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# Adv Algebra MATH 711

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This 260 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 711 at University of Michigan taught by Melvin Hochster in Fall. Since its upload, it has received 9 views. For similar materials see /class/231483/math-711-university-of-michigan in Mathematics (M) at University of Michigan.

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Math 711 Lecture of November 28 2007 Step 5 Reduction to the complete local case Now suppose that the result holds for ideals of height h of the form 1 ickR whenever k lt d Also suppose that R m K is a normal excellent local ring of prime characteristic p gt 0 of dimension d that I 1 xdR where 1 mm is a system of parameters for R and that n E 1 7 1 We next want to show that there is a counterexample such that R is also complete Over R we still have that n 6 IR If u e 1139 then there is a module finite extension domain T of E such that n 6 IT By the Theorem at the bottom of p 3 of the Lecture Notes from November 26 there is also a module finite extension S of R such that n 6 IS a contradiction Henceforth we may assume that our minimal counterexample is such that R is complete D For the next reduction we need the following fact Lemma Let A be a normal domain and let S be a domain ezrtension ofA generated by one element a that is integral over A so that S Let f fX be the minimal polynomial ofn over IC frac Then f has coe icients in A and S E Proof Let degf n Let be a splitting field for f over IC Let g be a monic polynomial over A such that 9a 0 Then fly working in It follows that every root p of f satisfies gp 0 Hence all of the roots of f in are integral over A We can write n f Pi i1 where the pi are the roots of f The coef cients of f are elementary symmetric functions of p1 pm and so are integral over A Since they are also in IC and A is normal the coef cients of f are in A ie f E Now suppose that h E AX is any polynomial such that 0 Then hlf working over Because f is monic we can carry out the division algorithm dividing h by f and obtaining a remainder of degree strictly less than n entirely over AX and the result will be the same as if we had carried out the dvision algorithm over Since the remainder is 0 when we carry out the division over ICX the remainder is also 0 when we carry out the division over AX Consequently h E fAX It follows that the kernel of the A algebra surjection AX a AM S such that X gt gt n is precisely fAX and the stated result follows E Step 4 Reducetion the case where R is complete and Gorenstein Choose a coe iicient field K for the complete counterexample R Then R is module finite over its subring 1 2 A K1xd which is regular and in paritcular normal Let S A domain module finite over a complete local domain is again local In S we still have that u E 3th WS since this becomes true when we make the module finite extension to R cf Problem 4 of Problem Set 4 Moreover since R is module finite over S we may identify R S so that we still have u Z 1 acdS Thus S also gives a counterexample By the preceding Lemma S g AXf where f is monic polynomial Since it is not a unit the constant term of f is in the maximal ideal of A It follows that S g as well Since is regular is Gorenstein We therefore have a minimal counterexample to the Theorem in which the ring is a complete local Gorenstein domain D Remark Until this point in the proof we have been concerned with keeping R normal In doing the reduction just above normality is typically lost But the remainder of the proof will be carried through for the Gorenstein case without any further refernce to or need of normality We shall soon carry through an investigation that requires the study of the tight closure of 0 in the injective hull of the residue class field of a Gorenstein local ring R m K which may also be thought of as the highest nonvanishing local cohomology module of the ring with support in m We shall therefore digress brie y to study some aspects of the behavior 0wm Comparison of nitistic tight closure and tight closure Let R be a Noetherian ring of prime characteristic 1 gt 0 When N Q M are modules that are not necessarily finitely generated we have a notion of tight closure N There is an alternative notion N fg M defined as follows fg i N M i NMO NQMOQM with MoN nitely generated As with tight closure studying this notion can be reduced to the case where N 0 and in this case fg i 0 M i 0M J MogM with Mo nitely generated It is not known whether under mild conditions on R these two notions are always the same There has been particularly great interest in the case where the module M is Math 711 Lecture of October 5 2007 More on mapping cones and Koszul complexes Let b B a A be a map of complexes that is injective We shall write d for the differential on A and 6 for the differential on 3 Then we may form a quotient complex Q such that Q71 Bnbnvln for all n and the differential on Q is induced by the differential on B Let C be the mapping cone of 15 Proposition With notation as in the preceding paragraph HnC E HnQ for all n Proof We may assume that every ml is an inclusion map A cycle in Qn is represented by an element 2 6 A71 whose boundary dnz is 0 in An1 n1Bn1 This means that dnz 71109 for some b E B714 Once we have specified 2 there is at most one choice of b by the injectiVity of 714 The boundaries in Qn are represented by the elements dn1An1 t nB T111187 171 n71Bn71 g dn1An1 71 A cycle in C71 is represented by a sum 2 69 U such that dn2 1 1 n71b 695717100 0 Again this element is uniquely determined by 2 which must satisfy dn E n1Bn1 b is then uniquely determined as 71 b where b E B714 is such that n1b Such an element b is automatically killed by 6711 since 157172571710 dn71 n71b dailanJ 07 and n2 is injective A boundary in C71 has the form dn1a71 nbn em This shows that d1 n71Bn71 g dn1An1 71 B71 as required D 2 Corollary Let g 1 icn E R be elements such that 7 is not a zerodiuisor on the R module M Let g 1 icn1 ie the result of omitting 7 from the sequence Then HA M E Hi for all i Proof We apply that preceding Proposition with A B IC M and bi given by multiplication by 7 in every degree i Since every term of IC M is a nite direct sum of copies of M the maps 15 are injective The mapping cone which is IC M therefore has the same homology as the quotient complex which may be identi ed with IC M 8 RicnR E IC R ER M R RicnR E IC R ER which is IC Mich and the result follows D We also observe Proposition Let b B a A be any map of complexes and let C be the mapping cone In the long e1act sequence HA Hc Hag k Hm the connecting homomorphism 871 is induced by 71 1 n1 Proof We follow the prescription for constructing the connecting homomorphism Let b E B71 be a cycle in B71 We lift this cycle to an element of C7 that maps to it one such lifting is 0 69 b the choice of lifting does not affect the result We now apply the differential in the mapping cone C to the lifting this gives 71n71 n71b 71n71 n71b 69 07 since b was a cycle in B71 Call the element on the right 04 Finally we choose an element of A71 that maps to 04 this gives 71 1 n1b which represents the value of 8n1b as required D Corollary Let g 1 icn E R be arbitrary elements Let 7 1 icn71 ie the result of omitting 7 from the sequence Let M be any R module Then there are short eacact sequences H 39E39 M gt a a HA M a AnnHFIQTM a 0 for every integer i Proof By the preceding Proposition the long exact sequence for the homology of the mapping cone of the map of complexes IC M EUCJET M has the form 71gtin 39 Eff M llf M P fixz M m FNETMM iiliM Since the maps given by multiplication by 7 and by 720 have the same kernel and cokernel this sequence implies the existence of the short exact sequences speci ed in the statement of the Theorem D The cohomological Koszul complex Notice that if P is a finitely generated projective module over a ring R 7 denotes the functor that sends N gt gt HomRN R and M is any module then there is a natural isomorphism HOHIRP g Pquot R M such that the inverse map 7 is defined as follows 7 is the linear map induced by the R bilinear map Bp given by Bpg gvu for g E Pquot u E M and v E P It is easy to check that 171P69Q m nQ and 2 that 71R is an isomorphism It follows at once that 3 an is an isomorphism for all n E N For any finitely generated projective module P we can choose Q such that P 69 Q E R and then since 7 69 g is an isomorphism it follows that 4 m is an isomorphism for every finitely generated projective module P If R is a ring M an R module and g 1 xn E R the cohomological Koszul complex IC M is defined as HOHlR Ko i R7 M and its cohomology called Koszul cohomology is denoted H The cohomological Koszul complex of R and it easily follows of M is isomorphic with the homological Koszul complex numbered backward but this is not quite obvious one needs to make sign changes on the obvious choices of bases to get the isomorphism To see this take the elements ujlwqji with 1 S j1 lt lt ji 3 n as a basis for K CZg R We continue to use the notation 7 to indicate the functor HomRi R We want to set up isomorphisms K271 E K that commute with the differentials Note that there is a bijection between the two free bases for K and 194 as follows given 1 S j1 lt lt ji lt 11 let M kni be the elements of the set 1 2n7j1jZ arranged in increasing order and let ujlym J correspond to ukl 7k denote as vjhm 7 which we shall also 71 4239 When a free R module G has free basis b1 bt this determines what is called a dual basis bl b for Gquot where b is the map G a R that sends bj to 1 and kills the other elements in the free basis Thus K24 has basis 1 However when we compute the 117 7J2quot value of the differential dzii on 1 vi while the coe icient of 1 hiil does turn out 117 71 to be zero unless the elements hl lt lt hi4 are included among the ji if the omitted element is jt then the coe icient of 1 hi 1 is dzii1 h j h1 7hz3971 031jid i1vh17m7hi717 which is the coe icient of vjlym J in dn Hwhlw Mil Note that the complement of j1 ji in 1 2 n is the same as the complement of h1 Mil1 in 1 2 71 except that one additional element jt is included in the latter Thus the coe icient needed is 71 1xjt where s71 is the number of elements in the complement of h1 Mil1 that precede jt The signs don7t match what we get from the differential in IC g R we need a factor of 715 1 t 1 to correct note that t 7 1 is the number of elements in j1 ji that precede jt This sign correction may be written as 715 1t 1 and the exponent is jt 7 1 the total number of elements preceding jt in 1 2 This sign implies that the signs will match the ones in the homological Koszul complex if we replace every v i by 71Zvi where E 21Ut 7 1 This completes the proof D This duality enables us to compute Ext using Koszul homology and hence Tor in certain instances Theorem Let g 1 icn be a possibly improper regular sequence in a ring R and let M be my R module Then EXtieR Ri M 2 HQ M 2 H7174 M TOISJRQR M 5 Proof Because the Koszul complex on the pi is a free resolution of R R we may use it to calculate ExtjR R M this yields the leftmost isomorphism The middle isomorphism now follows from the self duality of the Koszul complex proved above and we have already proved that the Koszul homology yields Tor when p is a regular sequence in R this is simply because we may use again that IC g R is a free resolution of R R D Depth and Ext When R a S is a homomorphism of Noetherian rings N is a finitely generated R module and M is a finitely generated S module the modules ExtN M are finitely generated S modules One can see this by taking a left resolution G of N by finitely generated free R modules so that Extjauv M Hj HomRG M Since each term of HomRG M is a finite direct sum of copies of M the statement follows If I is an ideal of R such that IM 31 M then any regular sequence in I on M can be extended to a maximal such sequence that is necessarily finite To see that we cannot have an infinite sequence 1 icn E I that is a regular sequence on M we may reason as follows Because R is Noetherian the ideals J7 1 xnR must be eventually constant Alternatively we may argue that because M is Noetherian over S the submodules JnM must be eventually constant In either case once JnM Jn1M we have that xn1M Q JnM and so the action of new on MJnM is 0 Since J7 Q I and IM 31 M we have that MJnM 31 0 and this is a contradiction since you is supposed to be a nonzerodivisor on MJnM We shall show that maximal regular sequences on M in I all have the same length which we will then define to be the depth of M on I The following result will be the basis for our treatment of depth Theorem Let R a S be a homomorphism of Noetherian rings let I Q R be an ideal and let N be a nitely generated R module with annihilator I Let M be a nitely generated S module with annihilator J Q S a The support ofN RM is VISJ Hence N RM 0 if and only ifISJ S In particular M IM if and only if IS J S b IfIM 31 M then there are nite ma1imal regular sequences 1 icd on M in I For any such ma1imal regular sequence Ext 3N M 0 ifi lt d and ExtN M 31 0 In particular these statements hold when N RI Hence any two mazrimal regular sequences in I on M have the same length c IM M if and only if Ext 3N M 0 for all i In particular this statement holds when N RI Proof a N R M is clealy killed by J and by I Since it is an S module it is also killed by IS and so it is killed by IS J It follows that any prime in the support must contain 6 ISJ Now suppose that Q E Spec S is in VISJ and let P be the contraction of Q to R It suf ces to show that N R MQ 31 0 and so it suf ces to show that Np RP MQ 31 0 Since I Q P Np 31 0 and NpPNp is a nonzero vector space over H RpPRp call it H5 where s 2 1 Mg maps onto MQQMQ N where A SQQSQ is a field If 2 l and we have H gt A But then we have N RMQ NP RP MQ HS RP Atg s XE H H ASt 5t7 0 as required The second statement in part a is now clear and the third is the special case where N RI Now assume that M 31 IM and choose any maximal regular sequence 1 zd E I on M We shall prove by induction on d that Ext 3N M 0 for i lt d and that EXtN M p 0 First suppose that d 0 Let Q1 Qh be the associated primes of M in S Let Pj be the contraction of Qj to R for l S j S h The fact that depthIM 0 means that I consists entirely of zerodivisors on M and so I maps into the union of the Qj This means that I is contained in the union of the P and so I is contained in one of the Pj called it Pjo P Choose u E M whose annihilator in S is Qjo and whose annihilator in R is therefore P It will suf ce to show that HomRN M 31 0 and therefore to show that its localization at P is not 0 ie that HomRP NpMp 31 0 Since P contains I AnnRN we have that Np 31 0 Therefore by Nakayama7s lemma we can conclude that NpPNp 31 0 This module is then a nonzero finite dimensional vector space over Hp RpPRp and we have a surjection NpPNp p Hp and therefore a composite surjection Np p Hp Consider the image of u E M in Mp Since AnnRu P the image 1 of u E Mp is nonzero and it is killed by P Thus AnnRPv PRp and it follows that v generates a copy of Hp in Mp ie we have an injection Hp gt Mp The composite map Np p Hp gt Mp gives a nonzero map Np a Mp as required Finally suppose that d gt 0 Let x 1 which is a nonzerodivisor on M Note that 2 zd E I is a maximal regular sequence on Since at E I we have that x kills N The short exact sequence 0 a M a M a MxM a 0 gives a long exact sequence for Ext when we apply HomR N 7 Because at kills N it kills all of the Ext modules in this sequence and thus the maps induced by multiplication by x are all 0 This implies that the long exact sequence breaks up into short exact sequences j 0 a Exth M a Extguv MxM a Extglm M a 0 We have from the induction hypothesis that the modules ExtN 0 for j lt d 7 l and the exact sequence above shows that ExtN M 0 for j lt d Moreover Ext WN 31 0 and dL1 shows that Ext71N is isomorphic with ExtN M The final statement in part b follows because the least exponent j for which say ExtRI M 31 0 is independent of the choice of maximal regular sequence 7 It remains to prove part lf IM 31 M we can choose a maximal regular sequence 1 icd on M in I and then we know from part b that ExtN M 31 0 On the other hand if IM M we know that IS l AnnRM S from part a and this ideal kills every Exta N M so that all of the Ext modules vanish D If R a S is a map of Noetherian rings M is a finitely generated S module and IM 31 M we define depthIM the depth of M on I to be equivalently the length of any maximal regular sequence in I on M or infj E Z Ext RI M 31 0 If IM M we define the depth of M on I as 00 which is consistent with the Ext characterization Note the following Corollary With hypothesis as in the preceding Theorem depthIM depthISM More over if PM is flat over R eg a localization of R then depthIRR R M 2 depthIM Proof Choose a maximal regular sequence in I say 1 xd These elements map to a regular sequence in IS We may replace M by Macl xdM We therefore reduce to showing that when depthIM 0 it is also true that depthISM 0 But it was shown in the proof of the Theorem above that that under the condition depthIM 0 there is an element a E M whose annihilator is an associated prime Q E Spec S of M that contains IS The second statement follows from the fact that calculation of ExtR commutes with at base change when the first module is finitely generated over B One may also use the characterization in terms of regular sequences D We also note Proposition With hypothesis as in the preceding Theorem let p 1 icn be gener ators ofI Q R IfIM M then all of the Koszul homology HA M 0 If IM 31 M then Hni M 0 ifi lt d and Hnd M 31 0 Proof We may map a Noetherian ring B containing elements X1 Xn that form a regular sequence in B to B so that Xi gt gt xi 1 S i S n For example we may take B RX1 Xn and map to B using the R algebra map that sends Xi gt gt mi 1 S i S n Let J X1 XnB Then depthIM depthJM and the latter is determined by the least integer j such that Ext3 BXB M 31 0 The result is now immediate from the Theorem at the bottom of p 4 D CohenMacaulay rings and lifting While preserving height Proposition A Noetherian ring R is Cohen Macaulay if and only if for every proper ideal I of R depthIR height Proof Suppose that R is Cohen Macaulay and let I be any ideal of R We use induction on height lf height I 0 then I is contained in a minimal prime of R and so 8 depthIR 0 Now suppose that height I gt 0 Each prime in Ass R must be minimal otherwise we may localize at such a prime which yields a Cohen Macaulay ring of positive dimension such that every element of its maximal ideal is a zerodivisor a contradiction Since I is not contained in the union of the minimal primes I is not contained in the union of the primes in Ass Choose an element 1 6 I not in any minimal prime of R and hence not a zerodivisor on R It follows that RiclR is Cohen Macaulay and the height of I drops exactly by one The result now follows from the induction hypothesis applied to IiclR E RiclR For the converse we may apply the hypothesis with I a given maximal ideal m of height d Then m contains a regular sequence of length d say 1 icd This is preserved when we pass to Rm The regular sequence remains regular in Rm and so must be a system of parameters for Rm killing a nonzerodivisor drops the dimension of a local ring by exactly 1 Hence Rm is Cohen Macaulay D We also note Proposition Let R be a Noetherian ring and let 1 icd generate a proper ideal I of height d Then there e1ist elements yl yd E R such that for every i l S i S d yi 6 xi gel1 xdR and for all i l S i S d yl yi generate an ideal of height i in R Moreover yl yd I and yd 5rd IfR is Cohen Macaulay then yl yd is a regular sequence Proof We use induction on d Note that by the coset form of the Lemma on prime avoidance we cannot have that x1 2 xdR is contained in the union of the minimal primes of R or else 1 xdR has height 0 This enables us to pick yl 1 A1 with A1 6 2 xdR such that yl is not in any minimal prime of R In case B is Cohen Macaulay this implies that yl is not a zerodivisor It is clear that 241102 xdR I The result now follows from the induction hypothesis applied to the images of 2 rd in Rle D Note that even in the polynomial ring Kac y z the fact that three elements generate an ideal of height three does not imply that these elements form a regular sequence 1 7 30y l 7 Z it gives a counterexample Proposition Let R be a Noetherian ring let p be a minimal prime of R and let 1 icd be elements off such that 1 iciRp has height i l S i S d Then there are ele ments 61 6d 6 p such that if yi xi 6139 13 d then yl yiR has height i l S l S d Proof We construct the 61 recursively Suppose that 61 6t have already been chosen t may be 0 lft lt d we cannot have that act p is contained in the union of the minimal primes of y1 yt If that were the case by the coset form of prime avoidance we would have that ict1R p Q Q for one such minimal prime Q Then Q has height at most t but modulo p all of 1 t1 are in Q so that height 6213 2 t l a contradiction D The following result will be useful in proving the colon capturing property for tight closure Lemma Let P be a prime ideal of height h in a Cohen Macaulay ring S Let 1 pkH be elements ofR SP such that 1 ickR has height h in R while 1 ick1R has height k 1 Then we can choose elements yl yh E P and 21 zk1 E S such that l yl yh 2 1 zk1 is a regular sequence in S 2 The images of 21 2k in R generate the ideal 1 ickR 3 The image of zk1 in R is n1 Proof By the first Proposition on p 8 we may assume without loss of generality that 1 ici generate an ideal of height i in R l S i S k We also know this for i k 1 Choose ZZ39 arbitrarily such that Z maps to obi l S i S k 1 Choose a regular sequence yl yh of length h in P Then P is a minimal prime of y1 yhS By the second Proposition on p 8 applied to the images of the of the Z in Syl yhS with p Pyl yhS we may alter the Z by adding elements of P so that the height of the image of the ideal generated by the images of 21 2 in Sy1 yh is i l S i S kl Since Syl yhS is again Cohen Macaulay it follows from the first Proposition on p 8 that the images of the 21 zk1 modulo yl yhS form a regular sequence But this means that yl yh21 zk1 is a regular sequence D Coloncapturing We can now prove a result on the colon capturing property of tight closure Theorem coloncapturing Let R be a reduced Noetherian ring ofprime characteristic p gt 0 that is a homomorphic image of a Cohen Macaulay ring Let 1 pkH be elements of B Let It denote the ideal 1 ictR 0 S t S k 1 Suppose that the image of the ideal k has height h modulo euery minimal prime of R and that the image of the ideal MAR has height h l modulo euery minimal prime of R Then a k ZR k1 Q b IfR has a test element 1 R n1 Q 1 ie n1 is not a zerodiuisor on RIg Proof To prove part a note that it suf ces to prove the result working in turn modulo each of the finitely many minimal primes of R We may therefore assume that R is a domain We can consequently write R SP where S is Cohen Macaulay Let h be the height of P Then we can choose yl yh E P and 21 zk1 in S as in the conclusion of the Lemma just above ie so yl yh 2 1 zk1 is a regular sequence in S and so that we may replace 1 xk1 by the images of the Z in R Since P has height h it is a minimal prime of J yl yhS and so if we localize at S 7 P we have that Math 711 Lecture of October 20 2006 We next prove illustrate the method of reduction to characteristic p by proving the Briancon Skoda theorem for polynomial rings over a field of characteristic 0 by that method We need Noether normalization over a domain and we first give a lemma Lemma Let A be a domain and let f E Abel zn Let N 2 1 be an integer that bounds all the exponents of the variables occurring in the terms of f Let b be the A automorphism ofAac1 icn such that 1 gt gt xi x272 for i lt n and such that it maps to itself Then the image of f under b when viewed as a polynomial in ion has leading term at for some integer m 2 l with a E A 7 Thus ouer Aa f is a scalar in Au times a polynomial in at that is monic Proof Consider any nonzero term of f which will have the form cax flac nag where 04 a1 an and ca is a nonzero element in A The image of this term under b is caac1 27 2 773112 mam my lynwgy and this contains a unique highest degree term it is the product of the highest degree terms coming from all the factors and it is Caga1g2a2 7Iy lan71zn Cagzna1Na2N2an1N 1 The exponents that one gets on at in these largest degree terms coming from distinct terms of f are all distinct because of uniqueness of representation of integers in base N Thus no two exponents are the same and no two of these terms can cancel Therefore the degree m of the image of f is the same as the largest of the numbers an a1N azN2 MANW1 as 04 a1 a runs through n tuples of exponents occurring in nonzero terms of f and for the choice 040 of 04 that yields m comic occurs in f is the only term of degree m and and cannot be canceled It follows that f has the required form D Theorem Noether normalization over a domain Let R be a nitely generated ezrtension algebra of a Noetherian domain A Then there is an element a E A 7 0 such that Ra is a module nite ezrtension of a polynomial ring Auk1 zd ouer Aa Proof We use induction on the number n of generators of R over A If n 0 then R A We may take d 0 Now suppose that n 2 l and that we know the result for algebras generated by n 7 l or fewer elements Suppose that R A61 bln has n generators If the 61 are algebraically independent over K then we are done we may take d n 1 and Z 61 l S i S n Therefore we may assume that we have a nonzero polynomial ail J E Abel xn such that f61 6 0 Instead of using the original by as generators of our K algebra note that we may use instead the elements 6176179W276276526179n1761 1676 where N is chosen for f as in the preceding Lemma With o as in that Lemma we have that these new algebra generators satisfy f ail 27 714 avg 71100 which we shall write as g We replace A by AM where a is the coef cient of it in g After multiplying by la we have that g is monic in at with coef cients in Aaac1 714 This means that 6 is integral over Aa6V1 171 R0 and so Ra is module finite over R0 Since R0 has n 7 l generators over AM we have by the induction hypothesis that R0b is module finite over a polynomial ring Aab21 zd1 Q R0b for some nonzero b E A and then Rab is module finite over Aab21 zd as well We can now prove Theorem generic freeness Let A be a Noetherian domain Let M be a nitely gen erated module over a nitely generated A algebra R Then there exists a E A 7 0 such that Ma is Aa free In particular there exists a E A 7 0 such that Ra is Aa free Proof Note that we may localize at an element repeatedly but finitely many times since one can achieve the same effect by localizing at one element the product of the elements used We use Noetherian induction on M and also induction on dim 1C A M where IC frac If a module has a finite filtration in which the factors are free the module is free By induction this comes down to the case where there are two factors N and MN When MN is free the short exact sequence 0 7 N 7 M 7 MN 7 0 is split so that M E MN 69 N We may take a finite prime cyclic itration of M and so reduce to the case where M RP We may replace R by RP and so assume that R M is a domain By the Noether Normalization Theorem for domains we may replace A by Al for a E A 7 0 and so assume that R is module finite over a polynomial ring R0 Abel ow over A We may then replace R by Bo Viewing R as a module over R0 This module has a prime cyclic filtration in which each factor is either Abel ow which is already free or a quotient Bi of it by a nonzero prime ideal and dim 1C AB lt n Thus for each Bi we can choose ai E A 7 0 such that BlL is Aai free and localizing at the product a produces a module with a finite filtration by free modules which will be itself free D We have the following consequence Corollary Let H be a eld that is nitely generated as a Z algebra Then H is a nite eld Hence the quotient of a nitely generated Z algebra by a maximal ideal is a nite eld Proof The second statement is immediate from the first statement To prove the first statement first suppose that H has characteristic 1 gt 0 The result that H is a finite 3 algebraic extension of ZpZ is then immediate from Hilbert7s Nullstellensatz or Zariski7s Lemma If not then Z Q H We can localize at one element a E Z 7 0 such that Ha H is Za free But if G is a nonzero free Za module and p is a prime that does not divide a then pG 31 G Thus pH 31 H contradicting that H is a field Discussion reduction of the Briancon Shoda theorem for polynomial rings over elds of characteristic 0 to the case of positive prime characteristic p Let K be a field of charac teristic 0 let f1 fn E Kac1 xd a polynomial ring over K let k be a positive integer and let 9 E I fk l We want to prove that g 6 f1 fnk and we assume otherwise The condition that g E Infk l implies that there is an equation gmilgm1 im0 where each ij 6 I k 1j I k 1j Thus for each j we can write ij as a sum of multiples of monomials of degree 11 k7 lj in f1 fn call the polynomials that occur as coef cients in all these expressions hl hN Let A be the subring of K generated over the integers Z by all the coef cients of 9 f1 fn and h hN Thus the elements 9 f1 f hl hN E Aac1 zd and if we let A f1 fnAac1 xn the equation holds in Aac1 xd so that g E frkil in Aac1 xn The idea of the proof is very simple we want to choose a maximal ideal M of A and take images in the polynomial ring Hz1 zd where H AM We will then be able to contradict the characteristic p Briancon Skoda theorem which will complete the proof for polynomial rings in equal characteristic 0 The only obstruction to carrying this idea through is to maintain the condition 9 13 after we kill M We can achieve this as follows Consider the sort exact sequence 0 gAhcl xdlIf Aac1 xdlIf Aac1 MBA11379 w 0 We can localize at a single element a E A 7 0 so that all terms becomes Aa free The first term remains nonzero when we do this since that is true even if we tensor further with K over Aa We may replace A by AG and so there is no loss of generality in assuming that all three modules are A free This means that the sequence is split exact over A and remains exact when we apply AM A 7 Moreover the first term remains nonzero Since AM has characteristic p we have achieved the contradiction we sought B Our next objective is to study multiplicities of modules on ideals primary to the maximal ideal of a local ring and connections with integral dependence of ideals Let M 31 0 be a finitely generated module over a local ring R m K and let I be an m primary ideal Recall that the function HileIn MI 1M where N denotes the length of N agrees with a polynomial in n for it gtgt 0 whose degree is the Krull dimension d of M which is the same as the Krull dimension of RAnnRM The leading e term of this function has the form J nd where e is a positive integer called the multiplicity 4 of M on I and which we denote e1 If I m we refer simply to the multiplicity of M In particular we may consider the multiplicity eR emR of R See the Lecture Notes from March 17 from Math 615 Winter 2004 Clearly we may alternatively de ne n1 eIM dlw If M 0 we make the convention that e1 0 Proposition Let R m K be local M a nitely generated R module ofKrull dimension d N a nitely generated R module and I J m primary ideals of R a If Qt Q AnnRM then eIM is the same as the multiplicity of M regarded as an PtQU module with respect to the ideal RQl i N 7 th b If dim N lt d then d lim c If Q J are m primary eJM S eIM d If dim M 0 eIltMgt W e If dim gt 0 then for any m primary ideal J of R eIJM eIM f IfM Q N where N is a nitely generated R module and M7 I N M then 51M d lim W In case dim lt d the limit is 0 g IfM has a nite ltration with factors Ni then eIM is the sum of the eINZ for those values ofi such that NZ has Krull dimension d Proof The statement in a is immediate from the definition since n1M IRQt 1M To prove part b simply note that Nm 1N is eventually a polynomial in n of degree dim N lt d For c note that if I Q J then 1 Q J 1 so that there is a surjection MI 1M a MJ 1M and MI 1M 2 MJ 1M for all n In the case of d n1M 0 for it gtgt 0 while 0 n0 1 To prove e choose a positive integer c such that C Q J Then JMI 1JM MI 1JM7 MJM S MI 1CM The last length is given for it gtgt 0 by a 6155 nd since substituting n c for n in a polynomial does not change its leading term This shows eIJM S eIM On the other hand dl MI 1JM 7 MJM 2 MI 1M7 MJM When we multiply by mid and polynomial with leading term 5 take the limit the constant term MJM yields 0 note that this argument fails when d 0 This shows that eIJM 2 eIM For part f note that by the Artin Rees lemma there is a constant e such that I CN M Q InM so that InCM Q I CN M Q InM Thus the limit is trapped between E M n1M d hm L61 ngtoo n nc1 d hm w ngtoo 11 Again the leading term of the Hilbert polynomial does not change when we substitute n c for n and so these two limits are both eIM when d dim M and 0 when dim lt d Finally for part 9 we may reduce by induction to the case of filtrations with two factors so that we have a short exact sequence 0 a N1 a M a N2 a 0 Then for each n we have a short exact sequence 0 a NlI 1M N1 a MI 1M a NgI 1N2 a 0 so that iMI 1M iN1I 1M m N1 N2I 1N2 We may multiply by dlnd and take the limit of both sides as n a 00 using part b and f of the Proposition D Corollary Let R m K be local M 31 0 nitely generated 0f Krull dimension d and I an m primary ideal Then MM Z 612 MpeIltRPgt PEAssM with dimRPd Proof If we take a finite filtration of M by prime cyclic modules and apply part g of the Proposition above the only primes P for which the corresponding cyclic modules RP make a nonzero contributions are those primes necessarily in Supp M such that dim RP d and these are the same as the primes in Ass such that dim RP d It therefore su ices to see for each such P how many times RP occurs in such a filtration A priori it is not even clear that the number cannot vary However if we localize at P all terms different from RP become 0 and the remaining copies of RPp g RpPRp give a filtratiion of Mp by copies of the residue class field of Rp Hence the number of times RP occurs in any prime cyclic itration of M is KR Mp D Remark In the statement of the Corollary we may write eMp instead of KR Mp where eMp is the multiplicity of Mp over Rp with respect to the maximal ideal by part d of the Proposition Math 711 Lecture of November 21 2005 We continue with our program of showing that an a irmative answer to Bass7s question follows from the intersection theorem of Peskine and Szpiro Lemma If R m K is a local ring andM a nitely generated module with AnnRM I then depthmR S depthIR dim S dim Proof For the right hand inequality choose P in the support of M such that dim RP dim Then depthIR dim S dim RP dim height P dim RP dim For the other inequality one may kill a maximal regular sequence in I both sides decrease by the same amount since M is unchanged Therefore we may assume that I has depth 0 and is contained in an associated prime P of 0 in R Then dim dim RP 2 depthmR D Lemma If N is a nitely generated nonzero R module such that idRN lt 00 then for every prime P in the support of N dim RP depthRp depthmR Proof We have that ldRPNp depthPRP RP h say Then with Hp RpPRp we have that Ext p HpNp 31 0 and so ukhK N 31 0 where h dim RP But then h h S idRN depthmR But by the preceding Lemma depthmR S depthPR dim RP S depthRp dim RP l H D Discussion the dual into the injectiue hull is independent of the choice of ring We want to establish the statement of the heading for module finite local extensions One of the main points is that when R m K a S n L is local map of local rings such that S is module finite over the image of R then ESL HomR S ER The latter is injective over S by the Corollary that is the next to last result in the Lecture Notes from March 19 Math 615 Fall 2004 We restate that result although we have changed the names of the rings to suit the current situation Corollary Let S be an R algebra let F be a flat S module and let E be an injectiue R module Then HomRF E is an injectiue S module We may apply this with E ERK and F S We get that E HomRS ERK is injective over S and by the adjointness of tensor and Hom we also get that Homsi E and HomRi E are the same functor on S modules for a given S module M HomSM E HomSM HomRS g HomRM 83 S E g HomRM E as required Note that every element of E is killed by a power of m and hence by a power of n It follows that E is a direct sum of copies of ESL But HomSLE E HomRL E E HomKL K as a K Vector space this has the same dimension as L and so as an L Vector space it must be isomorphic with L Thus E E ESL as claimed 1 2 Proposition Let R be a Noetherian ring and let M be a nitely generated R module a The support ofM is U Supp ExtlaM R j0 Moreuer if N maxdepthRp P is minimal in Supp then N Supp M U Supp EXtHM 3 j0 b IfR is a complete local ring of Krull dimension d and 7V denotes the dual into the injectiue hull of the residue class eld then d Supp M U Supp HhMV 30 Proof Clearly if Mp 0 then all of the EXtMRp vanish lt suf ces to show that if Mp 31 0 then some Extj p Mp Rp 31 0 for j S N It is enough to show this after localizing further at a minimal prime of Supp that is contained in P Therefore it suf ces to show this when P is a minimal prime of Supp But then we know that the first non vanishing Extj p Mp Rp occurs at the depth of Rp on the annihilator of Mp Since P is minimal in the support of R Mp has finite length over Rp and so its annihilator contains a power of PEP Thus the first non vanishing Extgp Mp Rp occurs when j is the depth of Rp and so j S N This completes the proof of part a For part b we may map a complete regular local ring S n K p RmK so that R E SJ Then Spec B may be identified with VJ Q Spec S as a closed set We may think of all of the modules involved as S modules instead with supports that are subsets of VJ and the issues are unaffected But now since g and since by the Discussion above it does not matter whether we take the dual into ESK or ERK we have MV Ext gij M S for every integer j by local duality and the result now follows from part a D We also note Proposition Let R be any complete local ring and let M be any nitely generated R module Then dim S i for alli 2 0 where 7V is the dual into the injectiue hull of the residue class eld Proof We use induction on the dimension of M If M has dimension 0 we have that Hg has finite length and therefore its dual has finite length as well while for i gt 0 we have that 0 Thus the result holds In the general case we may take a prime cyclic filtration of M and by taking the dual of the long exact sequence for local cohomology we may reduce to the case where M is 3 prime cyclic Since it does not matter over which ring we take the dual we may replace R by RAnnRM and so assume that M R is a complete local domain Then R is module finite over a regular local ring A and we may take local cohomology and duals over A instead of over R Again we may reduce to the case where R is a prime cyclic A module but since we may assume by induction that the result holds for dimension smaller than that of A we may even assume without loss of generality that R A M But then A 0 unless i dim A in which case anAV A has dimension i which su ices D We next observe Theorem Let R m K be a complete local ring let E ERK be an injectiue hull of the residue class eld and let 7V denote HomRi Let N be any nitely generated R module and let E denote the complea o EOaEl Ei m which we assume to be a minimal injectiue resolution ofN Let 8 be the comple1H3E Then a Each term of 8 is a nite direct sum of copies of E In fact 8 has the form O EHO EHl EHj where M ujm N Each map El a EHitl is given by a uZH gtlt Mi matria ouer R with entries in m The terms 8i are 0 forj lt k deptth while 8 31 0 The cohomology of the complea 8 is The complea HomRE 8 is the same as the complea HomRE E It is a complea consisting of nitely generated free modules Its cohomology is ExtE N It has the form AAA CL 0 VVVV O RMo Rul R j u where the matrices of the maps are the same as in part b and so have entries in m The complea 8 V is a complea of nitely generated free modules Its cohomology consists of the modules HznNV It has the form A H V OHRHOHRMHltRWltm It is the dual into R of the complea displayed in part e Proof For part a note that since Ass P the value of Hg on an injective module consists precisely of the direct sum of those copies of ERP such that P m Thus in fact HnEj 2 EW Quite generally the maps of modules b WEBVW j1 239 a WZ39 2 1 i1 4 are in bijective correspondence with the a gtlt b matrices of maps om where for all i j blJ a Wi Here the map b corresponding to the matrix om is such that its value on v E is 1JU7 z aj The statement in part b that the map is given by a pig by M matrix over R now follows from the fact that HomRE E E R We already know that because of the minimality of the resolution every element in the socle in EWt1 is killed by the map it follows at once that the entries of the matrix are in m Part c is then clear since it is equivalent to the assertion that M 0 for j lt d while Md 31 0 and M is the K vector space dimension of ExtK N Part d follows simply because may be viewed as the lth right derived functor of H3i The first statement in part e follows because every element of the image of map from E is killed by a power of m and so HomRE E7 2 HomR E The statement about cohomology then follows from the definition of Ext using injective resolutions of the second module while the remaining statements are immediate from parts a and b above and the fact that HomRE E g B Part f is immediate from the fact that HomRE E E R part d and the fact that on modules Q that are finite direct sums of copies of E we may identify the functors Q gt gt QV and Q gt gt HomR HomRE Q R In fact we have a natural identification of R with HomRE E and so it suf ces to give a natural isomorphism 6Q QV E HomRHomRE Q HomRE We can do this as follows Given f Q a E and g E a Q let 9Qltfgtlt9gt f 0 g E HomRE7 E Since this set up commutes with direct sums Q Q1 69 Q2 it suf ces to verify that 6Q is an isomorphism when Q E which is straightforward D Theorem Let R m K be complete local and let N be any nitely generated R module Let Id depthmR Let E be on lnjectlve hull of the residue class eld of R Then Ext 3E N 0 fort lt d Proof Let 7V denote the functor HomRi For any R modules Q and W we have that gt ExtieQWV 2 Term WV for if G is projective resolution of Q the latter is HAG R WV HiHomG R W since E is injective and by the adjointness of tensor and Hom this is 2 HiltHomRG HomRW E 2 Extng WV Math 711 Lecture of September 8 2006 Corollary If R is integrally closed in S then R normal then RH is normal t is integrally closed in SH If R is Proof The integral closure of RH in S H will be graded and so spanned by integral elements of SH of the form stk where s is homogenous Take an equation of integral dependence for st of degree say n on RH The coef cient of t is 0 and this gives an equation of integral dependence for s on R For the second part RH is integrallly closed in KH where K frac R and KH is integrally closed in KH frac RH since KH is a UFD D We next want to discuss integral closure of ideals Let R be any ring and let I be an ideal of R We define an element a of R to be integral over I or to be in the integral closure I of I if it satisfies a monic polynomial fz of degree n with the property that the coef cient of 2 is in It 1 S t S n We shall use the temporary terminology that such a monic polynomial is I special Note that the product of two I special polynomials is I special and hence any power of an I special polynomial is I special Let t be an indeterminate over R and let RIt denote the subring of the polynomial ring RH generated by the elements it for i E I This ring is called the Rees ring of I It is N graded with the grading inherited from RH so that the k th graded piece is It It follows easily that the integral closure of RIt in RH has the form RJ1tJ2t2Jktk where since this is an R algebra each Jk is an ideal of R We note that with this notation J1 I To see this note that if rt where r E R is integral over RIt satisfying an equation of degree n then by taking homogeneous components of the various terms we may find an equation of integral dependence in which all terms are homogeneous of degree n Dividing though by t then yields an equation of integral dependence for r on I This argument is reversible Exercise in this situation show that Jk is the integral closure of I39 We note several basic facts about integral closures of ideals that follow easily either from the definition or this discussion Proposition Let I be an ideal ofR and let u E R a The integral closure ofI in R is an ideal containing I and the integral closure ofI is b If h R a S is a ring homomorphism and u is integral over I then is integral over IS If J is an integrally closed ideal of S then the contraction of J to R is integrally closed 1 2 c u is integral over I if and only if its image modulo the ideal N of nilpotent elements is integral over IRN In particular the integral closure of 0 is N d The element u is integral over I if and only if for every minimal prime P of R the image ofu modulo P is integral over IRP e Every prime ideal of R and more generally every radical ideal of R is integrally closed f An intersection of integrally closed ideals is integrally closed g In a normal domain a principal ideal is integrally closed h If S is an integral e1tension of R then H R I Prgof That I Q I is obvious If r is in the integral closure of I then rt is iruegral over RIt But this ring is generated over RIt by the elements rlt such that r1 6 I ie such that rlt is integral over RIt It follows that RIt is integral over RIt and then rt is integral over RIt by the transitivity of integral dependence This proves a The first statement in b is immediate from the definition of integral dependence apply the ring homomorphism to the equation of integral dependence The second statement in b is essentially the contrapositive of the first statement The only if 7 part of c follows from b applied with S RN The if 7 part follows from d and so it will suf ce to prove The only if 7 part of d likewise follows from To prove the if 7 part note that the values of I special polynomials on u form a multiplicative system hence if none of them vanishes we can choose a minimal prime P of R disjoint from this multiplicative system and then no IRP special polynomial vanishes on the image of u in RP e follows from the second statement in c coupled with the second statement in b while f is immediate from the definition To prove g suppose that b is an element of R and a E E If I 0 it follows that a 0 and we are done Otherwise we may divide a degree n equation of integral dependence for a on bR by b to obtain an equation of integral dependence for ab on R Since R is normal this equation shows that ab E R and hence that a E bR Finally suppose that r E R is integral over IS Then rt is integral over SISt and this ring is generated over RIt by the elements of S each of which is integral over R It follows that SISt is integral over RIt and so rt is integral over RIt by the transitivity of integral dependence D Recall that a domain V with a unique maximal ideal in is called a valuation domain if for any two elements one divides the other This implies that for any finite set of elements one of the elements divides the others and so generates the same ideal that they all do together We shall use the term discrete valuation ring abbreviated DVR for a Noetherian valuation domain in such a ring the maximal ideal is principal and every nonzero element Math 711 Lecture of September 9 2005 We note that if M is a linear maximal Cohen Macaulay module over the local ring R m K and it E m 7 m2 is in suf ciently general position we shall make all this precise later when we return to a more detailed study of this topic then neither the multiplicity of M nor the least number of generators of M changes when we pass to MacM and RacR In the dimension 0 case the multiplicity of M is its length and so 1M S eM In the general case the multiplicity of M is the length of Macl icdM for a system of parameters in suf ciently general position one may need to enlarge the residue field to show that such a system of parameters exists In any case 1M S eM in general We shall next focus for a while on the problem of proving that ideals in usually polynomial rings are radical andor prime by working with large classes of such ideals This is the method of principal radical systems discussed last time It is based on two simple lemmas Lemma Let R be a Noetherian ring that is either local or N graded and let it E R be in the maximal ideal or be a form of positive degree Suppose that N is the nilradical of R that N is prime that it Z N and that RacR is reduced Then N 0 ie R is a domain Proof Suppose that u E N Since RacR is reduced we must have that u am for some i E R Since am 6 N at Z N and N is prime we must have that u E N Therefore N N By Nakayama7s lemma for local or graded rings N 0 D By applying this Lemma to RI in the situation below we obtain Corollary Let R be a Noetherian ring that is either local or N graded and let it E R be in the maximal ideal or be a form of positive degree Suppose that I is a homogeneous in the graded case proper ideal ofR with radical P where P is prime that it P and that P it is radical Then I P ie I is prime The next Lemma has various generalizations that may prove useful but we shall stick with the simplest case Lemma Let R be Noetherian let I be an ideal of R let J be the radical ofI and suppose that J Q P where P is prime Suppose that IacR is radical where it Z P and that icP Q I Then I J ie I is radical Proof Suppose that u E J Then it E I icR say u i our where i E I and r E R Then our uii E J Q P and so r E P Since icP Q I we have that our 6 I and so u i our Q I D We want to use these lemmas to prove the following result Theorem Let K be a eld let r and s be positive integers lett be an integer with l S t S minr s and let X be an r gtlt s matria of indeterminates ouer K Then ItX is a prime ideal ie is a domain The proof will take a while The idea is to include ItX in a much larger but finite family of ideals to which we can apply the lemmas above The ideals are typically radical 1 2 rather than prime The result is proved by reverse induction in that the largest ideals in the family are shown to be radical first The family has the property that for each ideal I in it that is not maximal in the family there is a larger ideal of the form I R in the family which will be known to be radical from the induction hypothesis We shall show first that the ideals ItX have radicals that are prime Thus once we show that they are radical it will follow that they are prime Note that if L is the algebraic closure of K and R is a K algebra R Q L K R and so to show that R is reduced or a domain it suf ces to show the corresponding fact for L K R Thus the problem we are discussing reduces to the case where K is algebraically closed and we assume this from here on This will enable us to take a naive approach to the material we need from algebraic geometry which will involve only basic facts about closed algebraic sets in af ne spaces A Note that showing that Rad ItX is prime is equivalent to showing that VItX is an irreducible closed algebraic set We think of points of A as corresponding to r gtlt s matrices over K Then VItX is precisely the set of r gtlt s matrices of rank 3 t 7 1 Proposition Let r s andt be as above Let A be an r gtlt s matria over a eld K Then A has rank 3 t 7 1 if and only ifA factors BO where B is an r gtlt t 7 l matria over K and O is a t 7 l gtlt s matria over K Proof We think of A as giving a linear map K5 7 KT where K5 is interpreted as s gtlt 1 columns The rank is at most t 7 1 if and only if the image has dimension 3 t7 l ie if and only if the map factors K5 7 Kh 7 KT where h S t 7 1 We may think of K E 1 as Kh EB K E l h and extend the map Kh 7 KT to the additional summand K E l h by letting it be 0 This gives a factorization K5 7 K E 1 7 KT for A which yields that A BO as required while any linear map with such a factorization obviously has rank at most t7 1 Corollary With notation as above VItX is irreducible Proof Think of AltT5t 1 g A3954 gtlt Aging as indexing pairs of matrices B C where B is r gtlt t 7 l and O is t 7 l gtlt s We have a map AltT5t 1 7 VItX that sends B 0 gt7 BC and by the preceding Proposition this map is surjective Since AltT5t 1 is irreducible and the image of an irreducible is irreducible VIt is irreducible D Of course this establishes that Rad ItX is prime For heuristic reasons we now carry through the proof that It is radical first for the case where t 2 Let Jk7h7aX Jkyhya denote the ideal generated by the entries of the first h rows of X the first k columns of X and the first a entries of the h 1 st row of X Here0 k s0 h rand0 a s lfhrorksallthevariableshave been killed and a 0 is forced We also abbreviate Jkyhyo Jkyh and Joyo Ja Note that Jk7h7a ka J07h7a If a S k Jk7h7a Jkyh Certain ideals have more than one description eg if h lt r Jk7h7s Jk7h170 We shall prove by induction that all of the ideals 2X Jk7h7aX are radical and prime if a 0 We assume the result for smaller matrices of indeterminates Evidently 2X JSW 5W is the ideal generated by all the indeterminates and is maximal We Math 711 Lecture of October 31 2007 Discussion local cohomology Let yl yd be a sequence of elements of a Noether ian ring S and let N be an S module which need not be nitely generated Let J be an ideal whose radical is the same as the radical of y1 ydS Then the dth local cohomology module of N with supports in J denoted Hf N may be obtained as lim k gt yf y N where the map from N Nk y1ydN to Nkh is induced by multiplication by 2h where z yl yd on the numerators If u E N ltu yf y gt denotes the image of the class of u in Nk in Hf N With this notation we have that W yl y gt lt2hu yf y hgt for every h E N If yl yd is a regular sequence on N these maps are injective We also know from the seminar that if S n L is a Gorenstein local ring and y yl yd is a system of parameters for S then HgyS H S is an injective hull for the residue class field L Sn of S over S In thesequel we want to prove a relative form of this result when R a S is a at local homomorphism whose closed fiber is Gorenstein Theorem Let R m K a S n L be a flat local homomorphism such that the closed ber SmS is Gorenstein Let dim R n and let dim SmS d Let a yl yd E n be elements whose images in SmS are a system of parameters Let E ERK be an injectioe hull for the residue class eld K Rm ofR over R Then E R HgyS is an injectioe hull for L Sn over S 7 In the case where the rings are of prime characteristic 1 gt 0 f ltE R 11595 raw R 11595 and ifu E E ands E S then M lt8 yi 7y gtq uq lt8q y lk yg gt Proof We first give an argument for the case where R is approximately Gorenstein which is somewhat simpler We then treat the general case Suppose that It is a descending l 2 sequence of m primary ideals of R cofinal with the powers of M We know that E limtRIt for any choice of injective maps RIt a RItH Let Qltyk ItS Jk where Jk yf y S For every k we may tensor with the faithfully fflat R algebra SJk to obtain an injective map Sthk a SQLHLk Since yl yd is a regular sequence on SItS for every It we also have an injective map Sthk a SQLMH induced by multiplication by z yl yd on the numerators The ideals Qltyk are n prinary irreducible ideals and as t k both become large are contained in aribtrarily large powers of 11 Once L Q m5 and k 2 s we have that ak Q msS 115 Q 115 Thus we have S R S R S E L 2139 7139 i i 21 139 L g S imtyk ak lmt7klt1t RJk imk imt It RJk R S N S N S N d hmk limtft R limk E R E R E R HaS We now give an alternative argument that works more generally In particular we do not assume that R is approximately Gorenstein Let E denote AnnEmt We first claim that that EM which we define as E R SJk is an injective hull of L over SM Ptmt R SJk By part f of the Theorem on p 2 of the Lecture Notes from October 29 it is Cohen Macaulay of type 1 since that is true for E and for the closed fiber of SJk since SmS is Gorenstein Hence Etyk is an essential extension of L and it is killed by Qltyk mtS Jk To complete the proof it suf ces to show that it has the same length as Styk Let M denote either Rmt or Et Note that M has a filtration with M factors each of which is K Rm Since SJk is R flat this gives a filtration of M R SJk with M factors each of which is isomorphic with K R SJk SmSJk Since Rmt Et it follows that SM and Etyk have the same length as required If t S t we have an inclusion E gt Ey and if k S 16 we have an injection SJk a SJk induced by multplication by Elf4 acting on the numerators This gives injections Et R Sk a Et 8 Sk since Sk is R flat and Et R Sk a Et R S since yl yd is a regular sequence on Et R S The composites give injections Etyk gt Etyk and the direct limit over If k is evidently E K HgyS The resulting module is clearly an essential extension of L since it is a directedTunion of essential extensions Hence it is contained in a maximal essential extension ESL of L over S We claim that this inclusion is an equality To see this suppose that u E ESL is any element Then it is killed by Qt Qltyk mtS Jk for any suf ciently large choices of t and k Hence it E AnnESLQ N which we know is an injective hull for L over SQl But Et R SJk is a submodule of N contained in E R HgyS and is already an injective hull for L over SQl It follows since they have the same length that we must have that E R SJk Q N is all of N and so u 6 Et R SJk Q E R HgyS To prove the final statement about the Frobenius functor we note that by the first problem of Problem Set 4 one need only calculate fg HgyS and this calculation is the precisely the same as in third paragraph of p l of theTLecture Notes from October 24 D 3 We are now ready to prove the analogue for strong F regularity of the Theorem at the top of p 5 of the Lecture Notes from October 29 which treated the weakly F regular case Theorem Let R m K a 511 L be a local homomorphism of local rings of prime characteristic p gt 0 such that the closed ber Sm is regular Suppose that c E R0 is a big test element for both R and S IfR is strongly F regular then S is strongly F regular Proof Let u be a socle generator in E ERltK and let y yl yd E n be elements whose images in the closed fiber SmS form a minimal set of generators of the maximal ideal nmS Let 2 yl yd Then the image of 1 in SltmS 241 ydS is a socle generator and it follows that u u 8 lt1y1 ydgt generates the socle in ESltL E R H5306quot Since c is a big test element for S it can be used to test whether 1 is in the tighticlosure of 0 in E R Hd ltS a This occurs if and only if for all 1 gtgt 0cltu lt1y1 ydgtq 0 in f ltE RH ltS and this means that cuq 8 lt1 24 yggt 0 in f ltE R H5306quot By part c of the Theorem on p 2 of the Lecture Notes from October 29 yl yd is a regular sequence on E R S from which it follows that the module f ltE R ltSltyi yg injects into f ltE R H5206quot Since S Sy f ygS is faithfully at over R the map f ltEgt a f ltE B R Sy f ygS sending in gt gt w 8 1 is injective The fact that cuq 8 lt1 24 yggt 0 implies that cuq 8 1g is 0 in f ltE 133 and hence that cuq 0 in R Since this holds for all 1 gtgt 0 we have that u E 0 a contradiction D The following result will be useful in studying algebras essentially of finite type over an excellent semilocal ring that are not F finite but are strongly F regular in many instances it permits reductions to the F finite case Theorem Let R be a reduced Noetherian ring of prime characteristic p gt 0 that is essentially of nite type over an ezrcellent semilocal ring B a Let B denote the completion ofB with respect to its Jacobson radical Suppose that R is strongly F regular Then B 83 R is essentially of nite type ouer B and is strongly F regular and faithfully at ouer R b Suppose that B A is a complete local ring with coe icient eld K Fir a p base A for K For all P lt A let RF AF A R We nay identify Spec RF with X Spec R as topological spaces and we let Zr denote the closed set in Spec R of points corresponding to primes P such that RE is not strongly F regular Then Zr is the same for all su iciently small P lt A and this closed set is the locus in X consisting of primes P such that Rp is not strongly F regular In particular if R is strongly F regular then for all P lt A RF is strongly F regular Proof a Since B a B is faithfully at with geometrically regular bers the same is true for R a B 83 R Choose c E R0 such that RC is regular Then we also have that Math 711 Lecture of December 5 2005 We have reduced the study of the vanishing conjecture to the case where A a R a S are complete local domains with A and S regular such that A a S and hence R a S is surjective and such that R is module finite over S If P and Q are the respective kernels of the maps A a S and R a S we also know that P is generated by part of a regular system of parameters for A and moreover that R A P Beyond that we may assume that M AI and then the vanishing conjecture asserts that I IQ IP We next want to reduce to the case where P has just one generator In the course of this reduction we lose the hypothesis that the the rings are complete local but we can get back to that case afterward The method we use to reduce to the case where P has just one generator is to replace A by the second Rees ring Ath v Q At lt where v 115 and R by PM RPRt 1 this is a homomorphic image of R A A and is therefore still module finite over A The quotient of A by vA is grpA a polynomial ring over AP and is still regular The quotient of PM by vR is RPReB PRPZR PjRPj1R 69 and this is an N graded ring In this ring RPR 7 QPR is a multiplicative system disjoint from the expansion of QPR and QPR becomes nilpotent if we localize at this multiplicative system because Q is a minimal prime of PR There is therefore a minimal prime of 11PM whose intersection with R is Q and which is necessarily graded Call this ideal Q Working with Q and vA we have from the vanishing conjecture for the rings A a A Q a A awe A vAgt that for all ideals of I of A with I IA IQ I IvA or IQ IA IvA This means that IQ A Q IvAth vl0 IP It follows that the general case of the vanishing conjecture reduces to the case where I is principal We have lost the condition that the rings A R and S be local and complete but we may repeat the earlier argument to reduce to this case again Thus the vanishing conjecture for maps of Tor is equivalent to the conjecture that when A15 is a principal prime of the complete regular local ring A such that AacA is regular and Q is a height one prime of a local domain R that is a module finite extension of A such that R A Q then for every ideal I of A IQ I Inc We next note that if W is a module over a regular local ring A and w E W then condition that the map A a W sending l gt gt w split is that for every ideal I of A the contraction of IW to A is I The condition is clearly necessary for if g W a A is a splitting and b gt gt bu E IW then applying 9 yields that bgbw gIW QIgW QIAI l For the converse we show that if 1 zd is a regular system of parameters for A it suf ces for all of the ideals It z i z to be contracted Both this condition and the splitting condition are unaffected by completion The condition yields that is injective for all t and we may take a direct limit to conclude that E a W A E is injective where E is the injective hull of the residue class field K of A This yields that HomAW E E a HomA E E is surjective and by the adjointness of tensor and Horn that HomAW HomAE a HomAE E is surjective By Matlis duality this means that HomAW A a A as surjective as required We can now complete the proof of the equivalence of the vanishing conjecture and the strong direct summand conjecture Since both imply the direct summand conjecture we may assume that the direct summand conjecture holds and this implies that for every ideal I of A IQ A Q IR A Q I and therefore IQ A IQ I But since IQ A Q Q A Ax we also have that IQ A IQ Ax IQ I Now Ax splits from Q if and only if IQ Ax Inc for all I By the calculation in the preceding paragraph this is equivalent to the condition that IQ I Inc for every I and we have already seen that this is equivalent to the vanishing conjecture for maps of Tor D There has been a partially successful metaconjecture that asserts that a result about regular rings should generalize to arbitrary Noetherian rings if the assumption of regularity is replaced by the hypothesis that certain modules have finite projective dimension One can generalize the vanishing conjecture for maps of Tor in this way Instead of assuming that A is regular one assumes that M is a Noetherian module of finite projective dimension over A and that ideals of depth at least k in A have height at least k in R modulo every minimal prime of R Under these hypotheses and rather weak conditions on the rings one can show that if G is a finite projective resolution of M over R then the cycles are in the tight closure of the boundaries in the complex G A R in degree one or more This implies that cycles become boundaries when one tensors further with a weakly F regular ring However we shall not attempt to give the best such result here but rather refer the reader to Hochster and C Huneke Phantom Homology Memoirs of the Amer Math Soc Vol 103 Number 490 1993 Amer Math Soc Providence RT Theorem 413 the case where T S considerably generalizes the result we discussed here From the point of view of this metaconjecture it is reasonable to ask whether when R is any Noetherian ring a module finite algebra extension R gt S such that deS lt 00 has the property that R a S splits This question was raised by J Koh in his thesis 1 H Koh The direct summand conjecture and behavior of codimension in graded extensions PhD Thesis University of Michigan 1983 The result is true when R contains a field of characteristic 0 but false in general otherwise counter examples were given in Juan 3 Velez in characteristic 2 and in mixed characteristic 2 cf J D Velez Splitting results in module nite extension rings and Koh s conjecture J Algebra 172 1995 pp 4547469 The proof in the equal characteristic 0 case depends on developing a notion of trace We want to assign a trace to every endomorphism of every nitely generated R module M of nite projective dimension at least when R is a Noetherian ring such that Spec R is connected We begin with the case where M is free In this case the trace TrRf Trf of f M a M may be de ned as the sum of the diagonal entries of a matrix representing f The matrix depends on the choice of a free basis for M but the trace does not because change of basis corresponds to replacing the matrix A by UAU l and the trace does not change It is clear that f gt gt Trf is an R linear map whose value on the identity map is the image of the rank of M in R In case B is Noetherian with Spec R connected and deM lt 00 we can de ne an R linear trace map HomRM M a R as follows Let W be the multiplicative system of all nonzerodivisors in R Consider the induced endomorphism W lM a W lM over W lR Then W lR is a semilocal ring whose localization at each maximal ideal has depth 0 Since a module of nite projective dimension over a local ring of depth 0 must be free it follows that W lM is locally free over W lR We want to see that the rank is constant Fix a nite projective resolution G of M by nitely generated projective modules The rank of Mp is the alternating sum of the ranks of the free modules in Gp Because Spec R is connected every projective module is locally free of constant rank It follows that the rank of Mp is independent of P for P corresponding to a maximal ideal of WAR Therefore W lM is free over WAR and we can de ne the trace of g E HomRM M as the trace of the map induced by g from W lM a W lM The di iculty is that we want the trace to be in R Q WAR not merely in W lR We can show that our trace is in R as follows The map f M a M lifts to a map b G a G We shall show that Trf is the alternating sum 2071jTrR j where h is the length of G Since the alternating sum is evidently in R this will prove what we need It su ices to prove the equality after localization at W Therefore we need only prove the result that if we have a map from an exact not merely acyclic complex of nitely generated free modules to itself the alternating sum of the traces of the maps is 0 the E correspond to the localizations of the GZ at W the base ring is now WAR and W lM is now included If the complex has length at most three say 0 a F2 a F1 a F0 a 0 where some of these may be 0 we have that F1 F2 69 F0 We can choose free bases for F2 and F0 and their union will be a free basis for F1 The matrix of bl then has block form A23 0 A0 Math 711 Lecture of December 9 2005 To conclude the proof of the syzygy theorem in characteristic p we shall make use of the following preliminary result that is independent of the characteristic Lemma Let M be nitely generated module of nite projective dimension over a Noe therian ring R Then M is a k th module of syzygies where k E N if and only iffor every prime ideal P of R either Mp is Ftp free or Mp has depth at least k on PRp Proof We first show that the condition is necessary Suppose that M is a kth module of syzygies of N which will also have finite projective dimension For any prime P such that Mp is not Ftp free we have that depthpRPMp depthpRPRp 7 dePMp depthpRPRp 7 dePNp 7 k depthpRPRp 7 dePN k depthpRPNp k 2 k as required To prove the converse suppose that for every prime P either Mp is free or Mp has depth at least k If k 0 there is nothing to prove Suppose that k 2 l and choose a basis f1 fh for HomRM R Then the maps f1 fh give a map M 7 Rh We shall show this map is injective and that the cokernel satisfies the condition to be a k 7 lst module of syzygies this will complete the argument by induction To show injectivity it will suf ce to show that given it E M 7 0 there is an R linear map f M 7 R such that at 31 0 for f must be an R linear combination of the fj and it follows that some fj is nonzero To prove the existence of f first replace it by a nonzero multiple whose annihilator is a prime P Then Mp has depth 0 over Rp Since k 2 1 it follows that Mp is free over Rp Hence we can certainly find a map Mp 7 Rp over Rp that is nonzero on xl This map has a multiple by an element of R 7 P that is the image of a map f M 7 R and at 31 0 It remains only to show that O RhM is a k 7 1 st module of syzygies Note that O has finite projective dimension If we localize at any prime P we have an exact sequence 07MP7R27CP70 There are two cases If Mp is not free then it has depth at least k Since Mp has finite projective dimenson over Rp the sum of the depth of Mp and its projective dimension is the depth of Rp Thus Rp has depth at least k and so does R2 The short exact sequence displayed above then implies that the depth of Op is at least k 7 1 One may see this from the long exact sequence for Extkp RpPRp In the remaining case where Mp is free over Rp we claim that the sequence splits so that Op is free as well This will complete the proof First note that HomRM Rp g HOHIRP Mp l It follows that the images of the fi generate the latter We change notation now and write R instead of Rp lt suf ces to prove that when M is R free and the fi generate HomRM R then the map o M 7 Rh given by u gt gt f1 fh splits Let b1 bs be a free basis for M Use this basis to identify M with R5 and let gi M 7 R be the map that sends bi to l E R and kills the other by Then h 9239 Z Tm fj 31 for each i where rm 6 R and rig is an s gtlt h matrix over B This matrix represents a map 6 Rh 7 R5 This is the required splitting since 945 9031 JAM whose 1th coordinate is h 2mebe 7 we 31 which is 0 if 1 31 i and is 1 if 1 i ie 67 bi bi as required D Proof of the syzygy theorem in characteristic 1 We now use the result on the depth of the order ideal and the lemma above to prove that if a kth module of syzygies M over a local ring R has finite projective dimension then either M is free or Bk embeds in M We use induction on k Let at be a minimal generator of M Then BMW has depth at least k or is the unit ideal It follows that there is a map M 7 R whose value on at is not a zerodivisor This implies that the annihilator of at in R is 0 and so the map R 7 M sending r gt gt rat is injective We then have a short exact sequence 07R7gtM7gtMR7gt0 MRac still has finite projective dimension If we can show that it is still at least a k 7 lst syzygy it will follow from the induction hypothesis that we have an injection Rk l gt MRac This map lifts to a map Rk l 7 M whose image G must be disjoint from Rae and so Rae G 2 R16 69 G is the required rank k free submodule of M To see that MRac is a k 7 1 st syzygy consider what happens when we localize at a prime P We consider two cases If P contains DM then Rp has depth at least k and so does Mp if it is free that holds because Rp has depth at least It follows that MRac has depth at least k 7 1 If P does not contain the order ideal then Rpac splits from Mp This implies that MRxp is free if Mp is and otherwise has at least the same depth as Mp and so has depth 2 k gt k 7 1 Thus MRac is a kth syzygy and the argument is complete D We next want to give an alternative proof of the syzygy theorem that is very close to the original proof of Evans and Grif th although what we prove is a bit stronger Their argument shows that the syzygy theorem follows from the improved new intersection Math 711 Lecture of October 25 2006 Proposition Let M be an R module Let a IfM has a nite ltration with factors N 1 S j S s and x is a nonzerodivisor on every N then MacM has a ltration with s factors Njach and Mx M has a ltration with ns factors there are n copies of every Njach 1 Sj S s b If 1 acd is a regular sequence on M and n1 nd are nonnegative integers then Mx1xgdM has a ltration by n1nd copies of Macl acdM Proof a By induction on the number of factors this comes down to the case where there are two factors That is one has 0 a N1 a M a N2 a 0 This has an isomorphic subcomplex 0 a ach a acM a acNg a 0 and the desired statement now follows from the exactness of the quotient complex It follows as well that Mx M has a filtration by the modules Njaanj and each of these has a filtration with n factors ackNjackHNj E For part b we use induction on d The case d 1 has already been handled in part a For the inductive step we know that Macffl 23711 M has a filtration by n1 nd1 copies of Macl xd1M and x xd is a nonzerodiVisor on each of these The result now follows from the last statement in part a with n nd D We next observe Lemma Let R be a Noetherian ring and let M be a nitely generated R module of di mension d gt 0 a M contains a mazrimum submodule N such that dim N lt d and MN has pure dimension d ie for every P 6 Ass MN dim RP d b Let W be a multiplicative system ofR consisting of nonzerodivisors and suppose that M and M are R modules such that W lM g W lM Then there e1ist e1act sequences OaMaMaOlaO andOaMaMaOgaOsuch thateach of01 andOg is killed by a single element of W c Let R m K be a complete local ring of dimension d and let M be a nitely generated faithful R module of pure dimension d Let 1 acd be a system of parameters for R IfR contains a eld there is a coe cient eld K Q R for R and M is a torsion free module over A KHacl acdll so that for some integer p gt 0 M and A become isomorphic when we localize at W A 7 In mi1ed characteristic there e1ists Cohen Macaulay ring A Q R containing 1 acd as a system of parameters such that A has the form Bf where B is regular and f 31 0 Moreover if W is the multiplicative system of nonzerodivisors in A then W consists of nonzero divisors of on M and W lM is a nite direct sum of modules of 1 the form W lBng where each gj is a divisor off In particular M 6 Bng is a Cohen Macaulay module over A of pure dimension d such that W lM and W lM are isomorphic as A modules Proof To prove a first note that since M has ACC on submodules it has a maximal submodule N of dimension less than d it may be 0 If N is another submodule of M of dimension lt d then d gt dim N 69 N 2 dim N N and so N N Q M contradicts the maximality of N Thus N contains every submodule of M of dimension lt d If MN had any nonzero submodule of dimension less than d its inverse image in M would be strictly larger than N and of dimension less than d as well b Since M Q W lM g W lM we have an injection M gt W lM Let ul uh be generators of M Suppose that ui maps to viwi 1 S i S h where vi 6 M and mi 6 W Let in w1wh Then M wM gt ZiRvi Q M The map W lM a W M that this induces is still an isomorphsim since in is a unit in WAR It follows that the cokernel 01 of the map M a M that we constructed is such that W401 0 Since 01 is finitely generated there is a single element of W that kills 01 An entirely similar argument yields 0 a M a M a 02 such that 02 is killed by an element of W c Let ul uh generate M Then the map R a Meah sending r gt gt rm ruh is injective It follows that Ass R Q Ass MGM Ass M so that R is also of pure dimension d Choose a field or discrete valuation ring V that maps onto a coef cient ring for B so that the residue class field of V maps isomorphically to the residue class field of R and let X1 Xd be formal indeterminates over V Then the map V a R extends uniquely to a continuous map B VHXl Xdll a R such that Xi gt gt obi 1 S i g d Let MB be the maximal ideal of B Since the map B a R induces an isomorphism of residue class fields and since RmBR has finite length over B the xi generate an m primary ideal of R R is module finite over the image A of B in R Moreover we must have dim A dim In the equal characteristic case where V K is a field we must have B A KHzl xdll Moreover M must be torision free over A since a nonzero torsion sub module would have dimension smaller than d Hence M and A where p is the torsion free rank of M over A become isomorphic when we localize at A 7 We suppose henceforth that we are in the mixed characteristic case We know that the ring A has pure dimension d It follows that A BJ where J is an ideal all of whose associated primes in B have height one Since B is regular it is a UFD Height one primes are principal and any ideal primary to a height one prime has the form 9 where g generates the prime and k is a nonnegative integer It follows that A BfB where f ff is the factorization of f into prime elements Let W be the multiplicative system consisting of the complement of the union of the ij The associated primes of A are the Pj fjA and these are also the associated primes of M Then W lA is an Artin ring and is the product of the local rings 14ij each of these may be thought of as obtained by killing f in the DVR obtained by localizing B at the prime ij M as a B module is then a product of modules over the various Apj each of which is a direct sum 3 of cyclic modules of the form Bij for l S s S kj Each of these is Cohen Macaulay of dimension d and the images the xi form a system of parameters since each of these rings is a homomorphic image of A We can now prove Theorem Let R m K be a local ring of dimension d and let 1 ird be a system of parameters for B Let I 1 xdR Let M be a nitely generated R module Then eM 0 ifdimM lt d and e M eIM ifdimM d Proof We have already proved that eIM 0 if dim lt d this is part c of the Lemma on p 5 of the Lecture Notes fTom October 23 Now suppose that dim d We may complete R and M without changing either multiplicity Let N be the maximum submodule of M of dimension smaller than d Then we may replace M by MN cf part d of the Lemma on p 5 of the the Lecture Notes from October 23 Thus we may assume that M has pure dimension d We may replace R by RAnnRM and so assume that M is faithful We view B as module finite over A as in part c of the preceding Lemma Since A contains 1 icd we may replace R by A and I by 1 icdA By parts b and c of the preceding Lemma there is a Cohen Macaulay A module M of dimension d such that each of M and M embeds in the other with cokernel of dimension smaller than d Thus by part e of the Lemma on p 5 of the Lecture Notes of October 23 we need only prove the result for M Hence we may assume that M is Cohen Macaulay But it follows from the part b of the Proposition at the beginning of this lecture that e MIM when M is Cohen Macaulay and we also know that eIM in this case B We next review the definition and some basic properties of the Koszul complex ICac1 J M where 1 xn E R and M is an R module We first consider the case where M R We let K1x1 J R be the free module G with free basis ill 71 As a module we let CZ1 J R be the free module iG which has a free basis with generators ujl uji jl lt lt ji The i differential is such that dui obi More generally the formula for the differential d is 239 duj1 A 39 39 39 A uji Zltleil jtu A 39 39 39 A 771 A ujt1 39 39 39 A ujzquot 151 Consider an N graded skew commutative R algebra A This is an N graded associative algebra with identity such that for any two forms of degree f g of degree h and k respec tively gf ilhkgf That is elements of even degree are in the center and multiplying 4 two elements of odd degree in reverse order reverses the sign on the product An R linear map d of A into itself that lowers degrees of homogeneous elements by one and satis es do dim indwmdv when u is a form is called an R derz vatz on of A Then G is an N graded skew commutative R algebra and it is easy to verify that the differential is an R derivation By the R bilinearity of both sides in u and 1 it suf ces to verify when u ujl ujh and v ukl uki with jl lt lt jh and k1 lt lt ki It is easy to see that this reduces to the assertion that the formula above is correct even when the sequence jl ji of integers in 1 2 n is allowed to contain repetitions and is not necessarily in ascending order one then applies to jl jh k1 ki To prove gtk note that if we switch two consecutive terms in the sequence j1 ji every term on both sides of changes sign If the jl ji are mutually distinct this reduces the proof to the case where the elements are in the correct order which we know from the definition of the differential If the elements are not all distinct we may reduce to the case where jt jt1 for some 15 But then ujl Au 0 while all but two terms in the sum on the right contain ujt 71th 0 and the remaining two terms have opposite sign Once we know that d is a derivation we obtain by a straightforward induction on k that if 111 vk are forms of degrees 11 ak then gtk gtk dvl vi 2ila1 39aquotlvj1 vjtil dvjt vjt 11 tz39 Note that the formula is a special case in which all the given forms have degree 1 It follows that the differential on the Koszul complex is uniquely determined by what it does in degree 1 that is by the map G a R where G is the free R module K1 R together with the fact that it is a derivation on Any map G a R extends uniquely to a derivation we can choose a free basis 1L1 un for G take the ac to be the values of the map on the ui and then the differential on IC1 acn R gives the extension we want Uniqueness follows because the derivation property forces gtk gtk to hold and hence forces to hold thereby determining the values of the derivation on an R free basis Thus instead of thinking of the Koszul complex Cxl acn R as arising from a sequence of elements 1 xn of R we may think of it as arising from an R linear map of a free module 6 G a R we might have written d1 for 6 and we write IC6 R for the corresponding Koszul complex The sequence of elements is hidden but can be recovered by choosing a free basis for G say 1L1 un and taking at 6011 l S i S n The exterior algebra point of view makes it clear that the Koszul complex does not depend on the choice of the sequence of elements only on the map of the free module G a R Different choices of basis produce Koszul complexes that look different from the sequence 5 of elements77 point of view but are obviously isomorphic In particular up to isomprphism permuting the elements does not change the complex We write IC1 xd M where M is an R module for ICx1 acn R 8 M The homology of this complex is denoted Hac1 acn Let g 1 xn and Let 1 ER We have the following comments 1 The complex is finite if M is not zero it has length n The ithe term is the direct sum of copies of M Both the complex and its homology are killed by AnnRM 2 2 The map from degree 1 to degree 0 is the map M a M sending 1L1 un gt gt 11 wnun The image of the map is IM and so Howl acd M E MIM A be V The map from degree n to degree n 7 l is the map M a M that sends it gt gt 11 7mm iacnun and so Hnx1 acn M E AnnMI 4 Given a short exact sequence of modules 0 a M a M a M a 0 we may tensor with the free complex IC 1 acn R to obtain a short exact sequence of complexes ICx1 acn M a ICx1 acn M a ICx1 acn M a 0 The snake lemma then yields a long exact sequence of Koszul homology HA M HA M llg M 221z M a H1 Mgt H g M a H1 MHgt MIM a MIM a MHIMH a 0 A UV V I kills H g M for every i and every R module M It su ices to see that and kills the homology the argument for at is similar Let 2 E CZg M and consider 2 u E CZ E Then dz an dz u flying Hence if z is a cycle dz an flying which shows that acnz is a boundary 6 Let g denote 1 IEWLL Let G Q G be the free module on the free basis 1L1 u 1 Then IC M may be identified with ltGgt R M g ltGgt M ma M This subcomplex is spanned by all terms that involve only 1L1 u 1 The quotient complex my be identified with IC g M as well one lets ujl ujiil u 8 w Math 711 Lecture of November 16 2005 Discussion 1 seeondproof of the direct 391 Wyu39wtwe in ehwwetw iotiep gt 0 Before proceeding further we give a different proof of the direct summand conjecture in positive prime characteristic p As before we may reduce to the complete case and so we may assume that the regular ring R has the form KHxl zdjj where K is a field and S is module finite over K Let L be a perfect field containing K eg the perfect closure or algebraic closure of K Then PM LHzl zdjj is faithfully at over R Since R a S splits if and only if HomRS R a HomRR R is onto an issue that is unaffected by applying PM ER 7 and since tensoring with a faithfully at algebra commtes with Hom when the first module is finitely presented we may consider whether PM a PM R S splits instead and so reduce to the case where K is perfect When K is perfect we have that Rq KHz i zgjj for every 1 p5 with e E N As earlier we may assume that S is a domain and in particular that S is torsion free over R It follows that S embeds in REM for some h and the image of 1 E S will have some nonzero coordinate Composing S gt Rh with the projection of REM on B using that coordinate we have an R linear map 15 S a R such that 151 31 0 and so we can choose 1 p5 such that 151 36 zgR R is free over Rq the monomials in the Q with every exponent less than 1 give a free basis When we expand the maximal ideal mq of Rq to R we get 36 zgR and so 151 Z qu is part of a free basis for R over Rq We may therefore choose an Rq linear map R a Rq that sends 151 gt gt 1 Then oo S a Rq is Rq linear and sends 1 gt gt 1 We may evidently restrict this map to Sq and so obtain an Rq linear map Sq a Rq that splits Rq gt Sq But we have a commutative diagram gist l 1 3 3 m where the horizontal isomorphisms are obtained by restricting the iterated Frobenius en domorphism Fe u gt gt M and the vertical maps are inclusions Since the map Rq gt Sq is split over Rq the isomorphic map R gt S is split over R D In any case we have now proved the direct summand monomial and canonical element conjectures in equal characteristic We next note the following fact Theorem If the local ring R has a big Cohen Macaulay module M then 71R 31 0 Proof What we need about M is that some system of parameters 1 zd for R is a regular sequence on M This hypothesis includes the condition that 1 zdM 31 M lf 71R 0 then for some t we have a map 15d of complexes 1C Cx i zg R a G where G is a free resolution of K such that Go R lifting Rz i z3R a K such that 15d 0 Choose 1 E M such that 1 z i zM but in kills the image it of v in z M This is possible because while z M is nonzero every 1 nonzero element is killed by a power of m and so each nonzero cyclic submodule has a non trivial socle We therefore have a map K 7 zM that takes 1 E K to u and we can lift this to the map G0 R 7 M that sends 1 gt7 1 Since G is free and CL IC x i xg M is acyclic this follows because z i xg is a regular sequence on M the map R 7 M sending 1 gt7 1 extends to a map of complexes G 7 KL The compostion of IC 7 G with this map G 7 CL gives a map 6 IC 7 CL such that 60 R 7 M sends 1 gt7 1 while 6d 0 We obtain another map C IC 7 CL simply by tensoring the map 60 R 7 M with the complex IC The maps 6 and C agree in degree 0 Since IC is free and K is acyclic 6 and C are homotopic and so there exists a map h ICdA 7 K such that Cd 7 6d h 0 6111 where 6111 Cd 7 ICd1 has a matrix in which each entry is inc for some j This implies that Cd1 7 6d1 v 7 0 v is in z i z M a contradiction since u is nonzero in z M D From this result we can conclude that the canonical element conjecture holds in dimen sion at most two as well as in equal characteristic since it reduces to the case of normal complete local domains which are Cohen Macaulay in dimension two By a dif cult result due to Ray Heitmann the direct summand conjecture holds in dimension three in mixed characeristic so that the canonical element conjecture is also true in dimension three We next want to show that the canonical element conjecture implies the improved new intersection conjecture Theorem Let R m K be a local ring of Krull dimension d such that me 31 0 Then the improved new intersection theorem holds for R that is if G is a nite complea of nitely generated free R modules of length n say 07Gn77G070 such that H0G 31 0 has a minimal generator u that is killed by a power ofm and HAG has nite length for i 2 1 then dim R d S n Proof Let i be a minimal generator of G0 such that 1 maps to u E M H0G Since a is killed by a power of m there exists a system of parameters 1 zd for R such that there is a surjection Rzl xd 7 Ru this lifts to a map R 7 G0 such that 1 gt7 1 Let Kat denote IC x i xg R for each t 2 1 Then we have the beginning of a map of complexess 0 7Gn7gt7gtG07gt0 1gt7uT 77gtR7gt0 all 7 a such that the vertical map R 7 G0 induces the specified map of the augmentations lt we were able to extend this to a map of complexes we would be able to do so if the upper row were acyclic for example we would have a contradiction when d gt n for the resulting map of complexes violates the canonical conjecture in the form given in part b of the first Proposition in the Lecture Notes of November 9 The point is that since d gt n the map at the dth spot is certainly 0 We cannot quite do this instead we shall show that one can give such a map of complexes when the bottom row is replaced by Kat for any suf ciently large value of t which is still enough to violate the canonical element conjecture We shall show by induction that we can carry out the construction of the map of complexes through the ith spot for all i 2 1 At every step we shall allow 15 to increase in constructing the next map Therefore we may assume for some i 2 1 that we have constructed a map 0 EGWEgtGi gtgtGo gt0 WT WT C531 KgltgtICgt gtu gt R 0 where 0 is the map 1 gt gt v For each h 2 l we have a standard map 6 ICEthgt a Kat which is the identity in degree 0 and in degree 1 is given by the diagonal d gtlt d matrix whose diagonal entries are the elements 51 xg In higher degree it is given by the exterior powers of the matrix that gives the map in degree 1 The only thing we need to know is that for every j 2 1 lm 6 Q mhICgt Let Z Ker GZ a G141 and Bi lm CZ1 a G1 Choose N such that mNHZGJ 0 Then mNZZ Q Bi By the Artin Rees Lemma we can choose 0 E N such that for all s 2 c msGi ZZ Q ms CZi Let h N 0 Then thZ Q mNZZ Q Bi We have a map ICt a G defined through the ith spot We get a composite map ICEthgt a K a G defined through the ith spot as well We claim that this map can be extended to one defined at the i 1st spot We have Gi1 Giil ml will K K2921 mi ngjh 5H1 K th 5 thjih To fill in the required map Kg a Ci it suf ces to show that for each generator b in a free basis for Kitjlh i6i6i1b is in Bi we can then choose 9 E Gi1 that maps to it and 9 will serve as the image of b under the map ICE a Gi1 that we are trying to construct Note that this element is a cycle its image in Gi1 can also be computed by traversing two other edges of the rightmost rectangle and 61416109 0 Because lm Q mhICEt we have that lm 15162 Q thi and so i6i6i1b E thi ZZ Q 3 as required This completes the proof of the inductive step and the result follows D We have already seen that the new intersection conjecture when H0G 31 0 has finite length and the intersection theorem if M N are finitely generated over R local and Math 711 Lecture of November 1 2006 Before attacking the problem of comparing symbolic powers of primes we want to discuss some techniques that will be needed One is connected with enlarging the residue class field of a local ring Proposition Let R m K be a local ring and let 6 be an element of the algebraic closure of K with minimal monic irreducible polynomial at 6 Let be a monic polynomial of the same degree d as f that lifts F to Let S Then S is module nite free of rank d and local ouer R Hence S is R flat The residue eld ofS is isomorphic with L KM and S has maximal ideal mS Proof S is module finite and free of rank d over R by the division algorithm Hence every maximal ideal of S must lie over m and the maximal ideals of S correspond bijectively to those of SmS l E 2 KM which shows that mS is maximal and that it is the only maximal ideal of S This also shows that SmS E D Discussion getting reductions such that the number of generators is the analytic spread Let R m K be local and I an ideal with analytic spread h One way of enlarging the residue field so as to guarantee the existence of a reduction of I with h generators is to replace R by Rt so that the residue class field becomes infinite For this purpose it is not necessary to enlarge B so that K becomes infinite One only needs that K have suf ciently large cardinality When K is finite one can choose a primitive element 6 for a larger finite field extension L the cardinality of the finite field L may be taken a large as one likes and a primitive element exists because the extension is separable Recall that the issue is to give one forms of B K R ngR that are a homogeneous system of parameters After making the type of extension in the Proposition one has because mS is the maximal ideal of S that L 83 ngSS g L 83 S R ngR g L R ngR g L K K R ngR If one makes a base change to K K B where K is the algebraic closure of K one certainly has a linear homogeneous system of parameters The coef cients will lie in L for any suf ciently large choice of finite field L Proposition Let R m K be any complete local ring Then R has a faithfully flat ezrtension S n L such that n mS and L is the algebraic closure of K IfR is regular then S is regular Proof We may take R to be a homomorphic image of T KHacl xdll where K is a field or of T VH1 icdll where V 7rV K is a complete DVR such that the induced map of residue class fields is an isomorphism In the first case let L be the algebraic closure of K Then T1 Lac1 xnll is faithfully at over T and the expansion of l 2 1 acdT to T1 is the maximal ideal of T1 Here faithful atness follows using the Lemma on p 2 of the Lecture Notes of October 18 becasuse every system of parameters for T is a system of parameters for T1 and so a regular sequence on T1 since T1 is Cohen Macaulay Then S T1 T R is faithfully at over R has residue class field L and m expands to the maximal ideal We can solve the problem in the same way in mixed characteristic provided that we can solve the problem for V if W 7rWL is a complete DVR that is a local extension of V with residue class field L then T1 WHxl acdll will solve the problem for T and T1 T R will solve the problem for R just as above We have therefore reduced to studying the case where the ring is a complete DVR V Furthermore if W 7rW L solves the problem but is not necessarily complete we may use W to give a solution that is a complete DVR Next note that if VA7rV K is a direct limit system of DVRs all with the same generator 7r for their maximal ideals such that the maps are local and injective then lim V is DVR with maximal ideal generated by 7139 It is then clear that the residue class field is lim AKA The reason is that every nonzero element of the direct limit may be viewed as arising from some V and in that ring it may be written as a unit times a power of 7139 Thus every nonzero element of the direct limit is a unit times a power of 7139 We now construct the required DVR as a direct limit of DVRs where the index set is given by a well ordering of the field L the algebraic closure of K in which 0 is the least element We shall construct the family V 7rKEL in such a way that for every A e L ueLrusAmegL This will complete the proof since the direct limit of the family will be the required DVR with residue class field L Take V0 V If A E L and VM has been constructed for M lt A such that for all M lt A VELZVSMQKHQL then we proceed as follows to construct VA There are two cases 1 If A has an immediate predecessor M and A 6 KM we simply let V VH while if A KM we take 6 A in the first Proposition to construct V 2 If A is a limit ordinal we first let V 7rV K limHlt VH If A is in the residue class field of V we let V V If not we use the first Proposition to extend V so that its residue class field is KA D To prove the theorem on comparison of symbolic powers in regular rings we shall also need some results on valuation domains that are not necessarily Noetherian In particular we need the following method of constructing such valuation domains Math 711 Lecture of October 5 2005 We shall denote by 1M the least number of generators of a nitely generated R module M By Nakayama7s Lemma if R m K is local or even quasilocal 1M dim Theorem Let R m K be a regular local ring of dimension n a K E K For every nite length R module M M and M have the same length M is automatically Cohen Macaulay Moreover 1M is dimKSocM which is the type of M Of course since M E M we also have that 1M dim SocM b Let M be a nitely generated Cohen Macaulay module of dimension d Then 1M is the type of M Proof For part a let 1 xn be minimal generators of R they are also a regular sequence Then K Ext K R g HomRK Rxl xnR HomRK K g K as required The statement that the lengths of M and M are equal is then immediate by induction if M has length 1 then M E K and we have already done this case Otherwise there is a short exact sequence 0 a N a M a Q a 0 where N is a proper nonzero submodule of M and then the length of M is the sum of the lengths of N and Q which are both nonzero and hence both less than the length of M We have a short exact sequence 0 a Qquot a M a N a 0 and so the length of M is the sum of the lengths of Qquot and N By the induction hypothesis these are the same as the lengths of Q and N which add up to the length of M and we are done For the remaining statement note that we have a short exact sequence d 6971 0 Anan M M where the map b sends it gt gt mu It is clear that the kernel of b is Anan Since the functor 7 is contravariant and exact on zero dimensional modules we obtain an exact sequence 0 lt Anan H M L MN where it is easy to see that 15 sends U1 un gt gt 21 xenon Thus it is clear that Coker bquot E Anan But the cokernel of is evidently MmM and since Anan has the same K Vector space dimension as Anan the result follows For part b note that if it E M is a nonzerodiVisor on M then MacM has the same minimum number of generators as M and its type is also the same as the type of M while g MacM and so the minimal number of generators and the type of are also unaffected By iterating we reduce to the case where M has finite length which we settled in part a D Theorem Let M be a nitely generated Coohen Macaulay module over a local R m K and let P be a prime ideal of R Then the type of Mp is less than or equal to the type of M 1 Proof We first consider the case where R is a homomorphic image of a regular local ring S We then replace R by S and m and P by their inverse images in S Thus we may assume without loss of generality that R is regular Then the type of Mp is the least number of generators of Mp E Mp and this is evidently at most the number of generators of M which is the type of M In the general case we consider the completion B of B Let Q be a minimal prime of PB lying over P in R The type of M is the same as the type of M over B a system of parameters for R is also one in B and the quotients will be isomorphic Since B is a homomorphic image of a regular local ring we have that the type of M equals the type of M and is greater than or equal to the type of MQ It therefore suf ces to show that the type of Mg is at least as large as the type of Mp The following lemma completes the proof with B Rp M Mp and o 1 Lemma Let M be a Cohen Macaulay module over a local ring B mBK and let B 7 C be a flat local homomorphism such that OmBO ls zero dimensional Then 0 8 M is Cohen Macaulay over 0 and its type is bigger than or equal to the type of M Proof Let 1 acd be a system of parameters for B Then it is also a system of param eters for C We replace B O and M with tensor products over B with Bdl dB and so assume that B and 0 both have dimension 0 lft is the type of M then Kt embeds in M as AnanB Applying O 83 7 yields the direct sum oft copies of OmBO as a submodule of 0 83 M which shows that the dimension of the socle in 0 83 M over 0 is at least the product of the type of M and the type of OmBO D The result below is true under various other hypotheses on R eg if R is excellent or a homomorphic image of Cohen Macaulay ring We shall not not need such great generality here Theorem Let R be a Noetherlan ring that is a homomorphz39c image of a regular rz39ng Let M be a nitely generated R module The set P E Spec R Mp is Cohen Macaulay ls Zarlshl open in Spec Proof We may replace R by the regular ring that maps onto it without affecting the issue Let I AnnRM After localizing at P IRp has pure height h We want to show that we can choose a E R 7 P such that Ma is Cohen Macaulay and we are free to localize at one element of R 7 P finitely many times to achieve this We do not change notation as we localize First choose a Z P but in all minimal primes of P that do not have height h After replacing R M by Ra Ma we may assume that I has pure height h Since R is regular M has finite projective dimension s Consider the modules Exti M R for 0 S l S s with l 31 h When we localize at P these finitely generated modules all become 0 and so there is a single element a Z P that kills them all Replace RM by PW Mm we may assume that ExtjeM R vanishes except when l h This implies that M is Cohen Macaulay To see this we may assume that we have localized at a single prime containing I Call the local ring obtained R m The vanishing of ExtM R for l gt h shows that deM S h by the Lemma below and so depthmM 2 dim R 7 h dim Rl dim The other inequality always holds D Math 711 Lecture of September 20 2006 We shall soon return to our treatment of the dimension formula which was stated in the Lecture of September 18 but we first want to make some additional remarks about the behavior of analytic spread Theorem Let K be a eld and T a nitely generated N graded K algebra with T0 K Let M be the homogenous maximal ideal ofT Let F1 Fs be homogeneous polynomials of the same positive degree d in T and let I F1 F5T Then 111ITM is the Krull dimension of the ring KF1 FS Q T and hence is the same as the mazrimum number of algebraically independent elements in KF1 FS ouer K Proof We shall show that KF1 FS g K TM ng TM We view the latter is K TM TMlITMtl 2 K T TiltllM and the ring on the right is the same as K T TIt because elements of T 7 M map to units in K and so already are invertible in this ring Note that K here is TM and so this ring is also the same as TItMTIt Now there is a map KF1 FS a TIt that sends gt gt th l gj S s To see that this is well defined note that the ideal of relations on the Fj over K is homogeneous Thus it su ices to see that if H E KY1 YS is a homogeneous polynomial of degree M such that HF1 Fs 0 then HF1t Fst 0 But the left hand side is t HF1 Fd t 0 0 We then get a composite map KF1 Fs a TIt a TItMTIt This map is clearly surjective since the image of T in the quotient is K and It is generated by the Et We need only prove that the kernel is 0 It is homogeneous let G be an element of the kernel that is homogeneous of degree h in F1 Fs Then G has degree hd in 1 zn If G is in the kernel then Gth is in MIhth and G 6 MW However all nonzero elements of this ideal have components of degree at least hd l in 1 mm a contradiction unless G 0 The final statement is a general characterization of Krull dimension in finitely generated K algebras D Remark This result gives another way to compute the analytic spread of the height one prime in a determinantal ring analyzed in the last Example beginning at the bottom of p 5 in the Lecture Notes of September 18 It is immediate that the analytic spread is n Example The Example discussed in the Remark just above shows that height one primes that have arbitrarily large analytic spread In a regular local ring a height one prime is 1 2 principal and so its analytic spread is 1 But there are height two primes of arbitarily large analytic spread Let X be an n gtlt n 1 matrix of indeterminates over a field K and let P be the ideal generated by the size n minors of X in the polynomial ring Then the analytic spread of P in KXM where M is generated by the entries of X is n l by the Theorem above for the minors of algebraically independent over K This is true even if we specialize the leftmost n gtlt n submatriX to be 2417 The minors are y and up to sign the products yn lxim l S l S These primes have height two the algebraic set of n gtlt n 1 matrices of rank at most 11 7 1 has dimension n2 n 7 2 On the open set where the first 11 7 1 rows are algebraically independent the space consisting of choices for the first 11 7 1 rows has dimension n 7 ln l the choices for the final row are linear combinations of the first 11 7 1 rows and are parametrized by An l giving dimension n2 71n 7 Recall that a map of quasilocal rings h R m 7 S n is called local if Q n The map of a local ring onto its residue class field is local while the inclusion of a local domain that is not a field in its fraction field is not local Proposition Let R m K be local a Ifh R m K 7 S n L is a local homomorphism and 1 Q m is an ideal of R then 1111 2 1111S b 1f1 and J are proper ideals of R then 1111 J 3 1111 111J c Let 1 and J are proper ideals of R If either 1111 or 111J is 0 then 1111J 0 1f the analyth spreads are positive 1111 J 3 1111 111J 7 1 Proof We replace R 7 S by Rt 7 St if necessary and the ideals considered by their expansions We may therefore assume the residue class fields are infinite For part a if 1 is integral over an ideal 10 with a 1111 generators then 15 is integral over 105 For part b simply note that if 10 is as above and J is integral over J0 with b 111J generators then 1 J is integral over 10 J0 which has at most a b generators To prove part c first note that the analytic spread of 1 is 0 if and only if 1 consists of nilpotents Thus if either a or b is 0 then 1J consists of nilpotents and 1111 J 0 as well Now suppose that both analytic spreads are positive and that 10 and J0 are as above Map the polynomial ring T ZX1 XaY1 Yb 7 Rso that X1 XaT maps onto 10 and Y1 YbT maps onto J0 Since 1J is integral over 10J0 it suf ces to show that 11110J0 S a b 7 1 Let M be the inverse image of m in T Then M is a prime ideal of T that is either X1 Xa Y1 YbT or pT X1 Xa Y1 YbT Let A TM Let I X1 XaA and j Y1 YbA Then we have an induced local map TM 7 R such that IR 10 and jR J0 By part a it will suf ce to show that 1111J S ab7 1 Let Qt denote the ideal X1 Xa Y1 Yb Q T and let 3 denote the ideal X1 XaY1 YbT There are two cases First suppose that M 2L Then 3 TM contains the rational numbers and may be viewed instead as the localization of the polynomial ring QX1 Xa Y1 Yb at X1 Xa Y1 Yb Then since the elements are forms of the same degree the Theorem above applies and the analytic spread is the transcendence degree of l S i S a l S j S b over Q But the fraction field of this domain is generated by the elements X1Y1 Xle and the elements XjXl 2 S i S 1 since which also shows that each Xj X1 is in the fraction field These b a7 1 elements are easily seen to be algebraically independent Exactly the same calculation of transcendence degree if Q is replaced by any other field H In the remaining case M Qt pT In this case note that since 3 and 3 1 are both free Z modules spanned by the monomials in X1 Xa Y1 Yb that they contain each B B 1 is free over Z It follows that p is not a zerodivisor on grg T Let H TM g ZpZ Then H T gran 2 H TpT ZPZ Z gruaT Let i TTpTgHX17 7Xa7 Yb anlz and g BT Because 1 is not a zerodivisor on grg T we have that ZpZ Z grgT E grgT and this is the ring whose dimension we need to calculate as in the proof of the Theorem above localization at M has no effect on this ring since the image of T 7 M consists of units in H We are now in the same situation as in the first case except that we are working with HX1 Xa Y1 Yb instead of QX1 Xa Y1 Yb D We are now ready to continue with our treatment of the dimension formula stated in the Lecture of September 18 Recall that we are assuming that R Q S are Noetherian domains with fraction fields f and g respectively that Q is a prime ideal of S lying over P in R that K RpPRp and that L SQQSQ We must show that height Q 7 heightP S tr deggf 7 tr degLK with equality of R is universally catenary Equality also holds if S is a polynomial ring over R Before beginning the proof we make the following observation Let P0 be a prime ideal of a local domain D In general dim DPO S dim D 7 height P0 while equality holds if D is catenary The inequality which is equivalent to the statement that dim D 2 height P0dim DPO follows from the following observation We can splice a saturated chain of primes of length H dim DPO ascending from P0 to the maximal ideal m of D corresponding to a chain of primes of length H in DPO with a chain of primes of length h descending from P0 to This yields a chain of saturated primes from m to 0 in D 4 that has length h k If moreover D is catenary then all saturated chains from m to 0 have the same length and this is dim D so that h k dim Proof of the dimension formula By adjoining generators of S to B one at a time we can construct a chain of rings 350 51 Sn such that for each i 0 S i S n we have that Si1 is generated over S by one element Let QZ Q S for each i Note that when R is universally catenary every Si is universally catenary It will su ice to prove the dimension formula whether the inequality or the equality for each inclusion S Q Si When we add the results each term associated with S for i different from 0 and 11 occurs twice with opposite signs The intermediate terms all cancel and we get the required result We henceforth assume that S RM where x need not be an indeterminate over B By replacing R and S by Rp and Rp R S we may assume that R P K is local We consider two cases according as whether at is transcendental or algebraic over R Case 1 x is transcendental over R Then the primes of S RM lying over P correspond to the primes of RPR 2 KM a polynomial ring in one variable There are two subcases Subcase la Q corresponds to the prime ideal 0 in KM ie Q PRM In this case SQ E Rx has the same dimension as B so that heigth height P We have that tr degQ l and L E Kx so that tr degLK l as well Since 0 171we have the required equality whether R is universally catenary or not Subcase lb Q is generated by PRM and a monic polynomial g of positive degree whose image 5 mod P is irreducible in The height of Q is evidently has height height P l a system of parameters for P together with 9 will give a system of parameters for RMQ The left hand side of the inequality is therefore 1 while the right hand side is 170 because L g K Again we have the required equality whether R is universally catenary or not Case 2 x is algebraic over B Let X be an indeterminate and map RX a BM S as R algebras by sending X gt gt x We have a commutative diagram l l RX 5 where the horizontal arrows are surjective and the vertical arrows are inclusions By hy pothesis the top horizontal arrow has a kernel which will be the principal ideal generated by a monic polynomial h of positive degree the minimal polynomial of at over f The kernel P0 of RX a S may therefore be described as hfX We claim that P0 is a height one prime of To see this we calculate RXpo Since R Q S P0 does not meet R and R 7 0 becomes invertible in Rpo Thus RPO is the localization of HX at Math 711 Lecture of December 1 2006 Recall the if R is a Noetherian ring of prime characteristic 1 gt 0 R is called F nite if F R a R makes B into a module finite algebra over FR rp r E R a subring of R that is also denoted RP When R is F finite the composition Fe R a R also makes B into a nite module over F5R we r e R a subring of R that is alternatively denoted R175 If R is F finite it is trivial that every homomorphic image of R is F finite The same holds for each localization WAR because inverting the elements in WP has the effect of inverting the elements of W If R is F finite so is RM if r1 rh span R over FR then the elements riicj l S i S h 0 S j lt p span RM over FRacp By induction any finitely generated algebra over an F finite ring is F finite and it is likewise true that any algebra essentially of finite type over an F finite ring is F finite A perfect field is obviously F finite and so a field that is finitely generated as a field over a perfect field is F finite it is a localization of a finitely generated algebra over a perfect field Thus if K is perfect each of the fields Kt1 tn is F finite where t1 t2 tn are indeterminates over K but the field Kt1 tn where we adjoin infinitely many indeterminates is not We note Proposition A complete local ring R m K of prime characteristic 1 gt 0 is F nite if and only if its residue class eld K is F nite Proof Since K Rm if R is F finite then K is Suppose that K is F finite and let c1 ch be a basis for K over Then R is a homomorphic image of a formal power series ring S KHacl xdll and it suf ces to show that S is F finite But the set of elements cjx llluwgd 0gjgh 0 ailtpfor0 i d spans S over FS JESll D This justifies the assertion in the Lecture Notes of November 29 that a complete local ring with perfect residue class field is F finite We next want to understand the behavior of the rank of 5R when R is a complete local domain with a perfect residue class field Note that when R is reduced of prime characteristic 1 gt 0 the three maps Fe R a R PJ E Q R and R Q Rlpg are isomorphic The isomorphism of Fe R a R with R175 Q R 1 2 follows from the fact that for a reduced ring R F5 is injective and F5R PJ E To understand the third map we need to de ne the ring Rlpg When R is a domain there we may take this to be the subring of an algebraic closure of the fraction field of R that consists of all the elements of the for rlpg for r E R In the general case one can show that there is an extension S of R unique up to canonical isomorphism such that the map R spa s E S In fact since R g PJ E via the map r gt gt rpg we think of77 R175 as R and take S to be B This means that when R is reduced we may think of 5R as Rlpg Theorem Let R m K be a complete local ring of Krull dimension d such that K is perfect Then for every e E N the torsion free ranh ofeR ouer R is pde Proof By the structure theory of complete local rings R is module finite over A KHacl xdll Let frac R and frac A IC The torsion free rank of Rlpg over R is the same as lpg l We have that clP5 IC 1PE IClPE KlPE IC and also clPE IC 1PE 21C so that a lp IclP KlP IC 1p 5 a 1C The map it a illp5 gives an isomorphism of the inclusion C Q with the inclusion Clp5 Q lpg so that clP5 IClPE 5 21C But then implies that clPE 5 Klp5 21C and the latter is the same as the torsion free rank over A of B AW 2 K1PEP5H Let yi avgp5 l S i S d Then B is free over A on the basis consisting of all monomials y fl ygd with 0 S ai lt p5 for l S i S d This free basis has cardinality ped pde as required D We are now ready to prove the existence of Hilbert Kunz multiplicities the result is stated on the first page of the Lecture Notes of November 29 but we repeat the statement Theorem Monsky Let M be a nitely generated module of dimension d ouer R m K where R has prime characteristic p gt 0 and let Qt Q m be m primary Then the Hilbert Kunz multiplicity eHKQ M ofM with respect to Qt ezrists and is a positive real number 3 Proof By the results of the Lecture of November 29 it suf ces to prove this when M R m K is a complete local domain with a perfect residue class eld Let 5Rmlpnl Yn T We shall prove that the sequence ynn is a Cauchy sequence This will prove that the sequence has a limit The fact that the limit is positive then follows from the lower bound in part c of the Lemma on p 2 of the Lecture Notes of November 29 The first key point is that 1R g Rlp has torsion free rank pd as an R module Thus 1R and 13de become isomorphic after localization at a nonzero element of the domain B By part b of the Lemma on p 4 of the Lecture Notes of November 29 there is a positive real constant C such that VMRWWWBMQHMWMSQWm for all n E N The leftmost term is pdeKQl By the Proposition at the top of p 4 of the Lecture Notes of November 29 fHKQ7 13X fHKQ7 RX 1 Thus becomes dele RX 7 farsz RX 1l S 019d 1 We may divide both sides by plt 1ld to obtain 0 7d Wnlninu OMWWW f Hence for all N 2 n m YNl S m Wn1llin1 7n2lml71v4 WNl Op d11p 1 p Op d p71 which shows that ynn is a Cauchy sequence as claimed D lt 1 1 177 S lt p p2 gt The proof of the Theorem of Monsky can be easily adapted to show more Theorem Let R m K by a local ring of prime characteristic 1 gt 0 let M be a nitely generated R module and let Qt Q m be an m primary ideal Then the Hilbert Kunz multi plicity with respect to Qt is additive in the sense that if one has a nite ltration ofM with factors Ni eHKQ M is the sum of the values of eHKQ Ni for those NZ of the same dimension as M Equivalently ifP is the set of necessarily minimal primes in the support ofM such that dim RP dim M then eHKQ M 2 RP MP5HKQ7 RP PEP Proof Additivity implies the formula because if one takes a prime cyclic ltration of M the only terms that contribute to the value of eHKQ M are those RP with P E P and the number of factors equal to RP is the same as KR Mp On the other hand it is easy to see that if one has then additivity follows because for every P E P KR Np is additive in N for modules N whose support is contained in Supp M It will suf ce to prove additivity after applying S ER 7 where S n L is complete local with L algebraically closed R a S is at local and n mS Hence we may assume without loss of generality that R is complete local with algebraically closed residue class field and it suf ces to prove that holds in this case We may replace M by M N where N is a maximal submodule of smaller dimension without affecting the issue Thus we may assume without loss of generality that M is of pure dimension We may replace R by RAnnRM without affecting any relevant issue so that the minimal primes of the support of M are those of B Let W be the multiplicative system that is the complement of the union of the minimal primes of R Exactly as in the argument on pages 5 and 6 of the Lecture Notes of November 29 we have M1 69 BM Q M such that each M has a unique minimal prime P E P in its support and localization at W induces an isomorphism We then have that 1313 gRPi for every t while Mipj 0 ifj 31 i We then have that h 5HKQ7 M 5HKQ7 M1 Mh 5HKQ7 M07 i1 and the formula will follow if we can show that it holds for all of the Mi We have thus reduced to the case where M has a unique minimal prime P in its support If we repalce M by 5M the Hilbert Kunz multiplicity with respect to Qt is multiplied by pde by the Proposition on p 4 of the Lecture Notes of November 29 The same is true for KR if Mp has a filtration by k copies of H RpPRp 8Mp has a filtration by k copies of 5H and the dimension of 81 over H is the same as the torsion free rank of 8RP over RP which is pde as required by the first Theorem on p 2 of today7s Lecture Notes Thus we may replace M by 5M for 6 gtgt 0 and so reduce to the case where RAnnRM is a domain Hence we can reduce to the case where R is a domain and M is torsion free over R If M has torsion free rank p over R we have already seen that eHKQ M p eHKQ R which is just what we need D Math 711 Lecture of November 9 2007 We note the following fact from field theory Proposition Let IC be a eld ofprime characteristic p gt 0 let be a separable algebraic ezrtension of IC and let f be a purely inseparable algebraic ezrtension of IC Then the map f K to the compositum Hf which may be formed within a perfect closure or algebraic closure of such that a b gt gt ab is an isomorphism Proof The map is certainly onto lt suf ces to show that f K is a eld every element has a qth power in A and so if the ring is reduced it must be a field and the injectivity of the map follows is a direct limit of nite separable algebraic extensions of IC and so there is no loss of generality in assuming that the is fnite over IC The result now follows from the second Corollary on p 4 of the Lecture Notes from September 19 or the argument given at the bottom of p 6 of the Lecture Notes from October 19 D This fact is referred to as the linear disjointness of separable and purely inseparable field extensions In consequence Corollary Let be a separable algebraic ezrtension of IC a eld of prime characteristic 1 gt 0 Then for every 1 p5 IC1q l Y Proof We need to show that every element of has a qth root in IC1q Since is a directed union of finite separable algebraic extensions of IC it suf ces to prove the result when is a finite separable algebraic field extension of IC Let M C d The field extension Clq Q lq is isomorphic with the field extension IC Q L Consequently lq Klq d also Since Clq Q IC1q Q lq to complete the proof it suf ces to show that IC1q Klq d as well But IC1q g Clq K A and so its dimension as a Clq vector space is the same as the dimension of as a IC vector space which is d D We next prove Proposition Let R be module nite torsion free and generically tale over a regular domain A of prime characteristic p gt 0 a R is reduced b For every 1 then map Alq AR a RA1q is an isomorphism Lihewise A AR a RA is an isomorphism c For every 1 RA1q is faithfully flat ouer R Moreover RA is faithfully flat ouer R l 2 Proof Let IC frac Then IC A R Ufa i where every i is a nite separable algebraic extension of IC a Since R is torsion free as an A module R Q IC A R Nil i from which the result follows b We have an obvious surjection Alq A R a RA1q Since R is torsion free over A each nonzero element a E A is a nonzerodivisor on R Since Alq is A flat this remains true when we apply Alq A i It follows that Alq A R is a torsion free A module Hence we need only check that the map is injective after applying IC A Q if there is a kernel it will not be killed The left hand side becomes Clq K Ufa Q and the right hand side becomes Ufa iUClq The map is the product of the maps Clq i a iUClq each of which is an isomorphism by the Proposition at the top of p l c This is immediate from part b since Alq is faithfully at over A for every 1 this is equivalent to the atness of Fe A a A Since A is the directed union of the Alq it is likewise at over A and since it is purely inseparable over A it is faithfully at Hence RA is faithfully at over R as well by part D Theorem Let R be module nite torsion free and generically etale over a regular do main A of prime characteristic 1 gt 0 Then there e1ist nonzero elements c E A such that chp Q RA1p For such an element 0 we have that c2R1q Q RA1q for all 1 and hence also that 02R Q RA Proof Consider the inclusion RA1p Q Rlp which is a module finite extension even Alp Q Rlp is module finite because it is isomorphic with A Q R If we apply IC A Q on the left hand side R becomes Ufa i and AV becomes Clp Hence the left hand side becomes h h h H micn 2 H zIClpl UziP 2391 2391 2391 by the Corollary on p l and the right hand side also becomes Ufa llp Hence if we take a finite set of generators for RV as an RA1p module each generator is multiplied into RA1p by an element 01 E A The product c of the ci is the required element Now suppose c E A0 is such that cR x Q R Let 1 1 Ce 01EHIPT and note that ce1 ccip for e 2 1 We shall show by induction on e that ceRlq Q RA1q for every e E N Note that CO c and this base case is given Now suppose that ceRlq Q RA1q Taking pth roots we have that cipRlpq Q RlpA1pq We multiply both sides by c to obtain CeHRlpq ccipRlpq g CRllepq g RA1pA1pq lepqL as required Since 1 1 1 1 17i liilt2 p p8 2 2e 02 is a multiple of Ce in Alq and in RA1q and the stated result follows E Also note Lemma Let R be module nite torsion free and generically tale over a domain A a IfA gt B is flat injective homomorphism of domains then B A R is module nite torsion free and generically etale over B In particular ifIC a is a eld EItBTLSZOTL and f is an tale ezrtension of IC then K f is an tale ezrtension of L b With the same hypothesis as in the rst assertion in part a if c E R is such that CROO Q RA then CB A R Q B A RB c Ifq is a minimal prime of R then q does not meet A and A gt Rq is again module nite torsion free and generically etale over A Proof a We prove the second statement first Since f is a product of finite separable algebraic extensions of IC we reduce at once to the case where f is a finite separable algebraic field extension of K and then f E CMg where x is an indeterminate and g is an irreducible monic polynomial of positive degree over IC whose roots in an algebraic closure of IC are mutually distinct Let g 91 95 be the factorization of 9 into monic irreducible polynomials over A These are mutually distinct and any two generate the unit ideal Hence by the Chinese Remainder Theorem 5 2c f 2 liclg H Niel9139 31 and every j is a finite separable algebraic extension of L It is obvious that B A R is module finite over A By the Lemma on the first page of the Lecture Notes from October 12 the fact that R is module finite and torsion free over A implies that we have an embedding of R gt A39h for some h Because B is A at we have injection B A R gt BENZ and so B A R is torsion free over B The fact that the condition of being generically etale is preserved is immediate from the result of the first paragraph above with IC frac A frac B and f IC A R b Every element of B A R1q E Blq A1q Rlq is a sum of elements of the form blq rlq while Cblq rlq blq crlq Since crlq E RA it follows that cltB A PM g B A Rgt1A HBWl g B A RgtlB l c q cannot meet AO because R is torsion free over A Thus q corresponds to one of the primes of the generic fiber IC A R which is a product of finite algebraic separable 4 field extensions of IC It follows that IC A Rp is one of these nite algebraic separable field extensions D We can now prove the result on test elements which we state again Theorem Let R be module nite torsion free and generically etale over a regular do main A Let c E R0 be any element such that CROO Q RA Then c is a completely stable big test element for R Proof Let Q be any prime ideal of R and let P be its contraction to A Then Ap a Rp satisfies the same hypothesis by parts a and b of the Lemma on p 3 and so we may assume that A is local and that Q is a maximal ideal of R We may now apply B A i where B By the same Lemma the hypotheses are preserved B becomes a product of complete local rings one of which is the completion of RQ The hypotheses hold for each factor and so we may assume without loss of generality that A m K a R frac m L is a local map of complete local rings as well Now suppose that H Q G are R modules and u E We may assume as usual that G is free This is not necessary but may help to make the argument more transparent We are not assuming however that G or H is finitely generated We know that u E Hence there is an element r of R such that ruq E qul for all 1 gtgt 0 Since r E R0 dim RrR lt dim Since RrR is a module finite extension of ArR A we must have that dim ArR A lt dim A and it follows that rR A 31 0 ie that r has a nonzero multiple a E A Then auq E qul for all 1 gtgt 0 We may take qth roots and obtain alq E RlqH for all 1 gtgt 0 We are using the notation RlqH for the expansion of H to Rlq R G ie for the image of Rlq H to Rlq R G Let ord denote any valuation on A with values in Z that is nonnegative on A and positive on m Then ord extends uniquely to a valuattion on A with values in Zlp such that ordb1q lqord b for all b E A0 and 1 To complete the argument we shall prove the following 7 Suppose that 671 is asequence of elements of A 7 0 such that 6714a E R H for all n and ord 6 a 0 as n a 00 Then CM 6 qul for all 1 and so u E This not only proves that c is a big test element it also gives a new characterization of tight closure which is stated as a Corollary of this proof in the sequel Moreover we will have proved that c is a completely stable big test element in every completed local ring of R and so it will follow that c is a completely stable big test element for R If the statement is false fix 1 such that GM Z qul For every rt we have burl E R X H Math 711 Lecture of November 28 2005 Theorem Let R m K be complete local and let N be a nitely generated R module of nite injectiue dimension Let d depthmR Let E be an injectiue hull of the residue class eld of R Then M ExtE N is a nitely generated module of nite projective dimension equal to d 7 deptth ouer R and such that Supp Supp Proof We use the same notations as in the Theorem stated on page 3 of the Lecture Notes for November 21 In particular 7V HomRi E E is a minimal injective resolution of N and 8 is H However now we know that idRN is finite and must agree with d depthmR By parts c and e of the Theorem mentioned ExtkE N is H HomRE 8 and HomRE 8 is a free complex of the form 0aRWwgtRMgt0 since ui 0 for i lt k By the last Theorem of the Lecture Notes of November 21 the cohomology vanishes except at the RM term Therefore this complex is a free resolution of its augmentation which is M ExtE N Moreover the matrices of the maps have entries in m so that it is a minimal resolution We know that uk 31 0 where k deptth and so deM d 7 k as claimed It remains to see that Supp Supp We know that Supp Uj Exta M R by part a of the Proposition at the top of the second page of the Lecture Notes from November 21 We may use the resolution to compute these Ext modules But when we apply HomRi R by part f of the Theorem cited we obtained the complex 8 V and its cohomology is NV Therefore Supp Uj Supp NV which by part b of the Proposition cited earlier is the same as Supp D We can now prove Theorem PeskineSzpiro The intersection theorem implies an a rmatiue answer to Bass s question ie it implies that if a local ring R m K has a nonzero nitely generated module N of nite injectiue dimension then R is Cohen Macaulay Proof We use induction on dim R the case where dim R 0 is trivial We may assume without loss of generality that R is complete By the preceding result R has a finitely generated module M such that deM lt 00 and Supp Supp We use induction on dim lf dim R 0 then R is Cohen Macaulay and we are done Choose P a minimal prime of R such that dim RP dim We consider two cases The first is where MPM E RP R M is zero dimensional ie of finite length Then dim R dim RP S deM by the intersection theorem 3 depthmR ie dim R S depthmR Since we always have the other inequality R is Cohen Macaulay and the proof in this case is complete Now suppose that dim gt 0 Then there is some prime Q0 other than the maximal ideal in Supp MPM and we can choose Q prime with Q0 Q Q C m such that l 2 dim RQ l ie such that Q is as large as possible with respect to the condition that it be strictly contained in m Since Q E Supp MPM we have that P Q Q and that Q E Supp M Supp N By the second Lemma on the first page of the Lecture Notes of November 21 we have that l dim RQ depthRQ depthmR since Q E Supp N with idRN lt 00 Since NQ is a nonzero module of finite injective dimension over RQ we have that RQ is Cohen Macaulay by the induction hypothesis Thus the displayed equation 1 becomes 2 1 dimRQ depthmR Since RP is a complete local domain we have that dim RP dim RQ height QP 1 dim RQPRQ Now P was chosen so that dim RP dim R and so we have dim R 1 dim RQPRQ Now RQ is a Cohen Macaulay ring and so all minimal primes yield quotients whose dimension is dim RQ Therefore we have 3 dim R 1 dim 3Q Comparing 2 and 3 we have that dim R depthm R and so R is Cohen Macaulay D Discussion The ezristence of big Cohen Macaulay modules versus the canonical element conjecture We have seen that if R has a big Cohen Macaulay module then 71R 31 0 However in terms of proving conjectures that do not refer to big Cohen Macaulay modules directly so far as I know every result that follows from the existence of big Cohen Macaulay modules follows from the canonical element conjecture It is therefore tempting to ask whether one can prove that if 71R 31 0 for a local ring R then R has a big Cohen Macaulay module So far as I know this is an open question Discussion Big Cohen Macaulay modules uersus big Cohen Macaulay algebras Similarly there do not seem to be results that follow from the existence of big Cohen Macaulay alge bras that do not follow as well from the canonical element conjecture if one thinks about such algebras one ring at a time77 However a statement known as the weakly functorial existence of big Cohen Macaulay algebras77 does yield something new the vanishing con jecture for maps of Tor this is a Theorem in the equal characteristic case and a conjecture in the mixed characteristic case We first want to discuss brie y what is known about the existence of big Cohen Macaulay algebras and then how this can be used to prove the vanishing conjecture for maps of Tor in equal characteristic as well as some consequences of the vanishing conjecture 3 Recall that if R is any domain R denotes the integral closure of R in an algebraic closure of its fraction field which is unique up to non unique isomorphism We may also characterize R as a domain integral over R that does not admit any proper integral extension domain A third characterization is that R is a domain integral over R such that every monic polynomial over R factors into linear factors over R We leave the verification of the equivalence to the reader We want to observe that if R a S is any map of domains then there is a commutative diagram R SJr l l R a S To see this note that R a S factors R a R Q S where F is the image of R in S The problem of finding the needed map R a SJr can be solved by filling in both missing arrows in the top row of the diagram R fr 5 l l l 39 R s so that the two squares commute This reduces the problem of finding a suitable map R a SJr to the separate cases where R a S is surjective and where R Q S In the latter case one can simply observe that the integral closure of R in SJr must be R since an algebraically closed field containing S will contain an algebraic closure of the fraction field of R In case B a S is surjective so that S RP note that since R is integral over B there is prime ideal Q of R lying over P It follows that RP injects into RQ The latter ring is a domain integral over RP since R is integral over R Moreover every monic polynomial factors into linear factors over RQ from which it follows that RQ E RP and we have the required map R a RQ E RP When we speak of the weakly functorz al existence of big Cohen Macaulay algebras for a class of local rings we mean that whenever R and S are local rings in the class and we have a local homomorphism R a S there exists a commutative diagram BEG ltgt l l RES such that B is a big Cohen Macaulay algebra over R and O is a big Cohen Macaulay algebra over S This is known in equal characteristic The positive characteristic 1 case was settled by M Hochster and C Huneke In nite integral extensions and big Cohen Macaulay algebras Math 711 Lecture of September 13 2006 In the proof of the preceding Theorem we used Chevalley7s Lemma Theorem Let M be a nitely generated module over a complete local ring R m Let be a decreasing sequence of submodules whose intersection is 0 Then for all k E N there exists N such that Mn Q mkM Proof For all h the modules Mn mhM are eventually stable we may consider instead their images in the Artinian module MmhM which has DCC and so we may choose nh such that Mn mhM Mn mhM for all n n 2 nh We may replace nh by any larger integer and so we may assume that the sequence nh is increasing We replace the original sequence by the Mnhh Thus we may assume without loss of generality that Mn mhM Mn mhM for all n n 2 h We claim Mk Q mkM for all k if not choose k and Uk 6 Mk imkM Now choose vk1 E Mk1 such that vk1 E Uk mod mkM and Recursively for all s 2 0 choose vks E Mks such that vks1 E vks mod mk5Mz this is possible because Mks Q Mks1 l mk5M This gives a Cauchy sequence with nonzero limit Since all terms are eventually in any given Mn so is the limit each Mn is m adically closed which is therefore in the intersection of the Mn D We have been assuming that valuation domains V are integrally closed It is very easy to see this if f is in the fraction field L of V but not in V then it lf is in the maximal ideal m of V Some maximal M of the integral closure V lies over m and so it is not a unit of V ie f Z V Thus V V It is also easy to see that if K Q L are fields and V is a valuation domain with fraction field L then V K is a valuation domain with fraction field K Moreover if V is a DVR then V K is a DVR or is K For the first statement each f E K 7 0 has the property that f or lf is in V and hence in V K as required Now suppose that V is a DVR and that it generates the maximal ideal Let W V K 31 K Each nonzero element of the maximal ideal m of W has the form nick in V where k is a positive integer Choose an element y of the maximal ideal of W such that k is minimum Then every 2 E m is a multiple of y in V and the multiplier is in W Thus m is principal It follows that every nonzero element m has the form uyt where t gt 0 since it is clear that the intersection of the powers of m is zero We next want to prove Theorem Let R be a Noetherian domain Then the integral closure PM of R is the intersection of the discrete valuation rings between R and its fraction eld L Proof Let f ba be an element of L not in PM where a b E R and b 31 0 It su ices to find a DVR containing R and not ba we may then intersect it with L Localize at a prime of R in the support of the R module PM RfR Since localization commutes 1 2 with integral closure we may assume that RmK is local Nonzero elements of R are nonzerodivisors in IS by atness and so the fraction field of R embeds in the total quotient ring of IS and we may view ba as an element of the total quotient ring of If I p is in the integral closure of aRp for every minimal prime p of IS then b is integral over IE If the equation that demonstrates the integral dependence has degree n we find that b E bn la bn ZaZ ban l a 13t and since Is is faithfully at over R this implies that b E bn la bn 2a2 ban l a R as well Dividing by a then shows that ba is integral over R a contradiction Thus we can choose a minimal prime p of IS such that b p is not integral over aISLp It follows that h6 is not integral over ISLp where the bars over the letters indicate images in Note that R injects into Thus the integral closure ITpf of p does not contain 56 and since it is module finite over ISLp by the last Theorem of the Lecture of September 11 it is a normal Noetherian ring Thus it is an intersection of DVR7s by part a of the last Theorem of the Lecture of September 8 and we can choose a DVR V containing ITpf and not 56 which is the image of ba so that V contains the isomorphic image of R but not the image of ba Now we may intersect V with the fraction field of R D Theorem Let R be any ring and let I Q J be ideals of R a r E R is integral over I if and only if there exists an integer n such that IrR 1 II rR Thus if J is generated over I by one element then J is integral over R if and only if there exists an integer n E N such that J7H1 IJ b IfJ 1 IJ a with n E N then Jnk IkJ for all k E N c If J7H1 IJ and Q Q J is an ideal and r E N an integers such that QT1 JQT then Qnf1 IQ T d If J is integral over I and generated over I by nitely many elements then there is an integer n E N such that J7H1 IJ If R is Noetherian then J is integral over I if and only if there exists an integer n E N such that J7H1 IJ e IfR is a domain and M is a nitely generated faithful R module such JM IM then J is integral over I IfR is a Noetherian domain then J is integral over I if and only if there is a nitely generated faithful R module M such that JM IM Proof Note that 1rR I r 1rt1 tr R Comparing the expansions for I rR 1 and II rR we see that the condition for equality is simply that r7l1 be in II rR rnI In and this is precisely the condition for r to satisfy an equation of integral dependence on I of degree n 1 This proves a We prove b by induction on k The result is clear if k 0 and holds by hypothesis if k 1 Assuming that Jnk IkJ for k 2 l we have that Jnk1 JnkJ IkJn1 Ik1Jn7 Math 711 Lecture of November 14 2005 We can further reduce all of the equivalent conjectures in mixed characteristic to the special case of the direct summand conjecture where the ring consists of formal power series over a coe icient ring Theorem The following are equivalent for complete local domains R m K of mixed characteristic p gt 0 and Krull dimension d 1 For every complete regular local ring R of the form VHacg zdjj where V is a coe cient ring in mixed characteristic a complete DVR of mixed characteristic p with ma1imal ideal pV R is a direct summand of every module nite ezrtension ring 2 Whenever R is a complete local domain H fR 72 0 3 For every complete regular local ring R of the form WHzg zdjj where W Vjplpgj e E N and V is a coe cient ring R is a direct summand of every module nite ezrtension ring In fact R is a direct summand of R Hence all of these 391quot are 1 39 39 t to the t t t that for every local domain of mixed characteristic p and Krull dimenson dimension d ma 72 0 Proof We have already seen that 3 is equivalent to the final statement and so it will su ice to show that 3 i l i 2 i 3 i l is clear since we may take e 0 in 3 To see that l i 2 let V be a coe icient ring of R and then R is module finite over A VHacg zd Then R A since R is module finite over A Since the maximal ideal mA of A is such that mAR is primary to m H fR 2 H MR 2 g1 Since A is a direct summand over A of S for every module finite extension S with A Q S Q R we have that Hg A injects into H S for every such S Taking the direct limit over all such S we find that Him A 72 0 injects into MESHMS 2 HiA1iLHsS HiAltAgt which is consequently not 0 It remains only to prove that 2 i In considering 3 it su ices to consider the case where the extension ring is an integral domain and so it certainly su ices to show that R is a direct summand of R Let 7r plpg denote the generator of the maximal ideal of W Let E H ilR which is an injective hull for the residue class field of R Then HjR 2 W R 11713 2 3 R E If this module is not zero its dual into E is not zero Choose any nonzero cyclic submodule RJ Then we have a nonzero surjection RJ a K and K gt E The composition gives a nonzero map RJ a E which extends to the whole module because E is injective But HomRR RE E 2 HomRR HomRE E 2 HomRR R 1 using the adjointness of tensor and Hom and Matlis duality Thus there is a nonzero map R a R Call the image of this map I and choose h maximum such that I Q nhR so that we can write I 7rhI0 where I0 is not contained in 7TB Since I 2 I0 as R modules and we have a surjection R a I we have a surjection R a I0 Since every ideal of R is m adically closed we have that 0 7TB 7rac vac VR N1 and this shows that for some N we have that I0 is not contained in 7r x97 avgR This means that we have an R linear map 6 R a R and u E R such that 6u Z 7r x97 x vR Consider the composition of 6 with multiplication by it thought of as a map R a R The composition 15 R a R is an R linear map such that 151 Z mac v x VR Let A0 WHm The maximal ideal is u 7r x97 avgAO and expands to to 11R 7r x97 avgR Thus 151 6 AiuA Since A has depth d over A0 which is regular A is free over A0 and 151 is part of a free basis for A over A0 Thus there is an AO linear map AJr a A0 that sends 151 to 1 6 A0 The composition with 15 produces an AO linear map 01 AJr a A0 that sends 1 gt gt 1 and so A0 Q AJr splits But since A is a module finite extension of A0 Ag A and it follows that A0 a Ag splits But there is a continuous W isomorphism AWac2 icdll A0 WHac V that sends obj gt gt for 2 S j S d This induces an isomorphism Ag 2 A Thus the inclusion of A0 gt Ag is isomorphic with the inclusion of A gt A Since the former splits so does the latter D Theorem The monomial conjecture holds in every local ring of positive prime character istic 1 Hence the canonical element conjecture holds for such rings as well Proof Suppose that 1 xd is a system of parameters for R m K such that 1dt 6 5M x1R for some t E N Taking qth powers where q 155 we find that 1 icdqt E x itq xgtqR for all 1 We can View H iR lith Raciv ic VR In this direcct limit system when we map Rx i xgR a Ric iqt xgqtR we multiply by x1acdqt which is 0 in the image Hence the image of each of the modules Rx i xgR in H iR is 0 and since the terms Rx i xgR are cofinal 3 in the direct limit system we find that Hf R 0 This is well known to be false In the interest of giving a relatively self contained treatment we deduce this fact from local duality We may replace R by its completion and then it is module finite over a regular local ring AmAK Then H7103 E H fAR and by local duality over A this is dual to HomAR A which is nonzero even if we tensor with the fraction field of A D We want to deduce the canonical element conjecture in equal characterisitc zero from the positive characteristic case We cannot do this from the direct summand conjecture because the needed implication only holds in residual characteristic p gt 0 but we can use the following result which is proved in the Lecture Notes for Math 711 Fall 2004 it is stated in the Lecture of December 3 and proved in the sequel using Artin approxi mation The version using only the condition in part a permits immediate reduction of many theorems for Noetherian rings containing a field to the case of local rings of af ne algebras over a finite field We assume this result here and we shall use it to deduce the canonical element conjecture for all local rings R such that Bred contains a field from the characteristic p gt 0 case which we have already handled Theorem Consider a family of nite systems of polynomial equations over Z such that each system in the family involves variables 1 xd and other variables Yi71 YZyh where both hi and the variables are allowed to depend on which system in the family one is considering Suppose that none of these systems has either a a solution in a nitely generated algebra over a nite eld such that the values of the icj generated an ideal of height d nor b a solution in a nitely generated algebra over a DVR of mixed characteristic p gt 0 such that the ideal generated by the values of the icj has height d Then no system in the family has a solution in a Noetherian ring in such a way that the values of the icj generate an ideal of height d Moreover a alone guarantees that there is no solution in a Noetherian ring containing a eld such the values of the icj generate an ideal of height d Note that once we have a solution in which the icj generate an ideal of height d we also have a solution in which the icj are a system of parameters for a local ring of dimension d simply localize at a height d minimal prime of the ideal generated by the obj Theorem The canonical element conjecture holds for all local rings containing a eld More generally it holds for any local ring R of Krull dimension d such that R has a quotient of Krull dimension d by a minimal prime P such that RP contains a eld Proof The final statement is immediate from the case where R contains a field We may therefore assume that R m K is such that R contains a field of characteristic 0 The existence of a counterexample implies that there is a system of parameters 1 icd for R together with a map from the Koszul complex IC ICic1 xd R to a free resolution G of K such that the induced map of augmentations is the canonical surjection Rxl icd a K and the map of free modules at the dth spot is 0 We may assume without loss of generality that Go R and that the map from R CO to G0 R is the Math 711 Lecture of December 10 2007 We have de ned an element of the homology or cohomology of a complex of nitely generated modules over a Noetherian ring R of prime characteristic p gt 0 to be phantom if it is represented by a cycle or cocycle that is in the tight closure of the boundaries or coboundaries in the ambient module of the complex We say that the homology or cohomology of a complex of finitely generated modules over R is phantom if every element is phantom A left complex is called phantom acyclic if all of its homology is phantom in positive degree Discussion behavior of modules of boundaries under base change Note the following fact if G i G is any R linear map in the application this will be part of a complex S is any R algebra and B denotes the image of d in G then the image of stage 59936 is the same as the image ltS R Bgt in S R G Note that we have a surjection G a B and an injection B gt G whence S RGgtS RG factors as the composition of S R G a S RB this is a surjection by the right exactness OfS R From these comments we obtain the following Proposition Let R a S be a homomorphism of Noetherian rings of prime characteristic p gt 0 for which persistence1 of tight closure holds Let G G GH be part of a compler and suppose that r is a phantom element of the homology H at G Then the image ofr in the homology H of S RGgtS RGgtS RGH is a phantom element Hence ifS is weakly F regular the image ofr in H is 0 1See the Theorem on p 3 and Corollary on p 4 of the Lecture Notes from November 7 l 2 Proof Let 2 E G represent 7 and let B be the image of G in G Then 2 E B and 1 2 represents the image of r in H By the persistence of tight closure for R to S 1 2 is in the tight closure over S of ltS R Bgt in S R G and by the discussion above this is the same as the image of S R GC The final statement then follows because we can conclude when the ring is weakly F regular that an element in tight closure of the module of boundaries is a boundary D There are many circumstances in which one can prove that complexes have phantom homology and these lead to remarkable results that are dif cult to prove by other methods We shall pursue this theme in seminar next semester For the moment we shall only give one result of this type However the result we give is already a powerful tool with numerous applications Theorem Vanishing Theorem for Maps of Tor Let A a R a S be Noetherian rings of prime characteristic 1 gt 0 where A is a regular domain R is module nite and torsion free over A and S is regular or weakly F regular and locally e1cellent2 Then for every A module M the map Torf M R a Torf M S is Ufor alli 2 1 This result is also known in equal characteristic 0 when S is regular but the proof is by reduction to characteristic 1 gt 0 It is an open question in mixed characteristic even in the case where S is the residue class field of R We discuss this further below Proof of the Vanishing Theorem for Maps of Tor Since M is a direct limit of finitely generated A modules and Tor commutes with direct limit it suf ces to consider the case where M is finitely generated If some Tor M R a Torf M S has a nonzero element 7 in the image this remains true when we localize at a maximal ideal of S that contains the annihilator of r and complete Thus we may assume that S M is a complete weakly F regular local ring and then persistence of tight closure holds for the map R a S Let m be the contraction of M to A Then we may replace A M and R by their localizations at m and so we may assume that A m K is regular local Let P be a finite free resolution of M by finitely generated free A modules To complete the proof we shall show that the complex R A P is phantom acyclic over R ie all of its homology modules in degree i 2 l are phantom These homology modules are precisely the modules Torf M R Hence for i 2 l the image of this homology in the homology of S R Rm P S A P 2The result holds when the completions of the local rings of R are weakly Fregular which is a conse quence of either of the stated hypotheses 3 is 0 by the Proposition at the bottom of p 1 But the ith homology module of the latter complex is Tor4 M S as required It remains to show that R A P is phantom acyclic Let h be the torsion free rank of R over A and let G Q R be a free A module of rank h Thus G 2 Ah as an A module Then RG is a torsion A module and we can choose c E A0 such that cR Q G Let 2 E R A Pi represent a cycle for some i 2 1 Let Bi be the image of R 8 Pi1 in R A Pi We shall show that czq E leq in flag A P for all 1 which will conclude the proof It will suf ce to show that for all e c kills the homology of f ltR A P in degree i 2 1 since 2 7 is an element of the homology and the module of boundaries in f R A is by another application of the Proposition at the bottom of p l the image of f 3i which is Big By 11 on p 3 of the Lecture Notes from September 12 we may identify THE A P g R A fjP By the atness of the Frobenius endomorphism of A the complex 72 P is acyclic Since G is A free we have that G A 72 P is acyclic and this is a subcomplex of R A fjP Because cR Q G we have that c multiplies R A fjP into the acyclic subcomplex G A fjP If y is a cycle in R A fjP in degree i 2 1 then cy is a cycle in G A fjP in degree i and consequently is a boundary in G A fjP But then it is a boundary in the larger complex B A f2P as well D We conclude by giving two consequences of the Vanishing Theorem for Maps of Tor Both are known in equal characteristic and are open questions in mixed characteristic Both can be proved by other means in the equal characteristic case but it is striking that they are consequences of a single theorem Theorem The Vanishing Theorem for Maps of Tor implies if it holds in a given charac teristic that direct summands of regular rings are Cohen Macaulay in that characteristic Before we can give the proof we need a Lemma that will permit a reduction to the complete local case Lemma Let R and S be Noetherian rings a Let S be a regular ring and let J be any ideal of S The the J adic completion T ofS is regular b Let R a S be a homomorphism and suppose that R is a direct summand of S Then for every ideal I of R the I adic completion ofR is a direct summand of the IS adic completion of S Proof a Every maximal ideal M of T must contain the image of J since if u 6 JT is not in M there exists a non unit u E M such that tuu 1 But 1 litu is invertible since 1tut u is an inverse Hence every maximal ideal T corresponds to a maximal ideal Q of S that contains J Note that S a T is at and Q expands to M Hence SQ a TM is faithfully at and the closed fiber is a eld It follows that the image of a regular system of parameters for S9 is a regular system of parameters for TM b Let f1 h E I generate I Let X1 Xh be formal power series indetermi nates over both rings Let 6 S a R be an R linear retraction Then RHXl Xhll is a direct sumnand of SHXl XhH we define a retraction S that extends 6 by letting 6 act on every coef cient of a given power series Let Qt be the ideal of RHacl xhll generated by the elements Xi 7 fi 1 S i S h Then there is an induced retraction SH1 hHQISHXM XhH RHXl The former may be identified with the IS adic completion of S and the latter with the I adic completion of R D Proof of the Theorem Let R be a Cohen Macaulay ring that is a direct summand of the regular ring S The issue is local on R and so we may replace R by its localization at a prime ideal P and S by Sp Therefore we may assume that R m K is local Second we may replace R by its completion E and S by its completion with respect to mS The regularity of S and the direct summand property are preserved by the Lemma just above Hence we may assume without loss of generality that R m K is complete local and then we may represent it as module finite over a regular local ring A with system of parameters 1 icd Let M Aicl icd Then the maps fi Tor M R a Tor M S vanish for i 2 1 Since S R 69 W over R over A is enough the maps fi are injective Hence TorfAx1 xd R 0 1 i S d This means that the Koszul homology Hi1 mm B 0 for i 2 l and by the self duality of the Koszul complex also implies that Extf4Aac1 icdA R 0 0 S i lt d It follows that the depth of R on 1 icdA is d dim A dim Hence R is Cohen Macaulay D Theorem The Vanishing Theorem for Maps of Tor if it holds in a given characteristic implies that the direct summand conjecture holds in that characteristic ie that regular rings are direct summands of their module nite extensions in that characteristic Proof Let A gt R be module finite We want to show that this map splits By part a of the Theorem on p 3 of the Lecture Notes from September 24 it suf ces to show this Math 711 Lecture of October 2 2006 Remark The quadratic transform of R m K along V is independent of the choice of regular system of parameters If x E m has minimum order in V the quadratic transform is the localization of at the contraction P of the maximal ideal of V If y also has minimum order then yx P and so acy E Rmxp Since uy for all u E m it follows that Rimtil Q Rim96h lf Q is the contraction of the maximal ideal of V to Rmy we have that elements of Rmy 7 Q are not in PRmacp and therefore have inverses in Rmxp Thus Rimtile Q Rimyelp The other inclusion follows exactly similarly D Remark The first quadratic transform T1 m1 K1 of the regular local ring R m K has dimension at most dim R and dim T dim R 7 tr degK1K In fact since R is regular it is universally catenary and the dimension formula holds Since the two fraction fields are equal the transcendence degree of the the extension of fraction fields is 0 Remark A local inclusion of valuation domains with the same fraction field must be the identity map For say that W mW Q V my with mW Q my If f E V 7 W then lf E W Since f Z W lf 6 mW But then lf 6 my which contradicts f E V D Lemma Let 1 xd be a regular sequence in the ring R with d 2 2 a ConstderT Rim1 Q R Then T E RX1X7ac2 Moreover 1 3 xd 2s a regular sequence in R21 b Consider T R2x1 godx1 Q R Then T RX2Xd1Xi7ii2 Moreover jTR ac liilT Proof a Since 1 2 is a regular sequence in R 1 xlX 7 2 is a regular sequence in RX killing 1 produces RaclRX and the image of the second element is 7x2 We claim that 1 is not a zerodivisor modulo xlX 7 2 for if gulf xlX 7 x2g in RX then 9 haul by the paragraph above Since 1 is not a zerodivisor in RX we find that f xlX 7 2h This means that RXx1X 7 2 injects into its localization at 1 which we may view as lexlX 7 2 Since 1 is a unit we may take X 7 avg1 as a generator of the ideal l 2 in the denominator and so the quotient is simply RM The image of RX1X 7 2 is then Ric2ac1 Q RM The last statement in part a follows because if we kill 1 in RX1X 7 x2 we simply get Pt1 The images of 3 icd form a regular sequence in this polynomial ring because that was true in Rxl 2 Part b now follows by induction on d as we successively adjoin avgx1 avg1 and so forth the hypothesis we need for the next fraction continues to hold The final statement is then immediate because the Jacobian matrix calculated from the given presentation is 1 times the size d 7 1 identity matrix D We also note Proposition Let R m K Q V be an inclusion of a regular local ring in a DVR let u E m be such that RuR is regular and let it E m have minimal order in V Let T be the rst quadratic transform off along V Then either 111 uac is a unit of T or else TulT is regular If the rst quadratic transform is a DVR it is always the case that 111 is a unit Proof Extend u to a regular system of parameters 8 for R If u itself has minimum order in V then uac is a unit of T If not then it is a unit of T times some element 1 6 S and ulT ulT if 11 uacl Hence we may assume without loss of generality that it 1 and u 2 in the regular system of parameters 8 Then 1 and avg1 are in the maximal ideal of T and to show that they form a regular sequence in T it suf ces to show that they form a regular sequence in the ring R1 g l d ll However mod x1 this ring becomes the polynomial ring Kf2 Ed and the image of avg1 is nghe quotient of T by this regular sequence is a localization of the polynomial ring Kfg fd and so is regular Hence chm1 is part of a regular system of parameters for T and so avg1 uac U1 is a regular parameter Note that in this case diT 2 2 so that if T is a DVR we must have that 111 is a unit We are now ready to prove the result mentioned at the end of the Lecture Notes for September 29 concerning finiteness of the sequence of quadratic transforms under certain conditions we follow the treatment in 8 Abhyankar Rami cation theoretic methods in algebraic geometry Annals of Mathematics Studies Number 43 Princeton University Press Princeton New Jersey 1959 Propositon 44 p 77 Theorem niteness 0f the quadratic sequence Let R m K Q V n L be a local inclusion of a regular local ring R of dimension d with fraction eld IC in a discrete ualuation ring Suppose that trdegLK d 7 1 Let T0 R and let Ti nilJ9 denote the ith quadratic transform of R along V so that for each i 2 0 Ti1 is the rst quadratic transform of Ti along V Then this sequence is nite and terminates at 3 some Th which means precisely that Th has dimension 1 Moreover for this value of h Th V IC so that V IC is essentially of nite type over R and the transcendence degree of the residue eld ofV IC over K is d 7 1 Proof Assume that the sequence is in nite By the dimension formula for every t dim dim R 7 tr degKZK so that tr degLKi tr degLK 7 tr degKiK d 7 1 7 tr degKiK dim R 7 tr degKiK 7 1 dim 7 1 Therefore at every stage we have that T Q V satsifies the same condition that R 7 V did Since the dimension is non increasing it is eventually stable and by replacing R by T for j gt 0 we might as well assume that the dimension of T is stable througout We call the stable value d and we may assume that d 2 2 It follows that every K is algebraic over K We have a directed in fact non decreasing union T of local rings inside V call the union W N L Here N is the union of the mi and L is the union of the K1 and so is algebraic over K We claim that W must be a valuation domain of IC If not choose it 6 IC such that neither it nor 115 is in W ie neither is in any Ti Write it yOzo where yo 20 E To B These are both in the maximal ideal of T0 if 20 were a unit we would have it E To while if yo were a unit we would have 115 E To If no is a minimal generator of mo of minimum order in V then it ylzl where yl yOuo and 21 zouo are in T1 We once again see that yl and 21 must both belong to m1 These have positive order in V but ordVy0 gt ordVy1 and ordV20 gt ordV21 We recursively construct yi and z in mi such that it yizi and ordVy0 gt gt ordVyZ while ordV20 gt gt ordVzZ At the recursive step let ui be a minimal generator of mi such that ordVuZ is minimum Then it yi1zi1 where yi1 yiui E Ti1 and 241 6 Ti As before the fact that it Ti1 and 115 Z Ti1 yields that yi1 and 2141 are both in nil1 as required and we also have that ordVyZ gt ordVyZ1 and ord gt ord V9141 This yields that ordVyZi is a strictly decreasing sequence of nonnegative integers a contradiction It follows that W is a valuation domain Since W Q V IC is a local inclusion of valuation domains with the same fraction field we can conclude from the third Remark on the first page of the Notes for this Lecture that W V IC Therefore V IC is essentially of finite type over R T0 and we have from the dimension formula that tr degLK dim R 7 dim d 7 1 2 1 contradicting our earlier conclusion that L is algebraic over K This contradiction establishes the result D Proof of the Key Lemma second step We recall the situation R is a regular domain with fraction field IC 1 is an element such that RvR is regular S is a reduced torsion free algebra essentially of finite type over R that is generically tale y E R 7 UR is such Math 711 Lecture of November 9 2005 We also note Proposition Let R m K be a local ring of Krull dimension d a The canonical element 71R is nonzero if and only if for every map b from a Koszul complea IC Cg R on a system of parameters g 1 mm to a free resolution G ofK lifting the canonical surjection a K the map bd R a Gd is nonzero b The the canonical element 71R is nonzero if and only if for every nonnegative left complea G of nitely generated free R modules of length at least d with augmentation M H0G 31 0 for every system of parameters g 1 mm of R and for every map b of the Koszul complea 1C Cg R to G such that the image in M of l E R K0 is a minimal generator of M the map bd R a Gd is nonzero Proof Part a follows from the fact that the near the dth spots the map of complexes looks likes this a Gd a syzdK gt Gd1 a T T Cd gt Kdil If the map bd is 0 we may use the composition of bd and the surjection Gd a syzdK to give a map to the truncated resolution which is 0 at the dth spot On the other hand given a map to the truncated resolution such that the map Cd a syzdK is 0 we can obviously use it to give a map to the free resolution in which the map C a Gd is 0 The condition in part b is obviously su icient from part a even if G is restricted to be a free resolution of K We shall show that a implies In the situation of b the image of l E R K0 in MmM is nonzero and so we can choose a surjection M a K so that this element maps to l E K This surjection M a K lifts to a map from G to a free resolution G of K The composition of the given map IC a G with this map G a G contradicts a D As a corollary of this result we have Proposition Let R m K a S n L be a map a local map of local rings of the same dimension such that mS is primary to ii If n5 31 0 then me 31 0 Proof The hypotheses imply that a system of parameters for R maps to a system of parameters for S Suppose that one has a map from the Koszul complex IC of a system of parameters of R to a free resolution of K over R such that the map at the dth spot is 0 Simply apply S ER 7 to contradict part b of the preceding Proposition over S D Corollary Let R m K be a complete local ring of Krull dimension d such that me 0 let ISL denote the completion of R let S be the quotient oflst by a minimal prime such that dim S d and let T be the normalization of S which is a complete normal local domain IfnR 0 then nR 0 n5 0 and nT 0 Hence if the canonical element conjecture holds for complete normal local domains then it holds for all local rings 1 2 Proof All of the maps R a E a S a T satisfy the conditions of the preceding Proposi tion D We shall now develop a variant of the Koszul complex which is always acyclic when the residual characteristic is p gt 0 and which we shall use to show that a family of special cases of the direct summand conjecture implies the canonical element conjecture In fact we shall show that this family of cases the canonical element conjecture the monomial conjecture and the direct summand conjecture are all equivalent Until further notice let R m K denote a complete local domain of residual char acteristic p gt 0 and let S denote either R which can be used either in the mixed characteristic case or in the case where R has characteristic p gt 0 or R which can be used in the characteristic p gt 0 case The property of S that we really need is that it be a quasilocal domain that is integral over R and that is closed under extraction of all p5 th roots Moreover in the mixed characteristic case it is convenient to assume that S contains all p5 th roots of unity If it E S is not zero we write ic x for the ideal U zlpES eEN Since S contains all p5 th roots of unity it does not matter which choice of gelp5 we make in describing the principal ideal gelp5 S Evidently gem is an increasing union of principal ideals of S which are free S modules and so every gem is a at ideal of S We note the following properties of these ideals Proposition Let notation be as in the preceding paragraph a For all nonzero ac y E S 96 W W 96 y x cm and these are all isomorphic with 18 S 240 1 flat ideal of S b For any nite set 1 xk of nonzero elements of S k k k HWJ39OO mlt75j gt H 9 7 j1 j1 j1 and all of these are isomorphic with 1M S 83 3 a flat ideal of S c For any nonzero it gem is a radical ideal of S Proof First note that if u E 3cm then illp5 E ic x for all e since sxlpn1p5 has the form slpgiclpn5 Part c is immediate since if MV 6 3cm by increasing N if necessary we may assume that it has the form p5 For part a if u E ic x 24 we also have that illp E ic x 240 and so u ulpu1pp 1 E x y and it is easy to see that this is the same as Math 711 Lecture of October 27 2006 Discussion the dz erenee operator Consider the ring QM of polynomials in one variable 11 over the rational numbers We de ne a Q linear function T from this ring to itself by TP01 P01 7 1 Note that T preserves degree and leading term We write 1 for the identity map on QM and A for the operator 1 7 T that sends P01 gt gt P01 7 P01 7 1 Note that A lowers degree by one if the degree is positive and kills scalars Moreover if the leading term of P01 is and where a E Q the leading term of AltP01 is adnd l which is similar to the behavior of the differentiation operator In particular if P01 has degree d AdP01 is the scalar dla where a is the leading coef cient of For each constant integer c 2 0 TCP01 P01 7 o By the binomial theorem for each h the operator M 7 1 7 739k 7 1 7 CDT 7 71 gt130 71 71kPn 7 k so that l 1 AkP01 P01 7 man 11i We also note Lemma If 0 7 Nb 7 7 Na 7 0 2s a bounded complex of modules of nite length the alternating sum of the lengths Za7li NZ is the some as Za7li HZN Proof Let B be the image of Ni in N and Z the kernel of NZ 7 Ni1 so that H HAN ZiBi Then we have short exact sequences 07gtZi7gtNi7gtBi717gt0 and 07Bi7gtZi7gtHi7gt0 for all t It will be convenient to think of our summations as taken over all integers l E Z this still makes sense since all but finitely many terms are zero and will permit a convenient shift in the summation index We then have Zlt7lgti ltHigt 7 Zlt7lgtiwltzigt 7 n3 7 Zlt7lgti ltzigt Zlt7lgtitllltBigt 7 i z z 271i Zz Z 1i 3i71 Z 1i zi 43271 Z 1i Ni D If R m K is local and 1 ard is a system of parameters then for any finitely generated R module M all the modules Hi0 xd M have finite length each is a finitely generated module killed by 0E1 acdR l 2 Theorem Serre Let R m K be a local ring of Krull dimension d and let M be a nitely generated R module Let I 1 xdR Then d 271Z Hi1acdM 20 is eIM ifdimM d and is 0 ifdimM lt d Proof By the Theorem at the end of the Lecture Notes of October 25 the subcomplex AS Whose ith term is In ilCi Where CZ CZ1 icd M is exact for all it gtgt 0 Call the quotient complex 92 The long exact sequence of homology coming from the short exact sequence o AE aic QEMgt0 shows that H4151 xd M E HiQ for all i if it gtgt 0 Let denote the Hilbert polynomial of M With respect to I Which agrees With MI 1M for all it gtgt 0 Then 271i Hix1796dM 239 is the same as ZeiyaHZQTHB for all n gt 0 and this in turn equals Darwin 239 by the Lemma just above Since 92 is the direct sum of copies of MI 1 iM for n gt 0 this is Which is AdHn by the formula in the Discussion at the beginning of this Lecture By that same Discussion this is also dl times the leading coe icient of Discussion mapping cones Let t A a B be a map of complexes of R modules so that we have a commutative diagram dv dv d A Bi 2 3271 2 1 ml WIT 5i1 523971 539 AZ39 2 Aiil 3 We de ne the mapping cone M to be the complex such that B 63 ill1 where the differential takes bi EB aFl gt gt dibi ili 1 i1ai1 EB 6i1ai1 It is easy to check that M is a complex Note that B is a subcomplex and the quotient is A1 the complex A but with the indices shifted so that the degree i term is 1414 It is straightforward to check that the mapping cone is a complex It is also straightforward to check that IC1 iod M is the mapping cone of the map o IC1 iod1 M a ICac1 xd1 M given by multiplication by god on each module We next observe that if 0AigtBgtOgt0 is a short exact sequence of complexes then the homology of the mapping cone M of A gt B is the same as HO The isomorphism is induced by MfBeA1 a3 a0 For a suitable choice of i iier is a cycle in Mi iff du i v Note that we automatically have 61 0 since 6v d v iddu 0 and o is injective The cycle is completely determined by u and it occurs in a cycle iff its image represents a cycle in 01 The module of boundaries is dBZ1 AZ Q Bi and obviously maps onto the module of boundaries in 01 D Corollary If god is not a zerodivisor on M then H4151 iod M 2 H4151 iod1 MiodM for all i Proof We apply the discussion of mapping cones when the map o is injective with A B IC1 xd1 M and o an The fact that 7 is not a zerodivisor on M implies that the map o is injective Note that 0 ICac1 xd1Mach The stated result is immediate Theorem Let R m K be local of dimension d and let x E m be part of a system of parameters generating a reduction of m Suppose that x is not a zerodivisor on M Then eM eMacM where MacM is viewed as a module over RacR Proof Let 1 iod be a system of parameters generating a reduction I of m where x ion Then the images of 1 iod1 generate a reduction J for macR in RacR Thus eM eIM and eMacM eJMacM and we may compute each of these as an alternating sum of lengths of Koszul homology But the correspondingly indexed Koszul homology modules are isomorphic by the preceding Corollary B Our next goal is to prove that under mild conditions rings of multiplicity l are regular We first need 4 Lemma Hironaka Let R m K be a local domain and let it E R 7 0 be such that it has a unique minimal prime P Suppose that RP is normal and that Rp is a discrete ualuation ring in which it generates the ma1imal ideal PRp Suppose also that the normalization S off is module nite ouer R which is true when R is complete and that every minimal prime Q of acS lies ouer P which is true ifR is universally catenary Then R is normal and P zR Proof Note that if R is universally catenary and Q is any minimal prime of acS in S if P is the contraction of Q to R the height of P must be one by the dimension formula R and S have the same fraction field and RP gt SQ is module finite so that the extension or residue class elds Rp a SQ is algebraic Since P contains ac we must have P P Since Rp is a discrete valuation ring it is normal and so S Q RP Hence Sp Rp is already local of dimension one and SQ is a further localization of dimension one It follows that SQ RP and that QSQ PRp Moreover since QSQ S Q we have that PR1 S Q and so only one prime Q of S lies over P We have that SQ is contained in the fraction field of RP and it is an integral exten sion Since RP is normal we must have that SQ RP and so every residue class in SQ can be represented by an element of B This implies that S Q R We can also see that acS Q we have that acS Q Q and to check Q Q acS it suf ces to show that this is true after localization at each minimal prime of icS since S is normal Q is the only such prime and QSQ PR1 xRp xSQ Since S Q R we now have that S acS B By Nakayama7s lemma this implies that S B so that R is normal Then P Q and Q acS icR D We shall say that a module M over a Noetherian ring R has pure dimension d M may be equal to R if for every associated prime P of M dim RP d An equivalent condition is that every nonzero submodule of M has dimension d Theorem Let R m K by a local ring The R is regular if and only ifR has multiplicity 1 and suppose R has pure dimension Proof If R is regular it is Cohen Macaulay and its multiplicity is the length when w we kill a regular system of parameters which is 1 Moreover R is again regular and so is a domain We therefore only need to show the If 7 part we assume that R has multiplicity 1 and P is of pure dimension and we need to prove that R is regular We use induction on dim lf dim R 0 then eR R 1 so that R must be a field and is regular We may replace R by P without affecting any relevant issue Then c act Z RpkmP P where P runs through all the associated primes of R each of which is minimal and such that dim RP dim R by hypothesis It follows that there is only one associated Math 711 Lecture of October 26 2007 It still remains to prove the final assertion of the Theorem from p 3 of the Lecture Notes of October 22 that if R is F finite and weakly F regular then R is strongly F regular Before doing so we want to note some consequences of the theory of test elements and also of the theory of approximately Gorenstein rings Theorem Let R m K be a local ring of prime characteristic p gt 0 a IfR has a completely stable test element then I is weakly F regular if and only if R is weakly F regular b If R has a completely stable big test element then 13 has the property that every submodule of every module is tightly closed if and only if B does Proof We already know that if a faithfully at extension has the relevant property then B does For the converse it suf ces to check that 0 is tightly closed in every finite length module over 1 respectively in the injective hull E of the residue class field over 13 which is the same as the injective hull of the residue class field over R A finite length module is the same as a finite length R module We can use the completely stable big for part b test element c E R in both tests which are then bound to have the same outcome for each element of the modules For a module M supported only at m 534M 2 fig 22 M g i R f M g f M a Proposition Let B have a test element respectively a big test element c and let N Q M be nitely generated respectively arbitrary R modules Let d E R0 and suppose it E M is such that cuq E qul for in nitely many values of 1 Then it E NITI Proof Suppose that duq E qul and that p81 ql lt 1 so that q qlqg Then duql 2 dquduq e qull q NM and it follows that for all qg du hl m e qud qml Hence duq1 6 qull in 51 whenever ql S 1 Hence if duq E qul for arbitrarily large values of 1 then duq E qul in feM for all g and it follows that cduq E qul for all 1 so that u E NITI D Theorem Let R be a Noetherian ring of prime characteristic p gt 0 a If every ideal ofR is tightly closed then R is weakly F regular b IfR is local and It is a descending sequence of irreducible m primary ideals co nal with the powers of m then R is weakly F regular if and only if It is tightly closed for all t 2 l l 2 Proof a We already know that every ideal is tightly closed if and only if every ideal primary to a maximal ideal is tightly closed and this is not affected by localization at a maximal ideal Therefore we may reduce to the case where R is local The condition that every ideal is tightly closed implies that R is normal and hence approximately Gorenstein Therefore it suf ces to prove For b we already know that R is weakly F regular if and only if 0 is tightly closed in every finitely generated R module that is an essential extension of K Such a module is killed by It for some t gt 0 and so embeds in ERIt E RIt for some t Since It is tightly closed in R 0 is tightly closed in RIt and the result follows D We next want to establish a result that will enable us to prove the final assertion of the Theorem from p 3 of the Lecture Notes of October 22 Theorem Let R m K be a complete local ring of prime characteristic p gt 0 IfR is reduced and c E R0 let 6 R a Rlq denote the R linear map such that 1 gt gt clq Then the following conditions are equivalent 1 Every submodule of every module is tightly closed 2 0 is tightly closed in the injective hull E ERK of the residue class eld K Rm of R 3 R is reduced and for every c E B there exists 1 such that the 6176 splits 4 R is reduced and for some c that has a power which is a big test element for B there exists 1 such that 6176 splits 5 R is reduced and for some c such that RC is regular there exists 1 such that 6176 splits Proof Note that all of the conditions imply that R is reduced We already know that conditions 1 and 2 are equivalent Let u denote a socle generator in E Then we have an injection K a E that sends 1 gt gt u and we know that 0 is tightly closed in E if and only if u is in the tight closure of 0 in E This is the case if and only if for some c E B0 respectively for a single big test element c E R0 cuq 0 in feE for all 1 gtgt 0 We may view 75 R a R as R Q Rlq instead Then feE is identified with Rlq R E and R acts via the isomorphism R E Rlq such that r gt gt rlq Then uq corresponds to 1 8 u and cuq corresponds to clq 8 u Then it 6 0 if and only if for every c E B0 respectively for a single big test element c E R0 the map K a Rlq RE that sends 1 gt gt clq u is 0 for all 1 gtgt 0 We may now apply the functor HomRi E to obtain a dual condition Namely u 6 0 if and only if for every c E R0 respectively for a single big test element c E R0 the map HomRR1q R E E a HomRK E is 0 for all 1 gtgt 0 The map is induced by composition with K a Rlq R E By the adjointness of tensor and Hom we may identify this map with HomRR1q HomRE E a HomRK E 3 This map sends f to the composition of K a Rlq R E with the map such that 5 1 gt gt fsv Since HomRE E E R by Matlis duality and HomRK E E K we obtain the map HomRR1q R a K that sends f to the image of fclq in Rm Thus u 6 0 if and only if for every 0 E R0 respectively for a single big test element 0 E R0 every f Rlq a R sends 01 7 into m for every 1 gtgt 0 This is equivalent to the statement that quc R a Rlq sending l a 01 7 does not split for every 1 gtgt 0 since if fclq a is a unit of R a lf is a splitting Note that if R a Rlq sending 1 gt gt clq splits then R a Rlq splits as well the argument in the Lecture Notes from September 21 see pages 4 and 5 applies without any modification whatsoever Moreover the second Proposition on p 5 of those notes shows that if one has the splitting for a given 1 one also has it for every larger 1 We have now shown that u 6 0 if and only if for every 0 E R0 respectively for a single big test element 0 E R quc R a Rlq sending l a 01 7 does not split for every 1 Hence 0 is tightly closed in E if and only if for every 0 E R0 respectively for a single big test element 0 E R0 the map 6 splits for some 1 We have now shown that conditions 1 2 and 3 are equivalent and that 4 is equivalent as well provided that c is a big test element Now suppose that we only know that c has a power that is a big test element Then this is also true for any larger power and so we can choose ql p51 such that 0 11 is a test element If the equivalent conditions 1 2 and 3 hold then we also know that the map R a Rlqql sending l gt gt 0 qu clq splits for all 1 gtgt 0 and we may restrict this splitting to Rlq Thus 1 through 4 are equivalent Finally 5 is equivalent as well because we know that if c E R0 is such that RC is regular then c has a power that is a big test element D Remark It is not really necessary to assume that R is reduced in the last three conditons We can work with Re instead of Rlq where Re denotes R viewed as an R algebra via the structural homomorphism f5 We may then define quc to be the R linear map R a Re such that l gt gt e The fact that this map is split for some some 0 E R0 and some 1 implies that R is reduced if 7 is a nonzero nilpotent we can replace it by a power which is nonzero but whose square is 0 But then the image of 7 is We 0 and the map is not even injective a contradiction Once we know that R is reduced we can identify Re with Rlq and c is identified with clq We want to apply the preceding Theorem to the F nite case We first observe Lemma Let R m K be an F mte reduce local ring Then lq 2 g 2 E R Rlq for all 1 p5 Math 711 Lecture of November 30 2005 In considering the vanishing conjecture for maps of Tor we note the following easy reductions in the problem 1 As already mentioned we may assume that M is finitely generated as an A module 2 If there is a nonzero element u in the image of Torf M R 7 Tor M S we can choose a prime M of S in the support of Sn Then the image of u in ToriAM SM 2 Torf M SM is nonzero and we may replace S by SM Henceforth we assume that S is local 3 We may the replace S by S which is faithfully at over S Note that Torf M g TorAM S S s z 4 Let m be the contraction of M to A Then we may replace A by Am M by Mm and R by Rm Note that because the elements of A 7 m are invertible when mapped into S TorAM S Tor Mm S Henceforth we assume that A is local 2 5 Instead of using A M R and S we may use A M E A A M R1 A A R and S noting that we have maps A 7 R1 7 S because S is complete Observe that the fact that R is torsion free over A is equivalent to the fact that R can be emebedded in a finitely generated free A module and it is then clear that R1 can be embedded in a finitely generated free A module Henceforth we assume that A and T are complete local rings 6 Since A is normal and R is torsion free and module finite over A the going down theorem holds for A 7 B Let Q be the kernel of the map R 7 S and let 73 be the contraction of Q to A Then there is a prime Q0 of R contained in Q and lying over 0 in A and we have maps A 7 R 7 RQO 7 S It follows that we may replace R by RQO which is a domain module finite over A Thus in considering the vanishing conjecture for maps of Tor it suf ces to consider the case where A R and S are complete local domains and R is a module finite extension of R while A and T are regular We next want to indicate two proofs of the vanishing conjecture in characteristic 1 gt 0 Proof of the vanishing conjecture for maps of Tor using tight closure theory Since the localization of R at A 7 0 is a finite dimensional vector space over A we can find a finitely generated A free module W Q R and an element 0 E A 7 0 such that CR Q W Now let G be a free resolution of M by finitely generated free modules let i 2 1 be an integer and let 2 be a cycle in GZ A B so that 2 represents an element of Torf M R We shall show that z is in the tight closure over R of the module of boundaries B within GZ A B First note that F G A R g FjG A R In one case we first map the entries of the matrices to R and then raise them to the p5 th power while in the other we raise them to the p5 th power and then map them to B This can also be said in a functorial way Let PM respectively A denote B respectively A as an algebra over itself 1 2 using the endomorphism Fe Then the diagram AER l l AER commutes both vertical maps are induced by F5 and the statement that we need is simply that PM A A A i may be naturally identified with PM R R A 7 By the associativity of tensor both are PM A 7 But FjG A R has the subcomplex FjG A W which is acyclic since F is exact over a regular ring FjG is acyclic and W is A free Now czq is a cycle and is in FjG 8 W since cR Q W and so czq is in Blpgl in F Gi 8 B This shows that z E Bquot But then when we apply iRS we find that the image of 2 becomes a boundary it is in the tight closure of the boundaries and over the regular ring S every finitely generated submodule of every module is tightly closed D We can replace the condition that S be regular by the condition that each of its com pleted local rings is weakly F regular In the locally excellent case this is simply the condition that S be F regular There are several other ways to generalize but we do not want to get too technical here Before giving the second proof we need a preliminary result Proposition Let S and A be rings a If S n L is regular local an S module O is faithfully flat if and only if C is a big Cohen Macaulay module for S If S n L is an Artin local ring an S module O is flat if and only ifO is free c IfO is faithfully flat over the local ring S n L with u E 07110 and we map S a C so that l gt gt u then the map S a O is pure ie it is injectiue and remains so when we tensor over S with any S module N d If 00 a O is a pure map ofA modules andB Q G are modules then the intersection of the image ofB A O in G A O with the image ofG A Co is the image ofB A Co in G A 00 e If 00 a O is a pure map of A modules and G is any complea of A modules then for alli the map HiG 00 a HiG 0 is injectiue f IfA a S is a ring homomorphism M is an A module and 00 a O is pure over S then Torf M CO a Torf M O is injectiue for all i Proof For part a only it 7 is clear For if it will su ice to show that for every prime P of S and for all i 2 l Tor SP O 0 This is clearly true for all i gt 0 since S is regular local Assume that the result is true for all integers gt i where i 2 1 We prove it for i Let 1 xh be a maximal S sequence in P Then P is an associated prime of 1 xhR and we have a short exact sequence 0 a SP a Sx1xhS a Q a 0 for some Q The long exact sequence for Tor yields the exactness of gt Tor 1Q O a Tor SQ O a Tor Sac1 ichS O a Q has a ltration by prime cyclic modules and so the leftmost term vanishes The rightmost term vanishes because 1 xh is a regular sequence on G Thus the middle term vanishes as well as required For part b note that we are not assuming that G is finitely generated first observe that we have the analogue of Nakayama7s lemma for any module D without finiteness hypothesis For if D nD then D MD for all t and for t gt 0 nt 0 Choose a vector space basis for 0710 and lift it to a set of elements for in uj E G Choose a free S module G with a free basis bj in bijective correspondence with the a and map G a G sending bj gt gt uj for every j Then S Glm 0 and so lm G G and we have a short exact sequence 0 a N a G a G a 0 for some N Since TorgL G 0 the sequence remains exact when we apply L 83 7 Since L lt8 G a L 8 G is an isomorphism by our construction of G we have that L 83 N 0 and so N 0 For part c it suf ces to show that the map N a N SG is injective for every finitely generated module N If u E N maps to 0 we can choose t gt 0 such that u Z NntN Therefore we obtain a counterexample with N killed by nt It therefore suf ces to prove that the map is pure after we apply SntS S i we can then tensor with N over SmtS The hypothesis of atness and that u 110 are preserved by the base change Thus we have reduced to the case where S is an Artin local ring But then G is free and the element a is part of a free basis so that S gt gt G splits For part d note that we have G A GO Q G A G Moreover GB A GO injects into GB 8 G which gives exactly what we need For part e let B be the image of Gi1 in Gi G lf 2 E G A Go is a cycle that maps to 0 in HAG A G then in G A G it is a boundary and hence in the image of B A G intersected with the image of G A GO But then by part d it is in the image of B A GO and so represents 0 in HAG A GO For part f we observe that if GO a G is pure over S then it is pure over A For if M is any A module M A 7 is the same as M A S 83 7 Thus we may assume that GO a G is pure over A The result is now immediate from part a applied to G a projective resolution of M over A D Remark Here is another approach to the proof of part e The result is obvious when GO is a direct summand of G as an A module But we can reduce to this case We can write G as Go 69 P Q where P is a free module and Q is a suitable submodule meeting GO in 0 Consider all maps GO a 01 where 01 has the form GO EDP1Q1 where P1 is the finitely generated submodule of P generated by a finite subset of the free basis and Q1 is some finitely generated submodule of Q Go 69 P1 Then GO a G is the direct limit of the maps GO a 01 as 01 varies Each GO a 01 is evidently pure and each cokernel 0100 is finitely presented by construction But then GO is a direct summand of Cl and the injectivity of the map of complexes is clear The stated result now follows by taking a direct limit D Proof of the vanishing conjecture for maps of Tor using the weakly functorial ezristence of big Cohen Macaulay algebras In fact while the arguments in the literature use algebras we shall carry the argument through using big Cohen Macaulay modules instead We need the following statement given a local map of complete local domains R m K a S n L of characteristic 1 there exist big Cohen Macaulay modules B over R and C over S and an R linear map B a C such that the image of B is not contained in 110 In this circumstance we can pick it E O 7 110 such that u is the image of an element y E B This yields a commutative diagram BEG l l RES where the vertical arrow on the left sends 1 gt gt y and the vertical arrow on the right sends l gt gt u Note that if we have weakly functorial big Cohen Macaulay algebras we may use these for B and O in the diagram with y l in B and u l in C We then get a commutative diagram Tor M B gt Tor M O Tor M R gt Tor M S Now B is a big Cohen Macaulay algebra over R and every system of parameters for A is a system of parameters for B so that B is a big Cohen Macaulay module over A Since A is regular part a of the Proposition shows that B is A flat Therefore the Tor module in the upper left corner of the diagram is 0 and so the composite maps from Tor M R a Torf M O are 0 The map S a O is pure by part c of the Proposition and so the vertical arrow on the right is injective by part It follows that the arrow on the bottom is 0 D Remark An alternative proof of the injectivity of the map Torf M S a Tor M O can be based on the fact that every at module is a direct limit of finitely generated free modules Suppose we think of O limj Oj where the Oj are free Then some Oj contains an element 1 that maps to u in C We call the index jg and work only with j 2 jo We then have a map S a Oj for all j 2 jo In each case the image of l is not in 110 or else it would be in 110 But then the image of l is part of a free basis for C so that S a Oj is split over S hence over A as well and so the induced map of Tor modules is injective D The vanishing conjecture for maps of Tor is a statement that can be formulated in terms of equations with a constraint that certain elements be parameters and so the equal characteristic 0 case follows from the positive prime characteristic 1 case Thus the vanishing conjecture for maps of Tor is a theorem in the equal characteristic case We want to show next that it is a very powerful tool We therefore explain how it can be used to prove several rather subtle results In particular we show how it can be used to prove l The direct summand conjecture 2 The conjecture that direct summands of regular rings are Cohen Macaulay Both of these conjectures remain open in mixed characteristic To see how the direct summand conjecture follows we note that is su ices to prove that if 1 ard is a system of parameters of a complete regular local ring A and R is a module finite extension domain of A then we do not have a relation d Emmi 71dt 0 j1 in R Consider the A module AI where I 5M mg 1 xdtA We take S Rm the residue class field of R Then the map TorfAI R a Tor AI K is supposed to be injective For any A algebra B TorfAf1fkAzB consists of the relations of f1 fk with coe icients in B modulo the relations spanned by the relations on f1 fk over A By hypothesis we have a relation r1 rd 71 on 1x1x1xdt whose image 0 0 71 in Tor RI K is nonzero since all relations on xi111acdt over A have coe icients in the maximal ideal of A This contradiction proves the direct summand conjecture To see how one may deduce that direct summands of regular rings are Cohen Macaulay suppose that R is a direct summand of S where S is regular lt su ices to show that each local ring of R is regular and by tensoring R a S with a suitable local ring of R we may assume that R m K is local Now replace R by 13 and S by its completion in the Math 711 Lecture of November 7 2007 Our current theory of test elements permits the extension of many results proved under other hypotheses such as the condition that the ring under consideration be a homomor phic image of a Cohen Macaulay ring to the case of excellent local rings or more generally rings for which we have completely stable test elements or completely stable big test ele ments depending on whether the result being proved is for finitely generated modules or for arbitrary modules Here is one example Theorem coloncapturing Let R m K be an ezrcellent reduced equidimensional local ring of prime characteristic p gt 0 and let 1 k1 be part of a system of parameters Let k 1 xkR Then 1 2R k1 1 In particular 1 R n1 Q 1 Proof Note that R has a completely stable test element c Suppose that 71ka E 1 but u E R 7 Ik This is also true when we pass to E which is a homomorphic image of a regular ring and hence of a Cohen Macaulay ring Therefore from the result on colon capturing from p 9 of the Lecture Notes from October 5 we have that u E Aggy whence cuq E I ql Ilqllst for all 1 and it follows that cuq E Ilqllst R M for all 1 Thus a E 1 Now suppose that 71ka E 1 Then uqZ1 E 1ql Q UPquot for all 1 and so CuqZ1 6 E for all 1 and cuq 6 E R avg Q 1515 by the result of the first paragraph applied to 36 xZH Hence c2uq E Ik ql for all 1 so u E 1 The opposite conclusion is obvious D A Noetherian ring is called locally ezrcellent if its localization at every maximal ideal equivalently at every prime ideal is excellent Corollary IfR is weakly F regular and locally ezrcellent then R is Cohen Macaulay Proof Both weak F regularity and the Cohen Macaulay property are local on the maximal ideals of R Hence we may assume that R is local Since weakly F regular rings are normal R is certainly equidimensional Since colon capturing holds for systems of parameters in R the result is immediate D We can also prove a global version of this Theorem above that is valid even in case the ring is not equidimensional We need one additional fact 1 2 Lemma Let R m K be an e1cellent local ring and let I be an ideal ofR that has height at least k modulo euery minimal prime of R Then IR has height at least k modulo euery minimal prime of R Proof If pi is a minimal prime of R pig is a radical ideal and is the intersection of certain minimal primes qZj The intersection of all qZj is the same as the intersection of the pig Since finite intersection commutes with at base change this is 0 Thus it will su ice to show that the height of I is at least k modulo every qZj To this end we can replace R by RPz Thus it is enough to show the result when R is an excellent local domain In this case E is reduced and equidimensional Any prime Q of E containing IE lies over a prime P of R containing I The height of Q is at least the height of Q0 where Q0 is a minimal prime of PIS Hence it su ices to show that if Q is a minimal prime of PIS then height Q height Since R and IS are equidimensional and catenary heightP dim R i dim RP dim 172 i dim PtPE Since the completion of RP is equidimensional dim PtPE dim PtQ Hence height P dim if i dim PtQ height Q as required D Theorem coloncapturing Let R be a reduced Noetherian ring ofprime characteristic p gt 0 that is locally ezrcellent and has a completely stable test element c This holds for ezrample ifR is reduced and essentially of nite type over an ezrcellent semilocal ring Let 1ack1 be elements of B Let It denote the ideal 1 actR 0 S t S k 1 Suppose that the image of the ideal Ik has height h modulo euery minimal prime of R and that the image of the ideal Ik1R has height kl modulo euery minimal prime of R Then 213 k1 Proof We first prove that Ik 2R ack1 Q I The stronger conclusion then follows exactly as in the Theorem above because c is a test element lf ick1u E Ik but u Z I we can choose 1 so that cuq Z IE This is preserved when we localize at a maximal ideal in the support of I cuqI ql We have therefore reduced to the case of an excellent local ring Rm By the Lemma above the hypotheses are preserved after completion and we still have cuq Z IquS IkSql where S is the completion of Rm Since c is a test element in S we have that u Z IkS Since S is a homomorphic image of a Cohen Macaulay ring this contradicts the Theorem on colon capturing from p 9 of the Lecture Notes from October 5 D We next want to use the theory of test elements to prove results on persistence of tight closure Persistence Let R respectively Rbig denote the class of Noetherian rings S such that for every domain R SP the normalization PM of R is module finite over R and has the following two properties 1 The singular locus in PM is closed 2 For every element 0 6 PM 7 0 such that R is regular 0 has a power that is a test element respectively a big test element in PM Of course Rbig Q R and Rbig includes both the class of F finite rings and the class of rings essentially of finite type over an excellent semilocal ring Theorem persistence of tight closure Let R be in R respectivelyih Rbig Let R a S be a homomorphism of Noetheriah rings and suppose that N Q M are nitely generated respectively arbitrary R modules Let u E NITI Then l u E S R NgtZ RM Proof lt suf ces to prove the result after passing to Sqj as qj runs through the minimal primes of S Therefore we may assume that S is a domain Let P denote the kernel of R a S Then P contains a minimal prime p of R Tight closure persists when we kill p because the element 0 used in the tight closure test is not in p Hence we may make a base change to Rp and so we may assume that R a S is a map of domains with kernel P It suf ces to prove that tight closure is preserved when we pass from R to RP since the injective map of domains RP gt S always preserves tight closure Henceforth we may assume that S has the form RP Choose a saturated chain of primes Then it suf ces to show that tight closure persists as we make successive base changes to RPl then to RPg and so forth until we reach RPh RP Therefore we need only prove the result when S RP and P has height one Let PM be the normalization of R and let Q be a prime of PM lying over P We have a commutative diagram RP PJQ l l RE where the horizontal arrows are module finite extensions and the vertical arrows are quo tient maps Tight closure is preserved by the base change from R to PM because it is an 4 inclusion of domains lf PM is regular then the element is in the image of the submodule and this is preserved when we pass to PUQ If not because PM is normal the de ning ideal of the singular locus has depth at least two and we can find a regular sequence b c in PM such that R and R are both regular Then b and c are not both in Q and hence at least one of them has nonzero image in RQ say that c has nonzero image For some s c5 is a test element respectively big test element in R and it has nonzero image in PUQ It follows that the base change PU a PUQ preserves tight closure and hence so does the composite base change R a PUQ Now suppose that the base change R a RP fails to preserve tight closure By the argument above the further base change RP gt PUQ restores the image of the element to the tight closure This contradicts the fourth problem in Problem Set 4 Hence R a RP preserves tight closure D Corolllary Let R a S be a homomorphism of Noetherian rings such that S has a com pletely stable respectively completely stable big test element c and suppose that N Q M are nitely generated respectively arbitrary R modules Let u E NITI Then l u E lt5 R Ngt Z RM39 Proof Suppose that we have a counterexample Then for some 1 01 uq Z ltS R Ngtq This continues to be the case after localization at a suitable maximal Q ideal of S and after completing the local ring SQ Hence we obtain a counterexample such that S n L is a complete local ring Let m be the contraction of n to R and let let R1 be the completion of Rm Then the initial instance of tight closure is preserved by the base change from R to R1 but not by the base change R1 a S This is a contradiction since R1 is in Rbig D Although the theory of test elements that we have developed thus far is reasonably satisfactory for theoretical purposes it is useful to have theorems that assert that specific elements of the ring are test elements not only that some unknown power is a test element For example the following result is very useful Theorem Let R be a geometrically reduced equidmensional algebra nitely generated over a eld K of prime characteristic p gt 0 Then the elements of the Jacobian ideal jRK that are in R0 are completely stable big test elements for R It will be a while before we can prove this We shall discuss the definition and properties of the Jacobian ideal in detail later For the moment we make only two comments First the Jacobian ideal defines the geometrically regular locus in R Second if R Kz1 xnlf is a hypersurface then jRK is simply the ideal generated by the images of the partial derivatives BfBzi in R For example suppose that K has characteristic p gt 0 with p 31 3 The Theorem above tells us that if R Kb y zlz3y323 then the elements 2 242 and 22 are completely stable big test elements 3 is iinvertible in This is not the best possible result the test ideal turns out to be all of m z y But it gives a good starting point for computing 7R The situation is essentially the same in the local case where we study Rm instead In this case once we know that 2 242 are test elements we can calculate the test ideal as 2 242 R 2y2 since this ring is Gorenstein and we may apply problem 4 of Problem Set 3 The socle generator modulo I 2 242 turns out to be 2122 and the problem of showing that the test ideal is mRm reduces to showing that the ideal 2 y2xyz2 is tightly closed in Rm The main point here is that the Theorem above can sometimes be used to make the calculation of the test ideal completely down to earth The proof of the Theorem above involves several ingredients One is the Lipman Sathaye Jacobian Theorem which we will state but not prove The Theorem is proved in J Lipman and A Sathaye Jacobian ideals and a theorem of Briangon Skoda Michigan Math J 28 1981 1997222 Moreover a complete treatment of the Lipman Sathaye argument is given in the Lecture Notes from Math 711 Fall 2006 See specifically the Lectures of September 25 27 and 29 as well as the Lectures of October 2 4 6 9 11 and 13 Another ingredient is the Theorem stated just below We shall say that an extension of a domain S of a field IC is etale if S is a finite product of finite separable algebraic extension fields of IC We shall say that an extension A a R of a domain A is generically tale if the generic fiber is e tale ie frac A A R is a finite product of finite separable algebraic extension fields of frac Theorem Let R be a module nite and generically tale ertension of a regular ring A of prime characteristic p gt 0 Let r E R0 be such that cR x Q RA Then 0 is a completely stable big test element for R We shall see that elements 0 as above exist and that the Lipman Sathaye Theorem can be used to find specific elements like this The reason that it is very helpful that cR x Q RA is that it turns out that RA E R A A is faithfully at over R since A is faithfully at over A This makes it far easier to work with RA than it is to work with R and multiplication by c can be used to correct the error in replacing R by RA We begin by proving the following preliminary result Lemma Let A mA K be a normal local domain and let R be a module nite ezrtension domain ofA that is generically tale over A Let ord be a Z ualued ualuation on A that is nonnegatiue on A and positive on mA Then ord ezrtends uniquely to A by letting orda1q lqord a for all a E A 7 The ezrtended ualuation tahes values in lePl39 Let m be a proper ideal of B Let d be the torsion free ranh off as an A module Let u E mRA A Then ord 2 ldl Proof We know that frac R is separable of degree d over frac A IC Let 6 be a primitive element The splitting field of the minimal polynomial f of 6 is generated by the roots of f and is Galois over IC with a Galois group that is a subgroup of the permutations on d the roots of f We may replace R by the possibly larger ring which is Math 711 Lecture of November 4 2005 We extend the de nition of local cohornolgy Let I be an ideal of a Noetherian ring R and let M be any R rnodule not necessarily nitely generated We de ne H M lirnt ExthI i M This is called the ith local cohomology module ofM with support in I HM lirnt HornRRIt M which may be identified with Ut AnnMIt Q M Every element of is killed by a power of I Evidently if M is injective then 0 for j 2 1 By a taking a direct lirnit over t of long exact sequences for Ext we see that if 0 7 M 7 M 7 M 7 0 is exact there is a functorial long exact sequence for local cohornology o 7 HM 7 HM 7 HM 7 7 HM 7 HM 7 H M 7 It follows that is the j th right derived functor of Hi In the definition we may use instead of the ideals It any decreasing sequence of ideals cofinal with the powers of I It follows that if I and J have the same radical then E for all i Theorem IfM is a nitely generated R module over the Noetherian ring R then 31 0 for some i if and only ifIM 31 M in which case the least integer I such that 31 0 is depthIM Proof IM M iff I AnnRM R and every element of every is killed by some power IN of I and by AnnRM their sum must be the unit ideal and so all the local cohornology vanishes in this case Now suppose that IM 31 M so that the depth d is a well defined integer in N We use induction on d If d 0 some nonzero element of M is killed by I and so 31 0 If d gt 0 choose an element at E I that is not a zerodivisor on M and consider the long exact sequence for local cohornology arising from the short exact sequence 07MLM7M M70 From the induction hypothesis 0 for j lt d 7 l and H Ii 1MacM 31 0 The long exact sequence therefore yields the injectivity of the map Hf1M L HEVM for j lt d 7 1 But every element of Hg1M is killed by a power of I and in particular by a power of x This implies that Hj1M 0 for j lt d 7 1 Since Hf M 7 Hf lgMxM 7 H M 2 is exact Hf RM which we know is not 0 injects into D Now let I f1 fn generate an ideal with the same radical as I Let C R denote the total tensor product of the complexes 0 a R a Rfj a 0 which gives a complex of at R modules 0gtRgt Rfj a a Rfjlfh a Rflfn a0 1 j1ltj2 Let C M C R ER M which looks like this 0gtMgt ij a 6 WI a m Mfrnfn a 0 j jlltj2 We temporarily denote the cohomology of this complex as It turns out to be the same functorially as H M We shall not give a complete argument here but we note several key points First HSMM Ker M a j ij is the same as the submodule of M consisting of all elements killed by a power of fj for every j and this is easily seen to be the same as Second by tensoring a short exact sequence of modules 0 a M a M a M a 0 with the complex C39q x R we get a short exact sequence of complexes This leads to a functorial long exact sequence for H347 These two facts imply an isomorphism of the functors and 7147 provided that we can show that 0 for j 2 1 when M is injective We indicate how the argument goes but we shall assume some basic facts about the structure of injective modules over Noetherian rings First note that if one has a map R a S and an S module M then ifg is the image of i in S we have H Hg This has an important consequence for local cohomology once we establish that theTtwo theories are the same see the Corollary below Every injective module over a Noetherian ring R is a direct sum of injective hulls ERP for various primes P ERP is the same as the injective hull of the reisdue class field of the local ring RP This we may assume without loss of generality that R m K is local and that M is the injective hull of K This enables to reduce to the case where M has finite length over R and then using the long exact sequence to the case where M K since M has a finite filtration such that all the factors are K Thus we may assume that M K The complex C f R is then a tensor product of complexs of the the form 0 a R a R a 0 and 0 aTR a 0 a 0 If we have only the latter the complex has no terms in higher degree while if there are some of the former we get a cohomogical Koszul complex Kglz ng where at least one gj 31 0 But then 91 gnK K kills all the Koszul cohomology Thus we get vanishing of higher cohomology in either case It follows that 7147 and are isomorphic functors and we drop the first notation except in the pToof of the Corollary just below 3 Corollary IfR a S is a homomorphism of Noetherian rings M is an S module and RM denotes M viewed as an R module via restriction of scalars then for every ideal I of PM HHRM HISM39 Proof Let f1 fn generate I and let gl 9 be the images of these elements in S they generated IS We have HfRM E HRM E HM E D We note that the complex 0 a R a Rf a 0 is isomorphic to the direct limit of the cohomological Koszul complexes IC39 ft R where the maps between consecutive complexes are given by the identity on the degree 0 copy of R and by multiplication by f on the degree 1 copy of R 7 note the commutativity of the diagram ft1 OER R gt0 in if ft OEREREO Tensoring these Koszul complexes together as f runs through f1 f we see that C f M hintCW a 5 M Hence whenever f1 fn generate I up to radicals taking cohomology yields HzltMgt eggtwf t M When R is a Cohen Macaulay ring of Krull dimension d and 1 ird is a system of parameters this yields H7103 hint Rx i R which was our definition of HR in this case We next recall that when R m K is a complete local ring and E ERK is an injective hull of the residue class field this means that K Q E and every nonzero submodule of E meets K there is duality between modules with ACC over R and modules with DCC if M satsi es one of the chain conditions then MV HomRM E satisfies the other and the canonical map M a MW is an isomorphism in either case In particular when R is complete local the obvious map R a HomREE is an isomorphism An Artin local ring R with a one dimensional socle is injective as a module over itself and in this case ERK R If R is Gorenstein and 1 ird is a system of parameters one has that each R Rx i R is Artin with a one dimensional socle and one can show that in this case ERK g HJ lIR Knowing this we can prove a local duality theorem for local cohomology when R is Gorenstein Theorem Let R m K be a Gorenstein local ring of Krull dimension d and let E H5203 which is also an injective hall for K Let M be a nitely generated R module Then for every integer j Extfe j M RV Proof Let 1 ird be a system of parameters for R In the Cohen Macaulay case the local cohomology of R vanishes for i lt d and so C R numbered backwards is a at 4 resolution of E Thus E Torfij M Let G be a projective resoultion of M by nitely generated projective R modules Then Extfe j M RV Hd j HomRG R E since E is injective HomRi E commutes with the calculation of cohomology The functor HomR HomRi R E is isomorphic with the functor 7 8 E when restricted to finitely generated projective modules G To see this observe that for every G there is an R bilinear map G gtlt E a HomR HomRG R E that sends g u where g E G and u E E to the map whose value on f G a R is fgu This map is an isomorphism when G R and commutes with direct sum so that it is also an isomorphism when G is finitely generated and free and likewise when G is a direct summand of a finitely generated free module But then Extfgj M RV E HdjG E E Torfij M E which is E as already observed D Note that if R is a finitely generated N graded K algebra of Krull dimension d such that R0 K a field and R Klej then with X ProjR we have that HZKXa 0x le1Rlo for i 2 1 This follows from the fact that if f1 fd is a homogeneous system of parameters for R then H nR0 is the cohomology of C39 X R0 and this complex with the first term dropped and degrees decreased by one is the same as the Cech complex for the sheaf OX with respect to the af ne open cover of X consisting of the ij Therefore the fact that the a invariant of X is negative implies the vanishing of Hd 1X OX If X and Y are Noetherian schemes over a field K and f g are coherent sheaves on X and Y respectively then one has the Kunneth formula HkltX x Y MK 9 2 69 HiltX f K HW g ijk This permits a geometric explanation of why Segre products with R tend not to be Cohen Macaulay in any instance where Hd 1X OX 31 0 For simplicity we only consider the case where R KR1 while R0 K Then R KK5 t has dimension d 1 Let Z ProjR K Ks t E X gtltK P If B were Cohen Macaulay then we would have H XAR Kst tj 0 where M is the homogeneous maximal ideal of R K Ks t and so Hd 1Z OZ 0 Then OZ OX K 9y and so by the Kunneth formula Hd 1Z OZ is a direct sum of terms one of which is Hd 1X OX K HOOP 911 2 Hd 1X OX K K E Hd 1X OX Thus R KK8 t cannot be Cohen Macaulay if Hd 1X OX 7 0 We next want to prove Reisner7s result stated earlier concerning when face rings K A are Cohen Macaulay We need some preliminaries We first show that the result can be reduced to the case where the field K has charac teristic p gt 0 Lemma Let R be a nitely generated N graded algebra over R0 Z Then Q 8 R is Cohen Macaulay if and only if ZpZ 8 R is Cohen Macaulay for all but nitely many positive prime integers p Proof By clearing denominators we may choose homogeneous elements f1 fd of positive degree such that the images of f1 fd form a homogeneous system of pa rameters for Q 8 B After inverting one nonzero integer a we may assume that Ba and the Koszul homology Hf1 fd Ba 2 Hf1 fd Ra consists of torsion free Za modules since the homogeneous components are finitely generated over Za all of these are Za free For any prime p not dividing a ZapZa g ZpZ Since the Koszul complex Cfl fd Ba and its homology consists of Za free modules the calcula tion of homology commutes with tensoring with any Za module Thus f1 fd is a regular sequence in Q 8 R if and only if its image in ZpZ 8 R is is a regular sequence for all p not dividing a The fact that f1 fd is a homogeneous system of parameters implies that Q 8 H0f1 fd R is a finite dimensional vector space over Q This implies that H0f1 fd Ra is a finite rank free Za module and then ZpZ ZH0f1fdRa H0f1 fd RpR will be finite dimensional over ZpZ for all p not dividing a D We need one more preliminary result Lemma Let R be a nitely generated N graded algebra with R0 K a eld and suppose that there are elements icj in the homogeneous maximal ideal m of R generating an ideal primary to m such that each RM is Cohen Macaulay Suppose also that R is of pure dimension d The for allj lt d has nite length Proof Let T be a polynomial ring in n variables such that R TI where I is ho mogeneous and let M be the homogeneous maximal of T Then all minimal primes of I have height n 7 d Moreover g H3Rm g H ARm is dual working over TM to Extfjij Rm TQ It therefore suf ces to prove that W ExtjRm TM has finite length Let Q strictly contained in M be any prime ideal containing I lo calizing at a prime not containing I kills R and therefore certainly kills W and let P QI Then WQ ExtjRp T9 and if height Q h this is dual over T9 to HgnijRp HIEECWRP 0 since Rp is Cohen Macaulay its only nonvanishing local cohomology module with support in its maxmal ideal occurs when the exponent is dimRpk7n7dhdin and hd7ngthj7nsincej ltd We are now ready to prove Reisner7s theorem Note that since we are working over a field it does not matter whether we use homology or cohomology in the statement of the result they are dual over K Theorem Reisner s criterion Let K be a eld and let A be a nite simplicial complea ouer K with uertices 1 xn Then KA is Cohen Macaulay if and only if for every link A of A the reduced simplicial cohomology with coe icients in K II AK vanishes for 0 S i lt dimA Here A itself is to be included among the choices for A as the link of the empty simplezr Proof Only the characteristic of the field matters in determining whether K A is Cohen Macaulay By applying the preceding Lemma to ZA we may assume that the field has characteristic p The Cohen Macaulay hypothesis implies that K A has pure dimension The same is true of Reisner7s criterion this comes down to verifying that A has pure Math 711 Lecture of September 18 2006 We have already noted that when R m K is a local ring and i Q m an ideal we may identify K R ngR 2 Rm Im1 12m12 I m ea S is called a standard graded A algebra if S is N graded with S0 A and the l forms S1 of S generate S as an A algebra If S is a standard graded K algebra where K is a field then R has a unique homogeneous maximal ideal m 6921 Sn the K span and even the span as an abelian group of all elements of positive degree We note as well that if RIt Q Rt is the Rees ring then RIt12t2lntnm RI RIthIt IRIt R il i ll IIgt13t2u1n1tnH 7 and it is quite straightforward to identify this with ng R Since R RI g RI it follows that K R ngR Rm R WM 912 Rlltl Rm R RM R Rift 2 K m Rift so that we may also view K R grR as K R RIt We give two preliminary results Recall that in Nakayama7s Lemma one has a nitely generated module M over a ring R m with a unique maximal ideal ie a quasilocal ring The lemma states that if M mM then M 0 By applying the result to MN one can conclude that if M is finitely generated or finitely generated over N and M N mM then M N In particular elements of M whose images generate MmM generate M if N is the module they generate we have M N mM Less familiar is the homogeneous form of the Lemma it does not need M to be finitely generated although there can be only finitely many negative graded components the detailed statement is given below First recall that if H is an additive semigroup with 0 and R is an H graded ring we also have the notion of an H graded R module M M has a direct sum decomposition M e M hEH as an abelian group such that for all hk E H Rth Q Mhk Thus every Mh is an RO module A submodule N of M is called graded or homogeneous if N a N m Mk hEH An equivalent statement is that the homogeneous components in M of every element of N are in N and another is that N is generated by forms of M 1 Note that if we have a subsemigroup H Q H then any H graded ring or module can be viewed as an H graded ring or module by letting the components corresponding to elements of H 7 H be zero In particular an N graded ring is also Z graded and it makes sense to consider a Z graded module over an N graded ring Nakayama s Lemma homogeneous form Let R be an N graded ring and let M be any Z graded module such that M 0 for all su iciently large n ie M has only nitely many nonzero negative components Let I be the ideal ofR generated by elements of positive degree IfM IM then M 0 Hence ifN is a graded submodule such that M N IM then N M and a homogeneous set of generators for MIM generates M Proof If M IM and u E M is nonzero homogeneous of smallest degree d then u is a sum of products itvt where each it E I has positive degree and every vt is homogeneous necessarily of degree 2 d Since every term itvt has degree strictly larger than d this is a contradiction The final two statements follow exactly as in the case of the usual form of Nakayama7s Lemma D Lemma Let S a T be a degree preserving K algebra homomorphism of standard graded K algebras Let m Q S and 11 Q T be the homogeneous mazrimal ideals Then T is a nitely generated S module if and only if the image of S1 in T1 generates an n primary ideal Proof By the homogeneous form of Nakayama7s lemma T is finitely generated over S if and only if TmT is a finite dimensional K vector space and this will be true if and only if all homogeneous components TmTls are 0 for 8 gtgt 0 which holds if and only if 115 Q mT for all s gt 0 Proposition Let R m K be a local ring IfI Q J Q m are ideals then J is integral over I if and only if the image of I in JmJ K R grR1 generates an n primary ideal in K R ngR where n is the homogeneous maximal ideal in T Proof First note that J is integral over I if and only if RJt is integral over RIt and this is equivalent to the assertion that RJt is module finite over RIt since RJt is finitely generated as an R algebra and hence as an RIt algebra If this holds we have that K R RJt is a finitely generated module over K R RIt and since the image of I generates the maximal ideal min S K R ngR E K R RIt the preceding Lemma implies that the latter statement will be true if and only if the image of I in JmJ K R ngR1 generates an n primary ideal in T K R ngR The proof will be complete if we can show that when T is module finite over S then RJt is module finite over RIt Let jl E Jdl jh E th be elements whose images in Jdl de1 thdeh respectively generate T as an S module We claim that 3 jdltdl jdhtdh generate RJt over RIt To see this note that the fact that these elements generate T over S implies that for every N N 7 N7dv 4 N J 7 1 23d mJ 1 i h such that dign For each xed N we may apply the usual form of Nakayama7s Lemma to conclude that JN N7dijdi 1 i h such that dign and so for all N we have JNtN INTditNTdi ditdi7 1 i h such that dign which isjust what we need to conclude that jdltdl jdh tdh generate RJt over RIt D The following fact is often useful Proposition Let K be an in nite eld V Q W vector spaces and let V1 Vh be uector subspaces of W such that V Q Ufa Vi Then V Q Vi for some i Proof If not for each i choose vi 6 V 7 Vi We may replace V by the span of the vi and so assume it is finite dimensional of dimension d We may replace Vi by Vi V so that we may assume every Vi Q V The result is clear when d 1 When d 2 we may assume that V K2 and the vectors 10 c E K 7 0 lie on infinitely many distinct lines For d gt 2 we use induction Since each subspace of V E Kd of dimension d 7 l is covered by the Vi each is contained in some Vi and hence equal to some Vi Therefore it suf ces to see that there are infinitely many subspaces of dimension d 7 1 Write V K2 69 W where W g Kd Z The line L in K2 yields a subspace L 69 W of dimension d 7 l and if L 31 L then L 69 W and L 69 W are distinct subspaces D Also note Proposition Let M be an N graded or Z graded module over an N graded or Z graded Noetherian ring S Then every associated prime of M is homogeneous Hence euery minimal prime of the support ofM is homogeneous and in particular the associated hence the minimal primes of S are homogeneous Proof Any associated prime P of M is the annihilator of some element u of M and then every nonzero multiple of u 31 0 can be thought of as a nonzero element of SP g Su Q M and so has annihilator P as well Replace u by a nonzero multiple with as few nonzero homogeneous components as possible If ui is a nonzero homogeneous component of u of degree i its annihilator JZ is easily seen to be a homogeneous ideal of S If Jh 31 JZ we can 4 choose a form F in one and not the other and then Fa is nonzero with fewer homgeneous components then n Thus the homogeneous ideals JZ are all equal to say J and clearly J Q P Suppose that s E P 7 J and subtract off all components of s that are in J so that no nonzero component is in J Let sa Z J be the lowest degree component of s and ab be the lowest degree component in n Then sa lag is the only term of degree a b occurring in so 0 and so must be 0 But then sa 6 Aims ub Jb J a contradiction D Corollary Let S be a standard graded K algebra of dimension d with homogeneous max imal ideal m Then there are forms L1 Ld of degree 1 in R1 such that m is the radical 0f L1 LdS Proof The minimal primes of a graded algebra are homogenous and dim S is the same as dim SP for some minimal prime P of R Then P Q m and dim S dim SP dim SPm S dim Sm S dim S so that dim S dim Sm height m If dim S 0 m must be the unique minimal prime of S and therefore is itself nilpotent Otherwise S1 cannot be contained in the union of the minimal primes of S or the Proposition just above would imply that it is contained in one of them and S1 generates m Choose L1 6 S1 not in any minimal prime and then dim SLl d 7 1 Use induction lf L1 Lk have been chosen in S1 such that dim SLl LkS d 7 h lt d choose Lk1 6 S1 not in any minimal prime of L1 LkS if S1 were contained in one of these m would be and it would follow that heightm S h a contradiction Thus we eventually have L1 Ld such dim S L1 LdS 0 and then by the case where d 0 we have that m is nilpotent modulo L1 LdS D We are now ready to prove the result that we have been aiming for Theorem Let R m K be local and J Q R an ideal Then any reduction I of J has at least anJ generators Moreover if K is in nite there is a reduction with 111J generators Proof The problem of giving i1 ia E J such that J is integral over i1 iaR is equivalent to giving a elements of JmJ that generate an m primary ideal of S K R ngR where m is the homogeneous maximal ideal of S Clearly we must have a 2 dim S anJ lf K is infinite the existence of suitable elements follows from the Corollary just above D Discussion If R m K is local and t is an indeterminate over R let Rt denote the localization of the polynomial ring RH at mRt Then R 7 Rt is a faithfully at map of local rings of the same dimension and the maximal ideal of Rt is mRt while the residue class field of Rt is lf J Q m anJ is the least number of generators of an ideal over which JRt is integral In fact Kt ngRtRt g Kt K K R ngR so that anJ anJRt 5 Remark If I and J are anyitwo ideals of any ring R IJ Q W There are many ways to see this Eg if r E I s E J and R a V is any homomorphism to a uation domain then r 6 IV and s 6 JV whence rs E IVJV IJV Thus rs E IJ for every such r and s and the elements rs generate IJ D We shall prove below that if R is local and J is any proper ideal then dim R dim ngR Assume this for the moment It then follows that dim K R ngR S dim We define the big height of a proper ideal J of a Noetherian ring to be the largest height of any minimal prime of J The height is the smallest height of any minimal prime of J We then have Proposition For any proper ideal J of a local ring R m K the analytic spread ofJ lies between the big height ofJ and dim Proof That 111I S dim R follows from the discussion above once we have shown that dim R dim ngR Let P be a minimal prime of J We want to show that heightP S 111J After replacing R by Rt if necessary we have that J is integral over an ideal I with a 111J generators Then J is contained in the radical of I In Rp we have that P is the radical of JRp since P is a minimal prime of J and so is contained in the radical of IRp Thus height P S a 111J as required D Corollary IfK is in nite every proper ideal J of a local ring R m K of Krull di mension d is integral over an ideal generated by at most d elements D Proposition Let R m K be local and J aproper ideal Then for every positive integer n the ideals J and J have the same analytic spread Proof The Rees ring of J may be identified with RJ t since t is an indeteminate over R and this is a subring of RJt over which the larger ring is module finite since the nth power of any element of Jt is in RJ t The injectivity is retained when we apply K ER 7 since tensor commutes with direct sum and the module finite property continues to hold as well It follows that K 8 ngR is a module finite extension of K R ngnR and so these two rings have the same dimension D In general if X is a matrix and B is a ring BX denotes the ring generated over B by the entries of X We frequently use this notation when these entries are indeterminates in which case B denotes the formal power series ring over B in which the variables are the entries of X If M rij is a matrix over a ring R and t is a nonnegative integer ItM denotes the ideal of R generated by the size t minors of M By convention this ideal is R ift 0 and is 0 ift is strictly larger than either of the dimensions of the matrix M Example Let I Q R m K and let r be a nonzerodivisor Then RIt RrIt in fact rt is algebraically independent of B so that there is an R isomorphism Rt a Rrt mapping t gt gt rt and this induces an R isomorphism RIM E RrIt It follows that 6 K R RIt E K R RrIt and so anI anIr Ir Q rR which has analytic spread one If I m or if I is m primary the analytic spread of I and of Ir is dim Thus the smaller of two ideals may have a much larger analytic spread than the larger ideal 1 2 quot39 an be a 2 gtlt n matrix of formal 1 242 yn indeterminates over K Let be the polynomial ring in the entries of X and let A This ring is known to be a normal ring with an isolated singularity of dimension n1 One can see what the dimension is as follows we may tensor with algebraic closure of K without changing the dimension this produces an integral extension and so we may assume that K is algebraically closed The algebraic set Z in A2 defined by I2X corresponds to 2 gtlt n matrices of rank at most one We can map A gtlt A1 to Z by sending u c to the matrix whose first row is u and whose second row is cu This map is not onto but its image contains the open set consisting of matrices whose first row is not 0 It follows that the dimension of Z is n 1 This ring is also known to be Cohen Macaulay Cf Hochster and J A Eagon Cohen Macaulay rings inuariant theory and the generic perfection of determinantal loci Amer J of Math 93 1972 102071058 The same properties hold if we localize A at the ideal generated by the entries of X and if we complete Call the local ring obtained S Let P be the prime ideal 1 xnR Then A SP is either the localization of Ky1 yn at yl yn or its completion In any case RP has dimension n so that P is a height one prime of S But its analytic spread is n In fact grpS has the form Au1 un where the ui satisfy the relations yiuj 7 yjui 0 If we kill only these relations we get a domain of dimension n 1 that maps onto grp Since grPS has the same dimension as S we have not proved this yet but will shortly the map onto grPS cannot have a nonzero kernel ie it is an isomorphism But then K 8 grPS E Ku1 un has dimension n Example Let K be a field and let X We next want to prove the assertion that the local ring R m K and the ring ngR have the same dimension In order to do so we review the dimension formula Recall that a Noetherian ring R is catenary if for any two prime ideals P Q Q any two saturated chains of prime ideals joining P to Q have the same length Localizations and homomorphic images of catenary rings are clearly catenary R is universally catenary if every polynomial ring in finitely many variables over R is catenary It is equivalent to assert that every algebra essentially of finite type over R is catenary A ring is called Cohen Macaulay if in each of its local rings some equivalently every system of parameters is a regular sequence Cohen Macaulay rings are catenary and therefore universally catenary since a polynomial ring over a Cohen Macaulay ring is Cohen Macaulay Regular rings are Cohen Macaulay and hence universally catenary A complete local ring is a homomorphic image of a regular ring and so is also universally catenary If f Q g are fields tr degQ denotes the transcendence degree over 9 over f Theorem dimension formula Let R Q S be Noetherian domains such that S is nitely generated over R and call the fraction elds f and g respectively Let Q be a prime ideal ofS lying ouer P in B Let K and L be the residue class elds of Rp and SQ Math 711 Lecture of November 6 2006 To nish our comparison of symbolic powers in a regular ring we shall make use of quadratic transforms also called quadratic transformations or quadratic dilatations in a more general context than in the proof of the Lipman Sathaye Jacobian Theorem Let R m K be a local domain with R Q V n a local map where V is a not necessarily Noetherian valuation domain The rst quadratic transform of R along V is the localization R1 m1 of at the contraction of n where x is any element of m such that icV mV This ring is again a local ring with a local map R1 a V The quadratic transform is independent of the choice of the element it To see this suppose that icV yV where y x E m Then 2415 6 is a unit in V so its inverse icy E R1 Since my it follows that Rm y Q R1 Moreover each element of Rmy that is invertible in V has an inverse in R1 so that if Q is the contraction of n to Rmy we have an induced inclusion map RmyQ a R1 An exactly symmetric argument gives the opposite inclusion As in our earlier situation we may take iterated quadratic transforms RCngCnggV Note that if m 1 xh then mV 1 ichV so that it may be chosen from among the obi Putting this together with the Lemma on p 2 of the Lecture Notes of September 29 we have Proposition Let R m K be regular local with 1 icd a regular system of param eters and suppose that R Q V n is local where V is a valuation domain If the ac are numbered so that ij Q 1V for allj gt 1 then the quadratic transform R1 is a localiza tion of the ring S Rpm1 rd1 which is regular of dimension d In particular R1 has dimension at most d Moreover SiclS g KX2 Xd where X is the image oficiicl 2 S i S d D Here is another important example Theorem Let R be a one dimensional local domain whose integral closure V n is local and module nite ouer B This is always the case ifR is a complete one dimensional local domain Let R 31 Rk QV be the sequence of iterated quadratic transforms Then for all su ciently large k Bk V Proof Since V is module finite over B it cannot have an infinite ascending chain of R submodules It follows that the chain R is eventually stable But if the maximal ideal of 1 R1 is not principal and has minimal generators yl yh with yl of least order in V then for somej gt 1 243241 6 V 7 Bi and 243241 6 Bi Therefore for suf ciently large i the maximal ideal of R1 is prinicpal But then R1 is a DVR and is a normal ring inside the fraction field of R and containing R It follows that R1 V We also note Theorem Let R m K be a local domain with R Q Vn a local inclusion where V is a valuation domain of the fraction eld of B Let q be a prime ideal of V lying ouer P 31 m in B Let R R1 R2Q39 Rk quot39QV be the sequence of quadratic trransforms off along V Let Pi be the contraction ofP to Bi Then PLP Q RlPl Q PL2132 Q Q RkPk Q Q VCl is the sequence of quadratic transforms of RP Q Vq along Vq Proof By induction on h it suf ces to see this when k 1 Let 1 xh generate the maximal ideal m of R with 1V mV Some it E m is not in n and since icV Q mV le 1 Z q and so 1 Z P Moreover 1Vq It follows that the quadratic transform of RP along Vq is the localization at the contraction of n of RPmEll where m is mP and E1 is the image of 1 in RP The stated result follows at once Note that we again have P1 31 ml the maximal ideal of R1 since x1 6 m1 7P1 D We next observe Lemma Let R m K be a regular local ring with algebraically closed residue class eld and suppose R Q V n is local where V is a valuation domain and Rm 7 Vn is an isomorphism Then there is a regular system of parameters 1 xd for R such that the rst quadratic transform is the localization of Elm1 god1 at the height d maximal ideal generated by 121 godx1 so that these elements are a regular system of parameters in the rst quadratic transform Proof Let 1 yg yd be one regular system of parameters for R such that 1V mV 11 contains 1 and so 11 lies over a prime ideal of R containing 1 Hence the quotient of Rmac1 by the contraction ofn is also a quotient of KY2 Yd where Y is the image of 241161 2 S i S d The resulting quotient domain imbeds embeds K isomorphically in K Vn and so is equal to K It follows that the contraction of 11 corresponds to a maximal ideal of KY2 Yd which must have the form Y2 7 c2 Yd 7 cd for elements c2 cd 6 K Therefore we may let xi yi 7 ciacl for each i 2 S i S d D Proof of the theorem on comparison of symbolic powers We want to show that if R is regular and P Q Q are prime then P Q Q for all n By considering a saturated chain of primes joining P to Q we immediately reduce to the case where the height if QP in 3 RP is one We may replace R by RQ and so we may assume that Q is m in the regular local ring Em and that dim RP 1 Suppose that R m K a S is a at local map where S is regular with maximal ideal mS Then it suf ces to prove the theorem for S for if P1 in S lies over P and we know the theorem for S we have PW g Pf g msw mnS and then PW g mnS m R m because S is faithfully at over R We may therefore replace R first by its completion and then by a complete regular local ring with an algebraically closed residue eld Hence from now on we shall assume that R is complete with residue class field K that is algebraically closed as well as that dim RP 1 We now introduce valuations Let V1 be a valuation domain of the fraction field of R whose maximal ideal contracts to P we may use for example order with respect to powers of PR to construct V1 Let W be the integral closure of RP which will be a discrete valuation ring because RP is a complete local domain of dimension one Since K is algebraically closed the residue class field of W is K Let V n be the composite valuation Then 11 lies over m and Vn K Moreover V has a prime 7 lying over P Now consider the sequence of quadratic transforms Each R1 has a prime Pi that is the contraction of q Now RPk is the kth quadratic transform of RP by the Theorem above and so for large k is the DVR W by the earlier Theorem Then RkPk is regular and so Pk is generated by part of a regular system of parameters We shall see in the sequel that PIE P Q m in this case Assuming this to complete the proof it suf ces to show that if a given R1 provides a counterexample where R0 R then so does its quadratic transform We might as well work with R and R1 Rim17 7dlMa where M is the maximal ideal 1 avgx1 md1Rmx1 Suppose f E R has m adic order H but order at least 11 l in Rp Since mn f Q BUR1 we have that f f E BUR1 Since 1 P1 fn f has the same order as f in R1p17 and since P1 lies over P this will be at least 11 1 It therefore will suf ce to show that f f has ml adic order at most 11 in R1 Since I llAA 1 already local it suf ces to show that f f Z Mn Suppose otherwise The ideal J17l1 is generated by elements p fH where M is a monomial of degree n l in 2 J and BUR1 U Math 711 Lecture of September 12 2007 In our treatment of tight closure for modules it will be convenient to use the Frobenius functors which we view as special cases of base change We first review some basic facts about base change Base change If f R a S is an ring homomorphism there is a base change functor S ER 7 from R modules to S modules It takes the R module M to the R module S R M and the map h M a N to the unique S linear map S R M a S R N that sends 5 8 u gt gt 5 for all s E S and u E M This map may be denoted 15 ER h or S R h Evidently base change from R to S is a covariant functor We shall temporarily denote this functor as BRHS It also has the following properties 1 Base change takes R to S Base change commutes with arbitrary direct sums and with arbitrary direct limits Base change takes R to S and free modules to free modules 4 Base change takes projective R modules to projective S modules Base change takes at R modules to at S modules Base change is right exact if M a M a M a 0 is exact then so is S RMgtS RMgtS RMHgt0 7 Base change takes finitely generated modules to finitely generated modules the num ber of generators does not increase 8 Base change takes the cokernel of the matrix rij to the cokernel of the matrix f 9 Base change takes RI to SIS 10 For every R module M there is a natural R lineaar map M a S 8 M that sends it gt gt 1 u More precisely R linearity means that ru gt gt grl 8 u 97 u for all 7 E R and u E M 11 Given homomorphisms R a S and S a T the base change functor BRHT for the composite homomorphism R a T is the composition BSHT o BRHS 1 Part 1 is immediate from the de nition Part 2 holds because tensor product com mutes with arbitrary direct sums and arbitrary direct limits Part 3 is immediate from parts 1 and If P is a projective R module one can choose Q such that PEBQ is free Then S R P BS R Q is free over S and it follows that both direct summands are pro jective over S Part 5 follows because if M is an R module the functor S R M 83 i on S modules may be identified with the functor M ER 7 on S modules We have S RM SU M RS SU M RM7 by the associativity of tensor Part 6 follows from the corresponding general fact for tensor products Part 7 is immediate for if M is finitely generated by n elements we have a surjection R a M and this yields S a S R M Part 8 is immediate from part 6 and part 9 is a consequence of 6 as well 10 is completely straightforward and 11 follows at once from the associativity of tensor products The Frobenius functors Let R be a ring of prime characteristic 1 gt 0 The Fmbemus or Peskz39ne Szpz m functor fR from R modules to R modules is simply the base change functor for f R a S when S R and the homomorphism f R a S is the Frobenius endomorphism F R a R ie F7 W for all 7 E R We may take the 6 fold iterated composition of this functor with itself which we denote f This is the same as the base change functor for the homomorphism Fe R a R where Fe7 rpg for all 7 E R by the iterated application of 11 above When the ring is clear from context the subscript R is omitted and we simply write f or 5 We then have from the corresponding facts above 1 feR R 2 f5 commutes with arbitrary direct sums and with arbitrary direct limits 3 4 5 feR R and 75 takes free modules to free modules 75 takes projective R modules to projective R modules 75 takes at R modules to at R modules 6 f5 is right exact if M a M a M a 0 is exact then so is feM feM a feM a 0 a es m e y genera e mo ues o m e y genera e mo ues e num er 0 7fetkfi39tl td dl tfi39tl td dl th b f generators does not increase 8 75 takes the cokernel of the matrix rij to the cokernel of the matrix 9 f5 takes RI to RIlqu By part 10 in the list of properties of base change for every R module M there is a natural map M a f5 We shall use M to denote the image of a under this map which agrees with usual the usual notation when M R R linearity then takes the following form 10 For every R module M the natural map M a feM is such that for all r E R and all u E M ruq rquq We also note the following given a homomorphism g R a S of rings of prime characteristic 1 gt 0 we always have that g o Ffa F o g In fact all this says is that grq grq for all r E B This yields a corresponding isomorphism of compositions of base change functors 11 Let R a S be a homomorphism of rings of prime characteristic 1 gt 0 Then for every R module M there is an identification S R f M E f S R M that is natural in the R module M When N Q M the map f5 N a feM need not be injective We denote that image of this map by qul or more precisely by Niall However one should keep in mind that qul is a submodule of feM not ofM itself It is very easy to see that qul is the R span of the elements of feM of the form M for u E N The module qul is also the R span of the elements a as u runs through any set of generators for N A very important special case is when M R and N I an ideal of R In this situation Ill3 is the same as M as defined earlier What happens here is atypical because F5R R for all e Tight closure for modules Let R be a Noetherian ring of prime characteristic 1 gt 0 If N Q M we define the tight closure NITI of N in M to consist of all elements a E M such that for some c E R0 cuq 6 N5 g feM for all 1 gtgt 0 Evidently this agrees with our definition of tight closure for an ideal I which is the case where M R and N I If M is clear from context the subscript M is omitted and we write Nquot for NITI Notice that we have not assumed that M or N is finitely generated The theory of tight closure in Artinian modules is of very great interest Note that c may depend on M N and even it However c is not permitted to depend on 1 Here are some properties of tight closure Proposition Let R be a Noetherian ring of prime characteristic 1 gt 0 and let N M and Q be R modules a NITI is an R module 4 b IfN Q M Q Q are R modules then N Q Mg and NITI Q N c If N Q M is any family of inclusions and N ANA Q AMA M then NM d IfR is a nite product of rings R1 gtlt gtlt Rn NZ Q M are Ri modules 1 S i S n M is the R module M1 gtlt gtlt Mn andN Q M is N1 gtlt gtlt N then NITI may be identify with N fvl1 gtlt gtlt Nn wn e IfI is an ideal ofR IN I Q IN VI f IfN Q M and V Q W are R modules and h M a W is an R linear map such that hN Q V then Q VV Q Proof a Let c c E R If cuq E qul for q 2 go then cruq E qul for q 2 go If dog 6 Nq for q 2 ql then ccu uq E qul for q 2 maxq0q1 b The first statment holds because we have that Ngl Q Mg for all 1 and the second because the map F5M a F5Q carries NE into N51 c is a straightforward application of the fact that tensor product commutes with direct sum and the definition of tight closure Keep in mind that every element of the direct sum has nonzero components from only finitely many of the modules d is clear note that R1 gtlt gtlt 1O Rf gtlt gtlt R3 e If c c E R0 ch E Ilql for 1 gtgt 0 and culql E qul for 1 gtgt 0 then ccfuq chcuq E Ilqlqul for 1 gtgt 0 and Ilqlqul INql for every 1 f This argument is left as an exercise B Let R and S be Noetherian rings of prime characteristic p gt 0 We will frequently be in the situation where we want to study the effect of base change on tight closure For this purpose when N Q M are R modules it will be convenient to use the notation ltS R Ngt for the image of S R N in S R M Of course one must know what the map N gt M is not just what N is to be able to interpret this notation Therefore we may also use the more informative notation ltS R NgtM in cases where it is not clear what M is Note that in the case where M R and N I Q R ltS R Igt IS the expansion of I to S More generally if N Q G where G is free we may write NS for ltS R NgtG Q S 8 G and refer to NS as the e1pansion of N by analogy with the ideal case Proposition Let R a S be a homomorphism of Noetherian rings of prime characteristic p gt 0 such that R0 maps into S In particular this hypothesis holds 1 if R Q S are domains ifR a S is flat or if S RP where P is a minimal prime of S Then for all modules N Q M ltS R NJT4gtM Q ltS R NgtM RM Proof lt suf ces to show that if u E Nquot then 1 8 u E ltS R Ngt Since the image of c is in SO this follows because cl uq l cuq E ltS R qulgt ltS R NW 5 The statement about when the hypothesis holds is easily checked the only case that is not immediate from the de nition is when R a S is at This can be checked by proving that every minimal prime Q of S lies over a minimal prime P of R But the induced map of localizations Rp a SQ is faithfully at and so injective and QSQ is nilpotent which shows that PR is nilpotent D Tight closure like integral closure can be checked modulo every minimal prime of R Theorem Let R be a Noetherian ring of prime characteristic 1 gt 0 Let P1 Pn be the minimal primes of B Let DZ RPi Let N Q M be R modules and let u E M Let Mi DZ R M MPZM and let NZ ltDZ R Ngt Let ui be the image ofu in Mi Then a E NITI over R if and only iffor all i l S i S n iii 6 over Di IfM R and N I we have that u E Iquot if and only if the image ofu in DZ is in IDi in Di working over Di for all i 1 i S n Proof The final statement is just a special case of the Theorem The only if 7 part follows from the preceding Proposition It remains to prove that if u is in the tight closure modulo every Pi then it is in the tight closure This means that for every i there exists Ci 6 R 7 Pi such that for all 1 gtgt 0 Ci aq E qul PiFeM since feMPZM working over DZ may be identified with f5 MPZfe Choose di so that it is in all the except Pi Let J be the intersection of the Pi which is the ideal of all nilpotents Then for all i and all q gt 0 i digm e qul JF5M since every diPZ Q J Then 0 2211 dici cannot be contained in the union of Pi since for all i the ith term in the sum is contained in all of the Pj except Pi Adding the equations yields cuq e qul JF5M for all 1 gtgt 0 say for all 1 2 go Choose ql such that qull 0 Then cql 1Lqu 6 quqll for all 1 2 go which implies that cquq E qul for all 1 2 qlqo B Let B have minimal primes P1 Pm and let J B Pn the ideal of nilpotent elements of B so that Bred RJ The minimal primes of RJ are the ideals PiJ and for every i BredPiJ g RPi Hence Corollary Let R be a Noetherian ring ofprime characteristic 1 gt 0 and let J be the ideal of all nilpotent elements of B Let N Q M be R modules and let u E M Then a E NITI if and only if the image ofu in MJM is in ltNJgt VIJM working over Bred RJ We should point out that it is easy to prove the result of the Corollary directly without using the preceding Theorem We also note the following easy fact Math 711 Lecture of November 22 2006 Our next objective is to exhibit linear maximal Cohen Macaulay modules for rings de ned by the vanishing of the minors of a matrix of indeterminates in two special cases one is the case of maximal minors and the other the case of 2 gtlt 2 minors We shall solve the second problem in two different ways one of which generalizes to the case of Segre products of standard graded K algebras each of which has a linear maximal Cohen Macaulay module Discussion rings de ned by the vanishing of the minors of a generic matrizr Let K be a field and let X denote an 7 gtlt s matrix of indeterminates over K where 1 S 7 S s Let denote the polynomial ring in the rs variables Xi The ideal generated by the size if minors ItX is known to be prime in fact is known to be a Cohen Macaulay normal domain This was first proved in Hochster and J A Eagon Cohen Macaulay rings invariant theory and the generic perfection of determinantal loci Amer J Math 93 1971 102071058 and was treated by two methods in the Lecture Notes from Math 711 Winter 2006 one is the method of principal radical systems adapted from the paper just cited and the other is via the method of Hodge algebras We shall assume the fact that these ideals are prime here An argument for the case if 2 is given in Problem 5 of Problem Set 5 in which the isomorphism of with the Segre product of two polynomial rings over K one in 7 variables and one in s variables is established We note that it is easy to see that when K is algebraically closed the algebriac set VItX Q A is irreducible this is the algebraic set of 7 gtlt s matrices of rank at most if 7 1 For any such matrix 04 the map K5 7 KT that it represents factors through Kt l eg through a t 7 1 dimensional subspace of KT containing the image of Oz and the factorization K5 7gt K7571 7gt KT enables us to write 04 y where Q is 7 gtlt t 7 1 and y is t 7 1 gtlt s Any matrix that factors this way has column space contained in the column space of Q which shows that we have a surjection Ayn gtlt AWUS a VItX This proves the irreducibility since the image of an irreducible algebraic set is irreducible and shows at least that Rad ItX is prime It is also easy to calculate the dimension of VItX and hence of Let p t 7 1 Consider the open set in W Q X such that the first p rows of X are linearly independent The open set U of choices y for these rows we may think of points y E U as p gtlt s matrices of maximal rank has dimension ps Each remaining row is a unique linear combination of the rows of y using p coef cients so that the last 7 7 p rows of the matrix can be written uniquely in the form in where n is an arbitrary 7 7 p gtlt p matrix This 1 2 gives a bijective map of AT pp gtlt U onto the dense open W Q X and so the dimension of VItX which is the same as dim W is r 7 pp ps pr s 7 p where p t 7 1 It also follows that the height of 1X is rs 7 pr s 7 p r 7 ps 7 p If t r the case of maximal minors the height is rs7r7lslrs7rs7sr7ls7rl If t 2 the dimension is r s 71 Discussion linear mazrimal Cohen Macaulay modules over a standard graded ring Let K be a field Recall that a standard graded K algebra is a finitely generated N graded K algebra R such that R0 K and R KR1 ie R is generated over K by its forms of degree 1 In dealing with the existence of linear maximal Cohen Macaulay modules over the local ring Rm of a standard graded K algebra at its homogeneous maximal ideal m 6921 R7 it is convenient to work entirely in the graded case Note that grmRmRm E grmR E B so that each of the local ring and the graded ring determines the other If N is a linear maximal Cohen Macaulay module over a local ring S n K then M grnN is a maximal Cohen Macaulay module over R grnS and R is a standard graded K algebra To check this it suf ces to to do so after replacing S by St so that the residue class field is infinite R is replaced by Kt K R and M by Kt K M which does not affect the Cohen Macaulay property But when the residue class field is infinite we can choose a minimal reduction I 1 QETS for m where r dim R dim S and 1 irr is a system of paramters and then grnN ngN g NIN K KX1 XT where KX1 XT is a polynomial ring Note that M grnN is generated in degree 0 MO NIN NmN Conversely if M is an N graded maximal Cohen Macaulay module over the standard graded K algebra R and M is generated by elements of equal necessarily smallest degree we shall refer to M as a linear mazrimal Cohen Macaulay module in the graded sense over R if eM 1M where eM is defined as the integer e such that nr71 agrees r 7 with the leading term of the Hilbert polynomial of M by which we mean the polynomial that agrees with dim KM for all it gtgt 0 Note that we can shift the grading on such an M so that it is generated in degree 0 This does not affect 1M nor eM This is precisely the condition for Mm to be a linear maximal Cohen Macaulay module over R In fact if we have a finitely generated graded module M over standard graded K algebra R then 1Mm dim KMmmMm dim 1M and a minimal set of generators of M as a module may be taken to consist of homoge neous elements Under the condition that M is generated in degree 0 grmRmM E M 3 and eRm Mm may be calculated from the Hilbert function of the associated graded mod ule which yields that eRm Mm eM If R has a system of parameters 1 irr consisting of linear forms which is automatic when K is infinite then we again have mM 1 xTM in the graded case since eM Mac1 QETM and 1M Discussion a linear homogeneous system ofparameters for the maximal minors Consider an r gtlt s matrix X of indeterminates Xi over K where r S s We can give a linear homo geneous system of parameters for as follows Let Dj be the diagonal whose entries are X17 X24 XWTTA where l S i S s 7 r l The linear homogeneous system of parameters consists of the elements below these diagonals there are rr 7 l2 such elements the differences Xk17jk7Xv71 these are all on the diagonal Dj asj varies l S j S s 7r l and the elements above all the diagonals again there are rr 7 l2 such elements The total number of elements is rr 7 l s 7 r 7 lr 7 l r 7 ls 7 l which is the dimension of the ring KXIT To check that the elements specified form a system of parameters it suf ces to check that the the maximal ideal is nilpotent in the quotient Note that because all elements below D1 are 0 and the image of D1 in the quotient has all entries equal to say 1 the image of X11 we find from the vanishing of the leftmost r gtlt r minor that it 0 so that 1 is nilpotent We can then prove by induction on j that all the elements on the diagonal that is the image of Dj these are all equal are nilpotent If we know this for all variables below the diagonal Dj by the induction hypothesis and it the image of X17 is the common image of the elements on D then the is nilpotent from the vanishing of the minor consisting of r consecutive columns beginning with the jth In fact the quotient of by this linear homogeneous system of parameters turns out to be isomorphic with Kac1 5T1AT where the numerator is a poly nomial ring and M x1 xsr1 This is left as an exercise see Problem 2 in Problem Set 5 It follows that the multiplicity of the ring is the number of monomials of degree at most r 7 l in s 7 r 1 variables which is We can now show Theorem With notation as above the ideal P generated by the r 7 1 size minors of the rst r 7 1 rows ofX is a linear mazrimal Cohen Macaulay module for R in the graded sense Proof The generators of P have equal degree and since P has rank one eP eR Clearly 1P eP Thus we need only see that P is maximal Cohen s r 7 l Macaulay when considered as an R module The key point is that P is a height one prime in R its inverse image in the polynomial ring has height s 7 r 7 l l s 7 r 2 one more than the height of ITX Moreover the quotient RP is Cohen Macaulay it is a polynomial ring over a ring obtained by killing minors of an r 7 l gtlt s matrix of 4 indeterminates and so its depth on the homogeneous maximal ideal of R is dim R 7 l The short exact sequence 0 7gt P 7gt R 7gt R P 7gt 0 now implies that P has depth equal to dim R as an R module D We next want to give a calculation of the multiplicity of the ring R KX12 when X is a matrix of indeterminates We already know from Problem 5 of Problem 7 2 Set 5 and Problem 6 of Problem Set 4 that the answer is r s We give an 7 7 alternative proof of this by a completely different method The idea of this method is the same as the idea of the proof that these rings are Cohen Macaulay via the technique of principal radical systems If 7 l the ring is a polynomial ring in s variables and the multiplicity is l which is correctly given by the formula We prove that the multiplicity is r j s I 2 by induction on the number of variables The idea is to kill one of the entries of the matrix say it 11 Since the ring is a domain 11 is not a zerodivisor and the resulting ring has the same multiplicity as R In this ring zljxil xllxij 0 in R for i j 2 l and so every prime ideal contains either all of the elements xlj or all of the elements z Since Pz1jlgjgsisaprimeandQzi1lgigrisaprimePananre precisely the minimal primes of xllR in R If we localize at P the elements 11 i 2 2 become invertible and the resulting ring is easily checked to be a field localizing at 12 produces a localization of a polynomial ring over The situation is the same if we localize at Q Thus CR CRP CRQ The former is the ring obtained by killing the 2 gtlt 2 minors of an 7 7 l gtlt s matrix of indeterminates and the latter by killing the 2 gtlt 2 minors of an 7 gtlt s 7 1 matrix of indeterminates The result now follows form the identity rs72 7 r7ls72 rs7l72 r i 1 7 r i 1 i 1 r i 1 Discussion linear mazimal Cohen Macaulay modules for Our first proof uses the fact that the ring has for t 2 l and endomorphism reminiscent of the Frobenius endomorphism To wit the K algebra endomorphism 7 that sends Xij gt gt ij for all i andj maps 2X into itself zijzhk 7 zikzhj 7 7 and the latter element is a multiplie of the former element If we think of R 2 KY1 YTKKZ1 ZS this endomorphism is induced by the K endomorphism of the polynomial ring KY1 szl ZS Math 711 Lecture of October 4 2006 Proof of the Key Lemma nal step We are now in the case where R m K is regular local of dimension d R m K Q VnL is local where V Q IC the fraction field of R and V is a DVR essentially of nite type over R In particular tr degLK d 7 1 and we have a finite sequence of quadratic transforms along V R7 my T07 m07K0 g T17 mlzKl g g Th7 mhz Kb V7 397 where h 2 l and dim di 2 2 fort lt h We must show that jVR Q UV where u E R is such that RuR is regular Let ui 6 mi have minimum order in V for 0 S i S h 7 1 Let U1 uuo 6 T1 Recursively so long as vi 6 Ti is not a unit we know by induction and the Proposition on p 2 of the Lecture Notes of October 2 that vi is a regular parameter in Ti and we may define UZ391 uiui E T141 The Proposition just cited shows that either UZ391 is a unit of Ti or a regular parameter For some smallest h S h uk is a unit of Tk for the same Proposition shows that if uh1 is not a unit of Th4 then uh is a unit of Th see the final statement of that same Proposition Then U UO Ul 1 0 le UO ul Uk7139Uk7 where uk is a unit of Tk and hence of V Thus UV 710711 uk71V for some h S h On the other hand the Corollary at the end of the Lecture of September 27 shows that jVR jTlTOJTQTl jThThil The last statement in part b of the Lemma on the first page of the Lecture Notes of October 2 shows that jTHlT is generated by ujiil Thus M Ma 7 inflv and each exponent dj 7 1 is at least one The inclusion jVR Q uV is now obvious from inspection of and D The module nite property for normalizations We shall now address the problem of proving that S is module finite over S Several of the details of the argument are left to the reader in Problem Set 2 The result we aim to prove is this Theorem niteness 0f the normalization Let S be torsion free generically e tale and essentially of nite type over a normal Noetherian domain R Suppose that the com pletion of every local ring ofR is reduced which holds ifR is either regular or excellent Then the normalization S ofS ouer R is module nite over S We first give some preliminary results Lemma Let R be a Noetherian domain and b a nonzero element such that Rb is normal a R is normal if and only if Rp is normal for every associated prime of b b Q E Spec R RQ is not normal is the union of the sets VP where P is an associated prime ofP such that Rp is not normal and so is Zarishi closed c If Rm has module nite integral closure for every ma1imal ideal in of R then R has module nite integral closure We refer the reader to problem 4 in Problem Set 2 Let S Ra1b1 ahbh where b1 bh are nonzerodivisors in B This is a subring of R5 where b b1 bh and each fraction can be written in the form aQb We may include bb among these fractions and so assume that some a b Since every R linear combination of the fractions is in S if I is the ideal of R generated by the a we have that S RIb where Ib i E I Q Rb Here I is an ideal containing b We next observe Theorem Rees Let R m K be a local ring such that 13 is reduced Let S be the ring obtained by adjoining nitely many fractions elements of the total quotient ring to R Then the normalization S ofS is module nite over S Proof If the ring itself is complete the second problem from Problem Set 1 shows that one may reduce to the case where R is a complete local domain and since we already know that the normalization of a complete local domain is normal we may even assume that R is normal We then want to prove that the normalization of S RIf is module finite over RI with f E I by the discussion just above This follows from the fifth and sixth problems of Problem Set 2 Now consider the case where S itself is not necessarily complete If the result fails then there is an infinite strictly ascending chain S S0 Q S1 Q S2 Q of algebras generated by fractions over R where Sj1 is module finite over Sj for j 2 0 But the chain 12Sj must be eventually stable since the complete reduced ring 13 will have finite normalization so that IstjH Istj for large j But this implies Sj1 Sj otherwise some fraction over R is an ISL linear combination of finitely many other such fractions but not an R linear combination of them and we can get a contradiction from this as follows We may use a common denominator r E R not a zerodivisor and write ar 230airwi where the ai E R and the E But then a 6 a1 as st R which implies a 6 a1 asR say a 20 aibi with bi E R and that means we can replace the by elements bi of R D We also need the fact that finite separable algebraic field extensions do not disturb the property of being reduced Lemma Let B denoted a reduced ring containing a eld IC and let be a nite seprable algebraic e1tension of IC Then K B is reduced Proof B is the directed union of nitely generated IC algebas B0 since is IC at KB is the directed union of the rings K B0 Therefore we may assume that B is nitely generated over IC and has nitely many minimal primes B embeds in its total quotient ring which is a nite product of elds Thus we may replace B by its total quotient ring it su ices to prove the result when B is a product elds It is easy to see that it su ices to consider the case where B is a eld and we may even enlarge B further to an algebraically closed eld 9 But where f is monic with distinct roots in Q and so K Q E This ring is isomorphic with a nite product of copies of Q by the Chinese Remainder Theorem since the roots of f are distinct D We are now ready to prove the main theorem stated earlier on the mo dule nite property for S over S Proof of the niteness 0f the normalizaion We rst replace S by a subring nitely gener ated over R of which it is a localization Since localization commutes with normalization it su ices to consider this subring Thus we may assume that S is nitely generated as an R algebra Second the integral closure of S is the product of the integral closures of the domains obtained by killing a minimal prime of S Thus without loss of generality it su ices to consider the case where S is a domain Each of the generators in a nite set of generators for S over R satis es an algebraic equation over R with leading coe icient 73 say and it follows that we may choose a nonzero element 7 E R the product of these leading coe icients such that Sl7 is integral over Rl7 The integral closure of the normal domain Rl7 in the fraction eld of Sl7 is the same as the normalization of Sl7 and is module nite over Sl7 by the rst Theorem on p 3 of the Lecture Notes of September 11 It follows that we may enlarge S by adjoining nitely many elements of its normalization and so obtain a domain with the property that Sl7 is normal for some nonzero 7 It then follows from the rst Lemma above that in order to prove that S has nite normalization it su ices to prove this for SQ for every maximal ideal Q of S Choose 5 E S such that it generates over IC Let P be the contraction of Q to R1 RM Then SQ is a localization of R1pS and S is generated over R1p by elements of its fraction eld Thus to nish the argument it su ices to show that the completion of R1p is reduced We may replace R by its localization at the contraction of P and so we may assume that R is local with reduced completion The completion of R1p is one of the local rings of the completion of R1 with respect to the maximal ideal of R Thus it su ices to show that this completion of R1 is reduced But this is R1 R B Q R B 2 K 1C R B and the result follows from the preceding Lemma because IC R B is reduced and is nite separable algebraic over IC D Sketch of the proof of the Jacobian theorem We are ready to tackle the proof of the Jacobian theorem but we rst sketch the main ideas of the argument and then ll in the details 00000000000000000000000000000000000000000000000000000000000000000000000000000000000I Math 615 Lecture of February 23 2007 The nal step the application of generic freeness We have the following Lemma If 0 a N a M a G a 0 is an ezmct sequence of D modules and G is D free then the sequence is split so that M E NEBG In this case for any D module or D algebm Q the sequence 0 a Q D N a Q D M a Q D G a 0 2s ezmct Proof To construct a splitting f G a M choose a free basis 8 for G and for every element b E 8 de ne fb to be an element of M that maps to b Exactness is preserved by Q D i becaue tensor product commutes with direct sum D We are now ready to complete the proof There are several exact sequences that we are going to want to preserve while pass ing to characteristic 1 gt 0 Since A has Krull dimension d and is module finite over KF1 Fd we know that F1 Fd are algebraically independent over K and hence over the smaller ring D This yields 1 0DF1Fd ADgtADDF1Fdgt0 where DF1 Fd is a polynomial ring over D After localizing at one element of D70 we may assume that all these modules are D free and henceforth we assume this We shall make a number of further localizations like this but only finitely many Note that localizing further preserves freeness So long as there are only finitely many localizations at one element D remains a finitely generated Z algebra Second we have 2 0gtAD a RD RDAD O We may assume that D has been localized at one more element so that the terms of the exact sequence above are D free For every j the ideal F1 FjA is contracted from R Kz1 xn This implies that the map AFl FjA a RFl FjR is injective This map arises from the map ADF177FjAD RDF177FDRD in two steps we may tensor over D with the fraction field f of D and then we may tensor over f Q K with K After we tensor with K we know that the map is injective Since 1 2 K is faithfully at in fact free over its subfield f is injective once we tensor with f Therefore the kernel if any is torsion over D Hence if we localize at one element of D 7 0 so that ADFl FjAD becomes D free the map is injective We may also localize at one element of D 7 0 so that the cokernel is free over D and therefore we have for every j an exact sequence RDF1FDRD 3 0 7ADF1 FjAD 7 RDF1FDRD 7 ADF177FjAD 70 consisting of free D modules Finally we have that CAFl EA 31 0 It follows that GADF1 FiAD is not a D torsion module since it is nonzero after we apply K D 7 Hence after localizing further at one element of D 7 0 we may assume that 4 0w CADFh JDAD ADFh 7EAD ADFh 7E GAD 0 is an exact sequence of free D modules such that the module GADF1 FiAD is not zero We now choose a maximal ideal M of D Then H DM is a finite eld and has prime characteristic 1 gt 0 for some p We write AW and RE for H D AD ADMAD and H D RD RDMRD g Hz1 zn respectively We use E to indicate the image l w of w in AW or RE By the preceding Lemma the sequences displayed in l 2 3 and 4 remain exact after applying H ED 7 From 1 we have an injection of HF1 Fd which is a polynomial ring into AH This shows that the dimension of Am is at least d Since the homogeneous maximal ideal of Am is generated by the Ej and these are nilpotent on the ideal F1 FdA we have that F1 Fd is a homogeneous system of parameters for AH From 2 we have an injection AE gt RE From 3 we have that F1 FjAE is contracted from R for every j From 4 we have 6 is not in F1 F AH although we still have that EFiH Elfl Ein39 in AK so that AE is not Cohen Macaulay This contradicts the positive characteristic version of the Theorem which we have already proved D Note we have completed the proof of the sharper form of the result on the Cohen Macaulay property for rings of invariants stated on p 4 of the Lecture Notes of February 16 in all characteristics now and consequently we have completed as well the proof of the Theorem stated in the middle of p 3 of the Lecture Notes of February 16 Remarks It might seem more natural to prove the Theorem stated in the mdidle of p 3 of the Lecture Notes of February 16 by preserving the Reynolds operator ie that the ring of invariants is a direct summand while passing to characteristic p It turns out that this is not possible as we shall see below What we actually did was to preserve finitely many specific consequences of the existence of the Reynolds operator namely the contractedness of the ideals F1 FjA from R while passing to characteristic p and this was suf cient to get the proof to work Consider the action of G SL2 K on X where X acid is a 2 gtlt 3 matrix of indeterminates that sends the entries of X to the corresponding entries of yX for all y E G We have already noted that the ring of invariants in this case is CA1 A2 A3 where Aj is the determinant of the submatrix of X obtained by deleting the jth column of X see the third Example on p 3 of the Lecture Notes of January 31 In this case A1 A2 and A3 are algebraically independent this is true even if we special the entries of the matrix X so as to obtain 1 1 y 72w 0 x y 7 where z y and z are indeterminates It is easy to descend the inclusion A RG CA1 A2 A3 Q X to an inclusion of finitely generated Z algebras one can take D Z and consider the inclusion ZA1 A2 A3 Q However this is not split after we localize at one integer of Z 7 0 nor even if we localize at all positive prime integers except a single prime p gt 0 The Reynolds operator needs the presence of all prime integers p 31 0 in the denominators Note that if the map were split after localizing at all integers not divisible by p we could then apply ZpZ 8 Q and get a splitting of the map ZpZA1 A2 A3 Q ZpZX But we shall see below that this map is not split At the same time we want to note that in the Theorem on generic freeness it is important that the algebras T are nested with maps To a T1 a T2 a a T5 The result is false if one kills a sum of submodules over mutually incomparable subalgebras or even a sum of such subalgebras Both our proof that ZpZA1 A2 A3 Q ZpZX does not split and our example of the fallure of generic freeness when the T are incomparable are based on looking at the same example Namely we consider the module mmm H Mmm MmM Mmm where X is the same 2 gtlt3 matrix of indeterminates discussed in the action of SL2 C above and D T0 Z Note that the numerator and the three summands in the denominator are all finitely generated Z algebras We shall see that Q 8 H is a nonzero vector space over the rational numbers Q and that H is a divisible abelian group ie that nH H for every nonzero integer n It follows that if we localizate at any nonzero integer n E Z H7 is nonzero and is not free over Zn If it were free over Zn it could not be divisible by p for any integer 1 that does not divide 11 since it is simply a direct sum of copies of Zn It remains to prove the assertions that Q 8 H 31 0 that pH H for every nonzero prime integer p gt 0 and that the map ZpZA1 A2 A3 a ZpZX is non split for every prime integer p gt 0 We first note that if Z1 Z2 Z3 are indeterrninates and B is any base ring then Bizh Z27 Zslzlzgz3 H B Z lt BZ17Z27Z3jZ2Z3 BZ17Z27Z3jZ1Z3 BZ17Z27Z3jZ1Z2 is nonzero in fact the numerator is the free B rnodule spanned by all Inonornnials ZTZSQZE3 where 11 a2 a3 6 Z and the denominator is the free B rnodule spanned by all such Inonornials in which one of the integers a1 a2 a3 is nonnegative Hence the quotient may be identi ed with the free B rnodule spanned by all Inonornials Z1 Z32 Z513 such that a1a2a3 lt 0 Since A1 A2 A3 are algebraically independent over C and hence over Q we have that HQ A1 A2 A3 HQ A is a nonzero vector space over Q We have a cornutative diagram HCC A 4 HCC A mm cm i l HltQ7 A HltQ7 A QA1A2A3 le The top row may be thought of as obtained from the bottom row by applying C Q 7 We next observe that because L CjAl A2 A3 Q CX is split by the Reynolds op erator for the action of SL2 C and the top row is obtained by tensoring this inclusion over CjAl A2 A3 with HC A the top arrow is an injection Since C is free and therefore faithfully at over Q the arrow in the bottom row is also an injection Thus HQ A QA17A27A3 ij is a nonzero vector space over Q and this is the same as the result of apply Q 8 i to ZleA1A2A3 ZleA2A3 ZleAlAs ZleAlm HZ7 A ZA17A27A3 Zle which is the module H described earlier Finally we shall show that H pH for every prirne integer p gt 0 and from this we deduce that ZpZA1 A2 A3 a ZpZX is non split for every prirne integer p gt 0 Note that HpH ZpZ 8 H lf ZpZA1 A2 A3 a ZpZX splits over ZpZA1 A2 A3 then by applying 7 ZpZ HZpZA we obtain in injection HZpZA a HpH The lefthand term is not zero and this will imply that HpH 31 0 Thus by showing that HpH 0 we also show that ZpZMAi A2 A3 ZpZle does not split The final step involves some explicit use of local cohornology theory We refer to to the Lecture of December 8 from Math 711 Fall 2006 which contains a concise treatment of the material we need here as well as further references but we give a brief description including one de nition of the functor Ext A detailed treatment of Ext is given in the Lecture Notes from Math 615 Winter 2004 There is a discussion of homotopic maps of complexes in the Lectures of February 2 and February 4 it is used to prove the independence of Ext from the choice of projective resolution in the definition below Ext itself is defined in the Lecture of March 22 from the same set of Lecture Notes First recall that if M N are modules over R the modules Extja M N are defined as follows Choose a free or projective resolution of M ie an exact complex gtPigtgtP0gtMgt0 such that the P are free or projective This complex will frequently be infinite Let P be the complex obtained by replacing M by 0 ie gtPZIgtgtP0gt0 Apply the contravariant functor HomRi N to this complex to obtain 0 a HomRP0 N a a HomRPZ N a Then Extja M N is the cohomology of the complex at the HomR Pi N spot this is still the kernel of the outgoing map at that spot modulo the image of the incoming map it is called cohomology because the maps increase the indices There are other definitions one may use an injective resolution of N instead for exam ple and there are formulations of the theory where neither projectives nor injectives are used ExtiM N is independent of the choice of the projective resolution up to canonical choice free isomorphism If M is held fixed ExtM N is a covariant functor of N If N is held fixed it is a contravariant functor of M The functor ExtM N may be identified canonically with HomRM N The elements of Extll3M N are in bijective correspondence with isomorphism classes of short exact sequence 0 a N a W a M a 0 the reason for the name Ext is that Exth M N classifies such extensions There are two long exact sequences associated with Ext lf 0 a N1 a N2 a N3 a 0 is a short exact sequence of R modules then there is a long exact sequence 0 a HomRM N1 a HomRM N2 a HomRM N3 a Ext M N1 a 2 Extng N1 a Extng N2 a Extng N3 a Ext1M N1 a Similarly if 0 a M1 a M2 a M3 a 0 is exact there is a long exact sequence 0 a HomRM3 N a HomRM2 N a HomRM1 N a Ext Mg N a a Extngs N a ExtM2 N a Extngl N a Ext1M3 N a Math 711 Lecture of September 15 2006 The next Theorem gives several enlightening characterizations of integral closure We first note Lemma Let I be an ideal of the ring R r E I and h R a S a homomorphism to a normal domain S such that IS is principal Then hr 6 IS Proof By persistence of integral closure hr E E But IS is a principal ideal of a normal domain and so integrally closed which implies that r 6 IS D Theorem Let R be a ring let I be an ideal of R and let r E R a r E T if and only iffor every homomorphism from R to a valuation domain V r 6 IV b r E T if and only iffor every homomorphism f from R to a valuation domain V such that the kernel off is a minimal prime P of R fr 6 IV Thus ifR is a domain r E T if and only if for every valuation domain V containing R r 6 IV Moreover it su ces to consider valuation domains contained in the fraction eld of RP c If R is Noetherian r E T if and only if for every homomorphism from R to a DVR V r 6 IV d IfR is Noetherian r E T if and only if for every homomorphism f from R to a DVR V such that the kernel of f is a minimal prime of R fr 6 IV Thus if R is a domain r E T if and only if for every DVR V containing R r 6 IV Moreover it su ces to consider valuation domains contained in the fraction eld of RP e IfR is a domain and I ul unR is a nitely generated ideal let 7 U U and let Ti be the integral closure of Si Then r E R is in I if and only ifr E ITZ for all i Let R be a Noetherian domain Then r E I if and only if there is a nonzero element c E R such that or E I for all n E N Note I0 is to be interpreted as R even if I 0 g Let R be a Noetherian domain Then r E I if and only if there is a nonzero element c E R such that or E I for in nitely many values ofn E N A H V Proof We first observe that in any valuation domain every ideal is integrally closed every ideal is the directed union of the finitely generated ideals it contains and a directed union of integrally closed ideals is integrally closed In a valuation dornain every finitely 1 generated is principal hence integrally closed since a valuation domain is normal and it follows that every ideal is integrally closed Now suppose that r E I Then for any map of f of R to a valuation domain V we have that r E W IV This shows the only if 7 part of a To complete the proof of both a and b it will suf ce to show that if fr 6 IV whenever the kernel of f is a minimal prime then r E I But if r Z I this remains true modulo some minimal prime by part d of the Proposition whose statement begins on the first page of the Lecture Notes for September 8 and so we may assume that R is a domain and that rt is not integral over RIt But then by the Corollary at the top of the third page of the Lecture Notes of September 11 we can find a valuation domain V containing RIt and not rt in the Noetherian case V may be taken to be a DVR by the second Theorem on the first page of the Lecture Notes from September 13 Then r E 21 ijvj with the ij 6 I and the vj E V implies rt Zyd jt zj E V since each ijt E It Q V a contradiction The fact that it suf ces to consider only those V within the fraction field of RP follows from the observation that one may replace V by its intersection with that field The proofs of c and d in the Noetherian case are precisely the same making use of the parnthetical comment about the Noetherian case given in the paragraph above To prove e first note that the expansion of I to S is generated by ui since uj and so IT uZTZ as well If r E I then r 6 IT by the preceding Lemma Now suppose instead that we assume that r 6 IT for every t instead Consider any inclusion R Q V where V is a valuation domain Then in V the image of one of the uj say ui divides all the others and so we can choose i such that S Q V Since V is normal we then have T Q V as well and then r E ITZ implies that r 6 IV Since this holds for all valuation domains containing R r E I Finally it will suf ce to prove the only if 7 part of f and the if 7 part of If I 0 we may choose 0 l and so we assume that I 31 0 Suppose that J I Br and choose h E N so that Jh1 IJh so that Jnh InJh for all n E N see parts a and c on the second page of the Lecture Notes from September 13 Choose 0 to be a nonzero element of Jh Then for all n or E Jnh InJh Q I as required Now suppose that c is a nonzero element such that or E I n for arbitrarily large values of 11 If r I we can choose a discrete valuation U such that the value of v on r is smaller than the value of v on any element of I then vr l S 0a for all u E I Choose 11 gt 00 Then mr n S vw for all w E I But if we take w or we have vw 00 mr lt mr n S vw a contradiction D For those familiar with the theory of schemes we note that the condition in part e can be described in scheme theoretic terms There is a scheme the blow up Y of X Spec R along the closed subscheme defined by I which has a finite open cover by open a ines corresponding to the af ne schemes Spec Si The normalization Y of Y ie the normalized blow up has a finite open cover by the open a ines Spec Ti I corresponds to a sheaf of ideals on X which pulls back locally via expansion to a sheaf of ideals j on Y The integral closure of I is then the ideal of global sections of j intersected with R Math 711 Lecture of October 29 2007 We want to study at local homomorphism R m K a S n L to obtain informa tion about S from corresponding information about the base R and the closed fiber SmS We recall one fact of this type from p 3 of the Lecture Notes from September 19 Proposition Let R m K a S n L be a at local homomorphism of local rings Then a dim S dim R dim SmS the sum of the dimensions of the base and of the closed ber b IfR is regular and SmS is regular then S is regular Part a is generalized in the Theorem given below We first want to note the following Lemma Let R a ring and S a flat R algebra a IfM is a nitely generated R module then AnnSS R M AnnRMS E S 8 AnnRM Hence if N N Q W are R modules with N nitely generated S RN S S R N N R xi5 b If is a nitely generated ideal of R M is any R module and N is a at R module then then ADDM RNI g ADDMI R Proof a Let ul u be generators of M Then there is a map 04 R a M that sends r gt gt rul run The kernel of this map is evidently AnnRM so that O ADDRM RiMn is exact Since S is at we may apply S ER 7 to get an exact sequence a 0 e s R AnnRM e s A s R M where the map ls a sends s E S to sl u1 l u Since the elements l uj generate S R M the kernel of S is AnnSS R M while the kernel is also S R AnnRM by the exactness of the sequence For the final statement one may simply observe that N R N AnnRN NN b Let f1 f be generators for I and consider the map y M a M such that u gt gt flu fnu Then O ADDMI MLMn 1 2 is exact It follows that M 0 AnnMI R N s M R N i M R N is exact where 6 y 8 1N sends it 1 gt gt flu 1 fnu 1 Therefore the kernel is AnnM RNL but from the kernel is also AnnMI R N D Comparison of depths and types for at local extensions In the Theorem below the main case is when N S and the reader should keep this case in mind When N is an S module we refer to NmN as the closed fiber of N over B When we refer to the depth of a finitely generated nonzero module over a local ring we always mean the depth on the maximal ideal of the local ring unless otherwise speci ed Theorem Let R m K a 511 L be a local homomorphism of local rings and let N be a nitely generated nonzero S module that is R at Let M be a nonzero nitely generated R module Then a dim N dim R dim NmN More generally dim M R N dim dim NmN b N is faithfully flat ouer R c If yl yk E 11 form a regular sequence on the closed ber NmN then they form a regular sequence on W RN for every nonzero R module W In particular they form a regular sequence on NIN for every proper ideal I of R and on N itself Moreouer Nyl ykN is faithfully at ouer R d depthM RN depthMdepthNmN In particular depthN depthR depthNmN e N is Cohen Macaulay if and only if both R and NmN are Cohen Macaulay More generally ifM is any nitely generated R module then M R N is Cohen Macaulay if and only if both M and NmN are Cohen Macaulay IfN is Cohen Macaulay then the type ofN is the product of the type off and the type of NmN More generally ifM R N is Cohen Macaulay the type ofM R N is the product of the type ofM and the type of NmN A H V IfM is U dimensional and NmN is U dimensional then the socle ofM R N is the same as the socle of Anan K NmN Proof We use induction on dim R to prove the first statement If J is the nilradical of R we may make a base change and replace R S and N by RJ SJS and NJN The 3 dimensions of R and N do not change and the closed fiber does not change Thus we may assume that R is reduced If dim R 0 then R is a field m 0 and the result is obvious lf dim R gt 0 then m contains a nonzerodivisor at which is also a nonzerodivisor on N because N is R at We may make a base change to RzR The dimensions of R and N decrease by one and the closed fiber does not change Hence the validity of the statement in a is not affected by this base change The result now follows from the induction hypothesis To prove the second statement note that M has a finite filtration whose factors are cyclic modules RIj and so M R N has a corresponding filtration whose factors are the RIj R N Thus we can reduce to the case where M RI But this case follows from what we have already proved by base change from R to RI b To see that a at module N is faithfully at over any ring B it suf ces to see that Q R N 31 0 for every nonzero R module Q If u E Q is a nonzero element then Ru gt Q and so it suf ces to see that Ru R N 31 0 when Ru 31 0 But Ru g RI for some proper ideal I of R and Ru R M g QIQ Thus it suf ces to see that IQ 31 Q for every proper ideal I of R Since any proper ideal I is contained in a maximal ideal it suf ces to see that mQ 31 Q when m is maximal in R In the local case there is only one maximal ideal to check In the situation here mN Q nN 31 N by Nakayama7s Lemma since N is finitely generated and nonzero c By a completely straightforward induction on k part c reduces at once to the case where k 1 Note that an important element in making this inductive proof work is that we know that Nle is again R at Write y for yl To see that NyN is R at it suf ces by the local criterion for atness p l of the Lecture Notes from October 1 to show that Torj3Rm NyN 0 The short exact sequence gt0 O NiN NyN O yields a long exact sequence part of which is 0 Torf Rm N Torf Rm NyN NmN L NmN where the 0 on the left is a consequence of the fact that N is R at Since y is not a zerodivisor on NmN we have that Torf Rm NyN 0 Faithful atness is then immediate from part Note that if M 31 0 we have that M R NyN 31 0 because NyN is faithfully at over B To show that y is not a zerodivisor on M R N we may use the long exact sequence for Tor arising from the short exact sequence displayed above when one applies M ER 7 The desired result follows because TorI M NyN 0 d We use induction on depthM depthNmN Suppose that both depths are 0 Because depthM 0 m is an associated prime of M and there is an injection Rm gt M We may apply 7 ER N to obtain an injection NmN gt M R N Since depthNmN 0 there is an injection 511 gt NmN and the composite Sn gtNmNeM RN 4 shows that depthM R N 0 as required If depthM gt 0 we may choose it E m not a zerodiVisor on M Then it is not a zerodiVisor on M R N because N is R at We may replace M by The depths of M and M R N decrease by 1 while the closed fiber does not change The result now follows from the induction hypothesis Finally suppose depthNmN gt 0 Choose y E 11 not a zerodiVisor on NmN Then we may replace N by NyN This module is still at over R by part The depths of N and NmN decrease by one while M is unaffected Again the result follows from the induction hypothesis e M R N is Cohen Macaulay if and only if dim M R N depthM R N and by parts a and c we have that dim M RN7depthM RN dim Mdim NmN7 depthMdepthNmN dim M7depthM dim NmNidepthNmN The last sum is 0 if and only if both summands vanish since both summands are nonneg ative This proves e f Choose a maximal regular sequence 1 xh E m on M This is also a regular sequence on M R N and so we may replace M by Macl ichM Both types are unaffected by this replacement Similarly we may choose yl yk E n that form a maximal regular sequence on NmN and replace N by Nyl yk N Hence we have reduced to the case where M and NmN both are 0 dimensional We next want to establish the final statement from which the calculation of the type will follow We have that AnnnM R N Q Annm M R N E Anan R N by part b of the Lemma from p 1 Since m kills the module Anan this tensor product may be identified with Anan R NmN E Anan K NmN Then AnnMn E K5 where s is the type of M and this module is NmN5 Clearly the socle is the direct sum of s copies of the socle in NmN whose dimension over L is the type of NmN and the statement about the product of the types is immediate Corollary Let R m K a S n L be a flat local homomorphism of local rings a S is Gorenstein if and only ifR and SmS are both Gorenstein b If is an irreducible m primary ideal in R and y1 yk E n have images that are a system of parameters in SmS then for every integer t S y i yiS is an irreducible m primary ideal of S Proof Part a is immediate from parts e and f of the preceding Theorem since the type of S is the product of the types of R and SmS and so the type of S will be 1 if and only if and only if both R and SmS are of type 1 For part b make a base change to RI which is 0 dimensional Gorenstein ring Thus SIS is Corenstein the images of yf yf are a system of parameters and hence a maximal a regular sequence Thus the quotient of SI by the ideal they generate is a 0 dimensional Gorenstein ring D We can now prove Math 711 Lecture of September 25 2006 Following Lipman and Sathaye l Lipman and A Sathaye Jacobian ideals and a theo rem of Briangon Skoda Michigan Math J 28 1981 1997222 we present the Briancon Skoda Theorem in a generalized form BriangonSkoda Theorem Lipman Sathaye version Let R be a Noetherian nor mal domain and let I0 be an ideal ofR such that ngOR is regular Let n 2 1 and let I be an ideal ofR generated over I by n elements say f1 f Let k 2 1 be any positive integer Then PIN 1 Q Ik The version stated in the Lecture of September 6 is the case where I0 0 and k 1 Notice that the result is non trivial even when n 1 where it states that all the powers of I are integrally closed We shall first explain how this result follows from the Lipman Sathaye Jacobian Theo rem although this will take a while and then focus on the proof of the latter We need an intermediate result Theorem LipmanSathaye Let B be a Noetherian normal integral domain and let u E B 7 0 be such that BuB is regular Let t denote the inverse ofu in the fraction eld of B Let f1 fn E B and let S Bf1t fut Let S be the integral closure ofS in its eld offractions Then un lS Q S We want to see that the second theorem implies the first We need some preliminary facts Lemma Let be any ring and I an ideal of R a The integral closure of the ezrtended Rees ring RItu in Rt u in degree k is t if k S 0 let Ik R That is the integral closure in Rt u is RvkRv2RvRYt t2thm IfR is a normal domain this is also the integral closure ofR in its fraction eld b Suppose that R is a Noetherian domain and that ngR is an integral domain or that its localization at every prime ideal is an integral domain Then every power ofI is integrally closed Proof a The integral closure in RIT u is Z graded The result for nonnegative degrees is clear In positive degree k if rtk is integral over RIt u it satisfies a monic polynomial of degree d for some d and the sum of the coef cients of tdk must be 0 Just as in the case of RIt this yields an equation establishing the integral dependence of r on II The final 1 2 statement follows because when R is normal so is Rt v and so Rt 1 must contain the normalization of RIt b b If 71 31 0 we may preserve this while localizing Since integral closure commutes with localization we may assume that R is local If I expands to the unit ideal there is nothing to prove Otherwise ngR is N graded over the local ring RI This implies that it is a domain this is left as an exercise in Problem Set 2 When ng R is a domain we can de ne a valuation on R whose value on a nonzero element r is the unique nonnegative integer h such that r E h 7 1h I is then the contraction of the nth power of the maximal ideal of a discrete valuation ring Again the details are left as an exercise in Problem Set 2 D Proof that the second theorem implies the Briangon Skoda theorem Let B RIot 1 Then BvB E ngOR is regular and B is normal by the Lemma above Then S Bf1t fnt RIt v is the extended Rees ring of I over R It fol lows that in degree n h 7 l S is nk 1t k1 The fact that v lS Q S implies that vn 1Snk1 Q Slk Iktk and so nk 1 Q 1 Until further notice R denotes a Noetherian domain with fraction field IC and S denotes an algebra essentially of finite type over R ie a localization at some multiplicative system of a finitely generated R algebra such that S is torsion free and generically tale over R by which we mean that IC R S is a finite product of finite separable algebraic field extensions of IC Note that may also be described as the total quotient ring of S We shall denote by S the integral closure of S in L We shall prove that S is module finite over S if R is regular and more generally If A and B are subsets of we denote by A L B the set a 6 uB Q A If C is a subring of and A is a O module then A L B is also a O module We shall write jSR for the Jacobian ideal of S over R If S is a finitely generated R algebra so that we may think of S as RX1 Xslfl fh then jSR is the ideal of S generated by the images of the size s minors of the Jacobian matrix ijBxi under the surjection RX a S This turns out to be independent of the presentation as we shall show below Moreover if u E S then jSuR jSRSu From this one sees that when S is essentially of finite type over R and one defines jSR by choosing a finitely generated subalgebra S0 of S such that S W lSO for some multiplicative system W of So if one takes jSR to be jSORS then jSR is independent of the choices made We shall consider the definition in greater detail later The result we aim to prove is Theorem Lipman Sathaye Jacobian theorem Let R be regular domain1 with frac tion eld IC and let S be an ezrtension algebra essentially of nite type over R such that S is torsion free and generically etale over R Let IC R S and let S be the integral closure ofS in L Then S mfgR Q S2 jSR 1We can weaken the regularity hypotheses on R quite a bit instead we may assume that R is a CohenMacaulay Noetherian normal domain that the completion of every local ring of R is reduced and that for every height one prime Q of S if P Q R then Rp is regular 3 Note that since jgR is an ideal of S we have that S Q S mfgT The statement that S Q SzngR implies that jSRS Q S ie that jSR captures the integral closure S of S all we mean by this is that it multiplies S into S We next want to explain why the Jacobian ideal is well defined We assume first that S is finitely presented over B To establish independence of presentation we first show that this ideal is independent of the choice of generators for the ideal I Obviously it can only increase as we use more generators By enlarging the set of generators still further we may assume that the new generators are obtained from the orginal ones by operations of two kinds multiplying one of the original generators by an element of the ring or adding two of the original generators together Let us denote by Vf the column vector consisting of the partial derivatives of f with respect to the variables Since Vgf gi ng and the image of a generator f in S is 0 it follows that the image of Vgf in S is the same as the image of gi when f E I Therefore the minors formed using Vgf as a column are multiples of corresponding minors using Vf instead once we take images in S Since Vf1 f2 Vfl Vfg minors formed using Vf1 f2 as a column are sums of minors from the original matrix Thus independence from the choice of generators of I follows Now consider two different sets of generators for S over R We may compare the Jacobian ideals obtained from each with that obtained from their union This it suf ces to check that the Jacobian ideal does not change when we enlarge the set of generators f1 fs of the algebra By induction it suf ces to consider what happens when we increase the number of generators by one If the new generator is f f51 then we may choose a polynomial h E RX1 XS such that f hf1 fs and if 91 9 are generators of the original ideal then 91 gh X5 7 hX1 Xs give generators of the new ideal Both dimensions of the Jacobian matrix increase by one the original matrix is in the upper left corner and the new bottom row is 0 0 0 l The result is then immediate from Lemma Consider an h l by s l matrizr M over a ring S such that the last row is 0 0 0 u where u is a unit of S Let M0 be the h by 5 matrix in the upper left corner of M obtained by omitting the last row and the last column Then 5M0 1M Proof If we expand a size 5 1 minor with respect to its last column we get an S linear combination of size 5 minors of M0 Therefore 51M Q 5M0 To prove the other inclusion consider any 5 by s submatrix A0 of M0 We get an s l by s l submatrix A of M by using as well the last row of M and the appropriate entries from the last column of M If we calculate detA by expanding with respect to the last row we get up to sign udetA0 This shows that 5M0 Q 51M D This completes the argument that the Jacobian ideal jSR is independent of the pre sentation of S over R We next want to observe what happens to the Jacobian ideal when we localize S at one or equivalently at finitely many elements Consider what happens when we localize at u E S where u is the image of hX1 XS 6 RX1 XS where we have chosen an Math 711 Lecture of October 7 2005 For the purpose of the next theorem we make the convention that the type of the 0 mod ule over a local ring R is S 1 It should be the vector space dimension of ExtE1K M 0 ie it should be 0 Theorem Let R be a homomorphic image of a regular ring and let M be a nitely generated R module Let t 2 1 be a xed integer Then the set P E Spec R Mp is Cohen Macaulay of type S t is Zarishi open in Spec Proof Let R SJ where S is regular Then Spec R is homeomorphic with the closed set VJ Q Spec S if we identify Spec R with VJ the locus we want in Spec R is the locus for Spec S intersected with VJ Thus it suf ces to consider the problem for S instead and we may assume without loss of generality that R is regular Let P be a prime of R such that Mp is Cohen Macaulay of type at most t lf Mp is 0 this will be true on a Zariski neighborhood of P and we assume Mp 31 0 By the preceding result we may localize at one element of RiP so that M will be Cohen Macaulay with annihilator of pure height h Then ExtM Rp E Extga Mp Rp M can be generated by t or fewer elements and by clearing denominators we may assume that these elements have the form Ltll utl where iii 6 Ext M R for all i If fewer than t generators are needed we may take some of the iii to be 0 Let N be the R span of the 1 Then ExtM RNP 0 and so we can localize at one element of R 7 P that kills ExtM RN After this localization we have that ExtM R N is generated by at most t elements and so for all Q Exth 1 2 ExthWQ RQ M has at most t generators But this implies that the type of Mg is at most t as required D Corollary Let R be a homomorphic image of a regular ring Then P E Spec R Rp is Gorenstein is Zarishi open in Spec Proof We may apply the preceding result with M R The fact that the type of Rp is at most one implies that it is exactly one D We have already proved for a local at homomorphism R m K a S n L of local rings that S is Cohen Macaulay respectively Gorenstein if and only if both R and SmS are Gorenstein We next want to give a global version of this result that also describes the behavior of the loci where these properties fail We treat the Cohen Macaulay and Gorenstein cases simultaneously by axiomatizing the properties we need 1 Recall that if R 7 S is a ring homorphism its fiber over P E Spec R is Hp R S where Hp RpPRp E frac RP the fraction field of RP Thus the fiber may also be described as R 7 P 1SPS The map S 7 R 7 P 1SPS induces an injection Spec R 7 P 1SPS 7 Spec S whose image is the set of prime ideals of S lying over P in R Thus the primes in the spectrum of the fiber are in bijective correspondence with the prime ideals of S that contract to P Theorem Let P denote a property of Noetherian rings such that 1 If a local ring R has P so does its localization at any prime 2 R has P if and only if its localization at every maximal ideal has P it then follows that all of its localizations have P 3 If R m K 7 S n L is a local map of local rings then S has P if and only if R has P Then the following statements hold a IfS is faithfully flat over R then S has P if and only ifR has P and every ber of R 7 S has P b IfR 7 S is at all of the bers have P and I is an ideal ofS such that VI is the set of primes of R that do not have property P then VIS is the set of primes ofS that do not have property P In particular these results hold when P is the property of being Cohen Macaulay and when P is the property of being Gorenstein Proof We first consider part a Assume that S has P For every prime P of B there is a prime Q of S lying over P Since SQ has P so does Rp Therefore S has P implies that R has P Each prime of the fiber R 7 P 1SPS corresponds to a prime Q of S lying over P and it su ices to show that every ring R 7 P 1SPSQ has P But this ring g SQPSQ which has P because S does Now assume that R and all fibers have P Let Q be a prime of S lying over P in R It su ices to show that SQ has P This is true because Rp and SQPSQ both have P the latter is a localization of R 7 P 1SPS To prove b let Q be a prime ideal of S and let P be its contraction to R Note that Q 6 VHS gt P E VI lf SQ has P so does Rp and so P Z VI and Q Z VIS lf Q 6 VHS then P E VI so that Rp does not have P and SQ does not have P D Note that when R is a Hodge algebra over K on H governed by E arbitrary base change on K produces a new Hodge algebra with the same data More precisely if K 7 K is any ring homomorphism PM K K R H is the image of H in PM under the map sending h gt7 1 8 h and 2 is the semigroup corresponding to 2 under the obvious isomorphism NH 7 NH then PM is a Hodge algebra over K on H governed by 2 The free basis of standard monomials for R evidently maps bijectively to a free basis for PM over K and the straightening relations for R map to the required straightening relations for PM In particular each fiber Hp K R is a Hodge algebra over a field Corollary A Hodge algebra over a Noetherian ring K is Cohen Macaulay respectively Gorrenstein if and only ifK is Cohen Macaulay respectively Gorenstein and each ber 3 is Cohen Macaulay respectively Gorenstein The same holds for any property of rings P satisfying the three conditions in the Theorem above The condition that each ber is Cohen Macaulay respectively Gorenstein is equivalent to the condition that for every eld H to which K maps H K R is Cohen Macaulay respectively Gorenstein Proof The Hodge algebra is a free over K on a basis containing 1 and is therefore faithfully at over K The result is immediate from part a of the Theorem just above The final statement follows from the fact that if P is the kernel of K a H the map to H factors through the fiber KpPKp The final statement now follows from the Lemma just following D Lemma Let B be a nitely generated H algebra Then B is Cohen Macaulay respectively Gorenstein if and only if B H E B has the speci ed property for every eld ezrtension H of H Proof B is faithfully at over B and so the if 7 part follows Now assume that B has the speci ed property The result will follow if each fiber is Gorenstein the fibers are then Cohen Macaulay as well Each fiber has the form 4 E L where L has the form BpPBp and so is a field finitely generated over H We proceed by induction on the number of generators of the field L over H If H Q LO Q L we have that H 8 L g 4 n L0 LJ L Therefore it suf ces to show that if C is a Gorenstein algebra containing a field L0 and L is s field generated over L0 by one element then D O LO L is Gorenstein There are two cases If L L0 where it is transcendental over L0 then O LO L is a localization of OM and this is Gorenstein since it is at over 0 with Gorenstein fibers lf L is generated by one element 6 over L0 and is algebraic let f be the minimal monic polynomial of 6 over L0 Then D g C LO LOWf g But OM is Gorenstein and the monic polynomial f is a nonzerodiVisor Thus the quotient is also Gorenstein D The following theorem gives that the defining radical ideal of the closed set of primes where a graded ring is not Cohen Macaulay or not Gorenstein is homogeneous We need a preliminary fact Lemma IfT is at over a reduced ring R and the bers are reduced then T is reduced Proof If T has a nilpotent element other than 0 we may localize at a minimal prime Q of its annihilator and if Q lies over P we may study RP a TQ instead Then TQ has depth 0 and so Rp has depth 0 Since this ring is reduced and local it must be a field But then TQ is the fiber over P a localization of the original fiber over P and is reduced D Theorem Let S Q Nh be a semigroup and let R be a Noetherian ring graded by S Suppose that the set of primes such that Rp is not Cohen Macaulay respectively not Gorenstein is closed Then the radical ideal I de ning this locus is homogeneous in the S grading Proof There is no loss of generality in assuming that S Nb we can enlarge S and define the new graded pieces to be 0 For each i 1 S i S h we can put a Z grading on Math 711 Lecture of September 28 2005 The following fact will prove critical Lemma Let R be a Hodge algebra with notation as above let ho be minimal in lndR and let I hOR Then for every positive integer j PR is the free K module spanned by those standard monomials he c 6 NH 7 2 such that cho 2 j ie such that the standard monomial has h as a factor Proof Since R is spanned over K by the standard monomials it is clear that j h R is spanned over R by the elements h hc for c 6 NH 7 Z We claim that each such monomial is either 0 or else is standard itself Let e denote the element of NH that is l on ho and 0 elsewhere If c je does not yield a standard monomial we have that c je is in E and so is the sum of a standard generator c and some element f 6 NH ie c je c f We claim that c must have a positive value on ho if not then the value of f on ho is at least j and we can write c c f 7 je which shows that c E E a contradiction Now consider a straightening relation for ha since c is positive on ho each nonzero term Auu on the right involveds a standard monomial u whose support contains an element of H call it hf with hf lt ho Since ho is minimal in lndR such an hf does not exist and it follows that ha 0 and so h hc 0 as well D We keep the notations of the Lemma in force with ho a xed minimal element of lndR If r is nonzero element of B there is a largest integer j 2 0 such that r E j by a degree argument for example and then r E 7 7 71 We call j the order of r with respect to I or with respect to ho and use ord1r ord r to denote this order It is clear that ord rs 2 ord rord This generalizes by induction to the case of a product of several elements If ord r j we write r for the element r Ij1 E IjIj1 Q ngR r is called the leading form of r Then if ord rs ord r ord s we have that rs rs Again this generalizes by induction to the case of a product of several elements We can now state an important consequence of the Lemma above Corollary With notation as above JV7391 has a free K basis the elements uquot where u is standard monomial whose ezrponent has valuej on ho Thus the elements uquot as it runs through all standard monomials are a free K basis for ngR The order of ho is 1 while for all h E H7h0 ord 0 Moreover for any standard monomial u with ezrponent c uquot HheH Wequot D Keeping these notations we next observe that we can give grlR the structure of a graded algebra not the one come from the fact that it is an associated graded ring as follows We simply let the graded piece in degree i be the K span of all the uquot such that u is a standard monomial of degree i in R We need to check that if u and u are standard monomials of degrees i and l respectively then uquot u is a K linear combination of elements v where v is standard of degree i l Suppose that it has order m with respect to I and v has order nu Then then MuM in ngR is the class of aid in ImmImm1 by the de nition of multiplication for ngR 1 2 But aid will be a K linear combination of standard monomials each of which is a multiple of hgnm and each of which has degree i l Say uu Z Aww wEM It follows that u u Z Aww wEMordIumm and since every in has degree i i we have proved what we need We now have that ngR is an N graded K algebra generated by Hquot h h E We use the bijective map h gt gt h to give a partial ordering on H and we let 2 be the image of 2 under the obvious isomorphism NH N H induced by the order isomorphism H E Hquot We claim that with the grading described just above ngR is a Hodge algebra over K on Hquot governed by 2 and that lndngR Q h h E lndR 7 To see this suppose that c E E is a generator that v is the element with this exponent in R and that w is the corresponding element in grI obtained from Hquot by using the same exponent Then we have a straightening relation gt0 v Z Auu uEM in R We want to show that from we can derive a corresponding relation in ngR and we need to compare the indiscrete parts There are two cases If e is positive on ho then the right hand side of must be 0 and we see that we have the relation in 0 in ngR No elements from Hquot are used on the right Now suppose that c is 0 on ho so that v is a product of elements of H of order 0 Then in ngR we have the relation gt0 w Z Auuquot uEM ord1u0 and all the elements of Hquot occurring as factors of the various uquot are in h h E lndR i hS which is exactly what we need We have therefore proved Theorem Let R be a Hodge algebra over K on H governed by 2 Let ho be minimal in lndR Then with Hquot 2 as above and the grading de ned on ng above so that degu degu when u is standard ngR is a Hodge algebra over K on Hquot governed by 2 hence with the same data as R Moreover lndngR Q h h E lndR 7 D We can now take the associated graded ring with respect to the ideal generated by a minimal element of the indiscrete part of the new algebra again and do so repeatedly Each repetition decreases the cardinality of the indiscrete part by at least one We must eventually reach the discrete Hodge algebra over K on H governed by E 3 Corollary Let R be a Hodge algebra ouer K on H governed by 2 Then there is a nite sequence of Hodge algebras ouer K with the same data each of which is the associated graded ring of its predecessor with respect to a principal homogenous ideal and such that the last element in the sequence is up to isomorphism the discrete Hodge algebra ouer K on H governed by E D We shall embark on a program of getting information about a Hodge algebra from information about the corresponding discrete Hodge algebra For the Cohen Macaulay property the following fact will be useful Note that if R m K is local Rt will denote the localization of the polynomial ring in one variable Rt at mRt which is faithfully at over R with closed fiber Theorem Let 73 be a property of Noetherian rings such that l R has property 73 if and only if all local rings ofR have property 73 2 Ifx is a nonzerodiuisor in m and RxR has property 73 then R has property 73 3 If Rt has property then B does The Cohen Macaulay property is an ezrample Let I be an ideal off such that ngR has property 73 Then RQ has property 73 for every prime ideal Q of R such that Q I is a proper ideal of R and in particular for every prime ideal ofR that contains I Before giving the proof we recall that if I is an ideal of R and t is an indeterminate then the Rees ring of I is RIt Q Rt which may also be written as RIt12t213t3 Let u lt in Rtt Rt lt The second Rees ring of I is RItu Q Rt lt which is easily seen to be Ru3Ru2RuRItI2t2I3t3 Then vRIt u Rv3Rv2RvII2tISt2I4t3 and we see that RIt ulu may be identified with RH1172t1213t2 1314t3 Here the powers of t are just place holders77 This ring is ngR Moreover since u is a unit in Rt lt it is evidently a nonzerodivisor in this ring and in RIt y We use b to denote lt here as a reminder that u is generally not a unit in RIt 1 since t is not in the ring We are now ready to give the proof of the theorem stated above Proof If QI is proper it is contained in a maximal ideal in of R Since RQ is a localization of Rm it suf ces to show that Rm has the required property Since ngRm Rm 2 ngR W Math 711 Lecture of November 8 2006 We have completed the proof of the theorem on comparison of symbolic powers of prime ideals in regular rings as soon as we have established Lemma Let P be a prime ideal of the ring R that is generated by a regular sequence 1 ick Then PW P for every integer n Proof Let u E R 7 P We need only show that u is not a zerodivsor on P Suppose ur E P with r P Choose h which may be 0 such that r 6 Ph 7 Pk evidently h lt n Then ur E P Q Pk This implies that the image of u in RP is a zerodiVisor on PhPh1 But by part d of the Proposition on p 2 of the Lecture Notes of October 23 PhPh1 is a free RP module with a free basis in bijective correspondence with monomials of degree h in variables X1 Xk Before proceeding further we want to record an import result on atness We first note Lemma Let M be an R module with a nite ltration such that it E R is not a zerodivisor on any factor Then it is not a zerodivisor on M Proof By induction on the number of factors it su ices to consider that case of two factors ie where one has a short exact sequence 0 a N1 a M a N2 a 0 If u E M is such that you 0 then the image of u in N2 must be 0 or else it will be a zerodiVisor on N2 But then u 6 N1 and so you 0 implies that u 0 D Next note that when R m K a S n L is local and M is an S module MmM is called the closed ber of M because it is the fiber over the unique closed point m of Spec In this case if we make a base change to RI where I Q m is an ideal of R R S and M become RI 515 and MIM respectively but the closed fiber does not change E In the result that follows the most important case is when M S Theorem Let R m K a S n L be a local homomorphism of local rings and let M be an S module that is R flat Then a dim dim R dim b If y E n is a nonzerodivisor on MmM then it is a nonzerodivisor on M and on MIM for every ideal I Q m of R Moreover if y E n is a nonzerodivisor on MmM then MyM is again flat over R If depthmR 0 then y E n is a nonzerodivisor on M if and only if it is a nonzero divisor on 1 2 c depthnM depthmnR depthnMmM Proof For part a we proceed by induction on dim lf dim R 0 then m is nilpotent and a holds even without the assumption that M is R at lf dim R 2 1 let Qt be the ideal of nilpotent elements in R and make a base change to RQL The dimensions of R and M do not change and the closed fiber does not change Thus we may assume that R is reduced But then m contains a nonzerodivisor at which is consequently also a nonzerodivisor on M because M is R at Make a base change to RzR By the induction hypothesis dim dim RzR dim Since dim dim 7 l and dim RzR dim R 7 l the result follows For part b suppose that y is not a zerodivisor on We want to show that y is not a zerodivisor on MIM Suppose y kills a nonzero element u of MIM We can choose N gt 0 so large that u Z mNMIM It follows that y kills the nonzero image of u in MI mNM and so there is no loss of generality in assuming that I is m primary In this case RI has a finite filtration in which every factor is copy of K Rm When we apply M ER 7 the fact that M is R at implies that MIM has a finite filtration in which every factor is a copy of M 8 Rm E By the Lemma above since y is not a zerodivisor on any of these factors it is not a zerodivisor on MIM as required To prove that M is R at it suf ces to show that Torfla N M 0 for every finitely generated R module N since every R module is a direct limit of finitely generated R modules Since a finitely generated R module N has a finite filtration with cyclic factors it follows that it suf ces to prove that TorRRI M 0 for every ideal I of B Let M MyM Starting with the short exact sequence O MiM MyM O we may apply RI ER 7 to get a long exact sequence part of which is Torf RI M TorfRI MyM MIM L MIM Since M is R at the leftmost term is 0 and since we have already shown that y is not a zerodivisor on MIM it follows that TorI RI MyM 0 for all I as required We next consider the case where depthmR 0 Then we can choose a nonzero element 2 E R such that 2m 0 ie 0 m R i R is exact Applying 7 ER M we have that 0 m R M a M i M is exact This shows both that m R M may be identified with its image which is mM and that Anan mM We have already shown that if y is a nonzerodivisor on MmM then it is a nonzerodivisor on M For the converse suppose it E M is such that ya 6 mM 3 We must show that u 6 mM But zyu E zmM 0 and so zu 0 ie u E Anan which we have already shown is mM as required To prove part c let 1 xh E m be a maximal regular sequence in R Since M is at we may make a base change to Rxl xhR Macl aohM Both sides of the equality we are trying to prove decrease by h since the closed fiber is unchanged Thus we may assume without loss of generality that depthmR 0 We complete the argument by induction on dim Since y E n is a nonzerodivisor on MmM if and only if it is a nonzerodivisor on M if one of these two modules has depth 0 on n then so does the other Therefore we may assume that depthnMmM gt 0 Choose y E n that is a nonzerodivisor on Then y is also a nonzerodivisor on M and MyM is again R flat Let MmM M We may apply the induction hypothesis to MyM to conclude that depthnMyM depthnMyM depthmR since MyM may be identified with the closed fiber of MyM Since depthMyM depthnM 7 l and i 7 depthnMyM depthTl 7 l the result follows E Lemma Let R m K to 511 L be a flat local map of local rings a Rt 7 St is flat where t is an indeterminate b I 7 g is flat Proof For a R ZZt 7 S ZZt is at by base change so that Rt 7 St is at and SM 7 St is a localization and so at Hence St is at over RM and the map factors Rt 7 Rt 7 St Whenver B is at over A and the map factors A 7 W lA 7 T T is also at over W lA This follows form the fact that for W lA modules 0 7 N gt M the map T W71A N 7 T W71A N may be identified with T A N 7 T A M To see this note that we have a map T A M 7 T W71A M and the kernel is spanned by elements of the form w lu 1 7 u 8 w lv But since w in invertible in T we can prove that this is 0 by multiplying by 1112 which yields um 1 7 u 8 um 0 This proves a To prove b note that it suf ces to prove that 0 7 N gt M a map of ISL modules remains injective after applying S 8 i in the case where N and M are finitely generated Given a counterexample we can choose it E EN that is not 0 and is killed when mapped into E M We can choose k so large that u Z 8 N and by the Artin Rees lemma we can choose 11 so large that m M N Q mkN Then there is a commutative diagram N gt M l i NmnM N gt Mm M 4 and we may apply S E i to see that the image of u in S E NmnM N is nonzero even if we map further to S NmkM but maps to 0 in S When applied to maps of nite length R modules the functor S 8 i preserves injectivity because R a S a S is at and S 8 7 and S ER 7 are the same functor on finite lengthAR modules V we have that R ER V g V since V is killed by m5 for some s and RmsR g Rms and so by the associativity of tensor Vg f lt ng RV D We can now make several reductions in studying Lech7s conjecture Theorem In order to prove Lech s conjecture that eR S eS when R m K a S nL is flat local and R has dimension d it su ces to prove the case where dim S dim R d R and S are both complete with in nite residue class eld S has algebraically closed residue class eld B is a domain and S has pure dimension Proof By the Lemma above and the final Proposition in the Lecture Notes of October 20 we can replace R and S by Rt and St and so assume that the residue class fields are infinite Likewise we can replace R and S by their completions We can choose a minimal prime Q of mS such that dim SQ dim SmS By the Lemma on p l of the Lecture Notes of October 30 we have that height Q height dim Since dim S dim SmS dim R we have that dim S dim SQ height By the Theorem on behavior of multiplicities under localization in complete local rings we then have eSQ S eS Thus if eR S eSQ we have eR S eS as well It follows that we may replace S by SQ and so we may assume that dim S dim R d We may have lost completeness but we may complete again By the second Proposition on p l of the Lecture Notes of November 1 we can give a local at map S nL a S 11 U such that S is complete 11 nS and L is algebraically closed Thus we may assume that S has an algebraically closed residue class field We can give a filtration of R by prime cyclic modules RPi l S i S h Then eR is the sum of the eRPi for those i such that dim dim Tensoring with S over R gives a corresponding filtration of S by modules SBS and eS is the sum of the eSPZS for those i such that dim SPZS dim Since dim S dim R dim SmS and for each i dim SPZS dim RPidim SmS the values of i such that dim dim R are precisely those such that dim SPZS dim Thus it suf ces to consider the case where R is a complete local domain If dim R dim S and R is a complete local domain then it follows that S has pure dimension We use induction on the dimension lf dim R 0 then dim S 0 and the result is clear Let R be the normalization of R R ER S is faithfully at over PM and still local the maximal ideal of PM is nilpotent modulo the maximal ideal of R Moreover S Q PM R S so that we may assume that R is normal Suppose that S contains an S submodule of dimension smaller than S say J and choose J maximum so that SJ has 5 pure dimension d Then B does not meet J and so injects into SJ Choose x E m 7 Then at is a nonzerodivisor in R and hence a nonzerodivisor on S and on J It is also a nonzerodivisor on SJ for any submodule killed by x would be a module over SxS and hence of smaller dimension It follows that 07J7S7SJ70 is exact and we get that 0 7gt JmJ 7gt SmS 7gt SJSJ 7gt 0 is exact Then dim JacJ S dim J 7 l lt dimS 7 l dimSacS Therefore SacS does not have pure dimension Because all associated primes of aoR have height one RxR has a filtration whose factors are torsion free modules over rings RPZ of dimension dim R 7 l where the P are the minimal primes of x By the induction hypoth esis every SPZS has pure dimension Since a finitely generated torsion free module over RPZ embeds in a finitely generated free module over RPi the tensor product of a finitely generated torsion free module over RPZ with S also has pure dimension Thus SacS has a filtration whose factors are modules of pure dimension and so has pure dimension itself This contradiction establishes the result B One approach to obtaining a class of local rings R for which Lech7s conjecture holds for every at local map R 7 S is via the notion of a linear mammal Cohen Macaulay module Recall that over a local ring R m K a module M is a Cohen Macaulay module if it is finitely generated nonzero and depthmM dim In particular M is Cohen Macaulay module over R if and only if it is Cohen Macaulay module over RI where I AnnRM Eg the residue class field K Rm is always Cohen Macaulay module over B By a maximal Cohen Macaulay module we mean a Cohen Macaulay module module whose dimension is equal to dim It is not known whether every excellent local ring has a maximal Cohen Macaulay module this is an open question in dimension 3 in all characteristics We write 1M for the least number of generators of the R module M If M is finitely generated over a local or quasi local ring R m K Nakayama7s lemma implies that 1M dim Note the following fact which has proved useful in studying Lech7s conjecture Proposition Let R m K be local and let M be a mammal Cohen Macaulay module Then eM 2 Proof We may replace R by ROS and M by ROS R M if necessary and so assume that the residue class field of R is infinite Let I 1 aod be a minimal reduction of m where d dim Then eM MIM 2 D Math 711 Lecture of October 1 2007 In the proof of the Radu Andr Theorem we will need the result just below A more general theorem may be found in Matsumura Commutative Algebra W A Benjamin New York 1 970 Ch 8 200 Theorem 49 p 146 but the version we give here will su ice for our purposes First note the following fact if I Q A is an ideal and M is an A module then Tor 14AI M 0 if and only if the map I A M a IM which is alway surjective is an isomorphism This map sends i u gt gt in The reason is that we may start with the short exact sequence 0 a I a A a AI a 0 and apply Q A M The long exact sequence then gives in part 0 Tor 14A M a Tor f AI M a I 34 M a M The image of the rightmost map is IM and so we have 0 a Tor f AI M a 84 M a M a o is exact from which the statement we want is clear Theorem local criterion for atness Let A a B be a local homomorphism of local rings let M be a nitely generated B module and let I be a proper ideal of A Then the following three conditions are equivalent 1 M is flat over A 2 MIM is flat over AI andI A M a IM is an isomorphism 3 MIM is flat over AI and Tor 14AI M 0 Proof The discussion of the preceding paragraph shows that 2 gt 3 and l i 3 is clear It remains to prove 3 i l and so we assume To show that M is at it su ices to show that if No Q N is an injection of finitely generated R modules then N0 A M a N A M is injective Moreover by the Proposition at the bottom of p 3 of the Lecture Notes of September 14 we need only prove this when N has finite length Consequently we may assume that N is killed by a power of I and so we have that IkN Q N0 forsomek Let N N0Ik iN forO S i S k TheNO Q N1 Q Q Nk N and it su ices to show that N A M a Nj1 A M is injective for each j We have now reduced to the case where Q Nj1Nj is killed by I From the long exact sequence for Tor arising from applying Q A M to the short exact sequence 0 Nj Nj1 Q 0 wehave Tor f Q M a N AM a N1 AM 1 2 is exact and so it suf ces to show that if Q is a nitely generated A module killed by I then Tor 14Q M 0 Since Q is killed by I we may think of it as a finitely generated module over AI Hence there is a short exact sequence 0gtZgtAIhgtQgt0 Applying i A M from the long exact sequence for Tor we have that Tor f AIh M Tor f Q M Z A M i AIh A M is exact By hypothesis Tor 14AI M 0 and so the leftmost term is 0 It follows that Tor 14Q M Ker 04 To conclude the proof it will suf ce to show that 04 is injective Hence it is enough to show that 7 A M is an exact functor on A modules Y that are killed by I For such an A module Y we have that Y AMg Y AI 141 AM Y AH AI AM gYlt Ai14UW and this is an isomorphism as functors of Y Since MIM is at over AI the injectivity of Oz follows D We want to record the following observation Proposition Let f A a B be a homorphism of Noetherian rings ofprime characteristic p gt 0 such that the kernel off consists of nilpotent elements ofA and for every element b E B there exists 1 such that bq E fA Then Spec f Spec B a Spec A is a homeomorphism recall that this map sends the prime ideal Q E Spec B to the contraction f 1Q of Q to A The inverse maps P E Spec A to the radical of PB which is the unique prime ideal ofB lying ouer P Proof Since the induced map Spec A a Spec AJ is a homeomorphism whenever J is an ideal whose elements are nilpotent and the unique prime of AJ lying over P E Spec A is PJ the image of P in AJ there is no loss of generality in considering instead the induced map Ared a Bred which is injective We therefore assume that A and B are reduced and by replacing A by its image we may also assume that A Q B Then A gt B is an integral extension since every b E B has a power in A and it follows that there is a prime ideal Q of S lying over a given prime P of A If u E Q then uq E A for some 1 and so uq E Q A P It follows that Q Q Rad PB and since Q is a radical ideal containing PB we have that Q Rad PB Therefore as claimed in the statement of the Proposition we have that Rad PB is the unique prime ideal of S lying over P This shows that Spec f is bijective To show that g Spec f is a homeomorphism it suf ces to show that its inverse is continuous ie that 9 maps closed sets to closed sets But for any b E B we may choose 1 so that bq E A and then gVbB Vqu g Spec A D The Radu Andr Theorem is valid even when R is not reduced In this case we do not want to use the notation Rlq Instead we let Re denote R viewed as an R algebra via the structural homomorphism Fe R a R We restate the result using this notation 3 Theorem Radu Andr Let R and S be F hite rings of prime characteristic 1 gt 0 such that R a S is flat with geometrically regular bers Then for all e Re R S a Se is faithfully flat Proof Let Te Re R S Consider the maps S a T8 a Se Any element in the kernel of S a Se is nilpotent It follows that this is also true of any element in the kernel of S a T5 Note that every element of Te has qth power in the image of S since r 8 sq rq 8 sq l rqsq It follows that Spec Se a Spec Te a Spec S are homeomorphisms Hence if Se is at over Te then it is faithfully at over Te It is easy to see that geometric regularity is preserved by localization of either ring and the issue of atness is local on the primes of Se and their contractions to T5 Localizing Se at a prime gives the same result as localizing at the contraction of that prime to S It follows that we may replace S by a typical localization SQ and R by Rp where P is the contraction of S ot R Thus we may assume that R m K is local and that R a S is a local homomorphism of local rings Evidently Se and Re are local as well and it follows from the remarks in the first paragraph that the maps S a T8 a Se are also local Let 7715 be the maximal ideal of Re of course if we identify Re with the ring R then 7715 is identified with the maximal ideal in of R We shall now prove that A Te a Se B is at using the local criterion for atness taking I meTe Note that since R a S is at so is Re a Re R S Te Therefore meTe g 7715 R S The expansion of I to B Se may be identified with meSe and since Re a Se as a map of rings is the same as R a S we have that Se is at over Re and we may identify meSe with 7715 8mg Se There are two things to check One is that BI is at over AI which says that Seme 8mg Se is at over Re R Sme R S The former may be identified with SmSe and the latter with Ke K SmS since Reme may be identified with Ke Since R is F finite so is K and it follows that Kw Klq is a finite purely inseparable extension of K Since the fiber K a K R S SmS is geometrically regular we have that Kw R SmS Ke K SmS is regular and in particular reduced Since it is purely inseparable over the regular local ring SmS we have from the Proposition on p 2 that Kltegt K SmS 2 SmSK1q is a local ring Hence it is a regular local ring We have as well that SmSe E SmS1q is regular since SmS is and is a module finite extension of SmSK1q Thus BIB SmS1q is module finite local and Cohen Macaulay over AIA SmSK1q which is regular local By the Lemma on p 8 of the Lecture Notes of September 8 BIB is free over AI and therefore at Finally we need to check that I A B a IB is an isomorphism and this the map t 7715 R S R RS Sltegt a 7715 8mg Se 4 The map takes u s EU to u 8 so We prove that b is injective by showing that it has an inverse There is an Re bilinear map me X 58 me R S R RS 58 that sends u u gt gt u 8 1 1 This induces a map 11 1 mltegt Rltegt 58 7718 R 5 Rltegt Rs 58 and it is straightforward to see that 1 o b sends u s u gt gt u l su u s u and that b o 1 sends u 8 u to itself D Note that if a Noetherian ring R is reduced and R a S is at with reduced bers over the minimal primes of R then S is reduced Because nonzerodivisors in R are nonzerodivisors on S we can replace R by its total quotient ring which is a product of fields and S becomes the product of the fibers over the minimal primes of R Hence if R a S is at with geometrically regular or even reduced fibers and R is reduced so is S This is used several times in the sequel Theorem If R a S is a flat map of F nite rings of prime characteristic p gt 0 with geometrically regular bers and R is strongly F regular then so is S Proof We can choose c E R0 such that RC is regular and then we know that there is an R linear map 6 Rl Y a R sending cl Y gt gt 1 Now RC a SC is at with regular fibers and RC is regular so that SC is regular as well By Theorem at the bottom of p 6 of the Lecture Notes of September 24 it su ices to show that there is an S linear map Sl Y a S such that cl Y gt gt 1 Let 6V 6 R ls Rlq R S a S so that 6V is an S linear map such that 6c1 7 1 1 By the Corollary on p 3 of the Lecture Notes of September 28 the inclusion Rlq R S a Slq which takes clq 1 to clq has a splitting Oz Sl Y a Rlq R S that is linear over Rlq R S Hence 04 is also S linear and 6V 0 04 is the required S linear map from Slq to S D We also can improve our result on the existence of big test elements now Theorem Let R be a reduced F nite ring of prime characteristic p gt 0 and let c E R0 be such that RC is strongly F regular Also assume that there is an R linear map RlP a R that sends l to c If S is F nite and flat ouer R with geometrically regular bers then the image of c3 in S is a big test element for S 3 In particular for every element c as above c is a completely stable big test element Hence every element c of R0 such that RC is strongly F regular has a power that is a completely stable big test element and remains a completely stable big test element after every geometrically regular base change to an F nite ring Proof By the Theorem at the bottom of p l of the Lecture Notes of September 26 to prove the result asserted in the first paragraph it suf ces to show that the image of c in S has the same properties because the map is at the image is in SO and so it suf ces to show that SC is strongly F regular and that there is an S linear map SlP a S such that the value on 1 is the image of c in S But the map RC a SC is at RC is strongly F regular and the fibers are a subset of the fibers of the map R a S corresponding to primes of R not containing 0 Hence the fibers are geometrically regular and so we can conclude that SC is strongly F regular We have an R linear map RlP a R that sends lgt gt c We may apply 7 ER 15 to get a map Rlp R S a S sending l to the image of c and then compose with a splitting of the inclusion Rlp R S a Slp to get the required map The statement of the second paragraph now follows because a localization map is geo metrically regular and F finite rings are excellent so that the map from a local ring to its completion is geometrically regular as well To prove the third statement note that whenever RC is strongly F regular there is a map RlP a R whose value on 1 is a power of c this is a consequence of the first Lemma on p l of the Lecture Notes of September 26 D Mapping cones Let B and A be complexes of R modules with differentials 6 and d respectively We assume that they are indexed by Z although in the current application that we have in mind they will be left complexes ie all of the negative terms will be zero Let b be a map of complexes so that for every n we have b7 3 a A7 and all the squares d An n A7171 ml wilT 571 3 B7171 commute The mapping cone Ciquot of b is defined so that C5 2 A7 69 B71 with the differential that is simply 1 on A7 and is 71 71on4 9674 on B71 Thus under the differential in the mapping cone an e bun H dnltanlt71 1 ltbn71e nimbnil If we apply the differential a second time we obtain dn71dnan71n71 n71bn7171n72 n726n71bn71gt696717267171ltbn71gt7 which is 0 and so we really do get a complex We frequently omit the superscript 91quot and simply write C for C2 Note that A Q C is a subcomplex The quotient complex is isomorphic with 3 except that degrees are shifted so that the degree n term in the quotient is B714 This leads to a long exact sequence of homology a HnA a HnC a n1B a n1A a One immediate consequence of this long exact sequence is the following fact Proposition Let b B a A be a map of left complezres Suppose that A and B are acyclic and that the induced map of augmentations H0B a H0A which may also be described as the induced map BOdoBl a AOd0A1 is injectiue Then the mapping cone is an acyclic left complezr and its augmentation is AOdA1 0BO D The Koszul complex The Koszul complex ICic1 icn R of a sequence of elements 1 icn E R on B may be defined as an iterated mapping cone as follows Let ICac1 R denote the left complex in which 1C1ac1 R Rub a free R module K06 R R and the map is such that ul gt gt x1 le we have 0 Rul H R a 0 Then we may define ICic1 icn R recursively as follows If n gt 1 multiplication by it in every degree gives a map of complexes ICac1 714 R i ICac1 714 and we let ICac1 icn R be the mapping cone of this map We may prove by induction that Kn 1 icn R is a free complex of length n in which the degreej term is isomorphic with the free R module on generators 0 S j S n Even more specifically we show that we may identify ICj 1 icn R with the free module on generators u7 indexed by the j element subsets o of 1 2 n in such a way that ifoi1 ij with l ll lt ltij n then J du7 27lt 1ituoit 151 We shall use the alternative notation mil24 for u7 in this situation We also identify uw the generator of ICO1 icn R with l E B To carry out the inductive step we assume that A ICac1 714 R has the specified form We think of this complex as the target of the map multiplication by ion and index its generators by the subsets of 1 2 n 7 1 This complex will be a subcomplex of ICac1 icn R We index the generators of the complex B which will be the domain for the map given by multiplication by ion and which is also isomorphic to Math 711 Lecture of December 4 2006 The next two results suggest that characteristic p techniques may be helpful in proving the existence of linear maximal Cohen Macaulay modules Let R be a ring Of prime characteristic p gt 0 The Frobenius closure IF of an ideal I Q R is r E R for some e E N r175 E filial Note that once this holds for one value of e it also holds for all larger values Alternatively IF is the union of contractions of I to R under the maps Fe R a R as e varies the union is increasing Note that I Q IF When R is Noetherian the contractions of I under the various F5 are the same for all 6 gtgt 0 Thus if J IF we can choose 6 gtgt 0 such that legl Q Ilpgl But since I Q J the opposite inclusion is obvious Hence for all e gt 0 IFW El Ilpgl Notice that when r 6 IF we have that lrp5 E Ilpgl for all e gt 0 so that IF Q I the tight closure of I in R Theorem D Hanes Let R m K be an F nite Cohen Macaulay local ring ofprime characterisitc p gt 0 Suppose that there exists an ideal I Q m generated by a system of parameters such that IF m Then for all su ciently large e 5R is a linear mazrimal Cohen Macaulay module ouer R Proof Choose any e such that mlpgl Ilpgl Since R is F finite 5R is a finitely generated module over R and obviously a maximal Cohen Macaulay module if 1 mm is a system of parameters in R they form an R sequence on 5R because mfg mg is a regular sequence on R Note that under the identification of 5R with R I 5R becomes Ilpgl and meR becomes mlpgl Since Ilpgl mlpgl we have that IER m 5R as required D Theorem Let R m K be any F nite Cohen Macaulay ring of prime characteristic p gt 0 Then R has a free ezrtension S such that R a S is local the induced map of residue class elds is an isomorphism and S has a linear mazrimal Cohen Macaulay module Proof Let 1 icd be any system of parameters for R Then for any su iciently large integer e E N we have that mlpgl Q I Let Z1 Zd be indeterminates over R let T RZ1 Zd and let S TJ where J is generated by the elements Z5 7 obi l S i S d Evidently S is module finite over R and so its maximal ideals all lie over m But 5 SmSgTZg7 1tdT a zero dimensional local ring with residue class field isomorphic with K Thus S is local with residue class field K Evidently S is free over R on the basis consisting of the images of all monomials Z ng with 0 S ai S p5 for l S i S d Thus S satisfies all of the requirements of the Theorem provided that we can show that it has a linear maximal Cohen Macaulay module 1 Let 21 zd be the images of Z1 Zd respectively in S Clearly 21 zd is a system of parameters for S since killing them produces Rxl acdR Since S is free over the Cohen Macaulay ring B it is Cohen Macaulay It will therefore suf ce to show that it satisfies the hypothesis of the preceding Theorem In fact the maximal ideal n of S is the Frobenious closure of 21 zdS The ideal n is generated by m and the 21 But mlp l Q 9517 7d Q 2 17 zzd p since at 255 in S while it is obvious that every 255 E 21 zd pgl D We next want to discuss some results concerning the existence of linear maximal Cohen Macaulay modules over Veronese subrings of polynomial rings This problem may seem rather special but the ideas used to solve the problem in dimension three for example can be used to prove the existence of approximately linear77 maximal Cohen Macaulay modules for standard graded domains over a perfect field of positive characteristic in dimension 3 and this circle of ideas has provided a substantial body of results on Lech7s conjecture for standard graded algebras Let K be field and let S be a standard graded K algebra By the tth Veronese subring S of S we mean 00 9 Sit 20 which may also be described as the K algebra KSt generated by St Clearly S is module finite over S since for every homogeneous element F of S Ft 6 S t Both Segre products and Veronese subrings arise naturally in projective geometry Let ProjR denote the projective scheme associated with a standard graded K algebra B This scheme is covered by open a ines of the form Spec Rpjo where F is a form of positive degree in B To get an open cover it suf ces to use finitely many F any set of homgeneous generators of an ideal primary to the homogeneous maximal ideal wil provide such a cover and we may take the to be one forms An important reason for studying Segre products is that ProjR K S E ProjR gtlt ProjS The Veronese subrings of S have the property that ProjSt ProjS for all t A specific homogeneous coordinate ring S for a projective scheme X over K which means that X ProjS gives an embedding of X in P by taking a degree preserving mapping of a polynomial ring KjXO X onto B so as to give an isomorphism of vector spaces in degree 1 The Veronese subrings of R turn out to give a family of different embeddings of X into projective spaces Although this is an important motivation for studying Veronese subrings we shall not need to take this point of view in the sequel Math 711 Lecture of September 21 2007 F nite rings Let R be a Noetherian ring of prime characteristic p gt 0 R is called F nite if the Frobenius endomorphism F R a R makes B into a module finite R algebra This is equivalent to the assertion that R is module finite over the subring FR rp r E R which may also be denoted RP When R is reduced this is equivalent to the condition that Rlp is module finite over R since in the reduced case the inclusion R Q Rlp is isomorphic to the homomorphism F R a R Proposition Let R be a Noetherian ring of prime characteristic p gt 0 a R is F nite if and only if Bred is F nite b R is F nite if and only and only if F5 R a R is module nite for all e if and only if F5 R a R is module nite for some e 2 l IfR is F nite so is every homomorphic image of R IfR is F nite so is every localization of R IfR is F nite so is every algebra nitely generated over R If R m K is a complete local ring R is F nite if and only if the eld K is F nite IfR is F nite so is the formal power series ring RHxl IfR is F nite so is every ring essentially of nite type over R IfK is a eld that is nitely generated as a eld over a perfect eld then every ring essentially of nite type over K is F nite Proof Parts c and d both follow from the fact that if B is a finite set of generators for R as FR module the image of B in S will generate S over FS if S RJ and also if S WAR In the second case it should be noted that FW 1R may be identified with W 1FR because localizing at w and a wp have the same effect For part a note that if R is F finite so is Bred by part c since Bred RJ where J is the ideal of all nilpotent elements Now suppose that I is any ideal of R such that RI is F finite Let the images of ill un span RI over the image of PtDP and let v1 vh generate I over B Let A Rpm Rpun Then R 14va th If we substitute the same formula for each copy of R occuring in an Rvj term on the right we find thtat R A ZRpuZvj ZRv vj 13971 ii 1 2 It follows that the n nh elements ui and uZuj span RI2 over the image of RI2p Thus RIZ is F finite By a straightforward induction RIZk is F finite for all k Hence if I J is the ideal of nilpotents we see that R itself is F finite For part b note that if F R a R is F finite so is the e fold composition On the other hand if F5 R a R is finite so is F5 S a S where S Bred Then we have S Q Slp Q Slq and since Slq is a Noetherian S module so is Slp Thus S is F finite and so is R by part a To prove e it suf ces to consider the case of a polynomial ring in a finite number of variables over R and by induction it suf ces to consider the case where S Likewise for part f we need only show that is F finite Let ul un span R over RP Then in both cases the elements nine 1 S l S n l S j S p 7 l span S over Sp Rpacp respectively PJ HQET H For g note that K Rm so that if R m K is F finite so is K If R is complete it is a homomorphic image of a formal power series ring KHacl xnll where K is the residue class field of B By part f if K is F finite so is B Part h is immediate from parts e and For part first note that K itself is essentially of finite type over a perfect field and a perfect field is obviously F finite The final statement is then immediate from part D A proof of the following result of Ernst Kunz would take us far afield We refer the reader to Kunz On Noetherian rings of characteristic 1 Amer J Math 98 1976 99971013 Theorem Kunz Every F nlte ring is excellent We are aiming to prove the following result about F finite rings Theorem existence of test elements Let R be a reduced F nlte ring and let 0 E R0 be such that RC is regular Then 0 has a power 0N that is a completely stable big test element This is terrifically useful Elements 0 E R0 such that RC is regular always exist In any excellent ring P E Spec R Rp is regular is open Since the complement is closed there is an ideal I such that VI P E Spec R Rp is not regular We refer to this set of primes as the singular locus of Spec R or of R Note that if R is reduced we cannot have I Q p for any minimal prime p of B because that would mean the Rp is not regular and RF is a field Hence I is not contained in the union of the minimal primes of R which means that I meets R lf 0 E I R then RC is regular primes that do not contain 0 cannot contain I Hence in a reduced F finite or any reduced excellent ring there is always an element r E R0 such that RC is regular and this means that Theorem above can be applied Hence Corollary Euery reduced F nite ring has a completely stable big test element It will take some time before we can prove the Theorem on existence of test elements Our approach requires studying the notion of a strongly F regular ring We give the definition below However we first want to comment on the notion of an F split ring De nition Fsplit rings Let R be a ring of prime characteristic p gt 0 We shall say that R is F split if under the map F R a R the left hand copy of R is a direct summand of the right hand copy of R If R is F split F R a R must be injective This is equivalent to the condition that R be reduced An equivalent condition is therefore that R be reduced and that R be a direct summand of R1 p as an R module ie there exists an R linear map 6 Rlp a 1 such that 61 1 Proposition Let R be a reduced ring of prime characteristic p gt 0 The following conditions are equivalent 1 R is F split 2 R a Rlq splits as a map of R modules for all 1 3 R a Rlq splits as a map of R modules for at least one value ofq gt 1 Proof l i Let 6 Rlp a R be a splitting Then for all 1 p5 gt 1 if q pe l we may define a splitting 65 Rl Y a Rlq by sealq solmy Thus the diagram Rlq L Rlq gT gT RlP L R commutes where the vertical arrows are the isomorphisms rlP gt gt rlq and r gt gt PM respectively Of course 61 6 Then 65 is Rlq linear and in particular R linear Hence the composite map 910920065R1 1a3 gives the required splitting 2 i 3 is clear Finally assume Then R Q Rlp Q Rlq so that a splitting Rl Y a B may simply be restricted to Rlp and 1 follows E Strongly Fregular rings We have de ned a ring to be weakly F regular if every ideal is tightly closed and to be F regular if all of its localizations have this property as well We next want to introduce the notion of a strongly F regular ring R for the moment we make this definition only when R is F finite The definition is rather technical but this condition turns out to be easier to work with than the other notions It implies that every submodule of every module is tightly closed it passes to localizations automatically and it leads to a proof of the Theorem on existence of test elements stated on p 2 Of course the value of this notion rests on whether there are examples of strongly F regular rings We shall soon see that every regular F finite ring is strongly F regular Let 1 S t S r S s be integers lf K is an algebraically closed field or an F finite field and X is an r gtlt s matrix of indeterminates over K then the ring obtained from the polynomial ring in the entries of X by killing the ideal 1X generated by the t gtlt t minors of X is strongly F regular and so is the ring generated over K by the r gtlt r minors of X this is the homogeneous coordinate ring of a Grassman variety The normal rings generated by finitely many monomials in indeterminates are also strongly F regular Thus there are many important examples In fact in the F finite case every ring that is known to be weakly F regular is known to be strongly F regular Conjecture Every weakly F regular F nlte ring is strongly F regular This is a very important open question It is known to be true in many cases we shall discuss what is known at a later point De nition strong Fregularity Let R be a Noetherian ring of prime characteristic 1 gt 0 and suppose that R is reduced and F finite We define R to be strongly F regular if for every 0 E B0 there exists qC such that the map R a Rlqc that sends 1 gt gt clqc splits overR That is for all c E B0 there exist qC and an R linear map 6 Rl YC a R such that 601 76 l The element qC will usually depend on c For example one will typically need to make a larger choice for op than for 0 Remark The following elementary fact is very useful Let h R a S be a ring homomor phism and let M be any S module Let u be any element of M Suppose that the unique R linear map R a M such that 1 gt gt u and r gt gt ru splits over R Then R is a direct summand of S ie there is an R module splitting for h R a S In fact if 6 M a R is R linear and 6a l we get the required splitting by defining s 6su for all s E S Note also that the fact that R a M splits is equivalent to the assertion that R a Ru 5 such that 1 gt gt u is an isomorphism of R modules together with the assertion that Ru is a direct summand of M as an R module We may apply this remark to the case where S Rlqc in the above de nition Thus Proposition A strongly F regular ring R is F split We also note Proposition Suppose that R is a reduced Noetherian ring of prime characteristic p gt 0 that c E R0 and that R a Rlqc sending 1 gt gt clqc splits ouer R Then for all g 2 gm the map R a Rlq sending 1 gt gt clq splits ouer R Proof lt su ices to show that if we have a splitting for a certain q we also get a splitting for the next higher value of 1 which is qp Suppose that 6 Rl Y a R is R linear and 6clq 1 We de ne 6V RlPq a Rlp by the rule 6rlpq 6r1q1P That is the diagram Rlpq 9quot 311 21 21 Rlq L R commutes Then 6V is Rlp linear and 6clpq 1 E Rlp By the Remark beginning on the bottom of p 4 R a Rlq splits and so R is F split ie we have an R linear map RlP a R such that M1 1 Then 0 6V is the required splitting D The following fact is now remarkably easy to prove Theorem Let R be a strongly F regular ring Then for every inclusion of N Q M of modules these are not required to be nitely generated N is tightly closed in M Proof We may map a free module G onto M and replace N by its inverse image H Q G Thus it su ices to show that H HE when G is free Suppose that u E We want to prove that u E H Since it 6 Hg for all 1 gtgt 0 cuq E qul Choose qC such that R a Rlqc with 1 gt gt clqc splits Then fix 1 2 qC such that cuq E qul Then the map R a Rlq sending 1 a clq also splits and we can choose 6 Rl Y a R such that 6c1q 1 The fact that cuq E qu gives an equation 77 cuq E rihg 2391 Math 711 Lecture of September 7 2007 Symbolic powers We want to make a number of comments about the behavior of symbolic powers of prime ideals in Noetherian rings and to give at least one example of the kind of theorem one can prove about symbolic powers of primes in regular rings there was a reference to such theorems in 5 on p 15 of the notes from the first lecture Let P be a prime ideal in any ring We define the nth symbolic power P of P as 7 E R for some 5 E R7Ps7 E P Alternatively we may define P as the contraction of Pan to R It is the smallest P primary ideal containing P If R is Noetherian it may be described as the P primary component of P in its primary decomposition While Pm P and PW P when P is a maximal ideal in general P01 is larger than P even when the ring is regular Here is one example Let at y z and 5 denote indeterminates over a field K Grade B Kb y 2 so that at y and 2 have degrees 3 4 and 5 respectively Then there is a degree preserving K algebra surjection R a Kt3 t4 255 g Kt that sends at y and z to t3 t4 and 155 respectively Note that the matrix 7 at y z X i y 2 362 t3 t4 t5 t4 t5 t6 The second matrix has rank 1 and so the 2 gtlt 2 minors of X are contained in the kernel P of the surjection R 7 KW 15 155 Call these minors f 2 7 242 g 3 7 yz and h 2433 7 22 It is not di icult to prove that these three minors generate P ie P f g We shall exhibit an element of Pm 7 P Note that f g and h are homogeneous of degrees 8 9 and 10 respectively is sent to the matrix Next observe that 92 7 fh vanishes mod xR it becomes 7yz2 7 7y2722 0 Therefore 92 7 fh am 92 has an 6 term which is not canceled by any term in fh so that u 31 0 Of course we could check this by writing out what u is in a completely explicit calculation The element 92 7 fh E P2 is homogeneous of degree 18 and x has degree 3 Therefore it has degree 15 Since at P and 11 6 P2 we have that u 6 Pa 1 2 But since the generators of P all have degree at least 8 the generators of P2 all have degree at least 16 Since degu 15 we have that u Z P2 as required Understanding symbolic powers is di icult For example it is true that if P Q Q are primes of a regular ring then P Q 6271 but this is somewhat di cult to prove See the Lectures of October 20 and November 1 6 and 8 of the Lecture Notes from Math 711 Fall 2006 This statement about inclusions fails in simple examples where the ring is not regular For example consider the ring RKUV W X Y ZUXVYWZK1L v w at y z where the numerator is a polynomial ring Then R is a hypersurface it is Cohen Macaulay normal with an isolated singularity It can even be shown to be a UFD Let Q be the maximal ideal generated by the images of all of the variables and let P be the prime ideal 1 11115 yzR Here RP E Then P Q Q but it is not true that Pm Q 622 In fact since 77115 yy wz E P2 and u P we have that x 6 Pa while at Z 622 which is simply Q2 since Q is maximal The following example due to Rees shows that behavior of symbolic powers can be quite bad even in low dimension Let P be a prime ideal in a Noetherian ring B Let t be an indeterminate over B When I is an ideal of R a very standard construction is to form the Rees ring RIt RIt1 t Q RH which is finitely generated over R if f1 fh generate the ideal I then Rift leuf M An analogous construction when I P is prime is the symbolic power algebra RPtP2t2P t g Rm We already know that this algebra is larger than RPt but one might still hope that it is finitely generated Roughly speaking this would say that the elements in P 7 P for su iciently large n arise a consequence of elements in P0 7 Pk for finitely many values of k However this is false Let R qx Y ZX3 Y3 Z3 where C is the field of complex numbers This is a two dimensional normal surface it has an isolated singularity It is known that there are height one homogeneous primes P that have infinite order in the divisor class group this simply means that no symbolic power of P is principal David Rees proved that the symbolic power algebra of such a prime P is not finitely generated over B This was one of the early indications that Hilbert7s Fourteenth Problem might have a negative solution ie that the ring of invariants of a linear action of a group of invertible matrices on a polynomial ring over a field K may have a ring of invariants that is not finitely generated over K M Nagata gave examples to show that this can happen in 1958 Analytic spread In order to give a proof of the result of Rees described above we introduce the notion of analytic spread Let R m K be local and I Q m an ideal When K is infinite the following two integers coincide 1 The least integer n such that I is integral over an ideal J Q I that is generated by n elements 2 The Krull dimension of the ring K R RIt The integer defined in 2 is called the analytic spread of I and we shall denote it anI See the Lecture Notes of September 15 and 18 from Math 711 Fall 2006 for a more detailed treatment The ring in 2 may be written as SK ImIEBI2mI2 EBI mI a Note that if we define the associated graded ring ngR of R with respect to I as Rel1212131n1w1 69 which may also be thought of as RItIRIt then it is also true that S E K R ng The idea underlying the proof that when K is infinite and h anI one can find f1 fh E I such that I is integral over J f1 fhR is as follows The K algebras S is generated by its one forms 1f K is infinite one can choose a homogeneous system of parameters for S consisting of one forms these are elements of ImI and are represented by elements f1 fh of I Let J be the ideal generated by f1 fh in R The S is module finite over the image of K 8 RJt and using this fact and Nakayama7s Lemma on each component one can show that RIt is integral over RJt from which it follows that I is integral over J Proof that Rees s symbolic power algebra is not nitely generated Here is a sketch of Rees7s argument Assume that the symbolic power algebra is nitely generated We now replace the graded ring R by its localization at the homogeneous maximal ideal By the local and homogeneous versions of Nakayama7s Lemma the least number of generators of an ideal generated by homogeneous elements of positive degree does not change It follows that P continues to have the property that no symbolic power is principal We shall prove that the symbolic power algebra cannot be nitely generated even in this localized situation which implies the result over the original ring R Henceforth R m K is a normal local domain of dimension 2 and P is a height one prime such that no symbolic power of P is prinicpal We shall show that the symbolic power algebra of P cannot be nitely generated over R Assume that it is nitely generated This implies that for some integer k PW Pk for all positive integers 11 Let I P0 The ring S RIt has dimension 3 since the transcendence degree over R is one The elements at y are a system of parameters for R We claim that there is a regular sequence of length two in m on each symbolic power J Pm To see this we take at to be the first term Consider JacJ If there is no choice for the second term then the maximal ideal m of R must be an associated prime of JxJ and we can choose 1 E J 7 acJ such that my Q xJ But then yv E xR and at y is a regular sequence in R It follows that v 11 with u E R 7 J Then mam Q acJ shows mu Q J But elements of m 7 P are not zerodivisors on J so that u E J a contradiction It follows that every system of parameters in R is a regular sequence on J J is a Cohen Macaulay module Thus if the symbolic power algebra is nitely generated any is a regular sequence on every P01 and therefore at y is a regular sequence in S It follows that killing 14 decreases the dimension of the ring S by two Since the radical of at y is the homogeneous maximal ideal of R we see that R S has dimension one This shows that the analytic spread of I is one But then I is integral over a principal ideal In a normal ring principal ideals are integrally closed Thus I is principal But this contradicts the fact that no symbolic power of P is principal D The notion of tight closure for ideals We next want to introduce tight closure for ideals in prime characteristic 1 gt 0 We need some notations If R is a Noetherian ring we use R0 to denote the set of elements in R that are not in any minimal prime of R If R is a domain RO R 7 Of course RO is a multiplicative system We shall use 6 to denote an element of N the nonnegative integers For typographical convenience shall use 1 as a symbol interchangeable with 195 so that whenever one writes 1 it is understood that there is a corresponding value of e such that q p5 even though it may be that e is not shown explicitly When R is an arbitrary ring of characteristic 1 gt 0 we write FR or simply F for the Frobenius endomorphism of the ring R Thus Fr rp for all r E R F or Fe indicates the eth iteration of FR so that F5 r rq for all r E R If R has characteristic 1 Ilql denotes the ideal generated by all 1 th powers of elements of I If one has generators for I their qth powers generate Ilql More generally if f R a S is any ring homomorphism and I Q R is an ideal with generators rEA the elements fr A generate IS De nition Let R be a Noetherian ring of prime characteristic 1 gt 0 Let I Q R be an ideal and let u E R be an element Then it is in the tight closure of I in R denoted Iquot if there exists c E R0 such that for all suf ciently large 1 CM 6 Ilql This may seem like a very strange definition at first but it turns out to be astonishingly useful Of course in presenting the definition we might have written for all suf ciently large e cape E Ilpgl 77 instead The choice of c is allowed to depend on I and u but not on 1 It is quite easy to see that I is an ideal containing I Of great importance is the following fact to be proved later Theorem Let R be a Noetherian ring ofprime characteristic 1 gt 0 IfR is regular then every ideal ofR is tightly closed If one were to use tight closure only to study regular rings then one might think of this Theorem as asserting that the condition in the Definition above gives a criterion for when an element is in an ideal that on the face of it is somewhat weaker than being in the ideal Even if the whole theory were limited in this fashion it provides easy proofs of many results that cannot be readily obtained in any other way We want to give a somewhat different way of thinking of the definition above First note that it turns out that tight closure over a Noetherian ring can be tested modulo every minimal prime Therefore for many purposes it suf ces to consider the case of a domain Let R be any domain of prime characteristic 1 gt 0 Within an algebraic closure L of the fraction field of R we can form the ring rlq r E R The Frobenius map F is an automorphism of L this is the image of R under the inverse of Fe and so is a subring of L isomorphic to R We denote this ring Rlq This ring extension of R is unique up to canonical isomorphism it is independent of the choice of L and its only R automorphism is the identity r has a unique qth root in Rlq since the difference of two distinct 1 th roots would be nilpotent and so every automorphism that fixes r fixes rlq as well Moreover there is a commutative diagram R Rlq ltgt H ire R P R where both vertical arrows are isomorphisms and F er rlq The reduced case When R is reduced rather than a domain there is also a unique up to unique isomor phism reduced R algebra extension ring Rlq whose elements are precisely all qthg roots of elements of B One can construct such an extension ring by taking the map BER to give the algebra map so that one has the same commutative diagram as in the domain case The proof of uniqueness is straightforward if 51 and 52 are two such extensions the only possible isomorphism must let the unique qth root of 7 E R in 51 correspond to the unique qth root of R in 52 for all 7 E R It is easy to check that this gives a well defined map that is the identity on R and that it is a bijection and a homomorphism In both the domain and the reduced case we have canonical embeddings Rl Y gt Rlq when 1 S 1 and we define R X U Rlq q When one has that GM rlff rhth one can take qth roots to obtain cm riqf1 mm Keep in mind that in a reduced ring taking qth roots preserves the ring operations We can therefore rephrase the definition of tight closure of an ideal I in a Noetherian domain R of characteristic 1 gt 0 as follows 7 An element u E R is in I iff there is an element 0 E R0 such that for all su iciently large 1 clqu E IRlq Heuristically one should think of an element of R that is in I S where S is a domain that is an integral extension of R as almost in I Note that in this situation one will have u flsl fhsh for f1 fh E I and 81 5h 6 S and so one also has it 6 ISO where SO Rsl sh is module finite over R 7 The condition is weaker in a way Rlq Q R is an integral extension of R but u is not necessarily in I R instead it is multiplied into I R by infinitely many elements clq These elements may be thought of as approaching l in some vague sense this is not literally true for a topology but the exponents lq a 0 as q a 00 Some useful properties of tight closure We state some properties of tight closure for ideals proofs will be given later Here R is a Noetherian ring of prime characteristic 1 gt 0 and I J are ideals of R We shall write Bred for the homomorphic image of R obtained by killing the ideal of nilpotent elements R m K is called equtdz39mensz39onal if for every minimal prime P of R dim RP dim An algebra over R is called essentially of nite type over R if it is a localization at some multiplicative system of a finitely generated R algebra If I J are ideals of R we define I R J 7 E R rJ Q I which is an ideal of R If J uR we may write I R u for I R uR Excellent rings In some of the statements below we have used the term excellent ring77 The excellent rings form a subclass of Noetherian rings with many of the good properties of finitely generated algebras over fields and their localizations We shall not give a full treatment in these notes but we do discuss certain basic facts that we need For the moment the reader should know that the excellent rings include any ring that is a localization of a finitely generated algebra over a complete local or semilocal ring The class is closed under localization at any multiplicative system under taking homomorphic images and under formation of finitely generated algebras We give more detail later Typically Noetherian rings arising in algebraic geometry number theory and several complex variables are excellent Here are nine properties of tight closure Property 2 was already stated as a Theorem earlier 1 I g 1 IIf1 Q J then 1 Q Jquot N If R is regular every ideal of R is tightly closed AAA co If R Q S is a module finite extension IS R Q Iquot 4 If P1 Ph are the minimal primes of R then u E R is in I if and only if the image of u in Dj RPj is in the tight closure of IDj in D working over D for 1 S j S h A UV V If u E R then u E I if and only if its image in Bred is in the tight closure of IRred working over Bred The statements in 4 and 5 show that the study of tight closure can often be reduced to the case where R is reduced or even a domain The following is one of the most important properties of tight closure It is what enables one to use tight closure as a substitute for the Cohen Macaulay property in many instances It is the key to proving that direct summands of regular rings are Cohen Macaulay in characteristic 1 gt 0 6 Coloncapturing If R m K is a complete local domain more generally if R m K is a reduced excellent and equidimensional the elements 1 xk zk1 are part of a system of parameters for R and k 1 xkR then 1 xk 2R k1 Q Of course if B were Cohen Macaulay then we would have k R zk1 Ik 7 Under mild conditions on R u E R is in the tight closure of I Q R if and only if the image of u in Rp is in the tight closure of IRp working over Rp for all prime respectively maximal ideals P of R The result holds in particular for algebras essentially of finite type over an excellent semilocal ring Tight closure is not known to commute with localization and this is now believed likely to be false But property 7 shows that it has an important form of compatibility with localization 8 If R m K is excellent Iquot I Property 8 shows that tight closure is determined by its behavior on m primary ideals in the excellent case 9 If R m K is reduced and excellent u E Iquot if and only if u is in the tight closure of R in R working over B These properties together show that for a large class of rings tight closure is determined by its behavior in complete local rings and in fact in complete local domains Moreover in a complete local domain it is determined by its behavior on m primary ideals We next want to give several further characterizations of tight closure although these require some additional condition on the ring For the first of these we need to discuss the notion of R for a domain R first The absolute integral closure R of a domain B Let R be any integral domain there are no finiteness restrictions and no restriction on the characteristic By an absolute integral closure of R we mean the integral closure of R in an algebraic closure of its fraction field It is immediate that R is unique up to non unique isomorphism just as the algebraic closure of a field is Consider any domain extension S of R that is integral over R Then the fraction field frac R is contained in the algebraic closure L of frac S and L is also an algebraic closure for R since the elements of S are integral over R and hence algebraic over frac 9 The algebraic closure of R in L is R Thus we have an embedding S gt R as R algebras Therefore R is a maximal domain extension of R that is integral over R this characterizes R It is also clear that R R When R R we say that R is absolutely integrally closed The reader can easily verify that a domain S is absolutely integrally closed if and only if every monic polynomial in one variable f E S factors into monic linear factors over S It is easy to check that a localization at any multiplicative system of an absolutely integrally closed domain is absolutely integrally closed and that a domain that is a homomorphic image of an absolutely integrally closed domain is absolutely integrally closed A monic polynomial over S P lifts to a monic polynomial over S whose factorization into monic linear factors gives such a factorizaton of the original polynomial over S If S gt T is an extension of domains the algebraic closure of the fraction field of S contains an algebraic closure of the fraction field of R Thus we have a commutative diagram 5 g T S gt T where the vertical maps are inclusions If R a S is a surjection of domains so that S g RP by the lying over theorem there is a prime ideal Q of R lying over P since R gt R is an integral extension Then R a RQ has kernel Q R P and so we have S E RP gt RQ Since R is integral over R RQ is integral over RP g S But since R is absolutely integrally closed so is RQ Thus RQ is an integral extenson of S and is an absolutely integrally closed domain It follows that we may identify this extension with S and so we have a commutative diagram R l aS l l l R a S where both vertical maps are inclusions Any homomorphism of domains R a T factors R a S gt T where S is the image of R in T The two facts that we have proved yield a commutative diagram l l l R gtgt S gt T where all of the vertical maps are inclusions Hence 10 Proposition For any homomorphism R a T of integral domains there is a commutative diagram R a T 1 1 R T where both uertical maps are inclusions D Other characterizations of tight closure For many purposes it su ices to characterize tight closure in the case of a complete local domain Let R m K be a complete local domain of prime characteristic p gt 0 One can always choose a DVR V tVV L containing R such that R Q V is local This gives a Z valued valuation nonnegative on R and positive on m This valuation extends to a Q valued valuation on R To see this note that R Q V V is a directed union of module finite normal local extensions W of V each of which is a DVR Let tW be the generator of the maximal ideal of W Then tV tGVWoz for some positive integer hW and unit 04 of W and we can extend the valuation to W by letting the order of tW be 1hW To construct V in the first place we may write R as a module finite extension of a complete regular local ring A mA By the remarks above it su ices to construct the required DVR for A There are many possibilities One is to de ne the order of a nonzero element a E A to be the largest integer k such that u 6 mg This gives a valuation because grmAA is a polynomial ring over K and in particular a domain Theorem Let R m K be a complete local domain ofprime characteristicp gt 0 u E R and I Q R Choose a complete DVR V mVL containing R m K such that R Q V is local Ezrtend the valuation on R given by V to a Q ualued ualuation on Rz call this ord Then it E 1 if and only if there ezrists a sequence of nonzero elements on E R such that for all n cnu E IR and ordc a 0 as n a 00 See Theorem 31 of Hochster and C Huneke Tight closure and elements of small order in integral extensions J of Pure and Applied Alg 71 1991 2337247 This is clearly a necessary condition for u to be in the tight closure of I We have Rlq Q R Q R and in so in the reformulation of the de nition of tight closure for the domain case one has clqu 6 Bl 7 Q IR for all su iciently large 1 Since one has 1 ordc1q 60rd c we may use the elements clq to form the required sequence What is surprising in the theorem above is that one can use arbitrary completely unrelated multipliers in testing for tight closure and u is still forced to be in 1 Solid modules and algebras and solid closure Ler R be any domain An R module M is called solid if it has a nonzero R linear map M a R That is HomRM R 31 0 An R algebra S is called solid if it is solid as an R module In this case we can actually find an R linear map 6 S a R such that 61 31 0 For if 60 is any nonzero map S a M we can choose s E S such that 60s 31 0 and then define 6 by 6u 60su for all u E S When R is a Noetherian domain and M is a nitely generated R module the property of being solid is easy to understand It simply means that M is not a torsion module over R In this case we can kill the torsion submodule N of M and the torsion free module MN will embed a free module Rh One of the coordinate projections 79 will be nonzero on MN and the composite M gtgt MN RhiR will give the required nonzero map However if S is a nitely generated R algebra it is oftem very dif cult to determine whether M is solid or not For those familiar with local cohomology we note that if R m K is a complete local domain of Krull dimension d then M is solid over R if and only 31 0 Local cohomology theory will be developed in supplementary lectures and we will eventually prove this criterion This criterion can be used to show the following Theorem Let R m K be a complete local domain Then a big Cohen Macaulay algebra for R is solid We will eventually prove the following characterization of tight closure for complete local domains This result begins to show the close connection between tight closure and the existence of big Cohen Macaulay algebras Theorem Let R m K be a complete local domain of prime characteristic p gt 0 Let u E B Let I Q R be an ideal The following conditions are equivalent 1 u E 1 2 There exists a solid R algebra S such that u 6 IS 3 There exists a big Cohen Macaulay algebra S ouer R such that u 6 IS Of course 3 i 2 is immediate from the preceding theorem Conditions 2 and 3 are of considerable interest because they characterize tight closure without referrring to the Frobenius endomorphism and thereby suggest closure operations not necessarily in characteristic p gt 0 that may be useful The characterization 2 leads to a notion of solid closure77 which has many properties of tight closure in dimension at most 2 In equal Math 711 Lecture of October 3 2005 Let S be an additive semigroup with identity 0 we are assuming that the semigroup operation is associative A ring R is S graded if it has a direct sum decomposition 3693 563 such that l E R0 and for all s t E S Rth Q Rst If R is S graded an R module M is said to be S graded if it has a direct sum decomposition M 565 M5 such that for all s t E S Rth Q M5t An element of some BS or M5 is said to be homogeneous of degree s or a form of degree s A submodule of N Q M where M is graded is called graded or homogenous if N 69W o M SES or if equivalently N is generated by homogeneous elements S is said to be a a cancellation semigroup if whenever su tu for s t u E S one has that s t S is said to have a linear order compatible with addition if it has a linear order S such that for all s t u E S ifs S t then su S tu Note that if s S t and u g u we have that su S tu since su S tu S tu Every subsemigroup of a cancellation semigroup is a cancellation semigroup and every subsemigroup of a semigroup with a linear order compatible with its addition has a linear order compatible with its addition one simply restricts the linear order on the larger semigroup In particular every subsemigroup of Z is a cancellation semigroup with a linear order compatible with addition since this is true of Z cancellation holds because Z is a group and for the linear order we may define a1 an 3 b1 bn if the two are equal or if there exists i l S i S n such that ai bi for i lt j while aj lt bj Proposition Let R be a Noetherian ring graded by a semigroup S such that S has can cellation and also has a linear order compatible with addition Let M be an S graded R module Then any associated prime ofM ie any prime of R that is the annihilator of an element of M is homogeneous Hence the nilradical Rad 0 ofR is homogeneous More generally the radical of a homogeneous ideal Qt is homogeneous Proof The final statement follows from the next to last statement applied to RQl while the next to last statement follows from the first statement because it implies that any minimal prime of R is homogeneous and an intersection of homogeneous ideals is homo geneous To prove the rst statement let u E M 7 0 have prime annihilator P Then every nonzero multiple of it has annihilator equal to P for if ru 31 0 then Pru 0 while if r 6 P rru 0 implies that rr 6 P and so r E P which implies that ru 0 a contradiction Among all those elements it having P as annihilator choose it so that the number k of its nonzero homogeneous components is minimum Then we can write it us1 ask where us 6 Msj for every j s1 lt lt sk every us 7 0 and k is minimum 1 We shall show that all of the ideals AnnRusj are homogeneous and that they are all equal Note that if u is a homogeneous element then 1 kills a sum of forms of distinct degrees if and only if 1 kills each of them this uses the fact that S is a cancellation semigroup It follows that the annihilator of a homegeneous element is a homogeneous ideal Now suppose that u is a homogeneous element of one of the ideals AnnRusj If 1 does not kill us for some i then on will be a nonzero multiple of u with fewer nonzero homogeneous components than u a contradiction This shows that all of the homogeneous ideals Ann Rusj are the same It follows that any element in Ann Rusk kills u and so is in We shall show that all of these are the same as P Consider any element r rtl rth of AnnRu where t1 lt lt th and each rtV E Rty We shall prove that every rtV kills ask This will show that r E AnnRusk and so P Q AnnRusk The point is that when one expands ru by the distributive law there is a unique highest degree term rth ask and so rthusk 0 Since rush 0 we can conclude that rtl l rthil also kills u and the fact that all the rtV kill ask now follows by induction on h Edample Some hypothesis on S is needed here Eg we may grade the polynomial ring Z2 y with S Z2 by letting ac have degree 0 and y have degree 1 The degree if the monomial ragb is 0 if b is even and 1 if b is odd In this ring ac y2 2 y2 is homogeneous but at y is not Thus the radical of the homogeneous ideal x2 242 is not homogeneous The problem is that while Z2 a group has cancellation it does not have a linear order compatible with addition The polynomial ring R Kac1 acn over any base ring K has an N grading also referred to as the grading by monomials such that R0117 an K201 2 The degree of xi is the lth standard basis vector ei for N The graded ideals with respect to this grading of R are precisely the ideals generated by monomials Corollary Let K be a eld and A a nite sz39mplz39cz39al complea with uertz39ces 1 xn Then KA is reduced Proof Since KA Kac1 nlIA it suf ces to show that A is radical Since the radical of IA is homogenous it suf ces to show that if a monomial u has the property that at 6 IA t 2 1 then u 6 IA But a monomial is in A if and only if the set of variables occurring with a positive exponent the support is not a face of A and this set is the same for at and u D Coupled with our earlier work this tells us the primary decomposition of IA it is the intersection of its minimal primes which are in bijective correspondence with the facets of A each is generated by the set of variables that is the complement of some facet Corollary Let R be a Hodge algebra ouerK on H governed by E and suppose that K is a eld and each generator ofE takes on only the values 0 and 1 on H Then R is reduced Proof There is a sequence of successive associated graded rings that begins with R and ends with the corresponding discrete Hodge algebra on K with the same data The con dition on the generators of 2 implies that the discrete Hodge algebra is a quotient of the 3 polynomial ring on variables corresponding to the elements of H by an ideal that is gen erated by square free monomials and so the discrete Hodge algebra is a face ring It is therefore reduced By induction on the number of associated graded algebras in the chain we see that each of them is reduced D Corollary IfR is an ASL over the eld K then R is reduced Proof In this case 2 is generated by functions on H that take the value 1 on two incom parable elements of H and are 0 elsewhere D Given a poset H there is an associated simplicial complex whose vertices are the el ements of H it consists of all subsets of H that are linearly ordered This simplicial complex is called the order complex of H The discrete Hodge algebra associated with an ASL R over K is the face ring over K of the order complex of H We want to make similar deductions for the Cohen Macaulay and Gorenstein properties In order to do so we need to study the Gorenstein property further and to do so we use the notion of Ext duals of Cohen Macaulay modules over regular rings We begin by reviewing some properties of Cohen Macaulay modules We first consider the case where the ring is local Proposition Let R m K be a local ring and let M be a nitely generated R module with Krull dimension d and annihilator I The following conditions are equivalent 1 M is Cohen Macaulay 2 depthmM d 3 M is Cohen Macaulay as an RI module 4 Some system of parameters for RI is a regular sequence on M 5 Every system of parameters for RI is a regular sequence on M 6 A sequence of elements ofR is a regular sequence on M if and only if its image in RI it is part of a system of parameters for RI 7 For every ideal J Q M the depth ofM on J is the same as the height of JRI in RI If these equiualent conditions hold then we also have a For every prime ideal P of R Mp is Cohen Macaulay ouer Rp b Ifacl ick E m are such that their images in RI are part of a system ofparameters then Macl ickM is Cohen Macaulay of dimension d 7 k c Euery nonzero submodule ofM has dimension d Thus ifP is an associated prime of M then RP has dimenson d Consequently M has no embedded primes and the associated primes ofM are the same as the minimal primes of I IfR is regular M is Cohen Macaulay and if and only if deM depthIR which is the same as the height of I since R is Cohen Macaulay Moreover in this case ifh is the height of I Ext 3M R uanishes ezrcept when i h Proof 2 is the definition of Cohen Macaulay and neither the depth nor dimension of M is affected when we replace R by RI Thus the first three conditions are equivalent ln proving the remaining equivalences we replace R by R I so that dim R dim M d The depth of M on an ideal only depends on the radical We now make use of the 4 results relating depth and Koszul homology from the Lecture Notes from February 18 from Math 615 Fall 2004 If 1 xd is one system of parameters then depthmM d if and only if depthltm17m 7 M d if and only if all of the higher Koszul homology of IC 1 xd M vanishes and this holds for a finitely generated module over a local ring iff 1 xd is a regular sequence on m Thus 4 i 3 i 5 i 4 is clear To see that 6 is equivalent to the others first note that since any regular sequence can be extended to a maximal regular sequence of length d all we need to show is that a maximal regular sequence on M must consist of parameters for R We know that if x is a nonzerodivisor on M then dim dim 7 1 Hence with N Rxl acdR we have that dim M RN dim Macl acdM dim M7d 0 which shows that M R N has nite length and therefore is supported only at m Now Supp M R N Supp Supp N Spec R Supp N Supp N and the fact that Supp Pt1 acdR implies that Rad 1 acdR m which in turn yields that 1 ard is a system of parameters for R Keep in mind that we have passed to the case where I 0 ie where M is faithful over R It is clear that 7 i 2 since we may take J m We postpone the proof of the converse implication until we have proved statement c about Cohen Macaulay modules To prove c suppose that M is Cohen Macaulay but has a nonzero submodule N of lower dimension Then it has a maximal such submodule Since the sum of two such submodules has the same property we may assume that N is maximum lf diM 0 there result is obvious and so we may assume that dim M gt 0 It is clear that M N has no submodule of dimension smaller than M or we could enlarge N In particular m is not an associated prime of MN and we can therefore choose at E M so as to avoid the associated primes of MN and the minimal primes of R It follows that x is a nonzero divisor on both M and MN The short exact sequence O N M MN o therefore remains exact when we apply RxR ER 7 and so we have an embedding NacN gt But dimNxN dimN 7 1 lt dimM 71 dimMxM and this produces a counter example in the Cohen Macaulay module M xM of lower di mension than M Note here that with N 31 0 we have NacN 31 0 by Nakayama7s Lemma The statement about associated primes follows because P is associated if and only if RP embeds in M Thus the associated primes of M are the same as the minimal primes of M and these minimal primes in Supp VI are the same as the minimal primes of I Thus the minimal primes of RI all have quotients of dimenson d dim RI With this hypothesis on RI we claim that the height of JRI is the same as the length of the longest sequence 1 ark in JRI that is part of a system of parameters We leave this statement as an exercise The equivalence of the first six conditions with 7 is now immediate To prove a again replace R by RI and P by PI Suppose that P has height h Then by the preceding paragraph there is part of a system of parameters 1 xh in P and these elements form a regular sequence on M Thus there images in Rp form 5 a regular sequence of length h on Mp and depthPRPMp 2 h dim Rp The other inequality always holds and so Mp is Cohen Macaulay over R p Part b is clear Now consider the case where R is regular Then M is Cohen Macaulay if and only if depthmM dim dim RI dim R 7 height Since deM depthR 7 depthm dim R 7 depthmM the Cohen Macaulay condition becomes that dim R 7 deM dim dim R 7 height I and this is eqivalent to the condition that deM height Note that since R is regular R is Cohen Macaulay and height I depthIR Finally Extj M R vanishes for i gt deM h and also for i lt depthAnnRMR depthIR h as well If R is Noetherian a finitely generated R module MN is called Cohen Macaulay if Mp is Cohen Macaulay over Rp for every prime equivalently for every maximal ideal P of R Evidently if W is a multiplicative system in R then W lM is Cohen Macaulay over W lR Note that if the annihilator of M is height unmixed and W lM 7E 0 then the annihilator of W lM is height unmixed in W lR one gets the expansions of the minimal primes of M that do not meet W When R is regular there is still a strong tendency for I AnnRM to be height unmixed ie for all minimal primes to have the same height Specifically Proposition Let R be a regular ring and let M be a Cohen Macaulay module ouer R with annihilator I If Supp M which is VI is connected then I is height unmizred ie all minimal primes ofI have the same height Proof Suppose that the minimal primes of I have more than one height Let P1 PS be the minimal primes of height h and let Q1 Qt be the minimal primes of other height Let J 1B and J 131 Qj Then VU VJ T VJ U VJ and VJ VJ are disjoint if a prime m were in both Mm a Cohen Macaulay module over Rm would have minimal primes DZRm and Qj Rm of different heights and we already know that this is impossible in the local case This shows that Supp is not connected D Now let M be a Cohen Macaulay module over a regular ring R such that Ann RM has pure height h For such a module M we de ne the Eat dual M of M as ExtM R Evidently M is a nitely generated R module We shall see that it is Cohen Macaulay with the same annihilator as M and that its dual is M The following two results summarize many of the important properties of the Ext dual 6 Theorem Let R be a regular Noetherian ring and let M M M be Cohen Macaulay modules with annihilators of pure height h Let N be Cohen Macaulay with annihilator of pure height h l a M is Cohen Macaulay with the same annihilator as M b There is a natural isomorphism M a M Thus up to isomorphism each ofM and M is the dual of the other c If0 a M a Ma M a 0 is ezaet then 0 MM a M a M a 0 is exact d OaMaMaN O is e1actthen0gtMgtM aNquot a0 is exact e Ifz is a nonzerodiuisor on M and xM 31 M then MacM is Cohen Macaulay with annihilator of pure height h l x is not a zerodiuisor on M and g f If W is a multiplicative system in R and W lM 7E 0 then it is Cohen Macaulay of pure height h and W lM W 1M Proof Part f follows from the fact that localization commutes with Ext when the first module is nitely presented We shall next prove Note that when one localizes at any prime M if M is not killed it becomes Cohen Macaulay with annihilator of pure height h Thus locally M is either 0 or of projective dimension h It follows that deM h Consider any projecitve resoluton of M of shortest possible length h where the GZ are finitely generated projective modules The higher homology of G is 0 while H0G M Note that for finitely generated projective modules G the natural map G gt GM that sends it to the map whose value on f E G is u is an isomorphism ie G is re exive Here G is HomRG R but in this case that agrees with the Ext dual R is easily checked to be re exive and the direct sum of two modules is re exive if and only if both are Thus finite rank free modules are re exive and hence so are their direct summands Moreover the dual of a finitely generated free module is free and so the dual of a finitely generated projective module is projective As in the local case Exti M R 0 except when i h this is obviously true when we localize at primes that do not contain I and it is true if P Q I by the local results The complex 0gtG3gtgtGZgtQ with the numbering reversed is therefore acyclic the higher homology is Extje M R for i lt h with augmentation M It is now clear that if we use this projective resolution to calculate M then we obtain the augmentation of the complex which is the double dual into R of G there is a natural isomorphism from G to its double dual This induces an isomorphism M Mi This is independent of the choice of length h projective resolution of M Given two such there are maps between them such that the composition in either direction is homotopic to the identity Dualizing into B provides maps of the dual complexes with the same property It is easy to check that the usual identifications of first M and then Math 711 Lecture of November 17 2006 Our next objective is to give a matrix factorization of a generic form over Z instead of ZE The idea is to replace the ring ZE by a ring of matrices over Z Discussion block form Let R be any associative ring with identity and let MAR denote the ring of n gtlt n matrices with entries of R which we may identify as usual with the ring of R linear edomorphisms of R thought of as n gtlt 1 column vectors over R The matrix 7 acts on the column y by mapping it to in The observation we want to make is that we may identify MkhR The naive way to make the identification is to partition each kh gtlt kh matrix into a k gtlt k array of h gtlt h blocks More conceptually we may think of the domain of an R linear map Rkh a Rkh as the direct sum of k copies of Rh ie as Rh k and we may think of the target of the map as Rhk the product of k copies of Rh Then the map is determined by its restrictions to the k direct summands Rh of the domain and the map from a particular summand Rh a Rhk corresponds to giving k R linear maps Rh a Rh one for each factor of the target module D39 39 39 I 39 in 39 3 variables over a matrix ring Let X1 Xk denote indeterminates both over R and over each matrix ring over R such that the X commute with one another with elements of R and with matrices over R Then we may identify the rings M7RX1 Xkl 2 MnRX1 Xk Given a finite linear combination Zh 7100 where each nab 6 MAR and each m is a monomial in the X1 we let it correspond to the matrix Eh ripyh Let 5 be a primitive dth root of unity Recall that its minimal polynomial is denoted lldt suppose that 6 ltIgtd which is the degree of lld and that 11d2 25 05712 3 1 00 where the c E Z If we take 1 5 2 55 1 as a basis for ZE then the matrix of multiplication by 5 on ZE is 0 0 0 0 700 1 0 0 0 701 6 0 1 0 0 702 7 0 0 0 1 70511 the companion matrix of lldt and ZE E ZM Q M5Z Notice that each of the powers L3 6 62 Qd l has an entry equal to 1 because 6139 maps the basis element 1 to the basis element 6139 0 S i S d 7 1 We then have Theorem Let d 2 2 and s 2 1 be integers and let f denote the degree d linear form over Z in sd variables given as f X11X12 39 39 X1d X51Xs2 39 39 Xsd Let 6 lt1gtd Then f has a matria factorization flgdsil a1 ad over RZXi7j1 i s1 j d of size 6sd 1 such that Ia va 1 S i S s 1 Sj S dR Proof We begin with the matrix factorization over RE which has size ds 1 given in the Theorem on p 5 of the Lecture Notes of November 15 We write X for the collection of variables Xi 1 S i S s 1 S j S d By the Discussions above we have an embedding of Zl llil gt MAME mam which will give a matrix factorization of f15Idsil with d factors in Mdsil M5R Under the identification Mdsil M5R Mgds71R this yields a matrix factorization for flgdsil whose entries are Z linear forms in the variables X In the factorization given in the previous Theorem every va occurred possibly with a coef cient 5 0 S k S 67 1 1n the new factorization 5 is replaced by a block corresponding to QkXZJ Since 6 has an entry equal to 1 the variable va occurs as an entry D Now that we have dealt with the generic case we can immediately get a corresponding result for any finitely generated ideal in any ring Theorem Let I be a nitely generated ideal of a ring R and let f 6 Id where d 2 2 Then for some integer N there e1ists a matria factorization fIN a1ad such that Ia I In fact there e1ists such a factorization in which 041 a2 ad and IOziI1 l d Proo Since 6 Id for some choice of elements it E I we can write 7 J f u1iquot39u1d us1quot39usd with all of the am 6 I We may assume all of the finitely many generators of I occur among the am by including some extra terms in which one of the factors is 0 We may then map ZXi7j 1 S i S s 1 S j g d a B so that va gt gt um Applying the homomorphism to the factorization for the generic form given in the preceding Theorem we obtain a factorization of f satisfying all but one of the conditions needed it need not satisfy the condition that 041042Old But we may satisfy the additional condition by increasing the size by a factor of d and taking all of the matrices to be cyca1 ad see the discussion on p 5 of the Lecture Notes of November 15 The following result will play an important role in our construction of linear maximal Cohen Macaulay modules over hypersurfaces 3 Theorem Let R m K be a local ring and let f E md be a nonzerodiuisor where d 2 2 Then for some s there is a matria factorization fIs a1ad such that 0 m for 1 S t S d Let G G0 R5 and let GZ be the image of rZJZ 041042Ozi so that GZ Q R5 as well and Gd fRs Let Mi GiGiH 1 S t S d Then all of the modules GiGd GGiH and Mi 0 S t S d 7 1 are mazrimal Cohen Macaulay modules over F RfR and all of them have nite projective dimension necessarily 1 over R Moreover s 0 S t S d 71 Proof Since fIs wit1 where it Oli1Old we have from the Discussion on the first page of the Lecture Notes of November 13 concerniiig Cohen Macaulay modules over hypersurfaces that if indicates images after applying R ER 7 then 71 31 ns 76 39 Rug 239 l jgel 1 is is acyclic and that 7 i 7 Ker it 1m rZJZ g Coker to g Coker 112 and the same holds with the roles of rZJZ and it interchanged The image of rZJZ contains fRs which is the image of with ie fRs Q GZ Q R5 and GifRs GiGd may be identified with lmE Thus every GiGd is a maximal Cohen Macaulay module over F of finite projective dimension over R Note as well that RdGi Coker rZJZ is a maximal Cohen Macaulay module over F of finite projective dimension over R We have that Gi1 lerJZH lmrJZozZH Q imR5 since Hal1 m and this is contained in miliiltR5gt mGi Hence Mi GiGiH is minimally generated over R by s elements and the short exact sequences 0 7 Mi 7 RsGi1 RsGi 0 show that every Mi is a maximal Cohen Macaulay over F of finite projective dimension over R as well A maximal Cohen Macaulay module over F that has finite projective dimension over R must have projective dimension 1 over R since its depth is d 7 1 D Proposition Let R m K be local and f E md 7 and Let f denote the leading form of f ie the image of f in ind7nd grmRd Suppose that N is a nitely generated R module N R is the most important case and that f is a nonzerodiuisor on grmN Let R RfR which has mazrimal ideal m mfR and let N NfN Then a f is a nonzerodiuisor on N b For every integer n 2 0 fN m N fmn dN c For every u E N 7 0 fg f u and the m adic order of fu is the sum of the m adic orders off and u 4 d gr g grmN fgrmNI Proof We first prove If u 31 0 say u E th 7 mh1N then it follows that fit Z mdh1N ie that fit 6 mdhN7mdh1N or else f u 0 a contradiction This proves both that ord fa d h and that fu f u Part a follows as well for if fa 0 then f u 0 To prove b note that if fit 6 mnN then d ord 2 n and so ord 2 n 7 d which shows that u E mn dN Finally to prove d note that WNm 1N 2 MN fNm 1N fN 2 miNmiN m m 1N fN Now if a on fv with a E m N and an E mn1N then fv a 7a E mnN and so i E mn dN by part Then m WW1W 2 m Nm 1N fm dN g lgrmNl flgrmNln2d g lgrmN fgrmNlm as required D Discussion To calculate the multiplicity of a local ring R m K or of an R module M one may work alternatively with the Hilbert function of grmR or the Hilbert function of grmM the lattter for example is de ned as dimKgrmM and is eventually a polynomial of degree dim 7 1 This function is the first difference of the Hilbert function If the Hilbert function of M has leading term gnr where r dim M then the leading term of the Hilbert function of grmM will be Ernr71 i 6 T71 rl 7 r 7 1 Corollary Let R m K be local and let f E m be such that its leading form f has degree d a If f is a nonzerodivisor in grmR then eRfR deR b IfN is a nitely generated R module and f is a nonzerodivisor on grmN then eN deNfN Proof lt suf ces to prove b which is more general In the notation of the preced ing proposition 5W may be calculated from the Hilbert function of grmN which is grN fgrN If the Hilbert function of grmN is Hn the Hilbert function for grN fgrN will be 7 Hn 7 d If the leading term of the polynomial de corresponding to is nf l the new leading term is inT Z since the n 7 L 21 2 1 7 Cn 7 d 1 has leading term cdr 7 lnr 2 for any constant c D polynomial on In order to prove the result we want on existence of linear maximal Cohen Macaulay modules we need to generalize the Theorem on p 3 to a situation in which we have tensored with a linear maximal Cohen Macaulay module N over R Math 711 Lecture of October 23 2006 One of our goals is to discuss what is known about the following conjecture of C Lech which has been an open question for over forty years Conjecture IfR a S is a flat local map of local rings then eR S eS This is open even in dimension three when S is module finite and free over R Note that one can immediately reduce to the case where both rings are complete Under mild conditions a local ring R of multiplicityAl is regularzA it suf ces if the completion R has no associated prime P such that dim RP lt dim Therefore the following result is related to Lech7s conjecture Theorem IfS is faithfully flat ouer R and S is regular then R is regular In particular if R m K a S n L is a flat local map of local rings and S is regular then R is regular Proof The second statement implies the first for if P is any prime of R then some prime Q of S lies over P and we can apply the second statement to RP a SQ to conclude that Rp is regular To prove the second statement let gt0 gtGngtgtG1gtGOgtRgtRmgt0 be a minimal resolution of Rm over R Then the matrix oa of the map GZ a Gi1 has entries in m for all i 2 1 Since S is R flat the complex obtained by applying S ER 7 namely quotquot S RGnquot quot S RG1 S RG0 Sm5gt0 gives an S free resolution of SmS over R Moreover the entries of the matrix of the map S 8 GZ a S R G141 are simply the images of the entries of the matrix oa in S these are in n and so the complex given in is a minimal free resolution of SmS over S Thus all of its terms are eventually 0 and this implies that all of the terms of are eventually 0 Hence K has finite projective dimension over R which implies that R is regular D Before treating Lech7s conjecture itself we want to give several other characterizations of e1 when I is generated by a system of parameters There is a particularly simple characterization in the Cohen Macaulay case We first recall some facts about regular sequences The results we state in the Proposition below are true for an arbitrary regular sequence on an arbitrary module However we only indicate proofs for the situation where R is local M is a finitely generated R module and 1 xd are elements of the maximal 1 ideal of R The proofs are valid whenever we are in a situation where regular sequences are permutable which makes the arguments much easier There is a treatment of the case where the regular sequence is not assumed to be permutable in the Extra Credit problems in Problem Sets 2 and 3 from Math 615 Winter 2004 It is assumed that M B there but the proofs are completely unchanged in the module case Recall that in all cases by Virtue of the definition the fact that 1 icd is a regular sequence on M implies that 1 icdM 31 M Proposition Let 1 xd E R let I 1 xdR and let M be an R module a Let t1 td be nonnegatiue integers Then 1 icd is a regular sequence if and only if xii gd is a regular sequence on M b If 1 icd is a regular sequence on M and a1 ad are nonnegatiue integers then x1 nagdw E 111 aegd1M implies that w 6 1 icdM c I 1 xd is a regular sequence on M 1 N are the monomials 0 degree n M M in 1 icd and w1wN are elements ofM such that ujwj E n1M then every wj 6 IM d If 1 icd is a regular sequence on M then ngM may be identi ed with 8131 Xd where the Xj are indeterminates and for nonnegatiue integers a1 ad such that 21 aj n the image of xi ng in I MI 1M corresponds to MIMXf1 Xd e If 1 icd is a regular sequence on M then MI 1M has a ltration in which the factors are 711 copies of MIM Proof a It su ices to prove the statement in the case where just one of ti is different from 1 we can adjust the exponents on one element at a time Since R sequences are permutable it su ices to do the case where only td is different from 1 and for this purpose we may work with Macl xd1M Thus we may assume that d l and the assertion we need is that act is a nonzerodiVisor if and only if it is Clearly if icw 0 then xtw 0 while if ictw 0 for t chosen as small as possible and w 31 0 then xxt 1w 0 b If all the ai are zero then we are already done If not we use induction on the number of ai gt 0 Since we are assuming a situation in which R sequences on a module are permutable we may assume that ad gt 0 Then dil 111 ad 7 aj l l V ad1 1dw7 icj w1acd wd 11 for elements wl wd E M Then 7 7 ad 11 adil a11 ad711 god 1 xd71w7dwd 1 acd71 M and since x111 x3611 gd is also a regular sequence on M we have that 11 adil a11 ad711 1 md71w7dwdex1 acd71 This yields that 11 adil a11 ad711 1 xd71weyc1 acd71 icdM providing an example in which the number of aj gt 0 has decreased This is a contradiction c 1x oneo e v at en in every 0 er k an in every monomia o e F39 fth M 1 3d Th 39 th M d39 391 fd gree n1 at least one xi occurs with exponent ll1 Thus Wu 6 3611 gogd1M and wj E M by part or eac monomia 1n 1 d we wri e or e correspon mg monomia dF h 39l 39X X 39tpf th d39 391 in 1 xd We define a map from IiM a MIM Z degmm by sending Mjwj gt gt Zj EU where 117 is the image of wj E M in MIM This map is well defined by part c and is obviously surjective The elements of n1M are precisely those elements of M which can be represented as Mjwj with every wj 6 IM and it follows at once that the kernel of the map is n1M e This follows at once from part d since we can initially use a filtration with factors IkMIkHM 0 S k S n and then refine it because each of these splits into a direct sum of copies of MIM such that the number of copies is the same as the number of monomials d of degree k in X1 Xd The number of monomials of degree at most d is n 2 D We next note Theorem If R m K is local of dimension d M is Cohen Macaulay of dimension d over R 1 xd is a regular sequence on M and I 1 xd then eIM M 1dM Proof By part e of the Lemma just above we have that MI 1M has a filtration such that 1 Every factor is E MIM d 2 The number of factors is ie is the same as the number of monomials of degree at most n in d indeterminates X1 Xd This gives the result for we then have d MI 1M n aMIM d d and the leading term of is D Theorem Let R m K be module nite over a regular local ring A such that 1 icd is a regular system of parameters of A and let M be an R module of dimension d Let I 1 xdR Then eIM is the torsion free ranh ofM over A Proof From the definition it does not matter whether we think of M as an R module or whether we think of it as an A module with maximal ideal n 1 acdA In the latter case if p is the torsion free rank of M as an A module we have an exact sequence of A modules 0 a A a M a C a 0 where O is a torsion A module so that dim O lt d It follows that eIM enM renA 0 r Aic1acdA r l r E Discussion If R is equicharacteristic we can always reach the situation of the Theorem above The mulltiplicity does not change if we replace R by But then we can choose a coe icient field K and the structure theorems for complete local rings guarantee that R is module finite over A KHacl icdll Q More generally Theorem Let R be module nite over a Cohen Macaulay local ring B such that 1 xd is a system of parameters for B Let M be an R module of dimension d Let I 1 xdR IfB is a domain eIM BIBp where p is the torsion free ranh ofM ouer B When B is not a domain if there is a short ezract sequence 0 a B a M a C a 0 with dim O lt d then eIM BIBp Proof MI 1M is independent of whether one thinks of 1 icd as in B or in R Thus we can replace R by B The result is then immediate from our results on additivity of multiplicity and the fact that when B is Cohen Macaulay eIB BI D We want to give a different characterization of multiplicities due to C Lech If E n1 nd is a d tuple of nonnegative integers and f is a real valued function of n we 5 write lim n r where r E R to mean that for all 6 gt 0 there exists N such that ngtltxgt for alli n1 might also write llim n r with the same meaning If g 1 Inln WHOO nd satisfying ni 2 N l S i S d we have that lf 7 rl lt 6 One icd is a system of parameters for Riwe temporarily de ne the Lech multiplicity e M to be We shall show that the limit always exists is 0 if dim lt d and with I 1 aMxil n1 z gd 71d lim EH00 795037 is eIM when dim d Lemma Let 1 and let M M and M be nitely generated R modules 1 a b c d We first prove xd d 2 1 be a system ofparameters for a local ring R m K Given n1 nd let 9671 If 3311 and let EM MIEMn1nd O M M MH O is exact then for any m primary ideal J M JM MJM iMJM M JM 0 S MJM i M JM S MJM Hence for all E 510 S 52M S 520 dM ie 0 EltMgt 7 510 51W Therefore if the three limits edist 65M S 6 M S 65M 65M If eM 0 and e M 0 then eM 0 IfM has a nite ltration with factors Nj we have that for any m primary ideal J MJM S Zj NjJNj Hence for all EM S Zj 1Nj and egM 0 whenever eNj 0 for all j If dim lt d then eM 0 If0 a M a M a M a 0 is ezract and dimM lt d then e M and e M e1ist or not alike and if they e1ist they are equal Math 711 Lecture of December 5 2007 From the local cohomology criterion for solidity we obtain Corollary A big Cohen Macaulay algebra or module B over a complete local domain R is solid Proof Let d dim R and let 1 icd be a system of parameters for B This is a regular sequnece on B and so the maps in the direct limit system Bacl xdB a aBx i B a are injective Since 0 31 Bacl xdB gt HimB Hg B B is solid B Our next objective is to prove Theorem Let R be a complete local domain Then there exists an R algebra B such that for every ring R1 with R Q R1 Q R such that R1 is module nite ouer R the following two conditions hold 1 B is a big Cohen Macaulay algebra for R1 2 For every pair of nitely generated Rl modules N Q M andu E NITI 1 u E ltB R Ngt in B R The proof will take a considerable effort The basic idea is to construct an algebra B with the required properties by introducing many indeterminates and killing the relations we need to hold The di iculty will be to prove that in the resulting algebra we have that mB 31 B Forcing algebras Let T ba ring it an h gtlt 1 column vector over T and let 04 be an h gtlt k matrix over T Let Z1 Zk be indeterminates over T and let I be the ideal generated by the entries of the matrix Z1 Zk By the forcing algebra which we denote HbrceaT of the pair 0 u 04 over T we mean the T algebra TZ1 ZklI In this algebra we have forced it to be a linear combination the coe icients are the images of the Z of the columns of 04 If M is the cokernel of the matrix 04 we have that 1 8 u 0 in HbrceaT T M Given any other 1 2 T algebra T such that 1 8 u 0 in T 8 M equivalently such that the image of u is a T linear combination of the images of the columns of 04 we have a T homomorphism rceaT a T that sends the Z to the corresponding coe iicients in T used to express the image of u as a linear combination of the images of the columns of 04 We shall say that rceaT is obtained from T by forcing o It will be technically convenient to allow the matrix 04 to have size h gtlt 0 ie to have no columns In this case the forcing algebra is formed by killing the entries of u Typically we are forcing it to be in the span of the columns of 04 When k 0 the span of the empty set is the 0 submodule in Th Now suppose that we are given a set E of pairs of the form u 04 where u is a column vector over T and 04 is a matrix over T whose columns have the same size as n We call the set E forcing data for T The size of u and of the matrix may vary By the forcing algebra Home T we mean the coproduct of the forcing algebras Home7 T as o varies in T One way of constructing this coproduct is to adjoin to T one set of appropriately many indeterminates for every 0 E Z all mutually algebraically independent over T and then impose for every 0 the same relations needed to form rceaT If E 01 on one may think of this algebra as gbrceai T 21 where the tensor product is taken over T When 2 is infinite one may think of 9b rceT as the direct limit of all the forcing algebras for the finite subsets 20 of E If E is forcing data for T and h T a T we may take the image of E to get forcing data over Tz one is simply applying the homomorphism h to every entry of every column and every matrix We write ME for the image of 2 under h Then 9brcehzT T T 9brceT We refer to the process of formation of rcehzT as postponed forcing we have in fact postponed the formation of the forcing algebra until after mapping to T In discussing forcing algebras we make the following slight generalization of the nota tions Suppose that E is a set of forcing data over S and g S a T is a homomorphism We shall also write Home T for 9b rcehzT Thus our notation will not distinguish between forcing and postponed forcing Note that if we partition forcing data 2 for T into two sets we can form a forcing algebra for the first subset and then form the forcing algebra for the image of the second subset We postponed the forcing process for the second subset But the algebra obtained in the two step process is isomorphic to 9b rceT One can also think of the formation of Home T as an infinite process Well order the set 2 Now perform forcing for one u 04 at a time Take direct limits at limit ordinals By transfinite recursion one reaches the same forcing algebra Home T that one gets by doing all the forcing in one step Algebra modi cations Let g S a T be a ring homomorphism which may well be the identity map and let P be a a set whose elements are nite sequences of elements of S We assume that if a sequence is in P then each initial segment of it is also in P Let Eng denote the set of pairs a 1 such that 1 ack1 is a sequence in P u E T and ack1u 6 1 ackT Thus if 1 ack1 were a regular sequence on T we would have that u 6 x1 ackT Let E be a subset of Eng We call 2 modi cation data for T over F We refer to 9b rceT as a multiple algebra modi cation of T over F We write AlgmodpygT for the forcing algebra Hbrcegm T and refer to it as the total algebra modi cation of T over P lf 0 is one element of Eng we refer to HbrceaT as an algebra modi cation of T For emphasis we also refer to it as a simple algebra modification of T We shall use iterated multiple algebra modifications to construct T algebras on which the specified sequences in P become regular sequences Note that if k 0 and mu 0 in T then we get an algebra modification HbrceaT in which 0 is a pair consisting of u and a l gtlt 0 matrix this algebra modification is simply TuT If we have a homomorphism h T a T and o a 1 3 we write ho for Ma 1 Note that if ack1u 6 1 xkT then ack1hu 6 1 kT Thus hang Q Smear With this notation if E is algebra modification data for T over F then ME is modifi cation data for T over F Algebra modifications are forcing algebras As in the general case of a forcing algebra we may talk about postponed modifications The construction of big CohenMacaulay algebras that capture tight closure Let R m K be a complete local domain Let P consist of all sequences in R that are part of a system of parameters in some ring R1 with R Q R1 Q R such that R1 is module finite over B Let E be the set of all pairs u 04 consisting for some ring R1 as above of an h gtlt 1 column vector over R1 and an h gtlt k matrix over a ring R1 such that u is in the tight closure over R1 of the column space of 04 Note that we know that whether this condition holds is unaffected by replacing R1 by a larger ring R2 such that R1 Q R2 Q R with R2 module finite over R Let B0 9b rceR lf B7 has been de ned for i 2 0 let Bn1 be the total algebra modi cation of Algmodp B7 of B7 over P Then we have a direct limit system B0 B1 Bn Bn1 3939 Let B limn B7 We shall prove that B is the required big Cohen Macaulay algebra Much of this is obvious Suppose that N Q M are finitely generated Rl modules and u E NITI over R1 Choose a finite presentation for MN over R1 so that MN is the cokernel of an h gtlt k matrix 040 over R1 The image of v in MN is represented by an element u E R f Then it is in the tight closure of the column space of Oz in R and it follows from the definition of B0 that u is a linear combination of the columns of Oz in B0 and hence in B It is likewise easy to see that if 1 ard is a system of parameters in R1 then it is a regular sequence on B Suppose that we have a relation k k1bk1 Z 954 211 on B Because B is the direct limit of the B we can find no such that BM contains elements that map to the bi The corresponding relation may not hold in BM but since it holds in B it will hold when we map to B for some n 2 no Thus we may assume that we have 51 k1 6 B such that xk1 k1 21 i i in B7 and every maps to bi when we map B a B By the construction of B7 we have that the image of k1 is in 1 xkBn1 We can then map to B to obtain that bk1 6 x1 xkB There remains only one thing to check that mB 31 B This is the most dif cult point in the proof This is equivalent to the condition that for some equivalently every system of parameters 1 ard in every R1 we have that 1 acdB 31 B To see this observe that if B B then 2B IB B and multiplying by I repeatedly yields by induction that ItB B for every t 2 1 If m1 is the maximal ideal of R1 then le Q ml which is contained in a power of le Hence mB B if and only if m1B B Likewise if 1 ard is a system of parameters for R1 generating an ideal I for some t we have m i Q I Q m1 and so m1B B if and only if B B If 1 xdB B then we have that 11bidbd for some finite set of elements of B We shall show that if this happens it happens for some algebra obtained from some choice of R1 by a finite sequence of forcing algebra extensions the first extension is the forcing algebra for a single pair u a such that u is in the tight closure of the column space of 04 over R1 The extensions after that are simple algebra modifications with respect to sequences each of which is part of a system of parameters for R1 Eventually we shall have to work even harder to make the problem more finite specifically we aim to replace these algebras by nitely generated submodules We then use characteristic p methods to get a contradiction However the first step is to get to the case where we are only performing the forcing procedure finitely many times With su iicient thought about the situation that one can do this is almost obvious but it is not quite so easy to give the argument formally Descent of forcing algebras Let g S a T be a ring homomorphism Let A be a directed poset and let SAEA be a directed family of subrings of S indexed by A whose union is S Let TQAEA be a directed family of rings such that lim T T and suppose that for every A E A we have a homomorphism g SA a T such that if A S u the diagram SLT l l Slgg gtTH l l Skg xgtTIJ commutes Thus g S a T is the direct limit of the maps SA a T Proposition Let notation be as above a Let E be forcing data over S Let E denote a subset of 2 such that all entries occurring are in S Suppose also that ifA S u then E Q EM and that the union of the sets EA is 2 Then 9b rceT is a direct limit of rings each of which is obtained from some TA by a nite sequence of such extensions of T each of which is obtained from its immediate predecessor by forcing one element of EA b Let P be a family of nite sequences in S closed under tahing initial segments and E Q SEAT De ne EA to consist of all elements of 2 whose entries are in TA whose corresponding sequence is in SA and such that if the sequence is 1 k1 and the element is u 1 3 then ick1u 6 1 xkT Then the algebra modi cation 9b rceT is a direct limit of rings each of which is obtained from TA by a simple algebra modi cation over a subset of P Proof a Let denote the poset each of whose elements is a pair A CD where I is a finite subset of EA The partial ordering is defined by the condition that A TD 3 u 1 precisely if A S u and I Q 11 There is an obvious map rce T a HbrcedTH Math 711 Lecture of September 29 2006 We prove the Corollary stated at the end of the Lecture of September 27 Proof If the ideals are different we may localize at a prime of R2 in the support of the module which is their sum modulo their intersection Thus we may assume without loss of generality that R2 is local We may replace R1 by its localization at the contraction of the maximal ideal of R2 and R0 by its localization at the contraction of the maximal ideal of R2 or of R1 the contractions are the same Thus we may assume that we are in the case where R0 gt R1 gt R2 are local and the homomorphisms are local By the Lemma near the bottom of the page 4 of the Lecture Notes of September 27 we have that R1 g R0X1 XmlpFl Fm for some m and prime 73 of R0X1 Xm Then leRO is generated by the image of 8 F Fm Likewise R2 R1Y1 Y5QG1 Gs 8G17 7G5 It follows that we can write Again jRQRl is generated by the image of R2 as a localization of R0X1 Xm Y1 YslFl Fm G1 Gs where the do not involve the Yi This means that the Jacobian matrix has the block form Mp N 0 MG where Mp is and MG is BGjBK We have that jRQRO is generated by the image of the determinant of this matrix No matter what N is this determinant is 8FFm BGG5 detMFdetMG 8X1 X 8Y Y as required D There are now three results whose proof are hanging one is the proof that S is module finite over S the second is the proof of the Key Lemma which is stated on p 3 of the Lecture Notes of September 27 and the third is the proof of the Jacobian Theorem itself We begin with the proof of the Key Lemma This will involve studying quadratic 1 transforms of a regular local ring along a valuation We first indicate our approach to the proof of the Key Lemma Proof of the Key Lemma step 1 We are trying to show that jgR Q uSC Assume the contrary Consider the primary decomposition of uSC Since we are assuming that S is module finite over S we still need to prove this 5 is a normal Noetherian ring and the associated primes of uS have height one We may choose such a prime Q such that jgR is not contained in the corresponding primary ideal in the primary decomposition of uSC Since the elements of S 7 Q are not zerodivisors on this primary ideal we also have that jSQR is not contained in SQ V Thus for the purpose of proving the Key Lemma we may replace 5 by V which is a discrete valuation ring and we may replace R by its localization at the contraction of Q to R In the remainder of the argument we may therefore assume1 that R m K is regular local We now digress to discuss quadratic transforms We first want to prove Lemma Let R m K be regular local with regular system of parameters 1 zd Then T fun1 ed1 Rmz1 is regular and the images of 2 rd are algebraically independent ouer K in Tle Kfg fd Proof If we localize T at a prime that does not contain 1 the resulting ring is a localiza tion of RM and is therefore regular For primes that contain 1 it suf ces to show that the localization is regular after killing 1 and this follows from the fact that Tle is regular even without localizing It therefore suf ces to prove that Tle is a polynomial ring First note that T k where u E m and this is an increasing union since xlmkac fH It is clear that Q T and the product of the jth and kth terms in the union is the j k th term Hence if there is a relation among the fj 2 S j S n we may lift it obtain a nonzero polynomial g whose nonzero coef cients are units of R we get them by lifting elements of K to R such that gx2ac1 rd1 out where t E We multiply both sides by iv where N 2 k is also larger than the absolute value of any negative exponent on 1 occurring on the left The left hand side becomes a nonzero homogeneous polynomial G of degree N in 1 zd whose nonzero coef cients are units The right hand side is in 1 aziV kmk Q z N kmk Q mN1 This gives a nonzero relation of degree N on the images of 1 mm in grmR a contradiction since when R is regular this is a polynomial ring in the images of the obj D De nition Let R m K be a regular local ring of Krull dimension d 2 2 and let suppose that R Q V is a local map to discrete valuation ring V Let ordV ord denote 1Note that in the re ned version of the Jacobian theorem7 it was assumed that RP is regular if P lies under a height one prime of S 7 so that we may make this reduction even in that case Math 711 Lecture of October 8 2007 Properties of regular sequences In the sequel we shall need to make use of certain standard facts about regular sequences on a module for convenience we collect these facts here Many of the proofs can be made simpler in the case of a regular sequence that is permutable ie whose terms form a regular sequence in every order This hypothesis holds automatically for regular sequences on a nitely generated module over a local ring However we shall give complete proofs here for the general case without assuming permutability The following fact will be needed repeatedly Lemma Let R be a ring M an R module and let 1 acn be a possibly improper regular sequence on M If ul un E M are such that n E 1417 0 11 then every uj 6 1 xnM Proof We use induction on n The case where n 1 is obvious We have from the de nition of possibly improper regular sequence that u 2311 acjuj with U1 un1 E M and so 2711 acjuj xnuj 0 By the induction hypothesis every uj xnuj 6 1 acn1M from which the desired conclusion follows at once D Proposition Let 0 M0 Q M1 Q Q Mh M be a nite ltration of M If 1 xn is a possibly improper regular sequence on every factor Mk1Mk 0 S h S hi1 then it is a possibly improper regular sequence on M If moreover it is a regular sequence on MMh1 then it is a regular sequence on M Proof If we know the result in the possibly improper case the final statement follows for if I 1 xnR and IM M then the same hold for every homomorphic image of M contradicting the hypothesis on MMh1 It remains to prove the result when 1 xn is a possibly improper regular sequence on every factor The case where h 1 is obvious We use induction on h Suppose that h 2 so that we have a short exact sequence 0gtM1gtMgtNgt0 1 and 1 icn is a possibly regular sequence on M1 and N Then 1 is a nonzerodiVisor on M for if mu 0 then 1 kills the image of u in N But this shows that the image of u in N must be 0 which means that u 6 M1 But 1 is not a zerodiVisor on M1 It follows that 0gtM1 M N o is also exact since it is isomorphic with the original short exact sequence Therefore we have a short exact sequence of quotients 0 M11M1 0 We may now apply the induction hypothesis to conclude that 2 7 is a possibly improper regular sequence on MiclM and hence that 1 xn is a possibly improper regular sequence on M We now carry through the induction on h Suppose we know the result for filtrations of length h 7 1 We can conclude that 1 xn is a possibly improper regular sequence on Mh1 and we also have this for MMh1 The result for M now follows from the case where h 2 Theorem Let 1 xn E R and let M be an R module Let t1 tn be integers 2 1 Then 1 icn is a regular sequence respectively a possibly improper regular sequence on M i xii icf is a regular sequence on M respectively a possibly improper regular sequence on Proof If M M then IkM M for all k If each of I and J has a power in the other it follows that M M iff JM M Thus we will have a proper regular sequence in one case iff we do in the other once we have established that we have a possibly improper regular sequence In the sequel we deal with possibly improper regular sequences but for the rest of this proof we omit the words possibly improper77 Suppose that 1 xn is a regular sequence on M By induction on n it will suf ce to show that 51152 xn is a regular sequence on M we may pass to 2 xn and and then apply the induction hypothesis It is clear that it is a nonzerodiVisor when 1 is Moreover Mic ilM has a finite filtration by submodules with factors M1M g MiclM l S j 3 t1 71 Since 2 7 is a regular sequence on each factor it is a regular sequence on Mic ilM by the preceding Proposition For the other implication it will suf ce to show that if 1 xj1x acj1 xn is a regular sequence on M then 1 xn is we may change the exponents to 1 one at a time The issue may be considered mod 1 icj1M Therefore it suf ces to consider the case j l and we need only show that if ic i 2 xn is a regular sequence on M then so is 1 icn It is clear that if 5 is a nonzerodiVisor then so is 1 By induction on n we may assume that 1 an is a regular sequence on M We need to show that if xnu 6161 xn1M then u 6 1152 icn1M lfwe multiply by oil we find that xnic i 1u 6 5152 icn1M and so 1571 t 951 u 101 202 957171071717 ie 1571 1 u 7 1u17 mm 7 7 71UWL1 0 By the induction hypothesis 1 an is a regular sequence on M and by the first part 571152 acn1 is a regular sequence on M By the Lemma on p l we have that u 7 1111 6 371202 acn1M and so u 6 1 acn1M as required D Theorem Let 1 xn be a regular sequence on the R module M and let I denote the ideal x1 xnR Let a1 an be nonnegatlue integers and suppose that u ul un are elements ofM such that 77 x311 ngj1uj ThenuEIM Proof We use induction on the number of nonzero aj we are done if all are 0 If ai gt 0 V 39 V 1 let y be Hj ix Rewrite as Zj 377 1 obj are again regular the Lemma on p 1 yields that yu 7 xiui E j j 31 and so yu E clM 38 j 31 Now ai 0 in the monomial y and there is one fewer nonzero aj The desired result now follows from the induction hypothesis D uj 7 at ya 7 0 Since powers of the If I is an ideal of a ring R we can form the associated graded rlng grim 7 RIeII2 e 691W 1 e an N graded ring whose kth graded piece is IkIk1 If f E h represents an element a E IhIh1 ngth and g E 1 represents an element b E IkI 1 ngRk then ab is the class of fg in IhkIhk1 Likewise if M is an R module we can form ngM MIM IMIZM IkMIk1M This is an N graded module over ngR in an obvious way with f and a as above if u E IkM represents an element 2 E IkMIkHM then the class of fu in IhkMIhk1M represents a2 lf 1 xn E R generate I the classes E 12 generate ngR as an RI algebra Let 6 RIX1 Xn 7 ngR be the RI algebra map such that Xi gt gt This is a surjection of graded RI algebras By restriction of scalars ng is also 4 a module over RIX1 Xn The RI linear map MIM gt ngM then gives a map 9M 1 PLIth 7an RI MIM ngM Note that 6R 6 If u E M represents in MIM and t1 tn are nonnegative integers Whose sum is k then X Xff lul H li1fful7 Where the right hand side is to be interpreted in IkMIk1M Note that 6M is surjective Theorem Let 1 xn be a regular sequence on the R module M and suppose that I 1 icnR Let X1 Xn be indeterminates over the ring RI Then gr1M g RIliXlz an 8121 MIM in such a way that the action of E 12 gIIRl1 0n ngM is the same as multi plication by the variable Xi In particular ificl icn is a regular sequence in R then ngR E RIX1 Xn in such a way that corresponds to Xi In other words ificl xn is a regular sequence on M respectively B then the map 6M respectively 6 discussed in the paragraph above is an isomorphism Proof The issue is Whether 6M is injective If not there is a nontrivial relation on the monomials in the elements With coe icients in MIM and then there must be such a relation that is homogeneous of say degree k Lifting to M we see that this means that there is an M 7 IM linear combination of mutually distinct monomials of degree k in 1 xn Which is in Ik1M Choose one monomial term in this relation it Will have the form x1 icfu Where the sum of the aj is k and u E M 7 IM The other monomials of degree k in the elements 1 xn and the monomial generators of 1k all have as a factor at least one of the terms x111 Join This yields that aj i n aj 1 7 Hjacj uinj uj j1 By the preceding Theorem u 6 IM contradictioning that u E M 7 IM D l Another description of the Koszul complex Let R be a ring and let p 1 xn E R In our development of the Koszul complex n we showed that CZg R has generators ujlji Where 1 S jl lt lt ji 3 n so that i the generators may the thought of as indexed by strictly increasing sequences of integers between 1 and n inclusive of length i We may also think of the generators as indexed by the i elernent subsets of 1 This means that with G IC1 R Ku1 Kun7 we have that I Z Kits R g G for alli E Z in such a way that Wind corresponds to ujl uji Thus the Koszul corn plex coincides with the shew commutative N graded algebra G A shew commutative N graded algebra is an associative N graded ring with identity such that if u and u are forms of degree d e respectively then on ildeuu The elements of even degree span a subalgebra that is in the center A graded derivation of such an algebra of degree 71 is a Z linear map 6 that lowers degrees by l and satisfies 6uu 6uu ildu6u when u and u or forms as above It is easy to check that the differential of the Koszul complex is a derivation of degree 71 in the sense specified Moreover given any R linear Inap G a B it extends uniqely to an R linear derivation of G of degree 71 If we choose a basis for G call it ul un and let it be the value of the map on ui we recover IC R in this way Maps of quotients by regular sequences Let g 1 icn and y yl yn be two regular sequences in R such that J y1ynR Q 1 R I It is obvious that there is a surjection RJ a RI It is far less obvious but very useful that there is an injection RI a RJ Theorem Let 1 icn and y1 yn be two regular sequences on a Noetherian mod ule M over a Noetherian ring R Suppose that Jy17 agnRQ17 Choose elements aij E R such that for all j yj 2211 aijaci Let A be the matria aij so that we have a matria equation yi yn 1 7A Let D detA Then D Q J and the map MIM a MJM induced by multiplication by D on the numerators in injectiue 6 Proof Let B be the classical adjoint of A so that BA AB DI where In is the n gtlt 11 identity matrix Then yl ynB 1 7AB x1 7D shows that D Q J The surjection RJ a RI lifts to a map of projective resolutions of these modules we can use any projective resolutions but in this case we use the two Koszul complexes IC g R and IC y R With these specific resolutions we can use the matrix A to give the lifting as far as degree 1 m R R gt Rx1xn gt 0 Al hi i Klg R R Ry1yn 0 Here we are using the usual bases for K1 R and Klg R It is easy to check that if we use the maps A KM R e 1C4 R for all i we get a map of complexes This means that the map R E KWQ R a 194 R E R is given by multiplication by D It follows that the map induced by multiplication by D gives the induced map ExtR1 xd M a Ext Ry1 yn We have already seen that these top Ext modules may be identified with Mxl M and Myl ynM respectively this is the special case of the Theorem at the bottom of p 4 of the Lecture Notes of October 5 in the case where i 11 Consider the short exact sequence O IJ RJ RI O The long exact sequence for Ext yields in part Extg l J M a EXWRI M a Ext RJM Since the depth of M on AnnRUJ Q J is at least 11 the leftmost term vanishes which proves the injectivity of the map on the right D Remark We focus on the case Where M R a similar comment may be made in general We simply want to emphasize that the identi cation of Extj RI R With RI is not canonical it depends on the choice of generators for I But a different identification can only arise from multiplication by a unit of RI A similar remark applies to the identification of ExtRJ R With RJ Remark The hypothesis that R and M be Noetherian is not really needed Even if the ring is not Noetherian if the annihilator of a module N contains a regular sequence 1 mm of length d on M it is true that Exti N M 0 for i lt d If d 2 1 it is easy to see that any map N a M must be 0 any element in the image of the map must be killed by 1 and AnnMacl 0 The inductive step in the argument is then the same as in the Noetherian case consider the long exact sequence for Ext arising When HomRN i is applied to o MLM Mle o The type of a CohenMacaulay module over a local ring Let R m K be local and let M be a finitely generated nonzero R module that is Cohen Macaulay ie every system of parameters for RI Where I AnnRM is a regular sequence on M It is equivalent to assume that depthmM dim Recall that the socle of an R module M is Anan HomRK It turns out that for any maximal regular sequence 1 mm on M the dimension as a K vector space of the socle in Macl icdM is independent of the choice of the system of parameters One way to see this is as follows Propostion Let R m K and M be as above with M Cohen Macaulay of dimension d over R Then for every ma1imal regular sequence 1 mm on M and for every i 1 g i g d I ExtK M 2 Extqu Macl xiM In particular for every mazrimal regular sequence on M the socle in Macl xdM is isomorphic to Extfe f M and so its K vector space dimension is independent of the choice mazrimal regular sequence Proof The statement in the second paragraph follows from the result of the first paragraph in the case Where i d By induction the proof that Ext K M 2 Extqu Macl aciM reduces at once to the case Where i 1 To see this apply the long exact sequence for Ext arising from the application of HomRK i to the short exact sequence O M M M lM O 8 Note that Extj K M 0 forj lt d since the depth of M on AnnRK m is d and that Ext K MiclM 0 for j lt d 7 l similarly Hence we obtain in part 0 Extfglar Mle ExtK M 3 ExtK M Since 1 6 m kills K the map on the right is 0 which gives the required isomorphism D Proposition Let M 31 0 be a Cohen Macaulay module over a local ring B Let 1 xn and y1 yn be two systems of parameters on M with 241 ynR Q 1 xnR and let A aij be a matria of elements of R such that yl y 1 yewl Let D detA Then the map Macl xnM a Myl ynM induced by multipli cation by D on the numerators carries the socle of Macl xnM isomorphically onto the socle ofMy1yM In particular if yi x5 1 S i S n then the map induced by multiplication by influwffl carries the socle of the quotient module Macl xnM isomorphically onto the socle of Proof By the Theorem on p 5 multiplication by D gives an injection Macl xnMt gtMy1 ynM which must map the socle in the left hand module injectively into the socle in the right hand module Since by the preceding Proposition the two socles have the same finite dimension as vector spaces over K the map yields an isomorphism of the two socles The final statement follows because in the case of this specific pair of systems of parameters we may take A to be the diagonal matrix with diagonal entries x571 icn D Frational rings De nition Frational rings We shall say that a local ring R m K is F rational if it is a homomorphic image of a Cohen Macaulay ring and every ideal generated by a system of parameters is tightly closed We first note Theorem An F rational local ring is Cohen Macaulay and every ideal generated by part of a system parameters is tightly closed Hence an F rational local ring is a normal domain Proof Let 1 ick be part of a system of parameters it may be the empty sequence and let I 1 ick Let 1 xn be a system of parameters for R and for every t 2 1 let Jt 1 ick xi xi d Then for all t I Q J and J is tightly closed so that Iquot Q J and I Q t Jt I as required In particular 0 and principal ideals generated by nonzerodivisors are tightly closed so that R is a normal domain by the Math 711 Lecture of October 9 2006 Lemma comparison of special sequences Let R be an in nite Cohen Macaulay Noetherian domain and let S be a torsion free generically etale R algebra essentially of nite type ouer B Let T be a localization of a polynomial ring in n uariables ouer R that maps onto S and let I be the kernel Let P1 PT be the minimal primes ofI in T Assume moreover that S and T are local Let g gl g and h hl h be two special sequences in I Then there is a nite chain of special sequences joining g to h such that for any two consecutive special sequences occurring in this chain the sequences of elements occuring di er in at most one spot Proof We first show that given two special sequences 91 g and h h and an integer i 1 S i S n we can choose it E T such that 91 9 remains special when gi is replaced by u and h hn remains special as well when hi is replaced by it Since regular sequences and hence special sequences are permutable in a local ring we may assume without loss of generality that i n Our first objective is to choose it such that for all j both sequences generate each IDijj Since P R 0 for every j we have for each j that IC Q ij To solve the problem we shall rst choose it with the required property in IC R T For every j let 0 denote the contraction of 917 79771 P2TP to IC R T and let D denote the contraction of any 7hn71 t P32ij to IC R T The O and D together constitute finally many vector spaces over the eld IC We claim that they do not cover IC R I Q IC R T for if they did then one of them would contain IC R I see the first Proposition on p 3 of the Lecture Notes of September 18 and this would contradict the existence of 9 if it were one of the O or the existence of h if it were one of the D Hence we can choose it 6 IC R I with the required property After multiplying by a suitable element of R 7 0 we may assume that u is in I and it will still have the required property since the multiplier is a unit in every Rpj Finally as in the proof of the Theorem on existence of su iciently many special sequences which is the last Theorem of the Lecture of October 6 we can choose 1 E I2 such that u u is not a zerodivisor modulo either 91 gn1T nor modulo hl hn1T this comes down to the assertion that u I2 is not contained in the union of the associated primes of the two ideals and by the Lemma on prime avoidance for cosets it su ices to show that I2 is not contained in any of them But this is clear because all the associated primes have height n 7 1 while I2 has height n Finally to prove the existence of the chain of special sequences we use induction on the number of terms in which the two sequences counting from the beginning agree Suppose 1 2 that gi hi for i lt j while sequences gj 31 hj j may be 0 here Then by the result of the paragraph above we may choose u such that the sequences 917 7 9j717u79j17 7 9n and 917 7 gj717u7hj17 7 hn are both special The first differs from 91 9 in only the jth spot and the second differs from h hn in only the jth spot as well By the induction hypothesis there is a chain of the required form joining these two and the result follows E The map I and the modules WSR Our next main goal is to construct the map I mentioned brie y in Step 3 of our Sketch of the proof the Jacobian theorem see p 4 of the Lecture Notes of October 4 We first need a lemma whose proof involves universal modules of differentials or Kahler differentials In the next seven paragraphs we assume only that IC is a commutative ring and that T is IC algebra A IC derivation of T into a T module M is a map D T 7 M such that l D is a homomorphism of abelian groups 2 FOE all 151152 6 T Dt1t2 t1Dt2 t2Dt1 and 3 For all e 6 IC Dc 0 Condition 2 implies that D is IC linear for Dct cDt tDc cDt 150 cDt Note that Dl l lDl lDl ie Dl Dl Dl so that condition 2 implies that Dl 0 In the presence of the other conditions 2 is equivalent to IC linearity for if the map is IC linear then for all e 6 IC Dc l cDl 00 0 There is a universal IC derivation from a given IC algebra T into a specially constructed T module QTK This is obtained as follows Let G denote the T free module whose basis consists of elements bt in bijective correspondence with the elements t of T Consider the T submodule H of G spanned by elements of the three forms 1 btlt2 7 btl 7 th for all t1 t2 6 T 2 bt1t2 7151th 7152th for all t1 t2 6 T and 3 bC for c 6 IC We write QTK for GH This is the universal module of di erentials or the module of Kahler di erentials for T over IC Note that we have a map d T 7 QTK that sends t gt gt b H the class of bt in GH The choice of elements that we killed we took them as generators of H precisely guarantees that d is a IC derivation the universal IC derivation 3 on T It has the following universal property given any T module N and a IC derivation D T a N there is a unique T linear map L M a N such that D L o d To get the map L on GH we de ne it on G so that the value on b is Dt this is forced if we are going to have D L o d One may check easily that H is killed precisely because D is a IC derivation and this gives the required map It is also easy to see that the composition of d with any T linear map is a IC derivation Thus for every T module N there is a bijection between the IC derivatons of T into N and HomTQTK N Given generators ti for T over IC the elements dti the index set may be in nite span QTK The value of d on a product of these generators is expressible in terms of the dti by iterated use of the product rule Given a polynomial ring ICXZ i E I over IC QTK is the free T module on the dXi and the universal derivation d is defined by the rule oF dF 8 dXi This formula is a consequence of the use of the iterated product rule and it is straightfor ward to check that it really does give a derivation Note that if U is a multiplicative system in T we have that QUaTK g U l TK Also observe that a IC derivation D T a M extends uniquely via the rule tu gt gt th 7 tDuu2 to a IC derivation U lD U lT a U lM so that the diagram U lT U lM l l T L M commutes The notations in the following Lemma are slightly different from those in our general setup Lemma Let N be a maximal ideal of T ICX1 Xn a polynomial ring over a eld IC Assume that ICX1 anN is separable eld edtension of IC Then 91 9 E NTN generate NTN if and only if the image of detngBaci in is not 0 Proof Consider the universal IC derivation d ICX a QMXWC the module of Kahler differentials which as noted above is the free T module generated by the elements Xm dX Of course if F E ICX then dF 218F8xjdacj The restric tion of d to N gives a IC linear map N a QMXWC and by the defining property of a derivation it sends N2 NQquK Thus there is an induced IC linear map of IC vector spaces SINN2 a C T QTK Both modules are vector spaces and it follows from the de ning property of a derivation that 6 is actually linear if t E T represents A 6 and u E N dtu tdu udt and the second term will map to 0 in T QTK Since TN is regular of dimension n JVJV2 is an n dimensional vector space over A The key point is that under the hypothesis that is separable over IC the map 6 is an isomorphism of vector spaces To see this observe that the map 6 sends the elements represented by generators gl g for NTN to the elements represented by the dgj and so it has a matrix which is the image of the matrix 893810 after mapping the entries to A Thus 6 is an isomorphism if and only if the Jacobian determinant det ng has nonzero image in L But this determinant generates JLK and so 6 is an isomorphism if and only if the Jacobian ideal of over IC is L But we may use any presentation of over IC to calculate JLK and so we may instead use ICZfZ where Z here represents just one variable and where f is a single separable polynomial The Jacobian determinant is then the value of fZ in which is not zero by virtue of the separability Thus 6 is an isomorphism Moreover we have already seen that if 91 97 are generators of NTN then the Jacobian determinant is not 0 in L But the converse is also clear because if 91 g are any elements of NTN they generate NTN if and only if their images in NTNNZTN NN2 span this vector space over by Nakayama7s lemma and this will be the case if and only if their further images in T QTK span that vector space over since 6 is an isomorphism But that will be true if and only if the images of the dgj span T QTK which is equivalent to the assertion that the images of the columns of the matrix 893810 after the entries are mapped to span an n dimensional space This in turn is equivalent to the nonvanishing of det 891815 in L We now return to our standard set of notations and assumptions as in Step 2 of the Sketch of the proof of the Jacobian theorem from p 3 of the Lecture Notes of October 4 Thus T is a localization of RX1 Xn that maps onto S with kernel I U is the complement in T of the set of minimal primes P1 PT of of I in T and I 11 Pi The B do not meet R and correspond to the minimal primes of IIC T The expansion of B to U lT is maximal ideal Mi corresponding to a maximal ideal of ICX1 Xn and has height n Here sz g ICX1 X lNi Corollary Let gl g E I If 91 g is a special sequence in I then the image y of detng8Xi is not a zerodivisor in S and so represents an invertible element of the total quotient ring of S Proof We may view as the product of the fields i where i is the fraction field of TPZ but may also be identified with ICX1 lt su ices to show that y does not map to 0 under a i for any i The fact that the image of y is not 0 in i follows from the preceding Lemma the separability of i over IC and the fact that for every i 91 gn generates Pipr which we may identify with MICX1 X lNi D We continue the conventions in the paragraph preceding the statement of the Lemma but because we shall let both S and its presentation vary we shall write 6 for the map T a S Math 711 Lecture of October 3 2007 Koszul homology We de ne the ith Koszul homology module Hiac1 acn M of M with respect to 1 xn as the ith homology module HiICx1 acn M of the Koszul complex We note the following properties of Koszul homology Proposition Let R be a ring and g 1 xn E B Let I Let M be an R module a HA M 0 2ft lt 0 or 2ft gtn b Ho M MIM c HWQ M AnnMI AnnRM kills euery Hiac1 xn CL D IfM 239s Noetherz39an so is its Koszul homology HA H For every t HA 7 2s a couartant functor from R modules to R modules g1f O M M MH O 2s a short ezmct sequence of R modules there is a long eract sequence of Koszul ho mology we llg MU HiQ M HiQ M i71 i M h Ifxl xn 2s a possibly improper regular sequence on M then HA M 0 i 2 1 Proof Part a is immediate from the definition Part 10 follows from the fact that last map in the Koszul complex from IC g M a ICO M may be identified with the map M a M such that 111 un gt gt 11 xenon Part c follows from the fact that the map 194 M a Kn1 M may be identified with the map M a M such that u gt gt x11 72U7 71 4107111 Parts d and e are clear since every term in the Koszul complex is itself a direct sum of copies of M To prove f note that if we are given a map M a MC there is an induced map of complexes IC R 8 M a IC R 8 M l 2 This map induces a map HA M a Hg M Checking that this construction gives a functor is straightforward For part g we note that u 0 a IC R R M a IC R R M a IC R m M a 0 is a short exact sequence of complexes because each Q g R is R free so that the functor Q g R ER 7 is exact The long exact sequence is simply the result of applying the snake lemma to This sequence can also be constructed by interpreting Koszul homology as a special case of Tor we return to this point later Finally part h is immediate by induction from the iterative construction of the Koszul complex as a mapping cone and the Proposition at the top of p 6 of the Lecture Notes of October 1 The map of augmentations is the map given by multiplication by 7 from Macl xn71M to itself which is injective because 1 icn is a possibly improper regular sequence D Corollary Let g 1 xn be a regular sequence on R and let I Then RI has a nite free resolution of length n ouer R and does not have any projective resolution of length shorter than n Moreover for every R module M Tor RI M 2 H4 M Proof By part f of the preceding Proposition IC g R is acyclic Since this is a free complex of finitely generated free modules whose augmentation is RI we see that RI has the required resolution Then by definition of Tor we may calculate Tor RI M as HZ IC g R RM which is precisely H43 To see that there is no shorter projective resolution of RI take M RI Then TorRI RI HWQ RI AnnRII RI by part c of the preceding Proposition If there were a shorter projective resolution we would have TorRI RI 0 Independence of Koszul homology of the base ring The following observation is immensely useful Suppose that we have a ring homomor phism R a S and an S module M By restriction of scalars M is an R module Let g 1 xn E R and let y y1y be the images of the mi in S Note that the actions of 1 and yi on MTare the same for every i This means that the complexes IC M and ICg M are the same In consequence Hal M g H g M for all j as S modules Note that even if we treat M as an R module initially in caclulating Hj g M 3 we can recover the S module structure on the Koszul homoolgy from the S module struc ture of M For every 5 E S multiplication by s is an R linear map from M to M and since HA 7 is a covariant functor we recover the action of s on HA Koszul homology and Tor Let R be a ring and let p 1 xn E B Let M be an R module We have already seen that if 1 xn is a regular sequence in R then we may interpret H4151 ion M as a Tor over R In general we may interpret HA M as a Tor over an auxiliary ring Let A be any ring such R is an A algebra We may always take A Z or A R If R contains a field K we may choose A K Let K X1 Xn be indeterminates over A and map B AX1 Xn 7 R by sending Xj gt gt obj for all j Then M is also a B module as in the section above and X1 Xn is a regular sequence in B Hence Proposition With notation as in the preceding paragraph Hin M 2 TorfBXB M Corollary Let g 1 xn E R let I QR and let M be an R module Then I hills H43 M for all i Proof We use the idea of the discussion preceding the Proposition above taking A B so that with X X1 Xn we have an R algebra map B g 7 R such that X gt gt 1 1 S i S n Then Iiiz M TOI BEB M When M is viewed as a B module every X 7 xi kiils M But X kills B B and so for every i both X 7 ac and X kill Tori3 B B It follows that every xi X 7 7 kills it as well and the result now follows from D An application to the study of regular local rings Let M be a finitely generated R module over a local ring R m A minimal free resolution of M may be constructed as follows Let b0 be the least number of generators of M and begin by mapping Rbo onto M using these generators lf Rbi Rbo gtM70 has already been constructed let bi be the least number of generators of Z Ker 041 and construct Oli1 Rbitl 7 Rbi by mapping the free generators of Ptle to a minimal set of generators of Z Q Rbi Think of the linear maps 04 i 2 l as given by matrices Then it is easy to see that a free resolution for M is minimal if and only if all of the matrices 04139 for i 2 l have entries in m We have the following consequence 4 Proposition Let R m K be local let M be a nitely generated R modules and let 04239 sz a Rbo Mao be a minimal resolution of M Then for all i Tor M K 2 Kbquot Proof We may use the minimal resolution displayed to calculate the values of Tor We drop the augmentation M and apply K ER 7 Since all of the matrices have entries in m the maps are alll 0 and we have the complex gKbi gmgKbO 0 Since all the maps are zero the result stated is immediate D Theorem AuslanderBuchsbaum Let R m K be a regular local ring Then every nitely generated R module has a nite projective resolution of length at most n dim Proof Let g 1 xn be a regular system of parameters for B These elements form a regular sequence It follows that K has a free resolution of length at most n Hence ToriM K 0 for all i gt n and for every R module M Now let M be a finitely generated R module and let Rbi Rbl Rbo R M O be a minimal free resolution of M For i gt n bi 0 because ToriM K 0 and so Rbi 0 for i gt n as required D It is true that a local ring is regular if and only if its residue class field has finite projective dimension the converse part was proved by J P Serre The argument may be found in the Lecture Notes of February 13 and 16 Math 615 Winter 2004 It is an open question whether if M is a module of finite length over a regular local ring R m K of Krull dimension n one has that dimKTorZM K 2 l The numbers dimKTor M K are called the Betti numbers of M If 1 xn is a minimal set of generators of m these may also be characterized as the dimensions of the Koszul homology modules H43 A third point of view is that they give the ranks of the free modules in a minimal free resolution of M The binomial coef cients are the Betti numbers of K Rm they are the ranks of the free modules in the Koszul complex resolution of K The question as to whether these are the smallest possible Betti numbers for an R module was raised by David Buchsbaum and David Eisenbud in the first reference listed below and was reported by Harthshorne in a Math 711 Lecture of September 7 2005 Our objective is to present a large number of open questions in commutative algebra to give an indication of the status of these questions as well as to develop many techniques of attack that may prove useful in addressing these questions At the same time we shall also present some of the consequences of af rmative answers to these questions proving the needed implications The extent to which these proofs will be given here depends somewhat on the accessibility of the proofs in other references in the literature All rings are assumed to be commutative associative with identity and all modules unital unless otherwise specified As an example of a very simple open question let X and Y yij be matrices of 2112 algebraically independent indeterminates over a field K Let It A denote the ideal generated by the size If minors of the matrix A the ring is deduced from the context Let K be a field let KX Y be the polynomial ring in the entries of X and Y over K we shall typically use this notation when dealing with one or more matrices and let R KX Yj11XY 7 YX It is an open question in general whether R is reduced a domain normal andor Cohen Macaulay These properties all hold when the matrices have size 3 or smaller We shall discuss methods that might be used to attack this problem and use them to prove that with K as above for an 7 gtlt 5 matrix of indeterminates X one has that is a Cohen Macaulay normal domain The method we shall focus on is that of principal radical systems Another approach that is sometimes useful is the theory of algebras with straightening law if time permits we will discuss that theory as well In both questions one can reduce at once to the case where K is algebraically closed Note that in R above the images of X and Y are generic commuting matrices and the algebraic set V11XY 7 YX consists of pairs of commuting n gtlt n matrices a natural object of study This algebraic set is known to be a variety ie to be irreducible by a result of M Gerstenhaber ln quite a different direction we shall consider several homological conjectures that are stated in or reduce to the case of a local ring and where it in fact suf ces to consider the case of a complete local ring One is the existence of big and small Cohen Macaulay modules and algebras The existence of big Cohen Macaulay algebras and the existence of small Cohen Macaulay modules both imply the existence of big Cohen Macaulay modules Big Cohen Macaulay modules and algebras are known to exist in equal characteristic and in mixed characteristic in dimension at most 3 Small Cohen Macaulay modules are not known to exist even in dimension 3 and the conjecture that they do exist at this point seems doubtful to me Another is the direct summand conjecture which asserts that a regular Noetherian ring is a direct summand of every module finite extension ring We shall prove the equivalence of this conjecture with several others including the canonical element conjecture the monomiol conjecture and with the improved form of the new intersection theorem and we shall show it can be used to deduce a conjecture of M Auslander an af rmative answer to a 1 question of Bass and a proof of the syzygy conjecture of Evans and Grif th The conjecture of Auslander is that a zerodivisor on a nitely generated module M 31 0 of nite projective dimension over a local ring R m K must be a zerodivisor on R Contrapositively a nonzerodivisor or by a trivial induction even a regular sequence in m on such a module M must be a nonzerodivisor respectively a regular sequence on R Bass7s question is whether if a local ring R m K possesses a finitely generated module M 31 0 of finite injective dimension must R be Cohen Macaulay We shall show how these follow from new intersection theorem which is explained below and hence how they follow from the direct summand conjecture The new intersection theorem and hence the two corollaries of it described above have been proved even in mixed characteristic by Paul Roberts using the Fulton MacPherson Baum intersection theory We shall also explain this proof but we assume certain facts about intersection theory without proof The Evans Grif th syzygy conjecture asserts in its simplest form that any k th module of syzygies over a regular local ring if not free needs at least k generators The hypothesis on the ring can be weakened The problem is still open in mixed characteristic in the same cases where the direct summand conjecture is open It can be deduced from the improved form of the new intersection theorem which is equivalent to the direct summand conjecture and we shall eventually explain that argument We want to explain in greater detail what the three conjectures monomial canonical element improved new intersection say as well as the new intersection theorem itself The monomial conjecture asserts that if 1 ard is a system of parameters for a local ring R of dimension d then for every positive integer If one has that ac i at 3 1R Both the new intersection and its improved version refer to a finite free complex G of finite rank free modules say 0gtGngtgtG1gtGOgt0 over a local ring R m Both conjectures assume that the homology modules HAG have finite length for i 2 l and both assume that the augmentation module H0G 31 0 In the intersection theorem it is assumed that H0G has finite length In the improved version the hypothesis is weakened it is only assumed that some minimal generator of H0G is killed by a power of m In both cases the conclusion is that dim R S n The hypothesis in the improved version is more technical and the resulting conjecture may not seem a lot stronger at first but it turns out to be apparently much stronger in its applications The new intersection theorem implies both Auslander7s zero divisor conjecture and an af rmative answer to Bass7s question and it also proves the Evans Grif th syzygy conjecture All of these conjectures are known if the ring contains a field but while the new intersection theorem has been proved by Paul Roberts in all cases the improved version like the direct summand conjecture remains open in mixed characteristic in dimension 4 or more The canonical element conjecture is a bit harder to explain One version is this Let R m K be local of Krull dimension d choose a free resolution of K over R by nitely generated free modules truncate this resolution to obtain an exact complex C of the form O syzdK Rbdil Rbl Rbo oy choose a system of parameters 1 xd for R and choose a lifting of the obvious map of augmentations Rxl xd a Rm K to a map from the Koszul complex IC 1 xd R to the complex C Then the conjecture asserts that no matter how the choices are made the map R Kdac1 acd R a syzdK is not zero Equivalently the element in syzdK that is the image of l E R is not zero Because the map of complexes is not unique the image of l in M syzdK is only determined mod 1 xdM Therefore an equivalent assertion is that the image of l in Macl ardM where M syzdK is not 0 Now consider what happens when we replace the system of parameters 1 ard by the system of parameters at i 102 There is a map of Koszul complexes Kol7 7 3 W Ko17 z di that is the identity on R and whose matrix mapping Rd a Rd in degree one is the diagonal matrix with diagonal entries x571 271 If one thinks of the Koszul complex as constructed using exterior powers the other maps are the exterior powers of the map in degree one and so the last map R a R is given by multiplication by the determinant of the matrix ie by multiplication by at wd By composing with the map already constructed from IC 1 xd R a C we obtain a map from IC 5 xg R a C The image of l is the product of the earlier image by 571 271 The modules Mt M form a direct limit system in which the maps Mt a Mt1 are induced by multiplication by 1 and on M and the direct limit is Hgmlw 7mdM E Hg We refer to the element 6 that is the image of 1 from any of the maps R ICdx i 3 R a M in the local cohomology module H isyzdK as the canonical element in HgsyzdK The canonical element conjecture asserts that E 31 0 If we take a different free resolution of K we get a different complex CL and a different dth module of syzygies M But there are induced maps of complexes between C and C in both directions whose compositions are up to homotopy the appropriate identity It follows that there are induced maps a H fM and HgM a H fM and these maps take the canonical element in to the canonical element in HgM and vice versa Therefore whether the canonical element vanishes is independent of the choice of free resolution of K There is a more invariant point of view that identifies E with an element of Tora K Hf R we shall discuss this later We shall also consider a strengthening of the direct summand conjecture the strong direct summand conjecture and its many consequences We shall likewise discuss the Buchsbaum Eisenbud Horrocks conjecture that the j th Betti number of a finite length module M 31 0 over a regular local ring R m K is at Math 711 Lecture of November 7 2005 A simplicial complex is constructz39ble if it is a simplex or recursively the union of two contructible complexes of dimension n intersecting in a constructible complex of dimension n 7 1 It is easy to verify that every constructible simplicial complex A satis es Reisner7s criterion over every field K by a straightforward induction that uses the Mayer Vietoris sequence The empty set is constructible and in dimension 0 every finite simplicial complex is constructible ln dimension 1 a simplicial complex is constructible if and only if it is connected ln dimension 2 things get more complicated A cylinder is not be constructible because the first cohomology group does not vanish A triangulation of a real projective plane is not Cohen Macaulay in characteristic 2 because the first cohomology equivalently homology group won7t vanish If we think of the real projective plane as a 2 sphere with opposite points identified it is clear that the fundamental group is Z2 On the other hand a trian gulation of a real projective plane gives a Cohen Macaulay face ring in every characteristic except two This means that the minimal resoluton of the corresponding defining ideal I Q Kjacl xn where n is the number of vertices generated by square free monomials depends on the characteristic Even its length depends on the characteristic since it must be longer in characteristic 2 the length is n 7 deptthjacl xnjI m is the homoge neous maximal ideal and the depth is equal to dim KjA dim A 7 l precisely when the ring is Cohen Macaulay and smaller otherwise We note that the Cohen Macaulay property for KjA only depends on the geometric realization lAl of A not on the face ring by a result of Munkres We shall not give complete details but we will give a sketch of the explanation One can define the reduced local cohomology of the space X at x with coef cients in K which we denote UXK as the direct limit over successively smaller open neighborhoods U of x of U YUK where YU is the quotient space obtained from X by collapsing X 7 U to a point One evidently need only take the limit over a family of choices for U that is cofinal in the set of all open neighborhoods of x The local cohomology will in general depend on the cohomology theory being used but when X is the geometric realization of A one can choose the neighborhoods U such that the spaces YU have finite triangulations and one can use simplicial cohomology throughout We will return to this point momentarily Note for example that when X is an n manifold near at we can choose arbitarily small neighborhoods U of x such that the closure D of U is an n ball in n space whose boundary is an n 7 l sphere and the spaces YU are n spheres Then KjA is Cohen Macaulay if and only if for all j lt dim and all x E lAl EU A 0 This turns out to be equivalent to Reisner7s criterion the key point is that the condition on vanishing of topological local cohomology is equivalent to the vanishing of the reduced cohomology of the various links A in degree lt dim To explain why we first recall that the suspension of the topological space X is the union of two cones over X identified only along X For a simplicial complex A there is a 1 simplicial version a cone over A with one additional vertex y can be obtained by taking all simplices of A together will all sets that are the union of simplex of A and Call this 09A The suspension is achieved as a simplicial complex namely OyA U OZA where y and z are two new vertices Let SX denote the suspension of X An easy application of the Mayer Vietoris theorem shows that K 0 for all i lt dim iff iSXK 0 for all i lt dim X 1 dim 500 We want to understand the local cohomology of lAl as a function of the point x of lAl chosen This turns out to depend only on the smallest simplex 039 E A such that x is an interior point of 0 The neighborhoods U of x can then be chosen arbitrarily small in such a way that the space YU is an iterated suspension of the link of 039 where the number of iterations is the same as the number of vertices of 039 The reduced cohomology then vanishes except in the highest dimension the dimension of A iff the same is true for the link A of 039 and this will complete the proof Recall that the closed star of z where x is a vertex of A consists of all simplices in A that contain at and their subsets The closed star is the cone with vertex at over the link of x The open star of x is the geometric realization of the closed star with the geometric realization of the link of x subtracted We want to verify the assertions of the preceding paragraph first considering the case where x is a vertex of A The open star U of x is an open neighborhood of x in lAl that contains arbitrarily small neighbhorhoods Ut homeomorphic with itself To see this note that the closed star is a union of rays emanating from x and terminating in a point A E A the link of x Use 15 6 01 to parametrize the points of the ray linearly with 0 corresponding to x and l to A Let A denote the set of all points parametrized by t gt 0 on these rays it is homeomorphic with A Let U be the union of and the A5 for 0 lt s lt t it is homeomorphic with the open star Given UgtQ with t lt 23 the effect of collapsing lAl 7 Us to a point is the same as the effect of collapsing UgtQ 7 Us to a point y and the resulting space can be thought of as the union of two cones over the link of at One consists of x and the points in A5 for 0 lt s S 12 The other consists of points A for 12 S s lt 15 together with the point 2 Thus YUt is homeomorphic with the suspension of the link of x for all t In fact all the YUt are homeomorphic with the space obtained by collapsing the link of x in the closed star of x Hence the reduced simplicial cohomology of the link vanishes in degree smaller than the dimension of the link if and only if the reduced simplicial local cohomology at x of lAl vanishes in degree smaller than the dimension of More generally if x is interior to 0 with vertices 1 5 we get a closed neigh borhood of x by taking the union of all lUl for 0 Q 039 which is an iterated cone over the geometric realization of the link A of 039 Call this iterated cone 00A and also call it V Let U be the set of points of V which when written as a convex linear combination of the vertices of a simplex involve all 1 zk with positive coef cients Then U is an open neighborhood of x in Let B be the union of the geometric realizaions of the simplices that do not contain at least one Q Then U V 7 B and YU may also be thought of as the result of collapsing B V 7 U Q V to a single point We want to show that x has arbitarily small open neighborhoods U such that YUt is homeomorphic with YU We also want to show that YU is homeomorphic with the result of iterated suspension of the link A of ac where the iteration is performed k times To construct the neighborhoods Ut simply note that for t gt 0 if 6 is the homothety centered at it that maps every ray from it into itself but multiplies distances from it by t then for t su iciently small if Vt Us and 3 denote the images of V U and B respectively then V is a closed neighborhood of ac and Us V 7 3 is an open neighborhood of it that may be made arbitrarily small Moreover the restriction of 6 maps V UB homeomorphically onto Vt Ut Bt Thus each YUt is homeomorphic to Yt for 0 lt t lt 1 It remains only to see that YU is the iterated suspension of the link A k times Note quite generally that when we take the cone over X and then collapse X to a point we get the suspension of X In our case If oj denotes o 7 obj we are collapsing the union of the 00 For j lt k let rj oj 7 Let Z be the result of collapsing the union of the 07 for j lt k in Oak By induction on k we may identify Z with the k 7 1 st iterated suspension of Finally YU may be identified with the result of collapsing Z in CM Z which is the suspension of Z as required D We next want to consider several interrelated conjectures and theorems for local rings We shall eventually prove the various relations We begin by stating several of these 1 The direct summand conjecture Let R be a regular ring and S a module nite ezrtension of R Then R is a direct summand ofS as an R module 2 The monomial conjecture Let R be any local ring and 1 icd a system of parameters for R Then for every nonnegatiue integer t ic i 39 g Z 1 x1R 3 The canonical element conjecture Let R m K be a local ring of dimension d andg 1 icd a system of parameters for B Let 9 denote an ezract sequence 07syzdK7Gd717gt7gtG07gtK7gt0 where the Gj are free so that this is a truncated free resolution of Here syzdK is a dth module of syzygies of K The canonical surjection Rxl xd 7 K lifts to a map b of complezres from the Koszul complea IC ICac1 icd R to the truncated resolution 9 0 syZdK Gdil G0 K 0 WT d7lT WT T 0 a R Cdnl gt R R R 0 where we have identi ed Cd g R g 1C0 For any such li ting bd 31 0 We shall eventually explain at length why this is called the canonical element conjecture Recall that if R is a domain R denotes the integral closure of R in an algebraic closure of its fraction field Like the algebraic closure of a field it is unique up to non unique isomorphism 4 4 Nonvanishing of local cohomology of R If R is a complete local domain of Krull dimension d then H R 31 0 5 Improved new intersection conjecture Let R m K be a local ring of dimen sion d and 0gtGngtgtG0gt0 a comple1 of nitely generated free modules such that H0G has a minimal generator that is killed by a power ofm and HAG has nite length for i gt 0 Then d S n It turns out that the conjectures stated in 1 through 5 are all equivalent when R has positive prime characteristic p and also when Rm has characteristic p gt 0 Formally they are also equivalent in equal characteristic 0 because all of them are true in that case These statements are all known to be true in the equal characteristic case and in mixed characteristic in dimension 3 3 The general case remains open They are also known to imply several other important results which we state below 6 The new intersection theorem Let R m K be a local ring of dimension d and 0gtGngtgtG0gt0 a complex of nitely generated free modules such that H0G 31 0 and HiG has nite length for all t Then d S n It is obvious that 5 i 7 The intersection theorem Let R m K be a local ring and M N nitely gener ated nonzero modules such that M R N has nite length Then dim N S deM Of course this is only interesting when deM lt 00 To see that 6 i 7 let 6 be a minimal free resolution of M so that its length is n deM and replace N by RI where I AnnRN Let G RI ER 6 Then the homology of this complex is killed by both I and AnnRM which means that it is nite length and so we can conclude that dim N dim RI S n deM D 8 Auslander s zerodivisor conjecture Let R be a local ring and M 31 0 a nitely generated module of nite projective dimension Let x be an element ofR that is a zero diuisor on R Then at is a zerodivisor on M Under the same hypothesis one can also state the contrapositive that a nonzerodivisor on M is a nonzerodivisor in R It follows that if M 31 0 has nite projective dimension then a regular sequence on M is a regular sequence in R 9 Bass s question Let R be a local ring that possesses a nitely generated nonzero module M of nite injectiue dimension Then R is Cohen Macaulay It is not di cult to deduce both 8 and 9 from Note that if R is Cohen Macaulay it does possess a nitely generated R module of nite injective dimension if g is a system of parameters the injective hull of K over R R turns out to be such a module We should note that 6 has been proved in all characteristics by Paul Roberts and so all of 6 through 9 are known in general We also mention 5 10 The EvansGrif th syzygy conjecture A kth module of syzygles of a nitely generated module over a regular local ring if not free has torsion free rank at least k This can be deduced from 5 and is known in equal characteristic but not in general in mixed characteristic We shall discuss all of this in detail later We next want to explain the source of the name canonical element conjecture77 for We note that even for a fixed choice 9 of truncated free resolution and a fixed system of parameters the map od is not unique but it is unique up to homotopy of maps of complexes This means that od can be changed to any map of the form od 1 hdd where 6d Cd a ICd1 g Rd has a d gtlt 1 matrix whose entries are up to sign the elements obj and h ICdA a syzdK Hence d1 is unique in syzdKx1 xdsyzdK Therefore we may phrase the condition instead by saying the for every system of parameters the image of 1 6 Cd is nonzero syzdKac1 xdsyzdK The condition becomes formally more dif cult to satisfy as the system of parameters generates a smaller ideal Suppose that A is d gtlt d matrix such that 241 yd 1 acdA where yl yd is also a system of paramters As in the discussion of in the Lecture of October 12 the exterior powers of A may be used to give a map of Koszul complexes ICy R a IC g R In degree d the map is multiplication by detA This map of Koszul complexes may be composed with the map from Cg R to g to obtain a map ICy R a 9 and the image of 1 E R ICdy R is detA times the image of 1 Q ICdi B This means that we are obtaining a wiell defined element me in Hz syzd Even when the ring is not Cohen Macaulay for any R module M Hf may be viewed as LmQMoM as g runs through all systems of parameters for R and it suf ces to consider hint x M for a single system of parameters g 1 xd where the map x M a Mac i1x1M is induced by multiplication by 1 and We refer to rm as the canonical element in Hf syszQ One other point should be made by passing to local cohomology we have avoided dependence on the choice of system of parameters and we have also avoided dependence on the choice of the map Cg R to 9 However the truncated resolution 9 is not uniquely determined and syzd K is only determined up to adding a free direct summand and isomorphism However given two different free resolutions of K one can lift the identity map on K to a map of complexes from each to the other The compositions of these maps of complexes are homotopic to the respective identity maps There are maps from the version of syzdK call it S arising from one resolution to the version call it Math 711 Lecture of September 21 2005 We continue to review some basic facts about simplicial complexes The conuea hull of a nite set 111 wk Q R is the smallest convex set that contains it it may also be characterized as r1vlrkuk0 ri 1r1rk1 Let e1 e be the standard basis for R The geometric realization lAl of a finite simplicial complex A With vertices 1 ic may be defined as the topological sub space of R obtained as the union of the convex hulls of the sets cl1 eik such that gel1 is a face of A Then the dimension of A is the same as the dimension of the real topological space We also recall the notion of the simplicial homology HA K of A With coef cients in a commutative ring K We form a free complex C such that Cj j 2 0 is the free K module on the free basis consisting of simplices of A of dimension j To define the differential we need only specify its value on a typical j simplex o gel1 JulH1 We do this by letting j1 d0 ZEDHW 9623 151 Then H A K is defined as the homology of this complex and is a topological invariant of lAl and in fact only depends on the homotopy type of lAl but the proof of this is outside the scope of these lectures We also define the reduced simplicial homology El A K as the homology of the com plex obtained by modifying the complex C above to include a term in degree 71 the additional term is the free module on one generator corresponding to the unique simplex of dimension 71 to Wit the empty simplex Q The value of d on every is 0 Note that EllA K E HiA K for i 2 l The rank of 710A K is one less than the rank of H0A The rank of the latter is the number of connected components of In particular When lAl is connected K0X 0 Proposition Let K be an integral domain and A a nite simplicial complea with uertices 1 ic There is a bijection between the facets o ofA and the minimal primes ofKA equivalently of IA in which the minimal prime corresponding to o is generated by the images of the variables not in o The quotient by the minimal prime corresponding to 0 may be identi ed with the polynomial ring ouer K in uariables corresponding bijectiuely with the uertices of 0 Hence if K is a eld dim dim A 1 More generally dim KA dim dim A l Proof The final statement about dimension follows from the analysis of the minimal primes and the fact that the dimension of Kac1 ick is dim h when K is Noe therian 1 We may analyze the minimal primes of I I A as follows Consider any prime P containing IA Note that for each subset of 1 xn that is not a face the product of the variables in that subset is 0 and so at least one of them is in P But a subset S of the vertices meets every subset that is not a face if and only if its complement is a face If the complement were not a face it would have to meet it a contradiction On the other hand if the complement of the face 0 fails to meet a set the set must be contained in o and therefore is a face Thus any prime P Q I contains the variables in the complement of a face and therefore the variables in the complement of a facet But the variables in the complement of a facet generated a prime Q that contains I each generator of I is divisible by one of variables in Q and there can be no smaller prime that contains I within Q since it would have to contain the complement of a larger face D Note that every quotient ring R of Kac1 icn obtained by killing square free mono mials has the form KA let A consist of all subsets of the variables the image of whose product is not 0 in R Think of 1 xn as coordinate functions on A where K is a field By a coordinate k plane we mean a k dimensional vector subspace of K A defined by the vanishing of a subset of the coordinate functions n 7 k coordinate functions will be needed Then with K a field VIA Q A is the union of the coordinate k planes where k may vary defined by the vanishing of the variables in the complements of the facets of A For example when n 3 and the facets of A are 1 2 1 3 and 2 3 VIA is the union of the three planes Vx1 V2 and V3 If the field is the real numbers and C denotes the standard simplex which is the convex hull of the standard basis e1 e then VIA O is the same as the geometric realization lAl of A We next want to describe Reisner7s criterion for when the ring KA is a Cohen Macaulay ring In order to do so we introduce the concept of a link within a simplicial complex Let A be a simplicial complex and o E A We define the link of o in A to be the simplicial complex 739 E A 739 0 Q and 739 U o E A One defines the link of a vertex in to be the same as the link of The link of Q is simply A The link of o gel1 may also be obtained recursively as follows take the link of mil in A to produce A1 At the the t lst stage take the link of gelH1 in At to produce At Then Ak is the link of o For example if A is the triangulation of the disk obtained from a convex polygon including its interior by joining an interior point x to each vertex on the boundary the link of x in A consists of the boundary it is one dimensional We shall eventually prove Theorem Reisner s criterion Let K be a eld and A a nite simplicial complex Then KA is Cohen Macaulay if and only iffor every A which is either A itself or a link of a simplem of A ifA has dimension h then HiA K 0 0 S i S h 71 Mentioning that A itself satisfies the condition is redundant here since it is the link of 0 but the statement about A is included for emphasis lf lAl is a manifold with boundary the condition on smaller links is automatically satisfied Although it is not immediately obvious the condition given is topological ie it depends only on We note that if A Math 711 Lecture of November 27 2006 Before giving a second proof of the existence of linear maximal Cohen Macaulay rno dules for KXIg and the extension of this result to the case of more general Segre products we want to note that the Segre product of two Cohen Macaulay rings need not be Cohen Macaulay even when one of them is a normal hypersurface and the other is a polynomial ring In fact let K be any field whose characteristic is different from 3 and let R KX Y ZlXS Y3 Z3 Kac y z and S Ks t where X Y Z s and t are indeterminates over K We shall show that T R K S is a three dimensional domain that is not Cohen Macaulay We have that T Kacs ys 2 8 act yt 215 Q Kx y z s t The equations 253 s3 ys3 0 and 2t3 03 yt3 0 show that 25 and 215 are both integral over D Kacs ys act 215 Q T The elements at y s and t are algebraically independent and the fraction field of D is Kxs ys ts so that dirn D 3 and D g KlXu X127 X217 X22lX11X22 X12X21 with X11 X12 X21 X22 mapping to 8 ysxt yt respectively It is then easy to see that ys act x3 7 yt is a homogeneous system of parameters for D and consequently for T as well The relation ZSXZDWS 7 W ZSVW 7 WWW now shows that T is not Cohen Macaulay for zszt Z act ysT To see this suppose otherwise The map Kl y y 2 8 tl Kl z y 2l that fixes Kx y 2 while sending s gt gt l and t gt gt l restricts to give a K algebra Inap Kacs ys 25 act yt 225 a Kx y lf zszt 6 act ysT applying this map gives 22 E at yKx y 2 which is false 7 in fact Kx y y E D 1 Segre products do have good properties that are important It was already noted that R K S is a direct summand of R K S This implies that every ideal of R KS is contracted from R K S In particular R K S is Noetherian and since it is N graded finitely generated over K This is quite obvious when R and S are standard since it is then generated by the products of elements in a basis for R1 with elements in a basis for S1 When R K S is normal so is R K S In particular this is true of the ring in the example above R KX Y ZlX3 Y3 Z3 is normal since it is Cohen Macaulay and the singular locus is the origin the partial derivatives of X 3 Y3 Z3 vanish simultaneously only at the origin and R K S Rst Discussion the dimension of the Segre product For any finitely generated N graded K algebras R and S with R0 K So we have that dim R K S dim R dim S 7 1 Each of R and S has a homogeneous system of parameters After raising the elments to powers we find that R is module finite over A KF1 FT where F1 FT form a homogeneous system of parameters of degree k and S is module finite over B KG1 GS where G1 Gs is a homogeneous system of parameters of degree k as well If K is infinite and R and S are standard we may even assume that k 1 here Then A K B where X is an 7 gtlt s matrix of indeterminates over K and so has dimension 7 s 7 l The result now follows because R K S is module finite over A K B To see this choose h gt 0 so that homogenous generators for R over A and for S over B have degree 3 h Let Vk be a K basis for Bk for k S h and let Wk be a K basis for Sk for k S h Then the finite set S of elements of the form 1 8 w where u E Vk and W 6 wk for some k S h generate R KS as a module over A K B To see this let F E R and G E S be given Then F is an A linear combination of elements in a fixed Vk with coef cients in A545 and G is a B linear combination of elements in a xed Wk with coef cients in B545 Here if t S h one may take k t and ift 2 h one may take k h It follows that every element of the form F 8 G is in the A K B span of S as claimed and elements of this form span R K S over K B Our next objective is to give a different proof that R has a linear maximal Cohen Macaulay module Again we consider the isomorphism g SKiY17 7Kl KKiZ17 7Z5 gKle 7Y39r7 Z17 7Z5 For every 6 E Z let T5 denote the K span of the monomials u E T such that dngM degzM 57 where dng u denotes the total degree of u in the variables Y1 YT and degZ M denotes the total degree of u in the variables Z1 Zs Then T5 is obviously an S module and T 635 T5 The following result is proved in 8 Goto and K i Watanabe On graded rings l Journal of the Mathematical Society of Japan 30 1978 1797213 3 Theorem With the above notatton T5 is a mammal Cohen Macaulay module of torsz39on free rank one over the Segre product Km m K K121 Zsl s 2 R7 Kle12ltX fors gt 6 gt 7r Proof The case where 6 0 is the statement that To S is Cohen Macaulay which we are assuming here We assume that 6 2 0 and proceed by induction on s The case where 0 2 6 gt 7r then follows by interchanging the roles of Y1 YT and Z1 Zs The case where s 1 is obvious Note that T5 Zng and that Zng is the ideal generated by the monomials of degree 6 in Y1Z1 YTZ1 which is P5 where P Y1Z1 YTZ1 P corresponds to the prime ideal of R generated by the variables in the first column The quotient is KX 12X where X is the r gtlt s 7 1 matrix obtained by omitting the first column of X this ring has dimension r s 7 1 7 1 r s 7 2 from which it follows that P is a height one prime of S To complete the proof it will suf ce to show that for 1 S 6 lt s SP5 is Cohen Macaulay the short exact sequence 07135757571970 then shows that depthmP5 depthmSP5 1 dim S We filter SP5 by the modules PkPk1 0 S k lt 6 Each of these is a module over SP and it suf ces to show that each is a maximal Cohen Macaulay module over SP We already know this when k 0 and so we may assume that 1 S k lt 6 We make use of the fact for every k E N Pk ZfT S This gives an injection of PkPk1 ZfTZf1T 2 TZlT 2 KY1 Y 22 Z5 T If we identify Pl Pl 1 as a submodule of T in this way the action of SP is obtained by identifying SP E KY1 YT KKZ2 Z5 Q T The generators of Pk map to the monomials of degree k in Y1 YT Thus we may identify PkP 1 with T Since 5 has been decreased by 1 and k lt 6 S s 7 1 the result follows from the induction hypothesis D Corollary thh notatton as in the Theorem above T54 is a linear mammal Cohen Maeaulay module over S g R Proof The generators of T54 have the same degree and since T54 is rank one eTs71 6ltT0gt r 2 which is the same as the number of monomials of degree 5 7 1 in Y1 YT D We can now prove 4 Theorem D Hanes Let K be an in nite eld Let R and S be standard graded K algebras that possess ltnear mammal Cohen Macaulay modules in the graded sense Then so does R K S Proof We may assume that M and N are the linear maximal Cohen Macaulay modules over R and S respectively and that they are generated in degree 0 Let Y1 YT be a homogeneous linear system of parameters for R where r dim R and let Z1 Zs be a homogenous linear system of parameters for S where s dim Let m and n be the respective homogeneous maximal ideals in R and S Then R is module finite over A KY1 YT and S is module finite over B KZ1 ZS Moreover R K S is module finite over A K B and R K S is module finite over A KB by the Discussion on the dimension of Segre products Since M is Cohen Macaulay it is A free and its rank 0 eM Similarly N is B free of rank d eN Note that mM Y1 YTM and that nN Z1 Z5N The action of any degree one form F of R on M 2 AC is an A linear map and can be thought of as being given by a c gtlt 0 matrix over A Since multiplication by F increases degrees by one the entries of each such matrix must be degree one forms of A Similarly the action of any degree one form G E S on N is given by a d gtlt d matrix of linear forms over B Consider the R K S module M K N We can consider it as Ac K Bd g Ted where TKiY17 7le KKlZ17 zzsl For 6 E Z we can define M K N5 as T5Cd Because the action of forms of R and forms of S preserves the bigrading on Ted coming from the Y grading and the Z grading on T every M K N5 is a module over R K S Since M K N5 is finitely generated even as a module over T0 AeKB g R KS it is finitely generated over R S For s gt 6 gt 7r it is maximal Cohen Macaulay over R K S since R KS is module finite over A KB and we know that it is maximal Cohen Macaulay over A K B To complete the proof we shall show that W M K N51 is a linear maximal Cohen Macaulay module over R K S It is maximal Cohen Macaulay and generated by elements of equal degree To complete the argument we shall prove that i 2 1W ed 7quot s 4W r 7 1 Include Y1 YT in a set of one forms F1 Fh that generate m and include Z1 Zs in a set of one forms G1 Gk that generate 11 Let M be the maximal Math 711 Lecture of September 19 2007 Earlier see the Lecture of September 7 p 7 we discussed very brie y the class of excellent Noetherian rings The condition that a ring be excellent or at least locally ex cellent is the right hypothesis for many theorems on tight closure The theory of excellent rings is substantial enough to occupy an entire course and we do not want to spend an inordinate amount of time on it here We shall summarize what we need to know about excellent rings in this lecture In the sequel the reader who prefers may restrict attention to rings essentially of finite type over a field or over a complete local ring which is the most important family of rings for applications The definition of an excellent Noetherian ring was given by Grothendieck A readable treatment of the subject which is a reference for all of the facts about excellent rings stated without proof in this lecture is Matsumura Commutative Algebra WA Benjamin New York 1970 Chapter 13 Before discussing excellence we want to review the notion of fibers of ring homomor phisms Fibers Let f R a S be a ring homomorphism and let P be a prime ideal of R We write Hp for the canonically isomorphic R algrebras frac RP RpPRp By the ber of f over P we mean the Hp algebra Hp R s 2 R i P 1SPS which is also an R algebra since we have R a Hp and an S algebra One of the key points about this terminology is that the map Spec Hp R S a Spec S gives a bijection between the prime ideals of Hp R S and the prime ideals of S that lie over P Q R In fact it is straightforward to check that Spec Hp R S is homeomorphic with its image in Spec It is also said that Spec Hp R S is the scheme theoretic fiber of the map Spec S a Spec This is entirely consistent with thinking of the fiber of a map of sets 9 Y a X over a point P E X as 941 Q E Y 9Q Pl 1 2 In our case we may take g Spec f Y Spec S and X Spec R and then Spec Hp R S may be naturally identi ed with the set theoretic fiber of Spec S a Spec If R is a domain the fiber over the prime ideal 0 of R namely frac R R S is called the generic ber of R a S If R m K is quasilocal the fiber K R S SmS over the unique closed point in of Spec R is called the closed ber of R a S Geometric regularity Let H be a field A Noetherian H algebra R is called geometrically regular over H if the following two equivalent conditions hold 1 For every finite algebraic field extension 4 of H H E R is regular 2 For ever finite urel inse arable field extension 4 of H H E R is re ular y p y p 7 g Of course since we may take 4 H if R is geometrically regular over H then it is regular ln equal characteristic 0 geometric regularity is equivalent to regularity using characterization When R is essentially of finite type over H these conditions are also equivalent to 3 K E R is regular for every field K 4 K E R is regular for one perfect field extension K of H 5 K E R is regular when K H is the algebraic closure of H These conditions are not equivalent to l and 2 in general because K E R need not be Noetherian We indicate how the equivalences are proved This will require a very considerable effort Theorem Let R a S be a faithfully flat homomorphism of Noetheriau rings If S is regular then R is regular Proof We use the fact that a local ring A is regular if and only its residue class field has finite projective dimension over A in which case every finitely generated module has finite projective dimension over A Given a prime P of B there is a prime Q of S lying over it It suf ces to show that Rp is regular and we have a faithfully at map RP a SQ Therefore we may assume that R P K a S Q L is a at local homomorphism and 3 that S is regular Consider a minimal free resolution of RP over R which a priori may be infinite Rb amp Rbn l a amp Rbo RP o By the minimality of the resolution the matrices 04339 all have entries in P Now apply S R 7 We obtain a free resolution sbn sbnel ampst S RSPs 0 where we have identi ed R with its image in S under the injection R gt S This resolution of SPS is minimal the matrices have entries in Q because R gt S is local Since S is regular SPS has finite projective dimension over S and so the matrices 04339 must be 0 for allj gtgt 0 But this implies that the projective dimension of RP over R is finite D Corollary If R is a Noetherian K algebra and L is an ezrtension eld of K such that L K R is regular in general this ring may not be Noetherian although it is if R is essentially of nite type ouer K because in that case L K R is essentially of nite type ouer L and therefore Noetherian then R is regular Proof Since L is free over K it is faithfully at over K and so L K R is faithfully at over R and we may apply the preceding result D Proposition Let R m K a S Q L be a flat local homomorphism of local rings Then a dim S dim R dim SmS the sum of the dimensions of the base and of the closed ber b IfR is regular and SmS is regular then S is regular Proof a We use induction on dim lf dim R 0 m and mS are nilpotent Then dim S dim SmS dim R dim SmS as required If dim R gt 0 let J be the ideal of nilpotent elements in R Then dim RJ dim R dim SJS dim S and the closed fiber of RJ a SJS which is still a at and local homomorphism is SmS Therefore we may consider the map RJ a SJS instead and so we may assume that R is reduced Since dim R gt 0 there is an element f E m not in any minimal prime of R and since R is reduced f is not in any associated prime of R ie f is a nonzerodiVisor in R Then the fact that S is at over R implies that f is not a zerodiVisor in S We may apply the induction hypothesis to RfR a SfS and so dim S 7 l dim SfS dim Rf dim SmS dim R i l dim SmS and the result follows b The least number of generators of Q is at most the sum of the number of generators of m and the number of generators of QmS ie it is bounded by dim Rdim SmS dim S by part a The other inequality always holds and so S is regular D Corollary Let R a S be a at homomorphism of Noetherian rings IfR is regular and the bers ofR a S are regular then S is regular Proof If Q is any prime of S we may apply part b of the preceding Theorem since SQPSQ is a localization of the fiber Hp R S and therefore regular Corollary Let R be a regular Noetherian K algebra where K is a eld and let L be a separable EItBTLSZOTl eld ofK such that L K R is Noetherian Then L K R is regular Proof The extension is at and so it suf ces to show that every Hp R L KR g Hp KL is regular Since L is algebraic over K this ring is integral over Hp and so zero dimensional Since L KR is Noetherian by hypothesis Hp KL is Noetherian and so has finitely many minimal primes Hence it is Artinian and if it is reduced it is a product of elds and therefore regular as required Thus it suf ces to show that Hp K L is reduced Since L is a direct limit of finite separable algebraic extension it suf ces to prove the result when L is a finite separable extension of K In this case L has a primitive element 6 and L g where g E is a monic irreducible separable polynomial over K Q Hp Let 9 denote the algebraic closure of Hp Then Hp K L Q Q K L and so it suf ces to show that 9 K L g 9 K Kielamid mid99M is reduced This follows because g is separable and so has distinct roots in Q E Theorem Let K be an algebraically closed eld and let L be any nitely generated eld e1tension of K Then L has a separating transcendence basis 8 ie a transcendence basis 8 such that L is separable over the pure transcendental ezrtension Proof If F is a subfield of L let F5ep denote the separable closure of F in L Choose a transcendence basis 1 icn so as to minimize L L where L Kx1 icnsep Suppose that y E L is not separable over Kx1 xn Choose a minimal polynomial for y over Kac1 xn Then every exponent on 2 is divisible by p Put each coef cient in lowest terms and multiply by a least common multiple of the denom inators of the coef cients This yields a polynomial Hac1 xn z E Kac1 such that the coef cients in Kac1 n are relatively prime and such that the poly nomial is irreducible over Kx1 By Gauss7s Lemma this polynomial is ir reducible in Kac1 icn It cannot be the case that every exponent on every obj is divisible by p for if that were true since the field is perfect H would be a pth power and not irreducible By renumbering the 1 we may assume that 7 occurs with an exponent not divisible by p Then the element 7 is separable algebraic over the field Kx1 icn1y and we may use the transcendence basis 1 icn1 y for L Note that acmy E Kx1 xn1ysep L which is therefore strictly larger than L Kac1 icnsep Hence L L lt L L a contradiction D We can now prove Theorem Let R be a Noetherian H algebra where H is a eld Then the following two conditions are equivalent 1 For every nite algebraic eld ezrtension H of H H E R is regular 2 For every nite purely inseparable eld EItBTLSZOTL H of H H E R is regular Moreover ifR is essentially of nite type over H then the following three conditions are equivalent to 1 and as well 3 K E R is regular for every eld K 4 K E R is regular for one perfect eld ezrtension K of H 5 K E R is regular when K H is the algebraic closure of H Proof We shall repeatedly use that if we have regularity for a larger field extension then we also have it for a smaller one this follows from the Corollary on p 3 Evidently 1 i But 2 i 1 as well because given any finite algebraic extension 4 of H there is a larger finite field extension obtained by first making a finite purely insep arable extension and then a finite separable extension The purely inseparable extension yields a regular ring by hypothesis and the separable field extension yields a regular ring by the second Corollary on p 4 Now consider the case where R is essentially of finite type over H Evidently 3 i 5 i 4 i 2 the last holds because any perfect field extension contains the perfect closure and this contains every finite purely inseparable algebraic extension and it will suf ce to prove that 2 i Let H denote the perfect closure U Hlq of H We first show that H E R is regular q Replace R by Rm Then B H ER Rm is purely inseparable over Rm consequently it is a local ring of the same dimension as Rm and it is the directed union of the local rings 4 E Rm as H runs through finite purely inseparable extensions of H contained in H All of these local rings have the same dimension call it d Let ul un be a minimal set of generators of the maximal ideal of B H E Rm and choose 4 suf ciently large that ul un are elements of A H R Rm Let J ul unA Since B is faithfully at over A we have that JB A J But JB is the maximal ideal of B which lies over the maximal ideal of A and so J generates the maximal ideal of A None of the generators is an A linear combination of the others or else this would also be true in B Hence ul un is a minimal set of generators of the maximal ideal of A Since A is regular n d and so B is regular Since the algebraic closure of H is separable over H it follows from the second Corollary on p 4 that 2 i To complete the proof it suf ces to show that if H is algebraically closed R is regular and L is any field extension of H then L E R is regular Since R a L E R is at it suf ces to show the fibers L E Hp are regular and Hp is finitely generated as a field over H Hence Hp has a separating transcendence basis 1 xn 6 over H Let K Mach aon Then L H HP H H17 z K HP Since Hp is a nite separable algebraic extension of K it su ices prove that L E K is regular But this ring is a localization of Lac1 ow and so the proof is complete D We say that a homomorphism R a S of Noetherian rings is geometrically regular if it is at and all the bers Hp a Hp R S are geometrically regular Some authors use the term regular for this property For those readers familiar with smooth homomorphisms we mention that if S is essen tially of finite type over R then S is geometrically regular if and only if it is smooth By a very deep result of Popescu cf D Popescu General Ne ron deslngularz39zatz39on Nagoya Math J 100 1985 977126 every geometrically regular map is a direct limit of smooth maps Whether Popescu7s argument was correct was controversial for a while Richard Swan showed that Popescu7s argument was essentially correct in R G Swan Ne ron Popescu deslngularlzatz39on Algebra and geometry Taipei 1995 1357192 Lect Algebra Geom 2 Int Press Cambridge MA 1998 Catenary and universally catenary rings A Noetherian ring is called catenary if for any two prime ideals P Q Q any two saturated chains of primes joining P to Q have the same length In this case the common length will be the same as the dimension of the local domain RQPRQ Nagata was the first to give examples of Notherian rings that are not catenary Eg in Nagata Local Rings lnterscience New York 1962 Appendix pp 20475 Nagata gives an example of a local domain D m of dimension 3 containing a height one prime P such that dim DP 1 so that 0 C Q C m is a saturated chain while the longest saturated chains joining 0 to m have the form 0 C P1 C P2 C m One has to work hard to construct Noetherian rings that are not catenary Nagata also gives an example of a ring R that is catenary but such that RM is not catenary Notice that a localization or homomorphic image of a catenary ring is automatically catenary R is called universally catenary if every polynomial ring over R is catenary This implies that every ring essentially of finite type over R is catenary A very important fact about Cohen Macaulay rings is that they are catenary Moreover a polynomial ring over a Cohen Macaulay ring is again a Cohen Macaulay ring which then implies that every Cohen Macaulay ring is universally catenary In particular regular rings are universally catenary Cohen Macaulay local rings have a stronger property they are equidimensional and all saturated chains from a minimal prime to the maximal ideal have length equal to the dimension of the local ring We shall prove the statements in the paragraph above We first note Theorem IfR is Cohen Macaulay so is the polynomial ring in n variables over R Proof By induction we may assume that n 1 Let M be a maximal ideal of RX lying over m in R We may replace R by Rm and so we may assume that R m K is local Then M which is a maximal ideal of RM lying over m corresponds to a maximal ideal of each of these is generated by a monic irreducible polynomial f which lifts to a monic polynomial F in Thus we may assume that M mRac Let 1 iod be a system of parameters in R which is also a regular sequence We may kill the ideal generated by these elements which also form a regular sequence in RX M We are now in the case where R is an Artin local ring It is clear that the height of M is one Because F is monic it is not a zerodivisor a monic polynomial over any ring is not a zerodivisor This shows that the depth of M is one as needed Theorem Let R m K be a local ring and M 31 0 a nitely generated Cohen Macaulay R module of Krull dimension d Then every nonzero submodule N ofM has Krull dimen sion d Proof We replace R by RAnnRM Then every system of parameters for R is a regular sequence on M We use induction on d If d 0 there is nothing to prove Assume d gt 0 and that the result holds for smaller d If M has a submodule N 31 0 of dimension 3 d 7 l we may choose N maximal with respect to this property If N is any nonzero submodule of M of dimension lt d then N Q N To see this note that N 69 N has dimension lt d and maps onto N N Q M which therefore also has dimension lt d By the maximality of N we must have N N N Since M is Cohen Macaulay and d 2 l we can choose it E m not a zerodivisor on M and hence also not a zerodivisor on N We claim that x is not a zerodivisor on M MN for if u E M 7 N and you 6 N then Rica Q N has dimension lt d But this module is isomorphic with Ru Q M since it is not a zerodivisor and so dim Ru lt d But then Ru Q N Consequently multiplication by it induces an isomorphism of the exact sequence 0 a N a M a M a 0 with the sequence 0 a N a xM a xM a 0 and so this sequence is also exact But we have a commutative diagram O gtN gtM gtM gt0 0NgtMgtMgt0 where the vertical arrows are inclusions By the nine lemma oriy an elementary diagram chase the sequence of cokernels 0 a NacN a MacM a MacM a 0 is exact Because it is not a zerodiVisor on M it is part of a system of parameters for R and can be extended to a system of parameters of length d which is a regular sequence on M Since it is a nonzerodiVisor on N and M dim NacN dim N 7 l lt d 7 1 while MacM is Cohen Macaulay of dimension d 7 1 This contradicts the induction hypothesis D Corollary If R m K is Cohen Macaulay R is equidimensional euery minimal prime p is such that dim Rp dim Proof If p is minimal it is an associated prime of R and we have Rp 7 R Since all nonzero submodules of B have dimension dim R the result follows E Thus a Cohen Macaulay local ring cannot exhibit the kind of behavior one observes in R KHz y y this ring has two minimal primes One of them p1 generated by the images of z and y is such that Rp1 has dimension 1 The other p2 generated by the image of z is such that Rp2 has dimension 2 Note that while B is not equidimensional it is still catenary We next observe Theorem In a Cohen Macaulay ring R if P Q Q are prime ideals of R then every saturated chain of prime ideals from P to Q has length height Q 7 height Thus R is catenary It follows that every ring essentially of nite type over a Cohen Macaulay ring is uni uersally catenary Proof The issues are unaffected by localizing at Q Thus we may assume that R is local and that Q is the maximal ideal There is part of a system of parameters of length h height P contained in P call it 1 xh by the Corollary near the bottom of p 7 of the Lecture Notes of Septermber 5 This sequence is a regular sequence on R and so on Rp which implies that its image in RP is system of parameters We now replace R by Rzl 5 when we kill part of a system of parameters in a Cohen Macaulay ring the image of the rest of that system of parameters is both a system of parameters and a regular sequence in the quotient Thus R remains Cohen Macaulay Q and P are replaced by their images which have heights dim R 7 h and 0 and dim R 7 h dim Pt1 xh We have therefore reduced to the case where R Q is local and P is a minimal prime We know that dim R dim RP and so at least one saturated chain from P to Q has length height Q 7 height P height Q 7 0 dim To complete the proof it will suf ce to show that all saturated chains from P to Q have the same length and we may use induction on dim Consider two such chains and let their smallest elements other than P be P1 and Pi We claim that both of these are height one primes if say P1 is not height one we can localize at it and obtain a Cohen Macaulay local ring S m of dimension at least two and a saturated chain p Q m with p PS minimal in S Choose an element y E m that is not in any minimal primes of S its image will be a system of parameters for Sp so that By p is m primary Extend y to a regular sequence of length two in S the second element has a power of the form ry u so that y ry u is a regular sequence and hence so is y it But then u y is a regular sequence a contradiction since u E p Thus P1 and similarly Pf have height one Choose an element f in P1 not in any minimal prime of R and an element g of Pf not in any minimal prime of R Then fg is a nonzerodivisor in R and P1 Pf are both minimal primes of my The ring Racy is Cohen Macaulay of dimension dim R 7 l The result now follows from the induction hypothesis applied to Rxy the images of the two saturated chains omitting P from each give saturated chains joining Placy respectively to Qacy in Racy These have the same length and hence so did the original two chains The final statement now follows because a polynomial ring over a Cohen Macaulay ring is again Cohen Macaulay D l Excellent rings A Noetherian ring R is called a G ring G77 as in Grothendieck77 if for every local ring A of R the map A a A is geometrically regular An e1cellent ring is a universally catenary Noetherian G ring R such that in every finitely generated R algebra S the regular locus P E Spec S Sp is regular is Zariski open Excellent rings include the integers fields and complete local rings as well as convergent power series rings over C and R Every discrete valuation ring of equal characteristic 0 or of mixed characteristic is excellent The following two results contain most of what we need to know about excellent rings Theorem Let R be an e1cellent ring Then every localization of R every homomor phic image of R and every nitely generated R algebra is excellent Hence every algebra essentially of nite type over R is excellent Theorem Let R be an e1cellent ring a IfR is reduced the normalization off is module nite over R b IfR is local and reduced then 13 is reduced c IfR is local and equidimensional then E is equidimensional d IfR is local and normal then 13 is normal For proofs of these results we refer the reader to Matsumura Commutative Algebra WA Benjamin New York 1970 as mentioned earlier Math 711 Lecture of September 14 2007 The following result is very useful in thinking about tight closure Proposition Let R be a Noetherian ring of prime characteristic 1 gt 0 let N Q M be R modules and let u E M Then a E NITI if and only if the image E ofu in the quotient MN is in O MN Hence if we map a free module G onto M say h G a M let H h 1N Q G and let i E G be such that hv u then u E NITI if and only ifv E Proof For the first part let c E R0 Note that by the right exactness of tensor products feMN g feMNql Consequently cuq E qul for all 1 2 go if and only if c q 0 in POWN for q 2 go For the second part simply note that the image of v in GH E MN corresponds to E in MN D It follows many questions about tight closure can be formulated in terms of the behavior of tight closures of submodules of free modules Of course when M is finitely generated the free module G can be taken to be finitely generated with the same number of generators Given a free module G of rank n we can choose an ordered free basis for G This is equivalent to choosng an isomorphism G E R R 69 69 R In the case of B one may understand the action of Frobenius in a very down to earth way We may identify feR g R since we have this identification when n 1 Keep in mind however that the identification of feG with G depends on the choice of an ordered free basis for G If u r1 B Brn E R then M 7 r With H E R qul is the R span of the elements M for u E H or for a running through generators of Very similar remarks apply to the case of an infinitely generated free module G with a specified basis b The elements b give a free basis for feG and if u rlbxl rsbxs then M r fbil rgbis gives the representation of M as a linear combination of elements of the free basis bi We could have defined tight closure for submodules of free modules using this very concrete description of M and qul The similarity to the case of ideals in the ring is visibly very great But we are then saddled with the problem of proving that the notion is independent of the choice of free basis Moreover if we take this approach we need to define NITI by mapping a free module G onto M and replacing N by its inverse image in G We then have the problem of proving that the notion we get is independent of the choices we make Our next objective is to prove that in a regular ring every ideal is tightly closed This depends on knowing that F R a R is at for regular rings of prime characteristic 1 gt 0 1 Eventually we sketch below a proof of the atness of F that depends on the structure theory for complete local rings of prime characteristic 1 gt 0 Later we shall give a different proof based on the following result which is valid without restriction on the characteristic Theorem Let R m K be a regular local ring and M and R module Then M is a big Cohen Macaulay module for R if and only z39fM 239s faithfully flat ouer R We postpone the proof of this result for a while it makes considerable use of the properties of the functor Tor However we do want to make several comments First note that it immediately implies that when R is regular F R a R is at In general R a S is at if and only if for every prime ideal Q of S with contraction P to R the map Rp a SQ is at l To see this note that for Rp modules M the natural map SQ R M a SQ RP M is an isomorphism because M a Rp RP M is an isomorphism and we have SQ Rp M g SQ Rp RP R M g SQ R M The latter is also SQ 83 S R If S is at over R since Sp is at over S we have that Sp is at over B On the other hand if N gt M is an injection of R modules and S RN a S R M is not injective we can localize at a prime Q of S in the support of the kernel This yields a map SQ R N a SQ R M that is not injective But if Q contracts to P we do have that Np a Mp is injective This shows that SQ is not at over Rp D Note that when S R and the map is F the contraction of P E Spec R is P Thus it suf ces to show that F is at on Rp for all primes P This is now obvious given the Theorem above any regular sequence equivalently system of parameters in R say 1 acn maps to 11 5 in R which is again a regular sequence Hence R is a big Cohen Macaulay algebra for R under the map F R a R and this proves that R is faithfully at over R We have the following additional comments on the Theorem Suppose that M is a module over a local ring R m K and suppose that we know that 1 xn is a system of parameters that is a regular sequence on M Let Qt 1 WR By the definition of a regular sequence we have that QlM 31 M We want to point out that this condition implies the a prz om stronger condition that mM 31 M The reason is that m is nilpotent modulo 2L Thus we can choose 5 such that m5 Q 2L If M mM we can multiply by mt to conclude that th mtmM mt1M Thus MmMm2Mth Then Mm5M Q QLM Q M and we find that QlM M a contradiction If M is faithfully at over R we have that Rm M MmM 31 0 so that mM 31 M Moreover whenever 1 xn is a system of parameters for B it is a regular sequence on R and the fact that M is faithfully at over R implies that 1 xn is a regular sequence on M This shows that a faithfully at R module is a big Cohen Macaulay module over R The converse remains to be proved We next sketch a completely different proof that F is at for a regular ring B As noted above this comes down to the local case We use the fact that a local map R a S of local rings is at if and only if the induced map If a S is at Hence by the structure theory of complete local rings we may assume that R KHacl xnjj is a formal power series ring over a field Since this ring is the completion of Kx1 acnjp where P 1 acn it suf ces to prove the result for the localized polynomial ring R Kjacl xn itself But FR K1115 x Thus allwe need to show is that Kpjxl xn Q K1xn is at We prove a stronger result R is free over FR in this case Since K is free over KP Kxff x is free on the same basis over Kpjxff x Thus we need only see that Kjacl xn is free over Kjacff acflj It is easy to check that the monomials x1 acar such that 0 S 1 S p 7 l are free basis D n We now fill in the missing details of the argument sketched above Proposition Let 6 R m K a S nL be a homomorphism of local rings that is local z39e 6m Q 11 Let Q be o nltely generated S module Then Q is flat over R if and only z39ffor every lnjectlve mop N gt M of nz39te length R modules Q R N a Q R M is z39njectz39ve Proof The condition is obviously necessary We shall show that it is suf cient Since tensor commutes with direct limits and every injection N gt M is a direct limit of injections of finitely generated R modules it suf ces to consider the case where N Q M are finitely generated Suppose that some n E S R N is such that u gt gt 0 in S R M It will suf ce to show that there is also such an example in which M and N have finite length Fix any integer t gt 0 Then we have an injection Nth m N Mth and there is a commutative diagram Q R N gt Q 813 M fl gl Q R Nth m N 4 Q R Mth 4 The image u of u in Q R Nth N maps to 0 under H by the commutativity of the diagram Therefore we have the required example provided that u 31 0 However for all h gt 0 we have from the Artin Rees Lemma that for every suf ciently large integer t th N Q th Hence the proof will be complete provided that we can show that the image of u is nonzero in Q RNth g Q RRmh RN g Rmh RQ RN g Q RNmhQ RN for h gt 0 But mhltQ R N Q nhltQ R N and the result follows from the fact that the finitely generated S module Q R N is n adically separated D We can now prove the following result which is the only missing ingredient needed to fill in the details of our proof that F is at Lemma Let R m K a S nL be a local homomorphism of local rings Then S is flat over R if and only z39fS is flat over R and this holds l S is flat over R Proof If S is at over R then since S is at over S we have that S is at over R Conversely if S is at over R then S is at over B because S is faithfully at over S if N gt M is injective but S R N a S R M has a nonzero kernel the kernel remains nonzero when we apply S 85 7 and this has the same effect as applying S R i to N gt M a contradiction We have shown that R a S is at if and only R a S is at If If a S is at then since R a Is is at we have that R a S is at and we are done It remains only to show that if R a S is at then E a S is at By the Proposition it suf ces to show that if N Q M have finite length then S N a S M is injective Suppose that both modules are killed by mt Since SmtS is at over Rmt if Q is either M or N we have that gQ Emt gmgo 2 Em olem Q 2 EM Q and the result now follow because S is at over R D l The following result on behavior of the colon operation on ideals under at base change while quite easy and elementary plays a very important role in tight closure theory Recall that when I Q R and R a S is a at homomorphism the map I R S a R ER S S is injective lts image is clearly IS the expansion of I to S Thus I R S may be naturally identified with IS when S is at over R Recall that if I and J are ideals of R then IzRJr RrJ I which is an ideal of R If J fl is principal we may write I R f for IR fR Math 711 Lecture of September 26 2005 Note that the straightening relations for a Hodge algebra are homogeneous if the terms on the right do not all have the same degree as v fix a degree different from that of v for which there are nonzero terms of that degree Then the sum of all the terms of that degree on the right must be zero a contradiction for the standard monomials must be linearly indepenendent over K lf c 6 Nb we write ch instead of ch so that the monomial he corresponding to c is HhEH hch No matter what the base ring K is there is a bijection between the semigroup ideals E Q NH and the nonzero monomial ideals of KH the polynomial ring with the elements of H as the variables This bijection takes 2 to the ideal which is the K span of the monomials he for c E 2 Conversely if I is any nonzero ideal generated by monomials in KH I is the K span of the set of monomials contained in I and the set of exponents of the monomials in I is a semigroup ideal in NH A minimal set of monomial generators for such an ideal I is unique and the exponents correspond to the generators of E in the sense defined earlier Since all of this holds when K is Noetherian it follows that the set of generators of E is finite Of course it is also possible to prove this without using any ring theory If H is a poset and h E H we define dim as the supremum of lengths t of strictly ascending chains h ho lt hl lt lt ht with h as the first element For example the dimension of a prime ideal P in the poset of prime ideals of a ring R is the same as the Krull dimension of RP Note that in a finite poset H if hl lt hg then dim hl gt dim hg since a longest chain starting with hg will yield a chain of length one more if we insert hl at the beginning We next note the following Theorem Let R be a Hodge algebra on H over K governed by 2 Then the standard monomials are a free K basis for R and R is the quotient of a polynomial ring in variables corresponding to the elements ofH by the ideal generated by the straightening relations More precisely if we introduce an indeterminate Xh over K for every element ofH then R is the quotient of KXh h E H by the ideal J generated by the straightening relations each translated into a relation on the indeterminates Xh for each ezrponent c that is a generator of Z we include one generator H Xi 7 Z M H Xi hEH uEM heH in J where c 6 NH 7 E is the ezrponent corresponding to u Proof Let d be the supremum of the sums 2 ch for c that is a generator of E and let hEH the weight of an element h E H be defined as d ldlmh Extend the weight function 1 2 to the exponent c by letting the weight of c be 2 czweightUL hEH Then for any two exponents 01 02 the weight of their sum is the sum of their weights Assume the result is false and that some monomial of degree N cannot be expressed as a linear combination of standard monomials as a consequence of the straightening relations Choose such a monomial whose exponent is of greatest weight This monomial 1 can be written as My where 1 corresponds to an exponent c that is a minimal generator of E It su ices to show that 1 is a linear combination of monomials of degree N corresponding to exponents of greater weight It will therefore be enough to prove this for 1 since multiplying by w will preserve the necessary qualities of the relation the weight of the exponent for each monomial is increased by the weight of the exponent chosen for w Consider the straightening relation UZuu uEM where each of the standard monomials u that occurs has the same degree as 1 It will su ice to show that the weight of the exponent 00 of each u that occurs is strictly larger than the weight of 1 Recall that v he Choose ho in the support of c of largest possible dimension call this dimension b Then the weight of c is Z Chltd1gtdimm hEH and whenever ch 31 0 we have that dim S b which shows that the sum is bounded by ltgt Z chltd1gtbltz chgtltd1gtbs dltd1gtblt ltd1gtb1 hEH hEH Now consider the weight of u ew occurring in the straightening relation Then 00 is supported at an element hl E H such that hl lt ho Then the weight of 00 is at least the weight of h which is d ldlmh1 Since dim h gt dim ho b this is 2 d Db which we showed in above is strictly larger than the weight of c as required D We shall say that two Hodge algebbras have the same data if their posets H and H are isomorphic say b H E H is the order isomorphism in such a way that the induced semigroup isomorphism NH 2 NH carries the semigroup ideal E for the first Hodge algebra isomorphically onto the semigroup ideal 2 for the second Hodge alagebra The simplest case is when H H and E 2 We define thez39ndz39screte part lndR of the Hodge algebra R over K on H governed by E to be the subset of elements h E H such that h is in the support of a standard monomial Math 711 Lecture of October 17 2007 We next want to prove the Theorem stated at the end of the Lecture Notes from October 12 Recall that A KHacl xnll and that P is cofinite in a xed p base A for K First note that it is clear that KHacl xnll a KeHacl xnll is faithfully at every system of parameters in the former maps to a system of parameters in the extension ring and since the extension is regular it is Cohen Macaulay Faithful atness follows from the Theorem at the top of p 2 of the Lecture Notes of September 14 Since a direct limit of at extensions is at it is clear that AF is at over A Since Keq Q K we have that Arq E Keqac mg 6 Kac1ic A Thus every A5 KeHacl xnll is purely inseparable over A and it follows that the union AF is as well Hence A a AP is local Note that the maximal ideal in each A5 is mAe x1 icnAe Every element of the maximal ideal of AF is in the maximal ideal of some A5 and so in mAe Q mAF Thus mAF x1 icnAF is the maximal ideal of AP The residue class field of AF is clearly the direct limit of the residue class fields K5 which is the union Us K5 KP this is the gamma construction applied to A K We next want to check that AF is Noetherian Note that AF is contained in the regular ring Kr acl xnll B and that each of the maps A8 a B is faithfully at Hence for every ideal I of A5 IB A5 I The Noetherian property for AF now follows from Lemma Let be a directed family of rings and injective homomorphisms whose direct limit A embeds in a ring B Suppose that for all i and for every ideal J of any Ai JB AZ I Then for every ideal I of A IB A I Hence ifB is Noetherian then A is Noetherian Proof Suppose that u E A I Q A and u 6 IB 7 IA Then a flbl fnb where f1 fn E I and b1 bn E B We can choose i so large that u f1 fn 6 AZ and let J f1 fnAZ Evidently u E JB AZ J and clearly J Q IA a contradiction For the final statement let I be any ideal of A Then a finite subset gl 9 E I generates IB Let I0 gl gnA Then I Q IB H A IOB A I0 Q I so that I IO U Since AF is Noetherian of Krull dimension n with maximal ideal 1 icnAF we have that AF is regular To complete the proof of the final Theorem stated in the Lecture Notes from October 12 it remains only to prove 1 2 Theorem AF is F mte Proof Throughout this argument we write K5 for K K1q A E P and A5 for KeHacl Let 61 6 be the nitely many elements that are in the p base A but not in P Let M be the set of monomials in Qip llp of degree at most 1 7 l in each element and let N be the set of monomials in xvip xip in each element Let of degree at most 1 7 l TMNMVZMEMV N We shall complete the proof by showing that T spans AF1p as an AF module First note that AF1p UltAe1p7 e and for every e AM KiPHxiP ailPH This is spanned over KelpHacl acdll by N Also observe that K5 is spanned over K by products of monomials in N and monomials in the elements Alqp for A E P and the latter are in K5 Hence Kelp is spanned byN over Ke1 and it follows that Kegpm aan is spanned by N over Ke1ac1 xnll A5 Hence Aip is spanned by T MN over A5 as claimed Note that AF Q KFHQCl xnll but these are not in general the same Any single power series in AP has all coe icients in a single K5 When the chain of fields K5 is infinite we can choose Ce 6 Ke1 7 K5 for every for every e 2 0 and then 2cm 6 KFHxH 7 Kim 50 Complete tensor products and an alternative View of the gamma construction Let R m K be a complete local ring with coe icient field K Q B When R A KHacl aonll we may enlarge the residue class field K of A to L by considering instead LHacl This construction can be done in a more functorial way and one does not need the ring to be regular Consider first the ring BL L K B This ring need not be Noetherian and will not be complete except in special cases eg if L is finite algebraic over K However RLmRL g L so that mRL is a maximal ideal of this ring and we may form the mRL adic completion of RL This ring is denoted L KR and is called the complete tensor product of L with R over K Of course we have a map R 7 RL 7 L KR Note that L R R A K N L KRl HtW lint L K In case B where g 1 xn are formal power series indeterminates this yields Kli ll N Klil N Ll l N EltL K g 1lt HtL K t1lt nt Lllgll which gives the result we wanted Now suppose that we have a local map R m K a S n K of complete local rings such that S is module finite over R ie over the image of R we are not assuming that the map is injective For every t we have a map RmtR a SmtS and hence a map L K RmtR a L K SmtS This yields a map limt L K RmtR a 11m L K SmtS The map RmS a SmS is module finite which shows that SmS has Krull dimension 0 It follows that mR is primary to n so that the ideals mtR are cofinal with the power of 11 Therefore the inverse limit on the right in is the same as limt L K RntR and we see that we have a map L KR a L KS We next note that when R a S is surjective so is the map L KR a L KS First note that BL a SL is surjective and that mRL maps onto nSL Second each element 039 of the completion of SL with respect to n can be thought of as arising from the classes modulo successive powers of n of the partial sums of a series 30313t such that 3 E ntSL mtSL for all t E N Since mtRL maps onto ntSL we can left this series to 70r1rt where for every t E N 7quot E mtRL and maps to 35 The lifted series represents an element of the completion of BL that maps to 039 Since every complete local ring R with coef cient field K is a homomorphic image of a ring of the form KHacl xn it follows that L KR is a homomorphic image of a ring of the form LHacl arnll and so L KR is a complete local ring with coef cient field L Next note that when R a S is a module finite not necessarily inj ective K homomorphism of local rings with coef cient field K we have a map L KR R s a L Ks since both factors in the ordinary tensor product on the left map to L KS We claim that this map is an isomorphism Since as noted above m5 is primary to n and both 4 sides are complete in the m adic topology it suf ces to show that the map induces an isomorphism modulo the expansions of mt for every t E N But the left hand side becomes L K R S g L K which is exactly what we need It follows that L KR is f39faithfully at over R we can represent R as a module finite extension of a complete regular local ring A with the same residue class field and then L KR L KA A B so that the result follows from the fact that L A is faithfully at over A With this machinery available we can construct RF when R is complete local with coef cient field K and P is cofinite in a p base A for K as Us K5 KR If R is regular this agrees with our previous construction If A R are complete local both with coef cient field K and A a R is a local K algebra homomorphism that is module finite not necessarily injective then we have Kelt gtKR Kelt gtKA A R for all e Since tensor commutes with direct limit it follows that RF 2 AP A R In particular this holds when A is regular It follows that RF is faithfully at over R Properties preserved for small choices of P Suppose that A is a p basse for a field K of characteristic p gt 0 We shall say that a property holds for all su iciently small co nite P Q A or for all P lt A if there exists P0 Q A cofinite in A such that the property holds for all P Q To that are cofinite in A We are aiming to prove the following Theorem Let B be a complete local ring of prime characteristic p gt 0 with cooe icient eld K let A be a p base for K and and let R an algebra essentially of nite type over B For T co nite in A let RF denote BF 83 R a IfR is a domain then RF is a domain for all P lt A b IfR is reduced then RF is reduced for all P lt A c IfP Q R is prime then PET is prime for all P lt A d If Q R is radical then RF is radical for all P lt A We shall also prove similar results about the behavior of the singular locus We first note 5 Lemma Let M be an R module let P1 Ph be submodules of M and let S be a flat R module Then the intersection of the submodules S R Pi for 1 S i S h is P1 Ph RM Here for P Q M we are identifying S R P with its image in S R M of course the map S R P a S R M is injective Proof By a straightforward indcution on h this comes down to the intersection of two submodules P and Q of the R module M We have an exact sequence O P Q Mi MPeBMQ where the rightmost map f sends it E M to u P 69 u Q Since S is R at applying S R Q yields an exact sequence 0 e s m P m Q a s m s m MP e s m MQ The rightmost term may be identified with 5 R MWS R P 69 5 R MS R Q from which it follows that the kernel of 15 8 f is the intersection of S R P and S R Q Consequently this intersection is given by S R P Q D We next want to show that part a of the Theorem stated above implies the other parts Proof that part 1 implies the other parts of the Theorem Part c follows from part a applied to PtP since iiPF BF 2 RP 2 RFPRF To prove that a i d let I P1 Pn be the primary decomposition of the radical ideal I where the P are prime Since BF is at over B RF is at over R Hence IRF which may be identified with RF R I is the intersection of the ideals RF R Pi 1 S i S h by the Lemma above By part a we can choose P cofinite in A such that every RF R Pi is prime and for this P RF is radical Finally c is part d in the case where I D It remains to prove part a Several preliminary results are needed We begin by replacing B by its image in the domain R taking the image of K as a coef cient ring Thus we may assume that B gt R is injective Then B is a module finite extension of a subring of the form KHacl ow with the same coef cient field by the structure theory of complete local rings We still have that R is essentially of finite type over A Moreover BF 2 AF AB from which it follows that RF 2 AF 1314 Therefore in proving part a of the Theorem it suf ces to consider the case where B A KHxl n and A Q R For each P cofinite in A Q K let r denote the fraction field of AF Let denote the fraction field of A Let Q be any field finitely generated over R that contains the fraction field of B To prove part a of the Theorem stated on p 4 it will suf ce to prove the following 6 Theorem Let K be a eld of characteristic 1 with p base A Let A KHxl ian and let AF and r be de ned as above for every co nite subset T of A Let Q be any eld nitely generated over A Then for all T lt A r 8 Q is a eld We postpone the proof of this result We first want to see just below that it implies part a of the Theorem stated on p 4 Beyond that we shall need to prove some auxiliary results first To see why the preceding Theorem implies part a of the Theorem on page 4 choose 9 containing the fraction field of R we can choose 9 frac R for example Since AF is A flat we have an injection AF A R gt AF A 9 Thus it su ices to show that this ring is a domain Since the elements of A 7 0 are already invertible in 9 we have that Q E frac A A 9 Since AF is purely inseparable over A inverting the nonzero elements of A inverts all nonzero elements of AF Moreover the tensor product of two frac A modules over frac A is the same as their tensor product over A Hence AP A o 2 AP A frac A A o 2 fracAF fmA o r can o It is now clear that Theorem above implies part a of the Theorem on p 4 In order to prove the Theorem above we need several preliminary results One of them is quite easy Lemma Let K be a eld of characteristic 1 gt 0 and let A be a p base for K The family of sub elds KF as T runs through the co nite subsets ofA is directed by Q and the intersection of these elds is K Proof K has as a basis 1 and all monomials 7 Ni A where t is some positive integer A1 At are mutually distinct elements of A and the 04339 are positive rational numbers in 0 1 whose denominators are powers ofp If u were in the intersection and not in K it would have a unique representation as a K linear combination of these elements including at least one monomial u as above other than 1 Choose A E A that occurs in the monomial u with positive exponent Choose T cofinite in A such that A Z P Then the monomial u is not in KP which has a basis consisting of 1 and all monomials as in such that the xj occurring are in T It follows that u KP D We shall also need the following result as well as part b of the Theorem stated after it Theorem Let be a eld of characteristic 1 gt 0 and let U be a nite purely inseparable ezrtension of Let be a family of elds directed by 2 whose intersection is L Then there edistsj such that for alli S j i 8 U is a eld Math 711 Lecture of November 5 2007 The following result is one we have already established in the F finite case We can now extend it to include rings essentially of finite type over an excellent semilocal ring Theorem Let R be a reduced ring of prime characteristic p gt 0 essentially of nite type over an ezrcellent semilocal ring B Suppose that c E R0 is such that RC is strongly F regular Then c has a power that is a completely stable big test element in R Proof If c is a completely stable big test element in a faithfully at extension of R then that is also true for R by part b of the Proposition at the bottom of p 8 of the Lecture Notes from September 17 The hypothesis continues to hold if we replace R by E 83 R and it holds in each factor of this ring We may therefore assume that R is essentially of finite type over a complete local ring A As usual choose a coef cient field K for A and a p base A for K Again the hypotheis continues to hold is we replace R by BP for P lt A and RF is faithfully at over R But now we are done since RF is F finite D We next want to backtrack and prove that certain rings are approximately Gorenstein in a much simpler way than in the lengthy and convoluted argument given in the Lecture Notes from October 24 While the result we prove is much weaker it does suf ce for the case of an excellent normal Cohen Macaulay ring and hence for excellent weakly F regular rings We first note Lemma Let M and N be modules over a Noetherian ring R and let it be a nonzerodiuisor on N Suppose that M is R free or much more generally that Ext3M N 0 Then RacR R HomRM N g HomRmRMacM NicN Proof The right hand module is evidently the same as HomRMacM NicN and also the same as HomR M NicN since any map M a NacN must kill icM Apply HomR M i to the short exact sequence o NLN NxNao This yields a long exact sequence which is in part 0 HomRM N L HomRM N HomRM NicN Ext3M N 0 and the result follows E 2 Theorem Let R m K be an excellent normal Cohen Macaulay ring or more gener ally any Cohen Macaulay local ring whose completion is a Cohen Macaulay local domain Then R is approximately Gorenstein Proof We may replace R by its completion and then R is module finite over a regular local ring A Q R Because R is Cohen Macaulay it is free of some rank h as an A module ie R 2 Ah Then to HomAR A is also an R module and is also isomorphic to Ah as an A module We shall see later that to is what is called a canonical module for B Up to isomorphism it is independent of the choice of A Then to is evidently also a Cohen Macaulay module over R We want to see that it has type one This only uses the Cohen Macaulay property of B it does not use the fact that R is a domain From the Lemma above we see that the calculation of to commutes with killing a parameter in A We may choose a system of parameters for A and R that is a minimal set of generators for the maximal ideal of A By killing these one at a time we reduce to seeing this when A K is a field and R is a zero dimensional local ring with coef cient field K We claim that in this case to HomKR K is isomorphic with ERK In fact to is injective because for any R module M HomRM to g HomRM HomKR g HomKM R R K g HomKM K by the adjointness of tensor and Hom This is in fact a natural isomorphism of functors Since HomKQ K is exact so is HomRQ to Thus to is a direct sum of copies of ERK But its length is the same as its dimension as a K vector space and this is the same as the dimension of R as a K vector space which is the length of R Thus to has the same length as ERK and it follows that to ERK We now return to the situation where R is a domain Since every nonzero element of R has a nonzero multiple in A we have that to is torsion free as an R module Thus if in is any nonzero element of to we have an embedding R a to sending 1 gt gt ii Let It z i zi where 1 xn is a system of parameters for R Then tw Ru must have the form th for some m primary ideal J of R Then RJt g RwItw Rw Q wItw Since wItw is an injective hull of the residue class field for RIt it is an essential extension of its socle Therefore RJt is an essential extension of its socle as well Consequently Jt Q R is irreducible and m primary It will now suf ce to show that the ideals J are cofinal with the powers of m By the Artin Rees Lemma there exists a constant integer a E N such that mNaw Ru Q mNRw me for all N But then JNa Q m since INa Q mN D We next want to prove some additional results on openness of loci such as the Cohen Macaulay locus The following fact is very useful 3 Lemma on openness of loci Let X Spec R where R is a Noetherian ring Then S Q X is open if and only if the following two conditions hold 1 IfPQQ andQESthenPES 2 For all P E S S W VP is open in VP The second condition can be weakened to 2quot For all P E S S contains an open neighborhood ofP in VP Proof It is clear that l 2 and 2quot are necessary for S to be open Since 2quot is weaker than 2 and it su ices to show that l and 2quot imply that S is open Suppose otherwise Since R has DCC on prime ideals if S is not open there exists a minimal element P of S that has no open neighborhood entirely contained in S For all primes Q strictly contained in P choose an open neighborhood UQ of Q contained entirely in S Let U be the union of these open sets the U is an open set contained entirely in S and contains all primes Q strictly smaller than P Let Z X 7 U which is closed It follows that Z has nitely many minimal elements one of which must be P Call them P P0 P1 Pk Then Z VP0 U VPk Finally choose U open in X such that P E U and U W VP Q S We claim that U U u U 7 W131 u u VPk is the required neighborhood of P It is evidently an open set that contains P Suppose that Q E U lf Q E U then Q E S Otherwise Q is in X 7 U WP u VP1 u UlPk and this implies that Q E VP But Q must also be in U and U W VP Q S D We can use this to show Theorem Let R be an ezrcellent ring Then the Cohen Macaulay locus P E Spec R Rp is Cohen Macaulay is Zarishi open Proof lt su ices to estabish l and 2 of the preceding Lemma We know 1 because if P Q Q then Rp is a localization of the Cohen Macaulay ring RQ Now suppose that Rp is Cohen Macaulay Choose a maximal regular sequence in PRp After multiplying by suitable units in RP we may assume that this regular sequence consists of images of elements 1 xd E P We can choose ci 6 R 7 P that kills the annihilator of gel1 in 4 Rxl mi 0 S i S d 7 1 Let c be the product of the Ci Then we may replace R by RC we may make nitely many such replacements each of which amounts to taking a smaller Zariski open neighborhood of P Then 1 xd is a regular sequence in P and is therefore a regular sequence in Q and in QRQ for all primes Q Q P Hence in considering property 2 it suf ces to work with R1 Rxl zdR whether RQ is Cohen Macaulay or not is not affected by killing a regular sequence Consequently we need only show that if P is a minimal prime of an excellent ring R then there exists 0 P such that RC is Cohen Macaulay We may assume by localizing at one element in the other minimal primes but not in P that P is the only minimal prime of B First we may localize at one element 0 P such that RPC is a regular domain because RP is an excellent domain the localization at the prime ideal 0 is a field and hence regular and so 0 has a Zariski open neighborhood that is regular Henceforth we assume that RP is regular It would suf ce in the argument that follows to know that it is Cohen Macaulay Finally choose a filtration P P1 2 2 PW 0 such that every PiPZH is killed by P We can do this because P is nilpotent If P 0 we may take Pi Pi l S i S 11 Alternatively we may take Pi Aninn i Second we can localize at one element 0 E R 7 P such that each of the RP modules PiPZH l S i S n 7 l is free over RP Here we are writing R for the localized ring We claim that the ring R is now Cohen Macaulay To see this suppose that we take any local ring of R Then we may assume that R m K is local with unique minimal prime P that RP is Cohen Macaulay and that P has a finite filtration whose factors are free RP modules all this is preserved by localization Let 1 xh be a system of parameters for R The images of these elements form a system of parameters in RP Then 1 zd is a regular sequence on RP But P has a finite filtration in which the factors are free RP modules and so does B one additional factor RP is needed Since 1 zd is a regular sequence on every factor of this filtration by the Proposition near the bottom of the first page of the Lecture Notes from October 8 it is a regular sequence on R Hence R is Cohen Macaulay D Remark It is also true that if M is a finitely generated module over an excellent ring R then P E Spec R Mp is Cohen Macaulay is Zariski open in Spec This comes down to establishing property 2 and we may make the same initial reduction as in the ring case killing a regular sequence in P on M whose image in PR is a maximal regular sequence on Mp Therefore may assume that P is minimal in the support of M and after one further localization that P is the only minimal prime in the support of M We may assume as above that RP is regular again R Math 711 Lecture of October 11 2006 We showed in the Lecture of October 9 that the map 1979 is injective We next want to show that its image WSR6g is indepedent of the choice of the presentation 6 and the choice of special sequence Q We first prove Lemma Let 3 be a ring J Q 3 an ideal and y elements ofJ that are nonzerodioisors in 3 Then 3 3 J g 243 B J 3 T 243 via the map that sends the class ofu 6 3 3 J to the class of an element 1 E 243 B J such that 1 ya Proof Given a 6 3 3 J we have since y E J that ya 6 3 and so ya 1 with v E 3 the choice of v is unique since is a nonzerodivisor in 3 We first want to see that v E 243 B J which means that if a E J then av E 243 Since a E J an b for b E 3 Then any yb and so av yb Since is not a zerodivisor this yields av yb as required Next note that if we change the representative of the class of u say to u 0 then yu 0 yu yc 1 yc v ye Since 1 changes by a multiple of y our map is well defined This establishes that we have a map 3 3 J 243 B J 7 7 3 243 of the form stated By symmetry there is a map y323J 3BJ i i 243 3 of the same sort By the symmetry of the condition yu 11 if the class of u maps to the class of i then the class of 1 maps to the class of u and vice versa This shows that the two maps are mutually inverse D Theorem The image of the map bag in is independent of the choice ofg and of the choice of 6 Proof To prove for a fixed presentation that the map is independent of the choice of special sequence suppose that we have two special sequences that yield maps with different images We can preserve the fact that the images W W are different while localizing at a suitable prime or even maximal ideal of T S is replaced by its localization at a corresponding 1 2 prime Simply choose the prime to be in the support of W W Thus there is no loss of generality in assuming that T and S are local The sequences in question remain special as we localize But then by the Lemma on comparison of special sequences from the beginning of the Lecture of October 9 we know that there exists a finite chain of special sequences joining the two that we are comparing such that any two consecutive sequences differ in at most one spot Thus we need only make the comparison when the two sequences differ in just one term and since the sequences are permutable we may assume without loss of generality that one of them is 91 g 92 g7 and the other is h 92 gm which we shall also denote hl hn We set up an isomorphism 91 gnTTI 2 an hnTlTI 91 gnT i an as follows Let B Rgg 97 let J Q B be the image of I ie Igg 9 Let x be the image of g and y the image of h We now have the isomorphism by applying the Lemma proved just above 0392 To complete the proof of the independence of the image from the choice of special sequence we note that the following diagram commutes 91gnTTI a h1hnTlTI 91 gnT hl 5 11 To see this one simply needs to see that if gt0 uhivgthgj 32 in T then H gtk 7 Y in where E i are the respective images of u and v in and y n are the respective images of the determinants of the two Jacobian matrices in ie that udet thBXi E v det ngBXi modulo I Slel By differentiating with respect to each Xj in turn and using the fact that all the g h and the gj are in I we see that because the terms not shown coming from the product rule have a coef cient in I th 7 vVg E 215339ng modulo I 32 Thus the matrix whose columns are th V92 Vgn and the matrix whose columns are 77 Wg thng V92 V971 32 are equal mod I By elementary column operations we may drop the summation term from the first column of the second matrix when we calculate the determinant Then we may factor it from the first column of the first matrix and 1 from the first column of the second matrix when we take determinants This yields E7 57 in A and follows It remains only to prove that the image of 1979 is independent of the choice of 6 T a S as well We first consider the case of a finitely generated R algebra S The choice of a presentation is equivalent to the choice of a finite set of generators for S over R We can compare the results from each of two different sets of generators with the result from their union and so it suf ces to see what happens when we enlarge a set of generators By induction it suf ces to show that the image does not change when we enlarge a set of generators by one element and so we may assume that we have 6 T RX1 Xn a S and an extension of 6 6TXn1 a S by sending XWH to 8 Let T TXn1 We can choose an element F E T such that F maps to s in S and it follows easily that the kernel I of 6V is 1 XWH 7 It also follows easily that if g gl 9 is special in I then 9 gl 971 with gn1 X7 7F is a special sequence in 1 The larger size 11 1 Jacobian matrix has the same determinant y as the size n Jacobian matrix of 91 9 with respect to X1 Xm and it is easy to check that there is an isomorphism T 917 797T1TI g 917 79n1TZTI 91 gnT 91 gn1T which is induced by the inclusion 91 gnTTI Q 91 gn1T T 1 Since the Jacobian determinants are the same we have a commutative diagram 917 797T1TI r 9179n1T1TI 91 gnT 91 gn1T C a C 11 and this yields that the images are the same We have now justified the notation WSR when S is finitely generated over R We leave it to the reader as an exercise to verify that if s is a nonzerodivisor in S then WS571 R WSR5 Once we know this by exactly the same argument we used to verify that the Jacobian ideal is independent of the choice of presentation for algebras 4 essentially of nite type over B it follows that WSR6 is independent of 6 when S is essentially of finite type over R For a given special sequence g it is obvious from the definition of 1979 that y multiplies the image of 1979 into S Q A Since the image is independent of the choice of special sequence and since by the Theorem on the existence of su iciently many special sequences at the end of the Lecture Notes of October 6 as the special sequence varies the values of y generate JSR we have Corollary WSR Q S2 jSR D The following result gives several properties of WSR that we will want to exploit Proposition Let S be generically tale torsion free and essentially of nite type over the Noetherian domain B Let W WSR a For any multiplicative system U in S WUilsR U lW b W is torsion free over S c For every prime ideal P of S if u u is part of a system of parameters for Sp then it is a regular sequence on Wp Thus W is S2 d If W Q W Q and Wp W for all height one primes of S and for all minimal primes of S that are also mazrimal ideals then W W e If R a S is a local homomorphism of regular local rings then jSR is principal and W S 1L jSR39 f If S is normal and Rp is regular for every prime ideal P of R lying under a height one prime ideal Q of S then W S L jSR Proof Part a is essentially the last part of 43 while b is evident from the fact that W Q by definition To prove c note that by a we may assume that S is local and that u u is part of a system of parameters We may choose a presentation zT a S and think of W as g gl gnTT Ig1 gnT where the sequence g1 g is special Let uo yo 6 T be representatives of u 1 Then uo I cannot be contained in the union of the minimal primes of g1 gn or else it will be contained in one of them by the Lemma on prime avoidance for cosets Since this will contain I it will be a minimal prime of I and contradicts the statement that u is part of a system of parameters in S TI Thus we can replace uo by an element ul representing it such that g1 g ul is part of a system of parameters for T Similarly yo I cannot be contained in the union of the minimal primes of g1 g u1T or else it is contained in one of them say Q Thinking modulo I we see that Q is a minimal prime of u in TI containing 1 a contradiction Thus we may choose u1u1 in T representing u 1 respectively and such that g1 g ul U1 is a regular sequence Clearly ul 111 form a regular sequence on Tgl gnT We claim Math 711 Lecture of October 24 2007 The action of Frobenius on the injective hull of the residue class eld of a Gorenstein local ring Let R m K be a Gorenstein local ring of prime characteristic 1 gt 0 and let 1 be a system of parameters Let It 3 3R for all t 2 1 and let a E R represent a socle generator in RI where I 1 1 R Let y 1 7 We have seen that E is an injective hull of K Rm over R where the map RIt a RItH is induced by multiplication by y acting on the numerators Each of these maps is injective Note that the map from RIt a RItk in the direct limit system is induced by multiplication by y acting on the numerators Let 6 E N be given We want to understand the module feE and we also want to understand the qth power map 1 gt gt 0 7 from E to feE lf 7 E R we shall write lt7 3 3gt for the image of 7 under the composite map R a RIt gt E where the first map is the quotient surjection and the second map comes from our construction of E as the direct limit of the RIt With this notation lt7 3 3gt ltykr 3k kgt for every k E N Since tensor products commute with direct limit we have that few limt femIt 11m RIt ql 11m RItq In the rightmost term the map from RItq a RIt1q Itqq is induced by multipli cation by tq acting on the numerators The rightmost direct limit system consists of of a subset of the terms in the system limt RIt and the maps are the same The indices that occur are cofinal in the positive integers and so we may idenitfy feE with E Under this identification if v lt7 3 3gt then 0 7 lt7 315 3gt We can now prove the assertions in the first paragraph of the Theorem on p 4 of the Lecture Notes from October 22 Proof that 0 is tightly closed m ERK for a weakly F T egular Gorenstem local ring Let R m K be a Gorenstein local ring of prime characteristic 1 gt 0 We want to determine when U ltu 1 gt is in 0 in E This happens precisely when there is an element 0 E R0 such that cvq 0 in feE for all 1 gtgt 0 But cvq ltcuq 3 gt which is 0 1 2 if and only if CM 6 Iq Ilql for all 1 gtgt 0 Thus 0 is tightly closed in E if and only if I is tightly closed in B This gives a new proof of the result that in a Gorenstein local ring if I is tightly closed then R is weakly F regular But it also proves that if I is tightly closed every submodule of every module is tightly closed In particular if R is weakly F regular then every submodule over every module is tightly closed D It remains to show that when a Gorenstein local ring is F finite it is strongly F regular We first want to discuss some issues related to splitting a copy of local ring from a module to which it maps Splitting criteria and approximately Gorestein local rings Many of the results of this section do not depend on the characteristic Theorem Let R m K be a local ring and M an R module Let f R a M be an R linear map Suppose that R is complete or that M is nitely generated Let E denote an injective hull for the residue class eld K Rm of R Then R a M splits if and only if the map E E R R a E R M is injective Proof Evidently if the map splits the map obtained after tensoring with E or any other module is injective it is still split This direction does not need any hypothesis on R or M For the converse first consdider the case where R is complete Since the map E R R a E R M is injective if we apply HomRi E we get a surjective map We switch the order of the modules in each tensor product and have that HOHIRR E E HOHIRM R E is surjective By the adjointness of tensor and Hom this is isomorphic to the map HomRM HomRE a HomRR HomRE By Matlis duality we have that HomRE E may be naturally identified with R since R is complete and this yields that the map HomRM R a HomRR R induced by composition with f R a M is surjective An R linear homomorphism g M a R that maps to the identity in HomRR R is a splitting for f Now supppose that R is not necessarily complete but that M is finitely generated By part b of the Theorem on p 3 of the Lecture Notes from Septmeber 24 completing does not affect whether the map splits The result now follows from the complete case because E is the same for R and for R and E 8 R ER 7 is the same as E ER 7 by the associativity of tensor D This result takes a particularly concrete form in the Gorenstein case 3 Theorem splitting criterion for Gorenstein rings Let R m K be a Gorenstein local ring and let 1 xn be a system of parameters for B Let u E R represent a socle generator in RI where I 1 acn let y 1147 and let It ac i xR for I 2 1 Let f R a M be an R linear map with f1 w E M and assume either that R is complete or that M is nitely generated Then the following conditions are equivalent 1 f R a M is split 2 For every ideal J of R RJ a MJM is injectiue where the map is induced by applying RJ R 3 For all t 2 1 RIt a MItM is injectiue 4 For all t 2 1 yt luw Z ItM Moreover if 1 xn is a regular sequence on M then the following two conditions are also equivalent 5 RI a RIM is injectiue 6 uw Z IM Proof 1 i 2 i 3 i 5 is clear The map RIt a MItM has a nonzero kernel if and only if the socle element which is the image of yt lu is killed and this element maps to yt luw Thus the statements in 3 and 4 are equivalent for every value of t and the equivalence 5 gt 6 is the case t 1 We know from the preceding Theorem that R a M is split if and only if E a E R M is injective and this map is the direct limit of the maps RIt a RIt R M by the Theorem on p 2 This shows that 3 i Thus 1 2 3 and 4 are all equivalent and imply 5 and 6 while 5 and 6 are also equivalent To complete the proof it su ices to show that 6 i 4 when 1 acn is a regular sequence on M Suppose ytuw E ac i Then mu 6 ac i xM 2M yt 1 xnM by the Theorem on p 3 of the Lecture Notes from October 8 D Remark If M S is an R algebra and the map R a S is the structural homomorphism then the condition in part 2 is that every ideal J of R is contracted from S Similarly the condition in 4 respectively is that It respectively I be contracted from S We de ne a local ring R m K to be approximately Gorenstein if there exists a de creasing sequence of m primary ideals I1 2 I2 2 Q It 2 such that every RIt is a Gorenstein ring ie the socle of every RIt is a one dimensional K vector space and the It are co nal with the powers of m That is for every N gt 0 It Q mN for all t gt 1 Evidently a Gorenstein local ring is approximately Gorenstein since we may take It ac i R where 1 xn is a system of parameters Note that the following conditions on an m primary ideal I in a local ring R m K are equivalent 1 RI is a Oidimensional Gorenstrein 2 The socle in RI is one dimensional as a K vector space 4 3 I is an irreducible ideal ie I is not the intersection of two strictly larger ideals Note that 2 i 3 because when 2 holds any two larger ideals considered modul I must both contain the socle of RI Conversely if the socle of RI has dimension 2 or more it contains nonzero vector subspaces V and V whose intersection is 0 The inverse iamges of V and V in R are ideals strictly larger than I whose intersection is I D If R itself has dimension 0 the chain It is eventually 0 and so in this case an approxi mately Gorenstein ring is Gorenstein In higher dimension it turns out to be a relatively weak condition on R Theorem Let R m K be a local ring Then R is approximately Gorenstein if and only ifR is approximately Gorenstein Moreover R is approximately Gorenstein provided that at least one of the following conditions holds 1 E is reduced 2 R is excellent and reduced 3 R has depth at least 2 4 R is normal The fact that the condition holds for R if and only it holds for E is obvious Moreover 2 i l and 4 i We shall say more about why the Theorem given is true in the sequel For a detailed treatment see Hochster Cyclic purity versus purity in excellent Noetherian rings Trans Amer Math Soc 231 1977 4637488 which gives the following precise characterization a local ring of dimension at least one is approximately Gorenstein if and only if R has positive depth and there is no associated prime P of the completion E such that dim l and EB embeds in Before studying characterizations of the property of being approximately Gorenstein further we want to note the folliwing Propositon Let R m K be an approximately Gorenstein local ring and let It be a descending chain of m primary irreducible ideals co nal with the powers of m Then an injective hull E ERK is an increasing union Ut AnnItE and AnnEIt g RIt so that E is the direct limit of a system in which the modules are the RIt and the maps are injective Proof Since every element of E is killed by a power of m every element of E is in AnnEIt for some t We know that AnnEIt is an injective hull for K over RIt Since RIt is 0 dimensional Gorenstein this ring itself is an injective hull over itself for K D This yields Theorem Let R m K be an approximately Gorenstein local ring and let It be a descending chain of m primary irreducible ideals co nal with the powers of m Let at E R 5 represent a socle generator in RIt Let f R a M be an R linear map with fl w E M Then the following conditions are equivalent 1 f R a M splits over R 2 For all t 2 l RIt a MItM is injective 3 For all t 2 l mu Z ItM Proof Since E ERK is the direct limit of the RIt we may argue exactly as in the proof of the Theorem at the top of p 3 D When is a ring approximately Gorenstein To prove a su icient condition for a local ring to be approximately Gorenstein we want to introduce a corresponding notion for modules Let RmK be local and let M be a finitely generated R module We shall say that N Q M is co nite if MN is killed by power of m The reader should be aware that the term cofinite module77 is used by some authors for a module with DCC The following two conditions on a cofinite submodule are then equivalent just as in the remark at the bottom of p 3 and top of p 4 1 The socle in MN is one dimensional as K vector space 2 N is in irreducible submodule of M ie it is not the intersection of two strictly larger submodules of M We shall say that M has small co nite irreducibles if for every positive integer t there is an irreducible cofinite submodule N of M such that N Q th Thus a local ring R is approximately Gorenstein if and only if R itself has small cofinite irreducibles Note the the question of whether R m K is approximately Gorenstein or whether M has small cofinite irreducibles is unaffected by completion there is a bijection between the cofinite submodules N of M and those of M given by letting N correspond to The point is that if N is cofinite in M MN is a finitely generated R module in fact it has finite length and M a MN is surjective since Mth E for all t so that N is the completion of N M Moreover when N and N correspond MN E MN since MN is already a complete R module In particular irreducibility is preserved by the correspondence We have already observed that Gorenstein local rings are approximately Gorenstein We next note 6 Proposition Let R m K be a local ring IfM is a nitely generated R module that has small co nite irreducibles then every nonzero submodule ofM has small co nite irre ducibles Proof Suppose that N Q M is nonzero By the Artin Rees lemma there is a constant c E N such that th N Q mt CN for all t 2 c If Mtc is cofinite in M and such that Mtc Q mtCM and MMHC has a one dimensional socle then Nt Mtc N is cofinite in N contained in th so that NNt is nonzero and has a one dimensional socle since NNt embeds into MMtc D Before giving the main result of this section we note the following fact due to Chevalley that will be needed in the argument Theorem Chevalley s Lemma Let M be a nitely generated module over a complete local ring RmK and let Mtt denote a nonincreasing sequence of submodules Then t Mt 0 if and only if for every integer N gt 0 there exists t such that Mt Q mNM Proof The if part is clear Suppose that the intersection is 0 Let VMV denote the image of Mt in MmNM Then the VMV do not increase as t increases and so are stable for all large t Call the stable image VN Then the maps MmN1M a MmNM induce surjections VN1 a VN The inverse limit W of the VN may be identified with a submodule of the inverse limit of the MmNM ie with a submodule of M and any element of W ltMt mNM Mt mNM Mt tN t N If any VMJ is not zero then since the maps VN1 a VN are surjective for all N the inverse limit W of the VN is not zero But VN is zero if and only if Mt Q mNM for all t gt 0 D The condition given in the Theorem immediately below for when a finitely generated module of positive dimension over a complete local ring has small cofinite irreducibles is necessary as well as suf cient we leave the necessity as an exercise for the reader The proof of the equivalence is given in Hochster Cyclic purity uersus purity in e1cellent Noetherian rings Trans Amer Math Soc 231 1977 4637488 Theorem Suppose that M is a nitely generated module over a complete local ring RmK such that dimM 2 1 Suppose that m is not an associated prime ofM and that ifP is an associated prime ofM such that dimRP 1 then RP EB RP is not embeddable in M Then M has small co nite irreducibles Proof We use induction on dim M First suppose that dimM 1 We represent the ring R as a homomorphic image of a complete regular local ring S of dimension d Because R is catenary and dimM l the annihilator of M must have height d 7 1 Choose part of a system of parameters 1 xd1 in the annihilator Now view M as a module over PM Sacl xd1 We change notation and simply write R for this ring Then R is a one dimensional complete local ring and R is Gorenstein It follows that R has small cofinite irreducibles and we can complete the argument by the Proposition on the preceding page by showing that M can be embedded in R Note that for any minimal prime p in R Rp is a zero dimensional Gorenstein ring In fact any localization of a Gorenstein local ring at a prime is again Gorenstein but we have not proved this here However in this case we may view Rp as the quotient of the regular ring Sq where q is the inverse image of p in S by an ideal generated by a system of parameters for Sq and the result follows To prove that we can embed M in B it suf ces to show that if W R then W lM can be embedded in WAR One then has M Q W lM Q WAR and the values of the injective map M gt WAR on a finite set of generators of M involve only finitely many elements of W Hence one can multiply by a single element of W and so arrange that M gt W lR actually has values in R But WAR is a finite product of local rings Rp as p runs through the minimal primes of R and so it suf ces to show that if p is a minimal prime of R in the support of M then Mp embeds in RP Now Mp has only 13Rj as an associated prime and since only one copy of Rp can be embedded in M only one copy of Hp Pup13Rj can be embedded in Mp Thus Mn is an essential extension of a copy of Hp Thus it embeds in the injective hull of the residue field of RP which since Rp is a zero dimensional Gorenstein ring is the ring Rp itself Now suppose that dimM d gt 1 and that the result holds for modules of smaller dimension Choose a maximal family of prime cyclic submodules of M say Rul Bus such that AnnRui is a prime 621 for every t and the sum N Rul B BRus is direct Then M is an essential extension of N if v E M it has a nonzero multiple N that generates a prime cyclic module and if this prime cyclic module does not meet N we can enlarge the family Since M is an essential extension of N M embeds in the injective hull of N which we may identify with the direct sum of the EZ ERRuZ Note that a prime ideal of B may occur more than once among the 62 but not if dim RQi l and Rm does not occur Take a finite set of generators of M The image of each generator only involves finitely many elements from a given Ei Let M be the submodule of EZ generated by these elements Then M Q Ei so that Ass Qi and M is an essential extension of RQi What is more M Q 6931Mi By the Propositiion at the bottom of p 5 it suf ces to show that this direct sum which satisfies the same hypotheses as M has small cofinite irreducibles Thus by we need only consider the case where M 6931 M as described We assume that for i S h Ass Mi with dim RQi l and with the 621 mutually distinct while for i gt h dim RQi gt 1 and these 62 need not all be distinct Now choose primes P1 Ps such that for every t dim RPZ 1 such that P1 P5 are all distinct and such that for all i P Q 62 We can do this for l S i S h the choice Pi 621 is forced For i gt h we can solve the problem recursively simply pick P to be any prime different from the others alreaday selected and such that P Q 621 and dim RPZ 1 We are using the fact that a local domain RQ of dimension two or more contains infinitely many primes P such that dim RP 1 To see this kill a prime to obtain a ring of dimension exactly two We Math 711 Lecture of September 19 2005 Our next objective is to prove the Cohen Macaulayness assertions in the statement of the second Theorem of the Lecture of September 12 The argument is entirely similar to what we did earlier in studying the ideal generated by the 2 gtlt 2 minors of a matrix of indeterminates We use reverse induction assuming the result that larger ideals of the form L Jsk X are Cohen Macaulay Suppose that a speci c prime of the form L Jsk is given Call the ideal P To show that is Cohen Macaulay it suf ces to show that the depth of on the ideal m generated by all the gelJ in is d dim we review the relevant facts about the Cohen Macaulay property below Let x 175k1 Since we already know that is a domain we have that x is a nonzerodivisor and so is Cohen Macaulay if and only if is and this may be described as J5k1X There are two cases If skl sk1 then IKX is a larger prime ideal of our family and so killing it gives a Cohen Macaulay ring by the induction hypothesis lf sk l lt sk1 then I is radical By the lemma on killing minors each of the variables 171 for sk 1 lt b lt sk1 kills IkXlsk1 Let 0 be the result of changing sk in a to sk 1 while leaving all other entries xed Let Q1 0 Jsk1X and Q2 L Jsk1 Both of these ideals are prime and we know that they have Cohen Macaulay quotients by the induction hypothesis This is also true for Q3 Q1 Q2 0 l szr1 Note that VP VQ1 U VQ2 by the lemma on killing minors since all of the ideals are radical we have that P Q1 Q2 Moreover KXQ1 has dimension d 7 1 among the numbers used in calculating the dimension sk has increased by one while all others including k have not changed Sim ilarly KXQ2 has dimension d 7 1 here only k has changed increasing by 1 Finally KXlQ3 has dimension d7 2 since in this case k has increased by l and sk has increased by one Since these are Cohen Macaulay in the short exact sequence O KleP a KlXQ1 KleQ2 leXlQs 7 0 the depths of the middle and right hand terms on m are d 7 l and d 7 2 respectively and so the depth of is d 7 l as required D We next review the basic facts about Cohen Macaulay rings 1 xn E R is called a possibly improper regular sequence on the R module M if 1 is not a zerodivisor on M and for every l l lt l S n x is not a zerodivisor on Macl xi11M It is called a regular sequence if moreover x1 xnM 31 M Let R 7gt S be a homomorphim of Noetherian rings I an ideal of R and M a nitely generated S module Then M is also an R module by restriction of scalars If M 31 M we de ne the depth of M on I or depthIM to be the supremum of lengths of regular sequences in I on M By a theorem this supremum is nite every regular sequence can be extended to a maximal one and all maximal regular sequences have the same length which is the l depth If M M the depth is de ned to be 00 It turns out the depthIM depthISM The most important case is where S R but it is often useful to have the greater generality available REFS By a theorem if d depthIM and N is any R module with annihilator I then Ext 3N M 0 for i lt d while Eict liaN M 31 0 In particular one may choose N RI and the long exact sequence for Ext may then be used to prove facts about the behavior of depth in short exact sequences For example if 0 a M a M a M a 0 is exact and depthIM lt depthIM then depthIM depthIM 1 We used this last assertion earlier when the depth of the middle term was d 7 l and the depth of the right hand term was d 7 2 A local ring R m K is called Cohen Macaulay if some equivalently every system of parameters is a regular sequence in the ring Regular local rings are Cohen Macaulay and so are their quotients by an ideal generated by part of a system of parameters the latter are called local complete intersections If R is a Cohen Macaulay local ring then each of its localizations at a prime is again Cohen Macaulay REFS A Noetherian ring R is called Cohen Macaulay if all of its localizations at prime equiv alently at maximal ideals are Cohen Macaulay local rings The quotient of Cohen Macaulay ring by a proper ideal generated by a regular sequence is again Cohen Macaulay All Noetherian rings of dimension 0 are Cohen Macaulay and all reduced Noetherian rings of dimension 1 are Cohen Macaulay as well Of course this includes the one dimensional domains By a theorem all normal Noetherian domains of dimension two are Cohen Macaulay REFS If a local ring R is a module finite extension of a regular local ring A then R is Cohen Macaulay if and only if it is free as an A module REF We want to focus on the graded case Let R be an N graded finitely generated algebra over a field K with R0 K Let m denote the unique homogeneous maximal ideal it 0 is spanned by all forms of positive degree and may also be described as 69R Let 711 dim R d By a theorem there is always a sequence f1 fd of forms in m such that Rad fl fd m Such a sequence is called a homogeneous system of parameters Note that if K is in nite and R is generated by R1 over K then these may be chosen in R1 In general it is necessary to use forms of degree larger than 1 If f1 d is a homogeneous system of parameters for R R is mo dule finite over its subring Kf1 fd and this subring is a polynomial ring ie f1 fd are algebraically independent over K The image of a homogeneous system of paramaters for R in Rm is a system of parameters for Rm Theorem R be an N graded nitely generated K algebra as above where K is a eld with R0 K and let m be the homogeneous ma1imal ideal ofR Suppose that dim R d The following conditions are equivalent 1 R is Cohen Macaulay 2 Rm is Cohen Macaulay 3 Some equivalently every homogeneous system of parameters for R is a regular se quence in R Math 711 Lecture of November 10 2006 Before proceeding further with our study of Lech7s conjecture we want to discuss a result known as the associativity of multiplicities whose proof uses Lech7s theorem on computing multiplicities using ideals generated by powers of parameters Before we prove this result we want to make some remarks on the behavior of limits of real valued functions of two integer variables these might as well be positive since we will be taking limits as the variables approach 00 either jointly or independently Let Cm n be such a function and suppose that Cm n mngtoo exists and also that the iterated limit lim lim Cm mace ngtltxgt exists as well Then they are equal If the joint limit is L then given 6 gt 0 there exists N such that for all m n 2 N jL 7 Cm lt 6 It follows that each for m 2 n Lm Cm n which we are assuming exists is such that jL 7 Lmj S 6 since all of the values of Cm n are at distance at most 6 from L for all m 2 N It also follows that the iterated limit must have a value that is at distance at most 6 from L Since this is true for all 6 gt 0 the iterated limit must be L Note however that when the joint limit exists it is possible that neither iterated limit exists Eg for m n 2 l we can let Gmn lminm n if m n is even and 0 if m n is odd For any fixed m the function takes on the values lm and 0 alternately for large n and the iterated limit does not exist On the other hand observe that we can define Gmn 1 if m 2 n and Gmn 0 if m lt n The iterated limits both exist but are different and the joint limit does not exist If alternatively we define Cm n 0 when m n and Cm n 1 when m 31 n both iterated limits exist and are equal but the joint limit does not Note that the Corollary on p 5 of the Lecture Notes of October 20 is a special case of the following Theorem namely the case where r 0 It is also used in the proof Theorem associativity of multiplicities Let M be a module of dimenson d over a local ring R of dimension d Let 1 ied be a system of parameters for R and let I 1 xdR Let r s be nonnegative integers such that rs d Let Qt 1 x R 1 2 and B n1 xdR Let P be the set of minimal primes of 1 aer such that dimMp dim RP d Then 61M Z PEP Proof Let th 71 21 and Bn x H 913 Consequently by Lech7s result on calculation of multiplicities which is the Theorem on p 3 of the Lecture Notes of October 25 E M th Bn M eIM lim gt mnace m rns In this instance we know that the joint limit exists The iterated limit exists as well since we have 1 5 M th 37 M M mM lim a lim gt lim 6 Qt mace m r nace n5 mace m r Since M and B have dimension d and 1 xd is a system of parameters so is 71quot xfl yer acg and MthM will have dimension 3 P will be a necessarily minimal prime of the support of MthM such that dim RP dim if and only if P contains Qt P contains Supp M and dim RP 8 Since dim s P must be a minimal prime of Qt Let Q denote the set of minimal primes of Qt such that dim RP 3 By the Corollary on p 5 of the Lecture Notes of October 20 egMthM Z eQ3RP RPMpthMp PEQ Note that since P is a minimal prime of Qt Qt expands to an ideal primary to the maximal ideal in Rp Consequently 6 2 e23 RP lim g RP Misthp PEQ lim mace We now see that the iterated limit exists Note that hm RAMPthMP mace m r is 0 if dim Mp lt 7 and Q1ltMpgt otherwise Therefore we need only sum over those minimal primes P of Qt in Supp such that dim Mp 7 and dim RP s The latter two conditions are equivalent to the condition dim Mp dim RP d given that P E Supp contains Qt since dim Mp S dim Rp S 7 and dim RP S dim s 3 However if P is a minimal prime of Qt we do not need to require that P E Supp M for if not Mp 0 and the term corresponding to P does not affect the sum Therefore the value does not change if we sum over primes P E P and this yields 61M Z 6BRP6Q1Mp PEP as required B Our next objective is to obtain some classes of local rings R such that Lech7s conjecture always holds for at local maps R a S Lech proved the result for the case where dim R S 2 However we shall first focus on cases that are given by the existence of linear maximal Cohen Macaulay modules and related ideas We shall say that an R module M has rahh p if there is a short exact sequence 0 a R9 a M a C a 0 such that dim O lt dim If M has rank p for some p we shall say that M is an R module with rahh When R is a domain M always has rank equal to the maximum number of elements of M linearly independent over B When R has a module M of rank p eM peR by the additivity of multiplicities The following result is very easy but very important from our point of view Theorem Let R m K be a local ring that has a linear mammal Cohen Macaulay mod ule of rank p Then for every flat local map R m K a S n L eR S eS Proof As observed earlier we may assume that mR is n primary It follows that S 8 M is Cohen Macaulay over S and also has rank p Then 1 1 1 act 7 eltMgt 7 7 24M 7 7 vlts R M s 7 elts M 7 eltsgt a P P P Therefore Lech7s conjecture holds for any ring R that admits a maximal Cohen Macaulay module M such that eM Consequently we shall focus for a while on the prob lem of constructing linear maximal Cohen Macaulay modules However we first want to point out the following idea which has also been used to settle Lech7s conjecture in important cases Suppose that R and S are as in the statement of the Theorem just above and that M is any Cohen Macaulay module with rank say p Then 1 1 1 EVltS RMVM eM CM CM eR 55 gas R M 2 Hence Math 711 Lecture of October 12 2005 Discussion Let R m K be a Cohen Macaulay ring of Krull dimension d We want to de ne an R module HR associated canonically with R For those familiar with local cohomology it will turn out that HR Hy R We shall attempt to avoid making use of any substantial knowledge of local cohomology theory The key point is that if 1 xd and y1 yd are two systems of parameters for R if x1 acdR Q yl ydR then there is a canonical map Rx1dR a Ryl yd This should not be confused with the obvious surjection Ryl yd a Rxl acd in the other direction in fact we shall eventually show that the maps Rx1dR a Ryl ydR are injective The map is constructed by choosing a d gtlt d matrix A rij such that Y XA where Y is the l gtlt d row matrix whose entries are yl yd and X is the l gtlt d row matrix whose entries are 1 acd The existence of A is entirely equivalent to the fact that each Q is an R linear combination of the elements yl yd The map Rxl xd a Ryl yd is then induced by multiplication by 6 detA acting on the numerators Let adjA denote the transpose of the cofactor matrix of A the classical adjoz39mf of A A standard identity yields that AadjA 61d where Id is d gtlt d identity matrix Since Y XA multiplying both sides on the right by adjA yields that YadjA 6X which shows that multiplication by 6 takes 1 xdR intO 241 ydR This shows that multiplication by 6 does induce a map Rxl xdR a Ryl ydR We next want to show that this map is independent of the choice of the matrix A We first recall some facts about the Koszul complex the point of view we shall take which is the exterior algebra point of view is discussed in the Lecture Notes of March 1 from Math 615 Fall 2004 d Consider the Koszul complex IC x1 xd R of 1 ard on R If G Ruj K16 acd R is the free module on the free basis 1L1 ud and the differential takes uj gt gt acj l S j S d then the differential on the whole complex is the uniqe extension of this map to an exterior algebra derivation on G The matrix A induces a map of Koszul complexes O AdG AdilG u gtRdLR gt0 m M TA in O AdG AdilG u gtRdR gt0 while if we replace A by another choice A such that Y XA with detA 6 we get another such map of complexes Since the top row is acyclic and the bottom row consists of projective modules the two maps of complexes are homotopic the needed facts about homotopy may be found in the Lecture Notes of February 2 and February 4 from Math 1

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