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Commuta Algebra

by: Mrs. Preston Lehner

Commuta Algebra MATH 614

Mrs. Preston Lehner
GPA 3.87


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This 25 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 614 at University of Michigan taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/231503/math-614-university-of-michigan in Mathematics (M) at University of Michigan.

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Date Created: 10/29/15
Math 614 Lecture notes Sep 19 2007 As a rst use of localization7 we will prove the following proposition7 that Vanishing of a module is a local property7 Proposition Let R be a ririg arid M ari R module Theri the following are equivalent 1 M 0 2 W lM 0 for every multiplicative set W Q R 3 Mp 0 for every p E Spec R 4 Mm 0 for every 111 E Proof 1 i 2 because 3 0 2 i 3 i 4 are trivial All that remains is to show that 4 i 1 Suppose Mm 0 for every maximal ideal m and take any 2 E M Then for any 111 we have M 07 and in particular 0 in Mm This means that there is some it E R m such that wz 07 and so it E 0 z7 and 0 2 g 111 So 0 z is contained in no maximal ideal of R7 and hence must be the unit ideal Thus7 0 1132 2 Sequences complexes and exactness De nition A sequence of R modules is de ned to be a set of R modules7 which can be indexed by some interval in Z along with a linear sequence of module homomorphisms connecting them For example d d d d M0 M14gtM0 igtM2OgtM11gtMm 1 171 172 1 d d d 171 MOM1EM26 and m aMz 2M1 lMo 0M4 are all sequences Any of the above sequences7 with the numbers going down as the arrows go to the right7 and dj Mj a Mj1 for each dj we may label Md for short A sequence Md is a complex if djodHL 0 whenever the composition is de ned In other words7 ldeJrL Q Ker dj A complex Md is exact at Mj if either 1 at least one of dj or dHl is not de ned7 or 2 lm dHl Ker dj A complex is exact if it is exact at Mj for all j where M is de ned Special kinds of exact sequences and functors The primary example of an exact sequence is the following Let N be an R submodule of the R module M Let j N a M be the inclusion map7 and 7139 M a MN the natural projection Then the sequence o NLMLMN o is exact ln general7 a short exact sequence is a 5 term exact complex with zeros at either ends 7 ie it has the form OaM QMLM aO Note that any short exact sequence is isomorphic to the sequence one gets from a module inclusion Namely7 the sequence above is isomorphic to the sequence 0 a Z39M a M a Mx39M a 0 A left exact sequence is an exact sequence of the form 0 a M a M a M A right exact sequence is an exact sequence of the form M a M a M a 0 De nition Let R7 T be rings A functor F RMod a TMod is additive if for any R modules MN and homomorphisms g 6 HomRMN7 we have Ffg Ff F9 and Ff Ff Note that if Md is a complex and F is an additive functor7 then the sequence is also a complex7 since we have Fdj o Fdj1 Fdj o dj1 F0 0 We say that an additive functor F is left eccact if it takes left exact se quences to left exact sequences iight esact if it takes right exact sequences to right exact sequences and eacact if it takes short exact sequences to short exact sequencesl In the previous lecture we have shown that ifj N gt M is an inclusion of modules then W lj is injective and W 1MN W lMW lN In the terminology of this lecture this says that the functor W l RMod a WARMod is eccact You will show in the exercises that if FGltp RMod a TMod is an adjunction where F and G are additive then F is right exact and G is left exact Hom Let A B be rings which are not necessarily commutative Then we have the notions of a left A module that is A acts on the left so we get an action 2 gt gt 12 along with the rule ad2 aa z with module category AMod a iight B module where B acts on the right so we get an action 2 gt gt 2b along with the rule 2bb zbb with module category ModB and an A B bimodule that is M is both a left A module and a right B module and the actions are compatible in the sense that azb azb for all a E A b E B 2 E M so we may write azb unambiguously with module category AMOdB Note that AMod E AModZ and ModB E ZModB are equivalences of categories If M N E AMod then the left A module homomorphisms H HomAM N from M to N form an Abelian group In general there is no natural left or right A action on this group However if M E AModB and N E AModC then HomAM N E BModC with bimodule action given by bfcm fmb c Indeed we have a functor HomA AModBOP gtltAM0dC a BModC Now x N E AModB Then there is a functor H HomAN7 AMod a BMod where the left B module structure on any HomAN P comes from the right B module structure of N We will show that H has a left adjoint T N B 7 BMod a AMod 1This is equivalent to saying that F applied to my exact sequence gives an exact sequence but we will not need the fact that these are equivalent conditions Tensor product For a left B module M we de ne the left A module N B M as follows Let Ngtlt M be the set theoretic product of N and P and F FN M the free Abelian group on N gtlt M Let G be the subgroup of F generated by all elements of each ofthe following three forms for all 771 771 E M 71 71 E N and b E B 71 771771 771 771 771 771 7171 771771 771 771 771 71b 771 771 b771 Then set N B M FG and denote the equivalence classes of 71771 with the tensor symbol as in 71 771 71771 G E FG This notation is ambiguous but universally used The group N 83 M has a left A module structure given by a71 771 171 771 To see that this is well de ned we de ne it rst on F that is a71771 171771 and we see that the action of any a on an element of G gives another element of G Hence we get an A action on N B M FG Tensor is leftadjoint to Horn We must de ne a natural isomorphism ppLP HomAN BM P a HomB M HomAN That is for any 9 N B M a P we must give a B linear map ltpg M a HomAN P so that in particular for every 771 E M g771 N a P is A linear We de ne p by ltpg g71 One must verify the following facts H For every 9 N B M a P and 771 E M g771 is A linear D Each ltpg is B linear C40 p is a group homomorphist 4 p is injective Cf p is surjective CT ltpMp is natural in M and P 2Actually we donlt need to prove this but it s useful to know We leave out the veri cations of parts 1 3 and 6 For part 4 suppose g 0 Then gn in 0 for every n E N and m E M But every element of N B M is of the form 21 71 mi so we have n m 21901148 0 0 Hence 9 0 For part 5 let h M a HomAN P be B linear and de ne77 g N B M a P by gn in To show that this is well de ned de ne it on basis elements of F as g nm hm and show that g vanishes on G as follows g n n m n hmn g nm g n m similarly for 71771 m and g nbm hmnb hbm g nbm Hence 9 is well de ned as the induced map from 9 Then we have gn X m hmn so that My h Corollary For an A B bimodule N the furictor HomAN 7 AMod a BMod is left eccact arid the furictor N Bi BMod a AMod is right eaact Proof By the above the tensor functor is left adjoint to the horn functor Thus by the exercise they have the required exactness properties D Note that the functors A A 7 AMod a AMod and HomAA7 AMod a AMod are both isomorphic to the identity functor on AMod For an A module M the isomorphism M a HomAAM is given by m gt gt a gt gt am and the isomorphism A 8 M a M is given by 21 a in gt gt 21 aimi Extension restriction and evaluation7 of scalars Now we return to the commutative ring setting Let g R a S be a ring homomorphism and N an S module Then there is a functor p SMod a RMod called restrictiori of scalars which sends N to N but with R action given by r z grz and it sends every S homomorphism of modules to itself If one considers S to be an S R bimodule via right action by 9 then p is isomorphic to the functor Hom5S 7 And we have Hom5S R M N i HomRM Hom5S N 2 HomRM pN so that S 83 7 RMod a SMod a functor we call eaterisiori of scalars is left adjoint to p Math 614 Lecture notes Sep 24 2007 hdorecnlIionL aQ andIExactnesscontkl Proposition 1 ft M gt M andj N gt N are R module inclusions then M N N M R N M R W R 1N R Proof We have short exact sequences S aMQMLMaO and I T NLNLW O and the right exactness of tensor product means we have the following com mutative diagram7 where the rows and column are exact M RN4L4W RN44w d e M RNM RNM R N60 h M R NH 0 0 By the exactness of the column7 M RN E Since 9 is surjective7 M 83 N lmg7 so that M 83 N 21 In particular7 lme Q lmg7 but 1119 1111539 1 this already follows from the surjectivity of b and g The exactness of the middle row means that g induces an isomorphism g MlinN lm 9 Moreover g 1lm e lmd lm f To see this rst take some 2 E g 1lm e Then by surjectivity of b there is some z with 92 ebp gdp so 27dx E Kerg lm f whence z 6 lm dlm f Conversely take some element dp fy E lmd lm f Then gdp gdx ebz 6 lm e so that dp fy E g 1lm e Thus under the isomorphism g we have 1lm e lmdlm f lm f Hence we have M RNglmgg M RNImf N M RN M NH 2 i i i 39 R lme lme lmdlmflmf lmdlmf The third isomorphism above is induced by g4 But d 1M Rj and fi R 1Nso we are done D The right exactness of tensor product will allow us to see what extension of scalars does to an R module at least in terms of its presentation We detour here to discuss this important topic Presentations Let M be any R module Then there is an R linear surjection from a free module onto R Say X is a set and RX is a free module that admits an R linear surjection 7T RX a M This map has some kernel say K which also admits an R linear surjection from a free module onto it say 7T RGay a K Let j K gt RX be the inclusion map and g j 0 7T RGay a RX the composition Then it follows easily that the following sequence is right exact RGay i RX L M a 0 1 A right exact sequence where the rst two modules are free ie a se quence in the form of is called a presentation of M We have shown that every R module has a presentation If X can be chosen to be a nite set we already know this is equivalent to M being nitely generated If both X and Y can be chosen to be nite sets we say that M is nitely presented Note that any map of nitely generated free modules with xed bases say 9 R a R can be represented uniquely by an m gtlt n matrix of elements of R in the following sense Let e1 en be the xed basis of R and e3 e be the xed basis of R If g is such a map let n7 1 S 239 S m 1 S j S m be such that for each 6739 gej rig6 Then the m gtlt 71 matrix A Tlj determines the action of 9 Moreover if we represent elements of R and R by column vectors where in R 11antr 22 176 and similarly for R then the usual matrix multiplication by A is exactly the action of 9 since 71 Z 7101 a1 j1 n n m 7 7 H 7 A39 i i E E 71707 51quot E a E magi 7L 397 397 397 397 a 171 771 771 171 2 WW 71 n n a1 E aj951 9 E am 9 11 11 an Conversely given an mgtltn matrix A Tlj ofelements of R and free modules R R with xed bases 6739 and 6 respectively there is a unique R linear homomorphism g R a R such that the action of 9 corresponds to the multiplication of column vectors by A in terms of the xed bases In the case of free modules that may not be nitely generated the pre ceding discussion can be carried through with no change except that each column77 of the matrix can have only nitely many nonzero entries That is given a free module map 9 RGay a R9 where the free modules have xed bases cghey and 616X respectively we have 96y ZRX Wyem which must be a nite sum and so for each y only nitely many of the my can be nonzero As above matrix multiplication by A my from column vectors in RGay which themselves have only nitely many nonzero entries of course to column vectors in RX which also have the niteness condition on nonzero entries corresponds exactly to the action of g and such matrix multiplication only makes sense anyway if each column of the matrix has only nitely many nonzero entries Hence we can consider any R module to be the cokernel of the action of such a matrix In particular if M is nitely presented then it is the cokernel of multiplication by a nite matrix of elements of B This point of view is very useful Extension of scalars applied to presentations Let p R a S be a ring homomorphism let M be an R module We have a presentation as in 1 above which is of course right exact Moreover we know that tensor product commutes with direct sum and that S RR E S by a natural isomorphism Hence we have the following commutative diagram with exact rows because extension of scalars is right exact S RRY RRX1 S RMHO ELu Eiv 9 WI SY SX S R M a 0 So to represent S 83 M as a cokernel it suf ces to understand what the action of g is Fix bases 6y for RGay and 6 for R9 and lets say that the R matrix of representing 9 in terms of these bases is TMmexyey We have vl re m Mmem and ul 8 my rky for all z E X y E Y and r E R Thus for any y E Y we have 9621 1 9u 16y 1 91 621 11 9 1 2 71115 Z 111 71115 Z ltp7 Cl6m 16X 16X 16X Hence the matrix representing 9 is exactly rmy which means that S 83 M is the cokernel of this matrix Thus the representation of M as the cokernel of an R matrix leads di rectly to the representation of S R M as the cokernel of an S matrix simply by applying p to each of the entries meXy6Y7 Flat and surjective base change on tensor products Let p R a S be a ring homomorphism and let M N be S modules Then in general M 83 N has two dz erent S module structures Let aM N M 83 N with the S module structure inherited from M and 6M N M 83 N with the S module structure inherited from N In symbols the S action on aM N is given by 5m 71 5m 8 n and the S action on 6M N is given by 5m 71 m 8 To see that this gives two different structures in general let R be any ring letS RM where z is an indeterminate over R and let p R a BM be the 4 obvious ring inclusion Let M N S Then it is clear that z 8 1 31 1 8 z in S 83 S However in aS S we have x 1 1 z 8 1 and in 6S S we have x 1 1 1 8 p so the S rnodule structures are distinct Frorn another point of view for any two S rnodules M and N one can form the tensor products M SN and M RN In general there is a surjection UMN M RN a M S N since the de ning relations on the tensor product over R form a subset of the de ning relations on the tensor product over S In particular rmn wrmn and mrn mltprn It happens that UMN is an isomorphism iff aM N 6M N However for some ring hornornorphisrns p R a S we have aM N 6M N ie UMN is injective for all pairs M N of S rnodules One such example is where S RI and p is the canonical surjection Then for any S rnodule M and r E R we have for any 2 E M that r2 r2 Soforanysr Sr Rm MandnENwehaveinM RNthat sm nfm nrm nm rnm rnm sn which shows that aM N 6M N Another example is when S WAR and p lW is the localization map For any r6 R w E W m E M andn E N we have m wrn m rn m rn r 7 w iw im in w w w r m mi ni rni i w w w w w w Proposition 2 Let p R a S be a ririg homomorphism with the property that URQ is an isomorphism for all pairs P Q of S modules Theri for arty pair MN of R modules we have S R M R N g 5 R M s 58 N Iri other words terisor product commutes with such a base charige Proof S RUVIltXgtRNgtg S SS RM RN S R 5 R M R N Via 055 g S 83 M 83 S 83 N by associativity of tensor S R M s S R N V V13 US RMS RN D Math 614 Lecture notes Oct 3 2007 The main topic for today is that of local properties But before we go into it the student may at this point be under the impression that the notions of free7 projective7 and 7 at7 are equivalent for modules over a ring This is not true as the following examples show The Z rnodule Q is at but not projective It is at because it is a localization of Z On the other hand if it is projective it has to be the direct surnrnand of a free Z rnodule 7 ie of a free abelian group So suppose 7r there is a free abelian group F and a pair of group hornornorphisrns F i Q 739 such that 7r oj 1Q Then we have ll Z lieu i1 where 61e is part of a xed free basis of F and a E Z Let m 1 max lail Then mj1m j1 2171 16 Hence the freeness of the basis implies that m divides each 1 but then lail lt m implies that each a 0 That is j1 0 so that 7Tj1 7T0 0 31 1 and 7r oj 7 1Q As for the other supposed irnplication if R A gtlt B is a product of nonzero rings then the R rnodule A RB is projective but not free It is projective because it is clearly a direct surnrnand of B On the other hand since the nonzero R ideal B kills A it follows that A is not R free Local properties First we note the following exercise worthy facts given without proof about any at base change Proposition Let F be a flat R module and let LM be R submodules of N We may think ofF 83 L as a submodule ofF 83 N and similarly for the other submodules of N With this identi cation 1 IfLQMthenF LQF M 2 F LMF LF M 3 F L MF L F M 4 ff M a Q is an R linear homomorphism then Ker1F R f F 83 Kerf as submodules of M and lrn1p 83 f F 83 lrnf as submodules of Q 5 S 83 F is aflat S module for any R algebra S Proposition Some Local Properties a Let N Q M andu E M If E Np for all P E 9R thenu E N b Let N g M Ipr Mp for all P e 9a then N M c Let S be a sequence of R module homomorphisms If Sp is a complep resp eccact for all P E 9R then so is 8 Hence the property of being a kernel a cohernel or an image can be checked locally at mamimal ideals d Let L M be submodules of N If Lp Q Mp for all P E 9R then L Q M e An R module F is flat if Fp is flat ouer RP for all P E f For any multiplicative set W W 1NR NW 1Pt g R is reduced ltgt WAR is reduced for all multiplicative sets W ltgt RP is reduced VP 6 Proof a lfu N then there is some maximal ideal P with P 2 Ann NRM I lf 6 Np then for some it E RP we have ion 6 N and hence wN Ru Q N so that w E I Q P which is a contradiction So Np b If N 31 M apply a to any n N c By repeated application we need only consider sequences ofthe form SLLMN We xsuchanS First suppose each Sp is a complex for all P E Then for all such P 1m9 o fP 1m9 o fp 1mgp o fp 1m0 0 so that lmg o f 0 whence g o f 0 Now suppose each Sp is exact We know that S is a complex so let x E Ker 9 Then for each P E 9R s I E Kergp Kergp lmfp lmfp so that by part a z 6 lm f Hence 8 is exact d We have 0 7 LP 7 LP 7 L Nmep N Lp NmLP for each P E 9R so that L N L which means that L Q N e Let j L gt M be an R module injection and consider the exact sequence j 1 j 0gtKgtF RL gtF RM Localizing at any P E 9R we get OSKPHF R LPigtF R MP OHKP HFP RP LPamp FP RP MP Since j is injective so is jp since RP is at over R and since Fp is at over RP 1FP 13ij is also injective Hence KP 0 for all P E 9R from which it follows that K 0 Thus j is injective so F is at over R f If 5 E W 1NR then without loss of generality xi 0 Hence xw ztwt 0 so that i E NW 1R Conversely suppose that 5 E NW 1R Then for some positive integer t we have ztwt zw 0 Then for some 1 E W we have vxt 0 so that lay 0 as well Thus in E NR which means that sw vxvw E W 1NR g Suppose R is reduced Then for any multiplicative set W we have NW 1R W 1NR W40 0 If RP is reduced for all P E 9R then 0 NRp NRp for all such P and hence NR 0 D Now we present a global7 version of the last proposition from the Oct 1 lecture Theorem Let F be a nitely presented R module The following are equiv alent a F is projective b F is at 0 F is loeally free in the sense that for all p E Spec R FF is Pip free d The same as C but for p E Proof We already know that projective modules are at If F is at then FF is at hence free over RF for all prime ideals p by the proposition from Oct 1 Finally suppose FF is free for all p E Let 7T R a F be a surjection from a nite free module We have an exact sequence HomRFR 71gt HomRF F a O a 0 For any p E 9R the fact that localization commutes with Horn for nitely presented modules along with the atness of FF means that we get the following commutative diagram with exact rows HornRF R HornRF F a 0 a 0 l H HomR FF Pig HomR FF FF Q 0 HZ HZ Hence Cp 0 for all p E 9R so that C 0 and 71 is surjective Hence there is some j F a R such that 1F 7nj 7r oj so that F is a direct summand of R and thus projective D Chain conditions Noetherian and Artinian For any partially ordered set E the following conditions are equivalent 1 Every nonempty subset of E has a maximal element Math 614 Lecture notes Oct 10 2007 The following lemma is of independent interest It is called prime avoid ance77 because it implies that if an ideal avoids7 each of a nite set of primes then it avoids7 their union as well Proposition Prime Avoidance lemma Let R be a ring I an ideal and let Q1 Qn be a hite set ofideals all but possibly two of which are prime such that for all 1 Sj S n I g Q Then I g UL Qj Proof There is nothing to prove if n 01 so we may assume it 2 2 By induction on n I is not contained in a union of a strictly smaller number of the Qj By renumbering if necessary Q3Qn are prime Let aj E Q I Ui Qi By minimality of the cover such as exist Let b H at tZSla1 12 Qt where the empty product is 1 and a a1 a2 1102b Clearly a a11 12b a2 Q1 and a a1 a21 alb Q2 Also for t 2 3 if a1 a2 6 Q then the factors of b are not in Qt so that 1102b Qt since Q is prime and hence a Qt If on the other hand a1 a2 Qt then at 6 Q is a factor of b which means that b 6 Qt so that a Qt Thus a E I UL Q as required D Corollary Let R be Noetheiiah and M a hitely generated R module Then the map p M a 69 MF FEASSM 1 giueri by the direct product of the localization maps is irijectiue Thus M 0 ltgt Mp 0 for all p 6 Ass M IfZM lt oo theri p is art isomorphism Proof Let AssM p1 pn where p 7 p for i lt j lf 2 0 then for all p E AssM we have that 0 2 g p so by Prime Avoidance we have 0 2 g UAss M Since this union is the set of zerodivisors of M there is some nonzerodivisor w of M such that wz 0 Hence 2 0 If Mp 0 for all p 6 Ass M then p is the zero map but it is injective so M 0 Suppose M lt 00 Then RAnnM is Artinian so the support of M matches the assassinator of M a nite set of maximal ideals Take any 111 E lfm AssM then m SuppM so M Mpm 0 for any p E Spec R and hence ltpm 0 a 0 is the identity map in this case If m 6 Ass M then for each component Mp of the direct sum we have Mpm 0 if p 31 m and Mpm m if p 111 So again we have that pm Mm a Mm is the identity map Hence since it is locally an isomorphism at maximal ideals we have that p is an isomorphism D Corollary Ari Artiriiari ririg is a direct product of riitely mariy local Ar tiriiari local ririgs Proof By the above we have R FEASSRRF HFEASSRRF and each RF is an Artinian local ring since it has only one prime ideal 131 D Primary decomposition ln nitely generated modules over Noetherian rings there is an analogue of the theory of unique factorization We represent any submodule N Q M as an iritersectiori of primary77 submodules of M in a way which is in a certain sense unique Namely for each prime p E AssMN one gets a p primary component which if p is minimal in AssMN is uniquely determined by MN De nition Let N Q M be a submodule and p E Spec R We say N is a p primary submodule of M if Ass MN lf Ass M p we say M is a p coprimary module A submodule N of M is primary if it is p primary for some p E Spec R We say N is an irreducible submodule if whenever N N are submodules of M and N N N either N N or N N Any representation of a submodule N Q M as an intersection N N1 N of a nite number of submodules N of M is called a decomposition of N The decomposition is irredundant if for any 1 S i S r N 31 i Ni If each of the N is irreducible we call the decomposition irreducible If each N is primary we call it a primary decomposition In an irredundant primary decomposition if each N is pi primary and p 7 13 for all i 31 j we call the decomposition a minimal primary decomposition of N and we say N is the P primary component of the decomposition Note If N N are p primary submodules of M so is N N This is because one can think of MN N as a submodule of EB Via the short exact sequence M MM M 0 7 7 7 7 0 N N N N NN Hence any irredundant primary decomposition can be replaced with a min imal primary decomposition by grouping those terms which are p primary for each p Theorem Primary decomposition Let R be a Noetherian ring and M a nitely generated R module a Any submodule ofM has an irreducible decomposition b Any irreducible submodule ofM is primary Hence any submodule has a primary decomposition c IfN 1N with each N pi primary is an irredundant primary decomposition of a proper submodule N C M then AssMN p17 7pr39 d Suppose N 121 N is a minimal primary decomposition where each N is pi primary and W C R is a multiplicative set such that pl W Q for 1 g i g t and p m W 7 o fori gt t Then W lN 121 WAN is a minimal primary decomposition of W lN in W lM where each WAN is W lp 7 primary e If N C M is a proper submodule and p is a minimal element of AssMN then the p prirnary component of N in any shortest pri mary decomposition ofN is Z1Np where in M a Mp is the local ization map Proof a If not let S be the set of submodules of M having no irreducible decomposition Let N be a maximal element of S Then N is not ir reducible so N N1 N2 where N 31 N1 and N 31 N2 But N1 and N2 properly contain N so by maximality they each have an irre ducible decomposition and gluing these together gives an irreducible decomposition of N b Suppose N is not primary Then there are two distinct primes 131132 6 AssMN Then there are submodules Rpl E LiN Q MN And for any 2 6 LAN we have p N 2 which means that Ll Lg N so N is not irreducible A O V We may assume N 0 and that the decomposition is minimal M is isomorphic to a submodule of 6917 MNi so Ass M g UAssMN p1 p i1 On the other hand pl 6 Ass M To see this let B g Nj which is nonzero by minimality of the decomposition Observe that B N B N1 OtBJO m N1 Hence Q 31 AssB Q AssMN1 p1 so pl 6 AssB Q AssM Repeating the argument for each i p1 p Q M g MNl A CL V Since localization commutes with intersection we get the equality Moreover for i gt t we have Ass W 1MN 0 so W lM WAN and these terms are removable from the intersection Fori S t AssW 1MW 1N Ass W 1MN W lpi as claimed and all these prime ideals are distinct e Saypp1 lfi Nthenaz NQN1forsomea R pbutais 1 F a nonzerodivisor on MN1 so 2 6 N1 Conversely NF N1p by part d since for allj 2 2 we have 13 Q p D Math 614 Lecture notes Oct 12 2007 Some dimension theory De nition Let R be a ring A chain of prime ideals is a nite sequence of proper containments P0 P1 mgpn where each pl 6 Spec R The chain above has length n We say the chain is saturated if there no proper re nement can be given by inserting another prime ideal The Krull dimension dim R ofR is de ned to be the supremum of the lengths of all chains of prime ideals of R The height of a prime ideal p is de ned to be htp dim RF For an arbitrary proper ideal I of R7 we de ne htI infhtp l I Q p E Spec R The dimension of an ideal I is dimI dim RI If M is an R module7 we de ne dim M dim R Ann M We say that R is edtendry if any containment p Q q of prime ideals has a saturated re nement7 and every such re nement of the given pair has the same length There are several things we can say immediately from the de nitions a If R is an integral domain or more generally if R has a unique minimal prime p7 then dimR is the supremum of the chains of primes where the smallest prime is p b If R7111 is local7 then dimR is the supremum of the chains where the largest prime is m 1Equivalently7 htp is the supremum of lengths of chains of prime ideals where the biggest prime in the chain is p C A CL V For any ideal I we have the inequality htI dim RI S dimR To see this rst assume I p is prime and let h htp and d dim Rp If h oo resp d 00 then the chains of primes contained in p resp containing p are unbounded and hence the chains of primes in R are unbounded If both numbers are nite then there exist chains of primes PoniCquot39CPhP corresponding to a chain in RF and PCIOCCI1Cquot39CCId corresponding to a chain in Rp so that gluing these chains together we get a chain in R of length d h Hence the supremum of all chains in R is 2 d h In the general case let p be a minimal prime over I such that dim Rp dim RI Then dimR 2 htp dimRp htp dimRI 2 htIdimRI Similarly if I Q q where q is a prime ideal then htq 2 htI htqI This follows from the other inequality by localizing at q Beyond these things there is not much we can say about height and dimension for ideals and modules over an arbitrary ring However if R is Noetherian we can characterize height in terms of number of generators up to radical Dimension in Noetherian rings For this section we make the standing assumption that R is Noetherian Theorem Krullls Principal Ideal Theorem PIT For any x E R and any prime ideal p which is minimal over a htp S 1 Proof First we prove it in the case where Rp is a local ring We may assume without loss of generality that htp gt 0 so that there is some prime q with q p Since p is minimal over it follows that z q Moreover 13 is minimal over 0 in the ring Rx and hence is Artinian Thus the following descending chain of ideals cannot be strictly descending qltz2qlt2gtltz2q gtltz2m Hence there is some n with q elm Now take any f 6 cl Then f gzr for some 9 E elm and r E B So we have xr fig 6 cl so that in Rq we have 6 anq but z q means that is a unit in Rq so that 6 anq and r 6 cf Thus f E qu elm We have shown that q zqm elm so that by the Nakayama lemma we have that q elm Now apply the Nakayama lemma in Rq to conclude that q 0 Hence q is minimal over the zero ideal which implies that htq 0 Since q p was arbitrary htp 1 In the general case pRp is minimal over xRp so htp ht pRp S 1 D Essentially as a corollary we get the much more powerful looking next theorem Theorem Krullls height theorem Let R be a Noetheriari ririg a u Q q are two prime ideals iri B there is a saturated ehairi of prime ideals starting with p arid eridirig with q b If 1n E R arid p E SpecR is miriimal over 1n theri htp S n Heriee htp lt oo e If E SpecR with htp n theri there ecoist elemerits 1zn E p such that p is miriimal over 1 zn Proof a If p q the single element comprises a saturated chain Oth erwise do the following inductive process Say we have a saturated chain PoniCquot39CPi of primes with p Q q Then pick z E q pi and choose a prime pi such that in 1313 marlp is minimal over pi Then htp1p 1 so the enlarged chain PoniCquot39CPiCPi1 3 Math 615 Lecture of January 5 2007 This course will deal with several topics in the theory of commutative Noetherian rings including the following 1 The theory of Grobner bases and applications a lot more about this momentarily 2 The structure theory of complete local rings One strategy in studying problems over Noetherian rings is to reduce first to the local case and then to the complete local case The structure theory of complete local rings can then be applied There are even deep theorems that permit one to pass from the case of a complete local ring to a finitely generated algebra over a field or complete discrete valuation ring Other techniques can be used to pass from a problem in such an algebra over a field of characteristic 0 to a corresponding problem over a field even a finite field of positive prime characteristic p 3 What can one do when a ring is not Cohen Macaulay In particular we will discuss the theory of Cohen Macaulay rings but will focus on techniques that show that all local rings are in some sense close to being Cohen Macaulay Although we shall discuss the subject in much greater detail later we give a brief discussion of Cohen Macaulay rings here so that we can explain the sort of theorem we want to prove Recall the a ring R is quasilocal if it has a unique maximal ideal m in this case we usually denote the residue class field Rm by K and refer to the quasilocal ring R m We reserve the term local ring for a Noetherian quasilocal ring Let R m K be a local ring of Krull dimension d This implies that there exist d elements 1 aod E m such that if I 1 aodR then Rad I m One cannot use fewer than d elements by the Krull height theorem Such a sequence of elements 1 aod is called a system of parameters A d tuple r1 rd is called a relation on 1 d d E 11 The relations are easily seen to be an R submodule of the free R module Rd There are some obvious relations the element 0 0j 0 i i 0 where obj occurs in the lth spot and 715 occurs in the j th spot is a relation The elements in the R span of these relations are referred to as trz39oz39al relations A local ring is called Cohen Macaulay if there is a system of parameters such that every relation on the parameters is trivial It then follows by a theorem that this is true for 1 every system of parameters By a theorem this property passes to localizations One then defines an arbitrary Noetherian ring to be Cohen Macaulay if all of its localizations at maximal ideals equivalently at prime ideals are Cohen Macaulay In certain graded cases one can give an alternative characterization as follows Let K be a field and R an N graded algebra ie R has a direct sum decomposition R 6920 Rn with l E R0 satisfying RmRn Q Rmn for all mm 6 N such that R is finitely generated over R0 K In this case it turns out that one can always choose forms F1 Fd of positive degree in R by raising the to various powers one can even arrange that they all have the same degree such that F1 Fd are algebraically independent over K and R is module finite over A KF1 Fd Of course A is isomorphic with a polynomial ring in d variables over K In this situation R is Cohen Macaulay if and only if R is free as an A module In higher dimension it is rare for modules over polynomial rings to be free In fact relatively few rings are Cohen Macaulay ln equal characteristic 0 one can start taking module finite extensions of a polynomial ring if the dimension is 3 or higher all suf ciently large such extensions fail to be Cohen Macaulay Examples Let S Kz y be the polynomial ring in two variables over the field K Let R Kz2xyy2 Q S One may take A Kic2y2 Q R Then R is free over A on the basis 1 avg and so is Cohen Macaulay On the other hand let R1 Kz2z3zyy Q S and let A1 Kz2y Q R1 Then R1 is module finite over A1 with minimal generators 1 3162 but is not free over A1 One has that yz3 7 x2zy 0 This relation on minimal generators shows that R1 is not A1 free and therefore not Cohen Macaulay Alternatively in the local ring of R1 at its homogeneous maximal ideal 2 y is a system of parameters and icy 7x3 is a non trivial relation on 2 y However many of the rings that arise in natural geometric situations such as complete intersections and rings defined by the vanishing of minors of a matrix of indeterminates are Cohen Macaulay Many problems become easier in Cohen Macaulay rings One of the results we are aiming to prove stated in a very special case helps to remedy the situation when the ring is not Cohen Macaulay Theorem Let R be a complete local domain ofprime characteristic p gt 0 Let 1 zd be a system of parameters for R and let r1 rd be a relation on 1 xd Then there is a complete local module nite ezrtension domain S of R such that the relation r1 rd becomes trivial over S This result has been known for well over a decade cf Hochster and C Huneke In nite integral extensions and big Cohen Macaulay algebras Annals of Math 135 1992 53789 Recently there have been improvements one can make all relations on all systems of parameters become trivial after just one module finite extension but new relations may 3 be introduced Beyond that very recently global versions of this theorem have been proved We shall discuss the situation in detail later in the course The Theorem above turns out to be false when R contains a field of characteristic 0 Nonetheless one can use the characteristic 1 results to prove important theorems in equal characteristic 0 We next want to begin our systematic treatment of the theory of Grobner bases Before doing so we shall review some facts about closed algebraic sets in K over an algebraically closed field K Review of the behavior of closed algebraic sets over an algebraically closed eld This section is meant as an overview of some basic results on closed algebraic sets over an algebraically closed field We give definitions and statements of some theorems but most proofs are omitted For a detailed treatment of this material the reader may consult R Hartshorne Algebraic Geometry Springer Verlag Graduate Texts in Mathematics 52 New York 0 Berlin 0 Heidelberg 1977 Chapter I There is also a complete discussion in the Lecture Notes from Math 614 Fall 2003 see particularly the Lectures of October 3 15 17 and 20 Let K be an algebraically closed field and let R Kac1 xn be a polynomial ring lf W Q R is any set VW v E K u 0 for all f E It is easy to see that if I is the ideal generated by W VI Moreover if f E Rad I ie fl 6 I for some integer k 2 1 then f also must vanish on VI and so VRad VW as well If X VI for some ideal I we say the X is a closed algebraic set in K or a Zarz39skz closed set in K In fact we have 1 K V0 and Q VR are closed algebraic sets 2 VI J VI U VJ for any two ideals I and J 3 VZ6A I xeA VOA for any family of ideals Abey The conditions above show that the closed algebraic sets are in fact the closed sets of a topology on K this is called the Zarz39skz topology Suppose that we are given an arbitrary set of points 73 Q K and we want to understand the Zariski closure f of 73 Since this will be the smallest closed set containinng 73 we want to find I as large as possible such that VI Q 73 But any element of I must vanish on VI which we want to contain 73 Therefore the largest ideal we can use is the ideal of all functions in Kac1 ow that vanish on 73 and this ideal defines f Note that if n l the closed sets in K are the finite sets and K itself In K2 one gets finite unions of points andor curves defined by one equation and K2 itself


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