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# SPECIAL TOPICS MATH 601

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Dimension Theory Mathematics 601 In this note we prove some of the standard results of commutative ring theory that lead up to proofs of the main theorem of dimension theory and of the Nullstellensatz An alternative approach to dimension theory is given in Chapter 11 of Atiyah and Macdonald AM who base their approach on the analysis of the Hilbert function of a local ring We need a couple of facts about localization The proofs are fairly straightforward and they can be found in Chapter 5 of Lemma 1 Let A Q B be rings with B integral over A N Ifb is an ideal ofB with a b N A then Bb is integral over Aa 16 US is a maltiplieatively closed subset of A then S lB is integral over S lA 93 If q E specB and p q A then q is a mapimal ideal of B if and only ifp is a mapimal ideal ofA Theorem 2 Incomparability Let A Q B be rings with B integral over A Ifq Q q are prime ideals ofB with q W A q A then q q Proof If p E specA we write Bp for the localization S lB where S Ap Let p q A Then Bp is integral over Ap by Lemma 1 Moreover qu Q q Bp are both prime ideals of B and each intersect with Ap in pAp Therefore by Lemma 1 qu is a maximal ideal so qu q Bp lntersecting down to B gives q q E Theorem 3 Let A Q B be rings with B integral over A Z Lying over Let p E specA Then there is a q E specB with q A p 2 Going up Let p1 Q p2 Q Q pn be prime ideals ofA and let ql Q qg Q Q qm be primes ofB with m lt n and qZ A p for each i Then there are primes qm qn ofB with qm Q qm Q Q qn andqi Ap for eachi Proof The extension EpAp is integral by Lemma 1 Let m be a maximal ideal of B Then m Ap is maximal in Ap by Lemma 1 so m Ap pAp However pAp A p Consequently ifqm Bthenq Ap For 2 by induction we can reduce to the case in 1 and n 2 Then we have a chain p1 Q p2 and a prime ideal all with all A p1 Since Bq1 is integral over Ap1 by Lemma 1 by 1 there is a Q E specBq1 with qgql Ap1 pgp1lfq2 x E B z ql E Q then qg E specB with ql Q qg and q2 A pg 1 To prove the following lemma we use ideas from in nite Galois theory We can avoid this by a Zorn7s lemma argument or by using ideas of integrality over ideals of a subring see Chapter 5 of Corollary 4 Let A Q B be ririgs with B iritegral over A Let q E specB with p q A Z htq S htp 2 If htp n theri there is a q E specB with htq n arid 01 A p 3 dimB dimA Proof For 1 suppose that htq n and let qo Q Q qn q be a chain of primes in B If p q A then there is a chain p0 Q Q pn p Moreover if the inclusions in the chain for q are all proper then the same holds for the chain for p by incomparability Therefore htp 2 n D To prove 2 suppose that htp n and suppose that p0 Q Q pn p is a chain of length 71 By the going up theorem there is a chain qo Q Q qn with q A p for each i Let q qn Then htq 2 n and q A p However by 1 we know that htq S n so htq 71 Finally for 3 we know that dimB suphtq q E specB S suphtq A q E specA S dimA by However dimB 2 dimA by 2 so the two inequalities together proves that dimB dimA We point out that our proofs appear to assume that all heights in question are nite By being careful with our arguments we see that our arguments will work even when heights are in nite Lemma 5 Prime avoidance lemma Let A be a ririg arid let p1 pn be prime ideals ofA arid let a be ari ideal ofA such that a Q Up Theri a Q p for some i Proof We prove the contrapositive by induction on n The case n 1 is clear Suppose that n gt 1 and that a Q U2 pi Then a Q Ui p for each j By induction for each j there is an d E a such that Q p for each i 31 j If for some j we have dj p then we are done If not then dj E p for all j Consider the element r L y E 12 i71zi1 n i1 We have y E a and y p for each i Hence a Q U2 pi 1 Lemma 6 Let KF be a normal eld extension let A be an integrally closed domain with quotient eld F and let B the integral closure ofA in K If q1 q2 E specB with ql A qg A then there is a o E GalKF with oq1 qg Proof If x E K with z integral over A then 0x is also integral over A for any 0 E GalKF so 0B B It is then easy to see that if q E specB then oq E specB for any 0 Also B F A since A is integrally closed in F Let L be the xed eld of G GalKF Then KL is Galois and LF is purely inseparable lf L 31 F then charF p gt 0 Let B B L the integral closure of A in L If p q A then there is a unique p E specB with p A p namely p B ldpn Epfor some Since q A p this uniqueness gives q B p Therefore we may assume that L F and so KF is Galois First suppose that K F lt 00 Then lGl lt 00 If x E qg then NKF95 H 095 6 Q2 G F P7 760 so NKF E ql hence o E ql for some 0 since ql is prime Thus z E o 1q1 Therefore qg Q Uaeaoml By the prime avoidance lemma qg Q oq1 for some 0 But since Clz PCI1 AUCI1 AUCl1 A7 incomparability gives qg oq1 We now consider the case where K F is in nite lf L is a Galois subextension of K with L F lt 00 set GLo Gloq1 L qg L The argument above shows that GL 31 Note that if N GalKL then 639 Nl L F lt 00 and o 6 GL implies that oN Q GL lf 7391 Tm are coset representatives of N in G with 739 6 GL then GL TiN Thus GL is closed in the Krull topology on G Also note if L Q L then GL Q GL lf L1L are nite Galois subextensions of KF 3 then the composite M of the L is also a nite Galois extension of F Then by previous remarks we get g 31 GM Q GL2 Therefore the collection of closed sets GL has the nite intersection property Since G is compact in the Krull topology see Theorem 21 of 11 in MC or Theorem 176 of Mo we see GL 31 Q where the intersection is taken over all sub elds L with L F lt 00 lf 0 is in this intersection then oq1 qg since K is the union of all these sub elds D Theorem 7 Going down Let A Q B be domains such that B is integral over A and A is integrally closed pr1 Q p2 Q Q pn are prime ideals ofA and if qm Q qm Q Q qn are prime ideals in B with qZ A p for each i then there are primes ql Q Q qm1 Q qm such that qi A p for each i Proof By induction we can reduce to the case in 1 and n 2 We then have a chain p1 Q p2 in specA and a qg E specB with qg A p2 Let F be the quotient eld of A and let L be the quotient eld of B Also let K be the normal closure of LF and let G be the integral closure of B in K By going up there are t1 Q t2 in specC with t A p for i 12 Also there is an t E specC with t B qg Then t A p2 t2 A so there is a o E GalKF with 0t2 t by the previous lemma Set ql 0t1 B Then ql Q qg 0t2 B and ql Aot1 At1 Ap1 D We now have the tools to approach dimension theory In order to prove the main theorem we need a lemma the Noether normalization lemma and Zariski7s theorem Lemma 8 Let A be a hialgebra and a domain and let p be a nonzero prime ideal of A Let t1 tn 6 A and let c E p with c 31 0 Suppose that the t7 are algebraically independent oyer h in Ap Then t1 tn c are algebraically independent oyer h in A Proof Suppose this is false Then there is a nonzero polynomial f E kp1zny with ft1tnc 0 Choose f with the degree in y minimal Write f 20 giyi with g E klz1zn and yd 31 0 If d 0 then f go 6 kp1pn is nonzero with ft1 tmc got1 tn 0 so gE 0 in Ap This is false since the are algebraically independent in Ap Therefore d gt 0 Now in Ap 0ft177tn70fE77 7590E77m as c E p Thus go 0 So f y Z3gi1yi Since c 31 0 and A is a domain fy evalu ated at t1 tn c is 0 contradicting minimality of d Thus t1 tn c are algebraically independent over k D If A is a hialgebra and a domain with quotient eld F let trdegkA trdegk F the transcendence degree of Corollary 9 Let A be an kealgebra and a domain and let F be the quotient eld of A Let p E specA Then trdegkA 2 trdegkAp htp Proof Take t1t E A such that E are algebraically independent in Ap over k Say p po 3 pl 3 3 pm 0 is a chain in specA Thus trdegkAp 2 n and htp 2 m Choose 01em with e E pi1 7 pi Assume that t1 p tn pel p eip E Ap are algebraically independent over k Then as pipi1 is prime in AleL and Ap1pp1 Api by the lemma t1 pi1e1 p are algebraically independent over k because 0 7g CH1 pi E piplH Thus by induction t1 pm em pm are algebraically independent over k But as pm 0 this means t1 em are algebraically independent over k Thus trdegk A 2 n m Hence trdegk A 2 trdegkAp htp D Recall that a k algebra A is an af ne k algebra if A can be generated by a nite number of elements as a ring extension of k That is A is af ne if A ka1 an for some a E A If A is an af ne k algebra and is an integral domain we say that A is an af ne k domain Theorem 10 Noether Normalization Let A be an a ne kedomaln Then there are t1 td E A algebraically independent over k with A integral over kt1 td Proof We prove this in the case that 00 The proof for nite elds is slightly di erent and we indicate the changes required for k nite after the proof Say A ka1 an We use induction on n If n 1 then either a1 is transcendental over k in which case we take t1 a1 and d 1 or al is algebraic over k in which case we take d 0 since A is then integral over k So assume that n gt 1 If a1 an are algebraically independent over k we can take t a for each 239 Thus suppose that they are not independent over k Then there is a nonzero polynomial f E kz1 zn with fa1 an 0 Say f 220 f where f is homogeneous of total degree 239 and fl 31 0 Let y z 7 em for t lt n where the e E k are to be determined Then f fy1 01x xn f y1 yn1xn Each f then has degree 239 in an Expanding f we see that the coef cient of xi is flel en1 1 Thus we can write 171 f fl517 77171ll 291917 7yn71lr 10 Choose the C so flel en1 1 31 0 possible as 00 and fl is a nonzero homogeneous polynomial Let b a 7 Gian for t lt n Then under the map b kz1 zn a A with a we have bi so 171 0 fa1 an flel en1ai Zgb1 bn1a i0 Dividing by flele1 shows that an is integral over the subring kb1b1 of A By induction there are t1td E kb1bn1 algebraically independent over k 5 with hb1bn integral over ht1td Thus A hb1 bn1an is integral over kt1td D For the case that h is nite we can use the same argument except we set y x 7 zfj for suitably chosen large integers ri Theorem 11 Zariski Let h be a eld and let A be an a ne k algebra that is a eld Then A is an algebraic extension of k Proof Suppose that A is an af ne h algebra that is a eld By Noether normalization there are t1 t E A such that t1 t are algebraically independent over h and A is integral over ht1 t The zero ideal of A is maximal since A is a eld Thus by Lemma 1 the zero ideal of ht1 t is maximal This forces ht1 t to be a eld However by the algebraic independence of the tkt1 t is isomorphic to a polynomial ring in r variables The only way it can be a eld is if r 0 Thus A is integral over k and so the eld A is algebraic over k D Theorem 12 Dimension Theorem Let A ha1 an be an a ne k7algebra and a domain Then 1 dimA trdegkA S n 2 Up 6 specA then trdegkA htp trdegkAp and so dimA htp dimAp 3 Every mawimal chain of primes in A has length dimA Proof Let F be the quotient eld of A Then F ha1 an so a1 an contains a transcendence basis of Thus trdeg F S ii If p E specA then htp trdeg Ap S trdeng S n by the corollary Say trdeng d Then htp S d for any prime p so dimA S d Let us refer to the rst part of 2 by 27 We rst prove 27 and do this by induction on d If d 0 then the inequalities above show that all terms in 27 are 0 so 27 holds Thus assume that d gt 0 and that 27 holds for d 7 1 By Noether Normalization A is integral over a subring A0 ht1td where the t are algebraically independent over k This is the same d as above since F is algebraic over the quotient eld of A0 Let B kt1td1 Q A0 Then A0 2 Since trdeng d 7 1 by induction 27 holds for B Let q E specB and p q B Then q 2 pBM a prime of BM as Blz pB g Bp We consider two cases Case 1 q pBlz Then htq 2 htp since given a chain p0 C p1 C C p p we have pOBM C C pTBM q Now BMq E Bpx so by the corollary and induction d trdegk BM 2 htq trdegkBq Z htp trdegkBp 1 trdeng1d so trdeg BM htq trdegkBxq Case 2 q 3 pBlzl Then htq gt htpBz 2 htp by the argument above Thus htq 2 1 htpBz So by case 1 htq 21 htpBz 21 htp We trdegkBllCI 2 htp 1 trdegkBllq 2 htltPgt1trdegkBP as Bp Q BMq So htq trdegkBq Z htp trdegkBp 1 trdegk B 1 d by induction By the corollary d trdegk BM 2 htCl trdegkBllQ7 so trdeg BM htq trdegkBxq Thus we have 27 for A0 Let 3 E specA Since AAO is an integral extension of integral domains and A0 klzl xd1 is integrally closed the going down theorem shows that htq3 htq3 A0 Also trdegk A B trdegkA0 J3 A0 since A B is integral over AO B AO so the quotient eld of A B is algebraic over the quotient eld of AO B A0 Thus htq3 trdegkA J3 htq3 A0 trdegkA0 J3 A0 trdegk A0 d trdegkA by what has been proven Therefore we have proved 27 To prove 1 note that we have seen that dimA S d trdegk A Let M be a maximal ideal of A By Zariski7s theorem AM is algebraic over k so 2 gives htM d But htM S dimA so we get dimA trdegkA htM for each M This proves As Ap is also an af ne kidomain for p E specA we see dimAp trdegkAp by 1 so dimA htp dimAp This nishes For 3 let p0 C C p be a maximal chain Therefore p is a maximal ideal of A We have 7 S d dimA We prove r d by induction on d If d 0 this is trivial Assume that d gt 0 Then A is not a eld since 0 is not maximal so 7 2 1 Thus p1 31 Now 0 131131 3 D pTpl is a maximal chain of length r 7 1 in Apl Since the original chain is maximal htp1 1 as there is no p with 0 C p C p1 So dimAp1 dimA 7 htp1 d 71 Induction gives r 71 d71 so r d This nishes the proof of the dimension theorem D We can use some of the results we have done to prove the Nullstellensatz We start off with a lemma which is the heart of the Nullstellensatz Lemma 13 Let k be an algebraically closed eld arid let I be a proper ideal ofkx1 xn Theri the zero set ZI is rioriempty Moreover ariy mamimal ideal ofkx1 an is of the form 1 7 a1zn 7 an for some a E k Proof Let M be a maximal ideal of klz1xn We write A kp1pn for conve nience Then AM is a eld and the natural inclusion k 7 A induces a eld injection k 7 AM We view AM as a eld extension of k However AM is an af ne k algebra since AM is generated by z1M xnM Thus by Zariski7s theorem AM is algebraic over k Since k is algebraically closed this forces AM k From this we can characterize M Let p A 7 k be the ring homomorphism that sends A to AM followed by an isomor phism AM 7 k Let a Notice that p is the identity on k Then Mp 7ai 0 so x 7 a E kerltp M Consequently M contains the ideal 1 7 a zn 7 an However this is a maximal ideal so M 1 7 a1 zn 7 an D We can now prove the rst statement ofthe lemma Let I be a proper ideal of klzl xn and let M be a maximal ideal that contains I Then M 1 7 ai pn 7 an for some a E k by what we have proved If P a1an then P E ZM Q ZI so ZI is nonempty Theorem 14 Nullstellensatz Let J be an ideal ofkx1 zn Theri IZJ Proof Let J be an ideal ofklzl zn The inclusion xJ Q IZJ is clear Also since the result is trivial if J klzl zn we assume that J is a proper ideal Let f E IZJ and let 2 be a new variable Consider the ideal J generated by J and 17zf in klp1 xmz An arbitrary element of this ideal is of the form 917zf 2 big where g g E klp1 pmz and each h E J Suppose that Q E ZJ lf Q a1an1 and P a1 an then hQ hP 0 for all h E J so P E ZJ But then 0 17zfQ 17an11fP 1 since f E IZJ This is impossible so ZJ is empty By the previous lemma J klz1xnz Thus we have an equation 917zfz hi9 1 where h E J and 99 6 klzl zmz By writing each 9 as a polynomial in z with coef cients in klp1 pn and by doing the same for higi we see that hi9 hgzi for some h E J By setting 2 fwe get an equation hz1zn1fi 1 in kp1zn1f Multiplying by f gives f Zhx1 zfmi i1 6 J since each h E J Therefore f E j as desired D References AM M Atiyah and I MacDonald Introduction to Commutative Algebra Addison Wesley Reading MA 1969 MC P McCarthy Algebraic Emenslzms 0f Fields Chelsea Publishing Co New York New York 1966 Mo P Morandi Field and Galois Theory Springer Verlag New York New York to ap pear Lecture Notes for Mathematics 601 Error Correcting Codes and Algebraic Curves Patrick J Morandi Fall 2001 Contents 1 Introduction to Coding Theory 11 Introduction to the Course 12 De nition of Error Correcting Codes 13 Parameters of a Code 14 Linear Codes 15 Error Correction 16 Bounds on Codes 17 Cyclic Codes 2 Introduction to Algebraic Geometry 21 Af ne Curves 22 Projective Varieties 23 The Function Field of a Curve 24 Nonsingular Curves 25 Curves over non Algebraically Closed Fields 3 Algebraic Function Fields and Discrete Valuation Rings 31 Discrete Valuation Rings 32 Discrete Valuation Rings of 33 Discrete Valuation Rings of Fk 4 Divisors and the RiemannRoch Theorem 41 Divisors of a Function Field 42 The Riemann Roch Theorem 43 Fields of Genus 0 and 1 5 Goppa Codes 51 Goppa Codes coming from qux1Fq 6 Examples of Function Fields 61 The Connection Between Points and Places 62 Connections Between Divisors 63 Elliptic Curves 64 Hermitian Curves 7 The Hasse Weil Theorem 71 The Riemann Zeta Function 72 Riemann Zeta Functions of Number Fields 73 Riemann Zeta Functions of Curves 35 35 43 45 48 50 1 Introduction to Coding Theory 11 Introduction to the Course This course will discuss error correcting codes and connections with algebraic geometry When coding theory began in the 1940s7 the main mathematical technique was linear algebra Later7 ring theory was used7 notably the theory of polynomial rings and quotient rings In the 1970s7 Goppa discovered a method for producing codes from algebraic curves7 and his class of curves gave some nice theoretical results in the theory of error correcting codes After spending some time on the basics of coding theory7 we will develop the theory of algebraic curves and algebraic function elds in order to de ne Goppa codes and prove some results about their error correction capability This will require spending some time on the basics of algebraic geometry7 allowing non algebraically closed base elds However7 since we will concentrate on curves7 we will be able to bypass much of the dif cult machinery of algebraic geometry In fact7 it is possible to work purely eld theoretically7 as does the book of Stichtenoth However7 I believe that this is too narrow a viewpoint7 and that thinking geometrically gives better insight One aspect where Goppa codes are bene cial is in nding asymptotic bounds on codes Without describing what this means7 it is necessary to be able to produce long codes One of the most important class of codes7 BCH codes7 can produce long codes but at the expense of requiring the use of increasingly large nite elds Goppa codes7 which can be viewed as a generalization of BCH codes7 get around this problem Finding long Goppa codes reduces the problem of nding algebraic curves with many rational points We will see that7 for any algebraic curve over a nite eld7 there is an analogue of the Riemann Zeta function There is also such an analogue for every algebraic number eld The Riemann hypothesis7 which states that the Riemann Zeta7s nontrivial zeros lie on the line Res i and which remains unproven7 was proved by Andre Weil for the Zeta functions associated to curves over a nite eld We shall see how this fact leads to a bound on the number of rational points of a curve7 and what implications this has for coding theory There is a website for this course The URL is mathnmsuedu pmorandimath601 12 De nition of Error Correcting Codes Let F be a nite eld and let F be the collection of all n tuples over F This is an n dimensional F vector space A code over F is a nonempty subset of F It does not have to be a subspace although all of our examples will be subspaces We will sometimes write elements of F as strings of elements without using commas or parentheses The elements of a code are called codewords7 and the elements of F are called words Before we give some notation7 we will follow the normal practice of writing E for the unique7 up to isomorphism7 eld with q elements Example 11 Let F F2 Then 07 1 is a code in F and 0011 is a code in F2 Likewise7 000111 is a code in F3 and 00000711111 is a code in F5 Example 12 Let 0 0 0 1 1 1 1 H 0 1 1 0 0 1 1 1 0 1 0 1 0 1 Then the kernel of H is a code in F7 This is called the Hamming code7 and was the rst real example of an error correcting code We will investigate this code further in a little while The important property of a code is its ability to correct errors Let us discuss this idea starting with an example In 1979 the Mariner 9 spacecraft took black and white pictures of Mars These pictures were created by using 64 shades of grey Each pixel of a photograph was assigned a shade of grey Each picture consisted of a 600 by 600 grid of pixels Thus7 to transmit one photograph7 the spacecraft had to send 360000 pieces of data7 each piece representing the color of a pixel Suppose that the shades of grey were represented by a number from 1 to 64 in binary Thus7 we could represent any color with a string of six binary digits If7 in transmission7 an error was made7 then NASA would incorrectly color the corresponding pixel Since electromagnetic activity can easily cause such errors7 this would be a problem NASA wanted an encoding system that would take received data7 perform some sort of test7 and determine if the received data was the same as that send7 and7 if not7 determine what was the actual transmitted data What they did was to encode each color as a string of 32 binary digits Of the 232 possible strings7 64 of them were valid codewords Suppose that a codeword is transmitted7 but errors are made One can recognize that an error is made if one sees that the received word is not a codeword The main principle of error detection7 Maximum Likelihood Detection MLD assumes that few errors are more likely than many errors Therefore7 the codeword that differs from the received word in the fewest number of components is the most likely transmitted codeword All error correcting schemes use this principle For example7 with the code 000111 if 101 is received7 then MLD would assume that 111 was transmitted However7 with 0011 if 01 or 10 was transmitted7 then MLD would not distinguish between 00 and 11 Even worse7 with the code 01 if a codeword is transmitted but an error is made7 the error cannot be detected because all elements of F are codewords To make more precise the notion of closeness of words7 we de ne a metric on F The function d F gtlt F a Z de ned by duv is equal to the number of components in which u and 1 vary In other words7 if is the t th component of a vector x then duv ui The function d is in fact a metric To see this7 note that the property duv 2 0 and duv 0 if and only if u 1 is clear Similarly7 duv d1u is clear The triangle inequality is not hard7 but is not completely obvious We need to show that if um w E F 7 4 then duw lt du2 d2w Then duw u 7 We note that 2 u 7 20 is a subset of 2 u 7 21 U 2 2 7 20 since if u 2 and 21 20 then u 20 Thus duw u S u U 2 2 7 S u 2 du2 d2w This metric will play a quiet but important role in coding theory 13 Parameters of a Code There are some important parameters attached to a code C over F Fq The rst called the length of the code is the value of n for which 0 is a subset of F The next which we will label k is de ned as k 10ng If C is a subspace of F then k dimFC and so k is the dimension of the code To de ne the nal parameter we need some preliminary concepts The third invariant is labeled d and is the distance of the code C It is de ned as d mindu2 u 7amp2 E C It should not be a problem that we are using d both for the metric and for the distance of a code These three parameters are then positive integers The parameters in order n k d of the codes of the rst example are 212 312 and 515 respectively With a little calculation we see that the parameters of the Hamming code are 743 Occasionally we will refer to a code with parameters 72 k and d as an n k d code 14 Linear Codes A code that is a subspace of F is said to be a linear code All of the codes we will consider in this course will be linear codes We will view the elements of F as row matrices There are some useful matrices attached to a linear code C Q F The rst is called a generator matrice It is a matrix G whose rows form a basis for C This is an k gtlt 72 matrix and its row space is equal to C The code C is then given in terms of G by C2G2 Fk The second matrix is called a parity check matrice This is a matrix H of full rank for which 0 is the right nullspace of HT That is u E C if and only if uHT 0 Alternatively u E C if HuT 0 Thus by rethinking about codewords as column matrices G is the nullspace of H It is then an n 7 k gtlt 71 matrix Moreover GHT 0 since xGHT 0 for all z E Fk To see the symmetry of these matrices we rst give some notation We will use for the usual dot product77 on F that is u 1 mm This is no longer an inner product since u u 0 can occur with u 31 0 for instance if F F2 and u has an even number of components equal to 1 In terms of matrix multiplication u 11 UZIT Mimicking what is done for inner product spaces we de ne the dual code CL of a code C by Giu Fuv0forallv G We claim that H is a generator matrix for GL and G is a parity check matrix for Gi To prove this we rst note the following matrix properties if A is an r gtlt 5 matrix then if Ax 0 for all z 6 F5 then A 0 and ii if yA 0 for all y 6 FT then A 0 These are both straightforward to prove As for the claim rst note that the row space of H is contained in Of For if z E Fn k then zH 11 HUT zvHTT z 0 0 for all 1 E G This implies that dimGf 2 n 7 k n 7 dimG However if u 6 Cf then 0 u xG uxGT uGTxT for all z E Fk Therefore uGT 0 and so GL is contained in the right nullspace of G Because G has rank k and is a k gtlt 71 matrix its nullspace has dimension n 7 k Thus dimGi S n 7 k and so both inequalities yield dimGi n 7 k Furthermore this equality shows that the row space of H is GL and that GL is the right nullspace of G This nishes the proof of the claim Example 13 Let G be the Hamming code Then G has parity check matrix H as de ned earlier A simple calculation shows that 1110000010101010011001101001 form a basis for G Thus we may take 1 0 0 1 1 1 1 1 OHOO 0 0 0 1 COOH OOHO 1 1 0 1 as a generator matrix for G The code GL is then the nullspace of G which has basis 011110011010011011010 Thus GL 0000000011110011010011011010101010111001100001111 Having a linear code allows us to give an alternative description of the distance of a code First of all we de ne the weight of a word to be wu du 0 the number of nonzero components of u Then since du1 wu 7 1 we see that d minwx z E Gx 31 0 By listing out the elements of the Hamming code it is easy to see that the smallest weight 6 of a codeword is 3 For the dual code to the Hamming code the listing of its elements shows that CL has distance 4 Example 14 The extended Golay code discovered by Golay the codiscoverer of the Ham ming code has generator matrix 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 0 0 0 1 0 1 1 0 G 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 1 1 0 0 0 1 This is a 12 gtlt 24 matrix It can be viewed in the form B l A with A and B both 12 gtlt 12 matrices In fact G 12 l A where OOOHHHOHHOH 0 1 1 1 1 1 1 1 1 1 1 1 OHOOOHHHOHHH HOHOOOHHHOHH HHOHOOOHHHOH OHHOHOOOHHHH HOHHOHOOOHHH HHOHHOHOOOHH HHHOHHOHOOOH OHHHOHHOHOOH OOHHHOHHOHOH OOOHHHOHHOHH H The Golay code is the code C where C is the row space of G The matrix G has rank 12 so the code has dimension 12 Thus there are 212 4096 codewords This code has distance 8 which can be veri ed in a hopelessly tedious manner by calculating the weight of all 4095 nonzero codewords or by proving some facts by induction about the rows of G One sees that the distance is at most 8 since the last row has weight exactly 8 The Voyager spacecrafts visited Jupiter Saturn Uranus and Neptune taking pictures of each planet and their moons The following picture was taken by Voyager 2 The photographs taken by these spacecrafts utilized the Golay code to encode the data representing the pictures A photograph consists of a rectangular grid7 and at each grid point7 or pixel7 a color is given For the Voyager spacecrafts7 the use of the Golay code allowed the photos to be made with up to 4096 colors Each color was represented as a codeword in the Golay code To send the information representing one picture7 each pixel was described by the codeword representing the color of that pixel This codeword was a 24 tuple of binary digits Since the Golay code has distance 87 it can correct up to three errors Thus7 each codeword transmitted could have up to three errors without losing any data 15 Error Correction We make formal the meaning of an error correcting code7 and we shall shortly see the connection between the distance and the error correction capabilities of the code De nition 15 A code C is s error detecting if whenever at least one but at most 5 errors are made in any codeword then the resulting word is not a codeword From this de nition and that of the distance of a code7 it is clear that a code of distance d can detect up to dil errors A more important de nition for us is that of error correcting De nition 16 A code C is said to be t error correcting if a word is a distance at mostt from some codeword then its distance from every other codeword is greater than t To help to understand this de nition7 suppose that a codeword i is transmitted and at most t errors are made7 meaning at most t components of the codeword are changed7 resulting in a word w If the code is t error correcting7 then since w is a distance of at most t from i then the distance from w to any other codeword is greater than t Therefore7 i is the closest codeword to w Using MLD7 we would correct w to i and thus recover the correct codeword Example 17 The code 01 cannot correct any errors7 nor can 0011 However7 000111 can correct 1 error It 111 is transmitted but one error is made7 the result has distance 2 from 000 The same is true starting with 000 The code 00000711111 can correct two errors Example 18 The Hamming code can correct one error Moreover7 there is a nice decoding algorithm7 if F F2 Suppose that i is a codeword7 and one error is made in transmitting 1 resulting in a word w Then w i ei where el is the i th standard basis vector of F7 Multiplying by the Hamming matrix H7 we get Hw Ho ei Hi Hei Hei Now7 an easy calculation shows that Hei is the i th column of H Therefore7 if we identify which column of H is Hw7 then we will determine i Once we know i7 we can recover i as i w ei This example illustrates an aspect of coding theory we would like codes that can correct errors and for which there is an ef cient decoding algorithm We know show the relation between the distance and the error correction capability of a code Theorem 19 Suppose that a code C has distance d Then G is t error correcting for t d 7 12j but is not t 1 error correcting Proof We give a metric theoretic proof of this fact Let t d 7 12j Suppose that w is a word a distance at most t from i We need to prove that w is a distance greater than t from any other codeword Suppose that u is another codeword lf dwu 3 t7 then consider the closed disks of radius t centered at i and u7 respectively Then it is in both disks By the triangle inequality7 we have dou S dou duw 3 2t lt d7 a contradiction to the de nition of d Therefore7 dwu gt t7 so the code is indeed t error correcting To see that C is not t 1 error correcting7 let um be codewords with duo d Note that d 2t 1 or d 2t 27 depending on whether d is odd or even If we change t 1 of the components of u that differ from those of i to make them equal to the corresponding components of i then we obtain a word it with duw t 1 but dwo duo 7 t 1 d 7 t 1 S t 1 This shows that C is not t 1 error correcting D From this theorem it is clear that the Hamming code is 1 error correcting since its distance is 3 If we have a linear code C with parity check matrix H7 we can use cosets to help decode Given a word w7 the syndrome of w is Hw Therefore7 the syndrome is 0 exactly when the word is a codeword The possible syndromes are in 1 1 correspondence with the cosets of 0 since 0 u C it if and only if Hu Hit It a codeword i is incorrectly transmitted as w7 then e w 7 i is the error word We have He Hit7 so e and it have the same syndrome Since G w C e7 we look in the coset C w for the smallest weight vector this is our error word e by MLD To use this coset decoding7 we need a list of syndromes and corresponding smallest weight error vectors These vectors are called coset leaders For example7 the following table would be the coset decoding table for the Hamming code Syndrome Coset Leader 000 0000000 001 1000000 010 0100000 011 0010000 100 0001000 101 0000100 110 0000010 111 0000001 The Maple worksheet eosetsmws available on the class website will produce this table 16 Bounds on Codes Suppose you have to transmit two pieces of information You could use the code 01 but that would give you no error correction You could use 000111 to be able to correct one error or up to i of the digits You could use 0000011111 and correct up to two errors or g of the digits In general by making the strings longer we can have better error correction However longer codewords mean more effort in sending receiving and decoding So we would like to have codes as short as possible with as good error correction as possible What restrictions are there Are there any relationships between the parameters 72 k and d There are several relationships all that give upper bounds or lower bounds for d To state some of these bounds we set Aqnd to be the largest size of a code of length n and distance d Much of coding theory has been to determine the values of this function and in determining the asymptotic behavior of Aqnd as a function of 6 dn as n 7 00 To help in some of the proofs we rst give a counting argument Consider the sphere of radius r centered at a word 2 To pick a word a distance 2 from 2 we need to change 2 of the components of 2 there are choices for these components Each component can be changed in q 7 1 ways since q Therefore there are q 7 1quot total words a distance of2 from 2 Therefore the sphere contains a total of 2170 q 7 1quot words There are many bounds on codes although we restrict our attention to just three The rst bound gives a lower bound of the size of a code with given parameters 72 and d Most bounds give upper bounds Proposition 110 GilbertVarshamov Bound q f39L Aqn7d Z 2201 139 Proof Let C be a code of maximal size of length n and distance d The spheres of radius d 7 1 centered at codewords then cover the entire space of words since if there is a word a distance of at least d from every codeword then we could add it to 0 without changing n or d If M is the size of this code then from our calculation of the size of spheres we have M 23 q 7 1quot 2 q which yields the bound D This bound says that there exists a code of length n and distance d and with M elements such that M 2 q 23 q 7 It does not say anything about arbitrary codes Proposition 111 Singleton Bound Let C be a code with parameters 72 k and d Then d S n 7 k 1 Therefore Aqn d S qn d Proof Consider the function p O 7 Fn d given by z1zn 1 znd1 In other words p removes the last d 7 1 components of a codeword Since every codeword of 10 C has weight at least d7 the function p is 1 1 Consequently7 0 S an dHl Since this is true for any code of length n and distance d7 we have Aqn7 d S lF L d l qn d Taking logarithms to the base q7 we get k S n 7 d17 or d S n 7 k 1 D If a code 0 satis es d n 7 k 17 then the code is said to be an MDS code maximum distance separable The following bound is also called the sphere packing bound Proposition 112 Hamming Bound ft d 712j then qn Aqltndgt Zi0 q 7 1y Proof Suppose that we have a code with length n and distance d7 and for which it has M codewords Let 1 be a codeword7 and consider the sphere of radius t centered at 1 As we have seen this sphere contains 220 q 7 1quot words Since the code can correct It errors7 the spheres of radius t centered at codewords are disjoint Therefore7 since there are q total words7 01 7 220 ELM 1V Since this is true for any such code7 we get the Hamming bound These last two codes give an upper bound7 in terms of n and d7 of the number of elements of any code with length n and distance d D The codes Goppa constructed from algebraic curves yield a better asymptotic lower bound than the Gilbert Varshamov bound By asymptotic bounds7 we mean the investigation of i 10gqAq 7 d lim sup n00 n The reason for considering asymptotic bounds comes from Shannon7s channel coding theo rem lf we de ne the information rate of a code to be kn7 then the theorem says that given any 8 gt 07 there is a code with information rate at least R for which the probability of in correct decoding of a received word is less than 8 However7 in order to make the probability of incorrect decoding low with a xed kn7 then n must be large This theorem is not stated quite correctly there is a limit on how large the information rate can be this limit is called the capacity of the channel A description of this can be found in Romans book 17 Cyclic Codes To de ne some of the important classes of codes we rst de ne cyclic codes First of all let 0 F 7 F be the shift map that is o1n xnx12n1 A code C is said to be cyclic provided that 00 C To give an alternate description of cyclic codes rst consider the F algebra of polynomials We may view F as a subspace of via the map p a0 an1 gt gt a0 alz an1x 1 Then F is mapped isomorphically onto the subspace of polynomials of degree less than n We can interpret o with respect to this embedding Since oa0 an1 an1 a0 an2 we have ltP 0007 7an2 ari71 aoz aniz cn l ari71 zao an72zn72 E a0 anq l an1x 1mod 71 Because of this if we view F g 7 1 then 0 corresponds to multiplication by x Therefore a code C Q 7 1 is cyclic if and only if zC G Since G is already an F vector space this additional condition is equivalent to that C be an ideal of 71 Now since is a PlD every ideal of 7 1 is principal and is generated by a divisor of x 7 1 Thus cyclic codes of length n are in 1 1 correspondence with divisors of x 7 1 Suppose that x 71 factors as zn71 Consider the code C 1 that is C is the code consisting of all polynomial cosets that are multiples of lf g go 91 zn k then gzzgxzk 1gz is a basis for C where W f 71 Therefore k n 7deggz We call g the generator polynomial of the code Let K be a nite extension eld of F If 041ozt E K let be the minimal polynomial over F of 04 and let g be the least common multiple of the Then g is the monic polynomial of least degree for which each 04 is a root We then can use g to build a cyclic code of length n where n is any integer for which g divides x 7 1 This code then consists of all polynomial cosets m for which p0439 0 for all i Note that for g to divide x 71 we need each a to be a root of x 71 that is we need a 1 Recall from the theory of nite elds the multiplicative group of any nite eld is cyclic A generator of the multiplicative group F is called a primitive elemerit of F If 04 is a primitive element of En then all powers of 04 are roots of x 7 1 and we can use powers of alpha to build a code of length 71 Note that n 1 must be a power of a prime for EL to exist Example 113 With a rearrangement of its digits the Hamming code is an example of a cyclic code To see this let 04 be a primitive root of F8 Since we may view F8 as F8 F2 z 1 we can view its elements as 3 tuples over F2 The nonzero elements 1 oz a6 of F8 then correspond to the columns of the Hamming matrix H By reordering the columns if necessary we view H 104 a6 where we think of these elements as 12 column vectors For example if 04 E then a0 100 a1 010 a2 001 a3 110 a4 011 a5 111 a6 101 Therefore we have 1 H 0 0 OHO 0 1 0 1 1 0 1 1 1 0 1 0 1 1 1 The condition for a vector 1 a0 a6 to be in the rearranged Hamming code is 00 01 0H11OL046 a0a1aa6a6 06 Therefore by viewing 1 as a polynomial U E FZM of degree at most 6 we see that 1 is a codeword if and only if 0a 0 In fact the generator polynomial is then the minimal polynomial of 04 which is 3 z 1 Example 114 BCH Codes Let n gm 7 1 and let 04 be a primitive element of the eld qu If e is a positive integer with e S n we can use the elements 04042 oze l to build a code This code is then the set of polynomial cosets m with pa pa2 10045 1 0 While we do not show it here this code has distance at least e and so can correct 1e 7 12 errors The integer e is called the designated distance of the code and is a lower bound of the actual distance Example 115 ReedSolomon Codes Let n q 7 1 and let 04 be a primitive element of Fq Then the code of all polynomial cosets p with pa pa2 1001 1 0 is called a Reed Solomon code By the claim of the previous example the distance of this code is at least d The generator polynomial is g x 7 ozx 7 ad l so k n 7 deggz n 7 d 1 Therefore by the Singleton bound the distance of the code is exactly d and the code is an MDS code Example 116 Suppose n 7 and d 4 The Reed Solomon Code we get from this data has k 4 The resulting code is We now consider some speci c examples of BCH codes Example 117 Consider F16 F24 1 the splitting eld over F2 of 4 1 A short calculation shows that any root 04 of z41 is a primitive element of F16 To see this recall that since F16 has order 15 Lagrange7s theorem implies that 04 is a primitive element provided that 043 31 1 and 045 31 1 Since the minimal polynomial of 04 is x4 z 1 which does not divide x3 71 or 5 71 neither of these polynomials has 04 as a root We consider the BCH code of all polynomials cosets having 04 042 a6 as roots Then 71 15 To calculate k we calculate the generator polynomial This is the least common multiple of the minimal polynomials of the six roots The minimal polynomial of Oz and 042 is x4 z 1 The minimal polynomial of 043 and 046 is 4 x3 2 z 1 and the minimal polynomial of 045 is x2 z 1 One way to nd these is to note that every element of F16 is a root of x16 7 x If you factor this into irreducible polynomials you need to check which irreducible has a given 0 as a root Alternatively you can nd the generator polynomial by using the Maple worksheet generatormws which is available on the course website In any case we get g x10 x8 5 4 x2 z 1 This has degree 10 so k n 7 deggz 5 This code can correct three errors since the designated distance is 7 Example 118 Consider the code built from the same eld F16 but with designated distance 5 Then the code comes from the roots aa2a3a4 In this case the generator polynomial is x8 7 6 4 1 and so 71 15 and k 15 7 8 7 By lowering the designated distance we have increased the dimension of the code 2 Introduction to Algebraic Geometry While the error correcting codes de ned by Goppa can be described purely in terms of elds they are better understood by also viewing them geometrically We will discuss the basic concepts of algebraic geometry and see how the study of nonsingular projective algebraic curves is equivalent to the study of algebraic function elds in one variable Furthermore by thinking of elds we will help to justify why we need to discuss projective curves instead of restricting only to af ne curves A brief treatment of the concepts of algebraic geometry needed for coding theory can be found in the book Codes and Curves by Walker To do algebraic geometry we need to use algebraically closed elds Recall that a eld k is algebraically closed if every nonconstant polynomial in has a root in k The fundamental theorem of algebra states that C is algebraically closed Neither R nor Q is algebraically closed as 2 1 has no root in either eld Nor is any nite eld F algebraically closed for if F 11 an then x 7 a1 7 an 1 has no root in F If F is an arbitrary eld then it has an algebraic extension that is algebraically closed this eld is called an algebraic closure of F Such a eld is unique up to isomorphism 21 Af ne Curves Let k be an algebraically closed eld We denote the polynomial ring over k in 2 variables by Algebraic geometry studies solutions to polynomial equations We de ne the a ne place A2k to be the set k2 of all pairs over k If it is not important to keep track of the eld k we will write A2 in place of A2k If P 11 6 A2 and f E kxy we will denote by fP the evaluation of f at P If f E kxy then the a ne curve f 0 is de ned to be Zf P e A2 fP 0 This set is sometimes called the zero set of f Example 21 The curve y 2 in A2 is a parabola In general a conic section is the zero set of a polynomial of degree 2 For instance 2 y2 1 is a circle and xy 1 is a hyperbola The curve yz 3 is sometimes called a cuspidal cubic curve Example 22 An elliptic curve is a curve given by an equation yz fx where f is a cubic polynomial with no repeated roots For example yz x3 7 z is an elliptic curve Example 23 A hyperelliptic curve is a curve given by an equation yz fx where f is a polynomial of degree at least 4 and with no repeated roots For example yz x5 7 1 is an elliptic curve over C or over any algebraically closed eld of characteristic not 5 Example 24 Suppose that k R a eld that is not algebraically closed There are no solutions to the equation 2 y2 1 0 over R so Zx2 y2 1 is empty However if k C then there are solutions including 20 It is to have solutions to polynomial 15 equations that the base eld is assumed to be algebraically closed If fzy is a polynomial over an algebraically closed eld k and if b E k then fxb is a polynomial in the one variable x If fxb is not constant then it has roots in k and so fpy 0 has solutions in A2k If X is an algebraic curve over k then the ideal of X is X f E kpy fP 0 for all P E X If fpy pxy51pnzy5quot is the factorization of a polynomial f into irreducible factors then it is easy to see that IZf p1 10 Moreover the coordinate ring of X is the quotient ring PX kzyIX One way of thinking about the coordinate ring is to consider it the ring of polynomial functions on X For two polynomials f and g induce the same function X a k precisely when f 7 g E X which is equivalent to the cosets and g being equal A topological space is said to be irreducible if it cannot be written as the union of two proper subcurves If f gh then Zf Zg U Zh From this if f is a squarefree polynomial it follows that Zf is irreducible if and only if f is irreducible Alternatively X is irreducible if and only if PX is an integral domain 22 Projective Varieties For students who have studied topology the construction of projective varieties parallels that of constructing the real projective plane We de ne an equivalence relation on A3 0 0 0 by de ning that abc a b c if there is a nonzero scalar A with a Aa b Ab and c Ac Geometrically the equivalence class of a point is the line through the origin that passes through the point We will write a b c for the equivalence class of abc and P2 will denote the set of all equivalence classes This is the projective plane Note that a b c represents a point in the projective plane only if at least one of the coordinates is nonzero Note that polynomial functions are not well de ned on points of projective space How ever we can get around this problem A polynomial f E klzyz is said to be homogeneous if every monomial of f has the same degree Alternatively f is homogeneous if there is an integer in with fAz Ay A2 Amfzyz for all A E k If f is homogeneous and P 6 P2 then the equation fP 0 is well de ned in other words if P N Q then fP 0 if and only if fQ 0 Therefore we can de ne zero sets of collections of homogeneous polynomials If f is a homogeneous polynomial then the projective curve f 0 is the zero set Zf P e P fP 0 We de ne a projective curve to be irreducible in exactly the same way as for af ne curves if f is a squarefree homogeneous polynomial then it follows that Zf is irreducible if and 16 only if f is irreducible We can de ne a coordinate ring for projective curves If X is a projective curve then X is de ned to be X ltf E klxyz f is homogeneous and fP 0 for all P E The homogeneous coordinate ring SX is the quotient ring klzyzIX As with af ne curves a projective curve is irreducible if and only if its ideal is a prime ideal and this happens exactly when SX is an integral domain Example 25 Consider the projective parabola X given by yz 2 If a b c E X then be 12 lf 0 31 0 then by dividing by c we may assume that c 1 Then 11 is on the af ne parabola y 2 On the other hand if c 0 then a 0 and b 31 0 for a b c to be a valid point Then by dividing by b we have the point 0 1 0 Thus we can think of X as the union of the af ne parabola y x2 and the extra point 0 1 0 Note that the af ne parabola is obtained by setting 2 1 in the equation yz 2 Moreover there is another af ne parabola inside X and this other curve contains 0 1 0 If we set y 1 then we have the equation 2 2 which again represents a parabola The point 0 1 0 is a point of this af ne parabola The purpose of considering this second af ne parabola is that any point of X is a point on some af ne curve in X Example 26 Let X be the projective elliptic curve yzz 3 7 22 The af ne elliptic curve yz x3 7 z is an af ne curve in X we can map it into X via ab gt gt a b 1 The image is X 7 0 1 0 if a b c E X and c 0 then the equation yzz x3 7 22 forces a 0 Then b 31 0 since a b c 6 P2 Thus we may divide by b to assume b 1 and so the only point on X for which 0 0 is 0 1 0 As with the previous example the af ne equation is obtained from the projective equation by setting 2 1 Example 27 In the previous examples we started with a projective curve and found an af ne curve inside it In this example we start with an af ne curve and produce a projective curve Let Y be the curve y 3 By adding a third variable 2 we can use y 7 3 to get the homogeneous polynomial y227x3 The af ne curve Y sits inside the projective curve 122 x3 as in the previous examples Similarly y2 1 x4 yields the homogeneous polynomial 1222 24 x4 What we are doing is adding enough copies of z to each monomial so that each has degree equal to the highest degree of a monomial of the original polynomial To have a formula for doing this if ay has degree d then zdfxzyz is the homogenization of fzy The resulting projective curve is called the projective closure of the af ne curve f 0 In general if Zf is a projective curve then any point of the form a b 0 of Zf is called a point at in nity lf gxy fy 1 then Zf is the union of the af ne curve 9 0 and the points of in nity The polynomial gxy is called a dehomogenization of fyz at the variable Note that if fhEyz is the homogenization of fEy then fEy fhEy 1 Con versely if gEyz is homogeneous of degree d and if fEy gEy 1 then 9061172 zdeg9 degffhy We do not in general recover gEy 2 by homogenizing gEy 1 since this polynomial may have smaller degree than the degree of gEyz For example if gEyz 22 E2 717 yz then gEy 1 1 E y has degree 1 while gEyz has degree 2 23 The Function Field of a Curve Suppose that an af ne curve X is irreducible Then its coordinate ring PX is an integral domain and so has a quotient eld which we denote by This is the eld kX gh gh E PXh 31 0 Let E and 17 be the images of E and y in PX The de nition of PX shows that PX That is PX is generated as a k algebra by E and 17 The eld kX then contains the eld kE7j generated by E and Since the quotient eld of an integral domain is the smallest eld containing the domain we see that kX Therefore kX is the eld extension of k generated by two elements um and subject to the relation fu1 0 To give other another view of the function eld suppose that X is the zero set of the irreducible polynomial fEy If we de ne an equivalence relation on the ring 5 awMay 971 E klnylh 6 f by gh g h if gh 7 g h E f Q kEy then kX g S N To verify this interpretation note that an arbitrary element of kX is of the form with gh E kEy where we continue to write bars to represent the image of a polynomial in PX When are gE and 3 equal in kX7 Since kX is the quotient eld of PX this occurs exactly when g Finally this occurs when M U or gh 7 g h E We may de ne a topology on a curve by de ning a set to be open if it is empty or if its complement is nite Thus proper closed sets are nite This is a special example of the Zariski topology on an algebraic variety curves are special examples of varieties We will view kX as the eld of rational functions de ned an open subset of the curve X a rational function is simply a function that can be represented as a quotient of polynomials Example 28 The function eld of the af ne parabola y E2 is the quotient eld of 7 E2 This ring is isomorphic to Mt the rational function eld in one variable t they are isomorphic via the map that sends t to E y 7 E2 Therefore the function eld is isomorphic to Example 29 The coordinate ring of the curve X Zy27E3 is kE yly27E3 This ring 18 is isomorphic to kt2t3 a subring of These rings are isomorphic via the map E gt gt t2 and gt gt t3 The function eld of this curve is then the quotient eld of kt2t3 which is Note that yz represents a rational function on X as does xZy Furthermore these functions agree since yz xZy in Therefore a rational function can be represented in more than one way as a quotient of polynomials Example 210 The coordinate ring of the elliptic curve yz 3 7 is klxyy2 7x3 7 We claim that its function eld is isomorphic to kt To see this note that the coordinate ring is ME 17 and so its function eld is That is the function eld is generated as an extension eld of k by E and Now we have 172 is 7 E Therefore is algebraic over ME and so Mfg g V 7 The element E cannot be algebraic over k so g kt the rational function eld in t Let fyz be an irreducible homogeneous polynomial and let X Zf We can also de ne a function eld of X Let N be the equivalence relation on the set gxyzhxyz g h E klxyz are homogeneous of the same degree h U 0 given by gh g h if and only if gh 7 g h E The function eld kX is then the set of equivalence classes under the natural operations Note that any fg E kX determines a well de ned function on an open subset of X if g h are both homogeneous of degree d then if A is nonzero then gltAaAbAc Ad9abc mac Wmth hltabcgt39 The function gh is de ned for all points P except when hP 0 We may then think of kX in much the same way as we think of the function eld of an af ne curve We now show that the function eld of a projective curve can be calculated by nding the function eld of an af ne curve Proposition 211 Let fxyz be an irreducible homogeneous polynomial Let X Zf be a projective curve arid ifU is the a rie curve Zfy 1 theri kX E Proof We will de ne a ring homomorphism p NU 7 kX and show that p is injective Thus p induces a homomorphism kU 7 kX which necessarily is injective We will then be done by showing that this map is surjective To de ne p write gy fxy 1 For May 6 kxy de ne b Note that if degh d then hzzyz hhzyzzd and this represents an element of This map is well de ned since if h 7 h E 9 then h h gl for some l E Then h zyz hxzyz By our hypothesis on f we have gzzyz fzyzzdegf Therefore gxzyzlzzyz 0 in kX by the de nition of Thus h h this proves that p is well de ned It is a simple exercise to see that p is a ring 19 homomorphism To prove injectivity suppose that h 0 Then hzzyz 0 in Note that hzzyz hhzyzzdegh For this to be zero in kX we have hhxyz fxyzmxyz for some in Dehomogenizing we get Mimi2 Mimi1 ay 1mzy 1 9zym9cy1 Therefore h 0 Thus p is injective For surjectivity every element of kX is rep resented by a quotient hxyzlyz of homogeneous polynomials of the same de gree Let their common degree be d Then hzyzlzyz zdhy 1zdlzy 1 hy 1lzy 1 This represents an element of MU and this element maps via p to hyzlxyz Therefore p is surjective and so kU E D Example 212 Let X be the projective parabola yz 2 The af ne parabola Y Zy 7 x2 is obtained from dehomogeniZing yz 7 2 Therefore kX MY 2 Similarly the function eld of y22 3 is kt and the function eld of the projective elliptic curve yzz x3 7 22 is kt t3 7 t There are important subrings of the function eld of a curve First if X is a projective or af ne curve then le ltp E kX p is de ned at every point of X This is called the ririg of regular functions on X and is the ring of globally de ned rational functions on X If X is af ne then one can show that lel g PX but that if X is projective then le k Next let P E X Then the local ririg of X at P is the set OpX ltp E kX p is de ned at P This is the ring of regular functions de ned locally at P For ltpP to be de ned p must be de ned in an open neighborhood of P since if p gh then P Zh and so p is de ned on the open neighborhood h 31 0 of P For convenience we will typically write Op in place of OpX It is a local ring its unique maximal ideal is MppEOpiltpP0 A short exercise shows that Mp is an ideal of Op Furthermore if p 6 Op Mp then we may write p fg with fP 31 0 Then gf 6 Op is an inverse for p Therefore since every element outside Mp is a unit Mp is the unique maximal ideal of Op If Y Zf is an af ne curve then OpY E kY hP 31 0 Proposition 213 Let P be a point of ari irreducible a rie curve X Zf arid let mp g E lel gP 0 Theri Op lelmp 20 Proof By de nition kX Q Op Moreover everything in kX mp is invertible in Op so kXmp Q Op Conversely let p 6 Op Then we may write p gh with gh E kX by the de nition of Since p is de ned at P we have hP 31 0 Therefore h mp so gh e kmmp Thus Op kmmp D It is not hard to show that if P ab then mp is generated by the images in kX of the polynomials z 7 a and y 7 b Let X Zf be an irreducible projective curve and let Y Zg be the af ne curve obtained by dehomogeniZing f We have seen that kX If P E Y the de nition of local ring then shows that OpX OpY Therefore the local ring at a point can be determined from the previous proposition Example 214 The line X Zy has function eld We determine the local rings of points on X Let P 10 6 X Note that PX g Mt and kX is the quotient eld of this ring The isomorphism kX kt sends the image of a polynomial ay to ft0 We claim that OpX ktta the localization of Mt at the maximal ideal t 7 a For gzyhxy E OpX if ha0 31 0 Therefore ht 0 is not divisible by t7 a and so gt0ht0 E ktta The reverse inclusion is easy since gt 0ht0 E ktta is de ned at P since It 7 a does not divide ht 0 24 Nonsingular Curves To have a complete correspondence between function elds and projective curves we must restrict to nonsingular curves Suppose that fyz is a homogeneous polynomial It de nes a projective curve X in P2 and we refer to it as a plane curve By recalling formulas for tangent lines from calculus we say that the curve X is nonsingular at a point P E X if at least one of the three partial derivatives afdx afdy or 6f62 do not vanish at P If X is nonsingular at every point P then we say that X is a nonsingular curve Similarly an af ne curve 9 0 is nonsingular at a point P if dgdx and 396y do not both vanish at P and the curve itself is nonsingular if it is nonsingular at all its points Example 215 The af ne parabola y x2 is nonsingular for the partial derivatives of y 7x2 are 72x and 1 neither vanish simultaneously Similarly Zy 7 x3 and Z2 y2 71 are nonsingular However Zy2 7 x3 is singular at the origin Example 216 Consider the projective parabola yz 2 The partial derivatives are 72z 2 y they only simultaneously vanish at 000 which does not represent a point in P2 Therefore the parabola is nonsingular However the projective closure of the cubic curve y 3 is X Zy22 7 3 Here the partial derivatives are 73 22 2yz which vanishes at 0 1 0 E X Therefore this curve has a singularity Note that the singularity occurs at the point at in nity Example 217 The elliptic curve X Zy227x3x22 is nonsingular if the characteristic of k is not 2 To see this7 the three partial derivatives of 122 7 x3 22 are 73z2 22 2127 and y2 2oz It is easy to see that all three partials vanish only at z y z 07 and so there is no point on X for which this happens ln fact7 if y2 fx is any elliptic curve7 where f is a cubic polynomial with no repeated roots7 then this curve is nonsingular For7 a point P Lab on the curve satis es b2 fa The partials of y2 7 f are 7f x and 2y for these to vanish at P7 we must have b 0 and f a 0 Then fa f a 07 and this cannot happen since a would then be a multiple root of Example 218 Any hyperelliptic curve is nonsingular7 as long as the characteristic of the base eld is not 2 The argument of the previous example works word for word to prove that a hyperelliptic curve is nonsingular By making use of some theorems from Math 5827 we can now begin to give the connection between function elds and curves We rst note that a commutative ring is said to have dimension 1 if all nonzero prime ideals are maximal It is not an obvious result7 but the coordinate ring of an af ne curve has dimension 1 Also7 the local ring at a point on any curve also has dimension 1 The general result is that the geometric dimension of an af ne algebraic variety is equal to the ring theoretic dimension of its coordinate ring7 and that these are also equal to the transcendence degree over k of the function eld Next7 if X is a curve7 then OpX is a local Noetherian domain of dimension 1 for any P E X We have already remarked that its dimension is 1 It is Noetherian because the coordinate ring of an af ne curve is a quotient ring of The coordinate ring is then Noetherian7 and since OpX is a localization of the coordinate ring7 it is also Noetherian We have seen above that OpX is a local ring Let A be a local domain with maximal ideal M7 and set F AM7 a eld Then A is a discrete valuation ring if and only if it is Noetherian7 dimension 17 and dimFMM2 1 Note that F k for A OpX this can be seen indirectly in the proof below Theorem 219 Let X Zf be an irreducible curve Then P E X is nonsingular if and only if OpX is a discrete valuation ring of Proof By the remarks above7 it is enough to prove that P is nonsingular if and only if dimkMpM 17 where Mp is the maximal ideal of Op Let M x 7 my 7 b7 a maximal ideal of De ne a map 6 M 7 k2 by 69 13 2703 We note that 6 yields a vector space isomorphism MMZ 2 k2 The polynomial f yields the 1 dimensional subspace V M2 M2 of MMZ7 and 0V has dimension 1 if and only if P is nonsingular7 and it has dimension 0 otherwise Recall that kX and Op kXmp where 111 g E lel gP 0 Then mpmizv 1V1f1 42ffg MM2f The maximal ideal Mp of Op is then mpOp and so MADME g rapmi by a Math 582 exercise Thus MiaM e minmi Mf M MMZ v by one of the homomorphism theorems of vector space theory Therefore MADME has dimension 1 if and only if P is nonsingular Example 220 The curve X Zy2 7 x3 is singular at the origin The coordinate ring of X is kt2t3 and so the local ring at the origin is klt2t3tzta This is a proper subring of the discrete valuation ring ktt By another characterization of discrete valuation rings a local Noetherian domain of dimension 1 is a discrete valuation ring if and only if it is integrally closed The ring kt2t3t2ta is not integrally closed since It is integral over it t satis es the polynomial T2 7 t2 However t kltz t3t2t3 which shows that this ring is not integrally closed 25 Curves over nonAlgebraically Closed Fields Classical algebraic geometry uses algebraically closed elds in order to have solutions to polynomial equations However in several situations including coding theory we need to work with arbitrary elds Let k be an arbitrary eld and let K be an algebraically closed extension eld of k An af ne curve over k is a curve Zf Q A2K such that ay E Similarly a projective curve over k is a curve Zf Q P2K such that fyz E kxyz An af ne curve Zf over k is said to be absolutely irreducible if ay is irreducible in Kley An analogous de nition holds for projective curves If X Zf is an af ne curve over k then its ideal over k is Xk X Hey gxy 6 May gP 0 for all P E X and its k coordinate ring is 1 Xk klnylHXkl This is a subring of the coordinate ring PX If X is absolutely irreducible then the function eld KX exists we de ne the k function eld kX to be the quotient eld of PXk Therefore we see that kX is the set of all quotients of polynomials 9x y hxy where gh g h if gh 7 g h E fkzy We similarly can de ne the k function eld of an irreducible projective curve We also can de ne a k version of the local ring of a point which we also denote by OpXk This is the ring OpXk ltp E kX ltpP is de ned OpX This notation does not show the dependence on k but this will not give us a problem since we will not consider this concept for two elds at a time The characterization of nonsingularity in terms of Op can be made relative to k as follows le is a curve over k then P E X is nonsingular if and only if Op is a discrete valuation ring The proof of this is a moderately standard use of facts about discrete valuation rings7 and we won7t give it here A point P Lab on a curve X is said to be a k mtional point if ab 6 k We write Xk for the set of all k rational points of X Thus7 X XK To simplify terminology7 we will often refer to a K rational point as simply a point7 and use the terminology rational point to note a point whose coordinates are in k or some other sub eld of K A curve over k is nonsingular if it is nonsingular as a curve over K That is7 X is nonsingular if every point P E A2K lying on X is nonsingular As we will see7 a major problem in working with codes coming from algebraic geometry is to nd curves over E with lots of rational points Example 221 Let n 2 3 be an integer7 and consider the projective curve x y z Fermat7s last theorem says that there are no Q rational points of this curve having all nonzero coordinates Example 222 The projective curve X Zx2 y2 22 over R has no R rational points However7 it has lots of CC rational points7 such as 1 0 239 Example 223 The projective curve X Z2 yz over R is not absolutely irreducible even though 2 y2 is irreducible in Rhy since over C the polynomial 2 y2 factors as x W x 7 ty Therefore7 the k curve Zf need not be absolutely irreducible even if f is irreducible over k Example 224 We will see near the end of the course that Weil7s proof of the Riemann hypothesis for curves over a nite eld E yields a bound on the number of rational points This bound7 which is in terms of the genus g of the curve and of the size q of the base eld7 says that the number N of rational points satis es the inequality UV q1l S 29 We will de ne the genus in the rst chapter of Stichtenoth the proof of this bound is done in Chapter 5 Let X be the curve over qu given by qur1 1 This is called the Hermitian curve over E12 It turns out that the genus of this curve is g qq 7 127 and that N 1 13 the largest possible given the bound above 24 3 Algebraic Function Fields and Discrete Valuation Rings Let Fk be a eld extension Then F is said to be an algebraic function eld in one oaiiable over k if there is an element x E F that is transcendental over k and for which F lt oo Recall that ifs is transcendental over k then the eld is isomorphic to a rational function eld in one variable over k We will not distinguish between transcendental elements and variables from now on These elds are precisely the function elds of algebraic curves de ned over k We have indicated without full proof why the function eld of a curve is such a eld For the converse if F is generated over by a single element y then since y is algebraic over kz there is an irreducible polynomial pt E for which py 0 By clearing denominators we may view pt E t and the equation py 0 yields an equation fy 0 with f a polynomial in two variables over k Then F is the function eld of Zf Q A2 We note one simple but important property of algebraic function elds in one variable which we state as a proposition Proposition 31 Let Fk be an algebraic function eld in one oaiiable ft 6 F is tran scendental over k then F lt 00 Proof There is an x E F transcendental over k and with F lt 00 We will be done by proving that z is algebraic over kt since then lt 00 and then F i kW lFi ktllkt9 i MN m kltzgtnklttgtltzgt kw l is nite since both terms are nite To show that z is algebraic over kt we know that t is algebraic over Therefore there are E with tnfn1xt 1 f0z 0 By writing each as a quotient of polynomials and then clearing denominators we have an equation of the form anxt an1xt 1 a0x 0 Viewing this equation as a polynomial in z with coef cients in kt we conclude that z is algebraic over 1 This proposition could also be proved by using the notion of a transcendence basis A nite transcendence basis of a eld extension Kk is a set t1 tn of elements of K such that t1 is transcendental over k and tZJrL is transcendental over kt1 t for each i 2 1 and for which K is algebraic over kt1 tn The number of elements of a transcendence basis is uniquely determined and is called the transcendence degree of Kk For an algebraic function eld in one variable Fk the transcendence degree is 1 lft is transcendental over k then t is a transcendence basis for Fk by the uniqueness of transcendence degree Therefore every element of F is algebraic over kt including the element x of the proof of the proposition Moreover F is nitely generated over k so it is also nitely generated over thus as F is algebraic over kz we see that F lt oo 25 Let Fk be an algebraic function eld in one variable If k is the algebraic closure of h in F then k is called the eld of constants of Recall that k is the set of all elements of F that are algebraic over k We will see the reason for calling elements of k constants once we interpret elements of F as functions Now we prove that k is a nite dimensional extension of k This is a special case of a theorem from eld theory says that if Fk is a nitely generated eld extension then the algebraic closure of h in F is a nite extension of k Proposition 32 Let Fk be an algebraic function eld in one variable If k is the eld of constants of Fh then H k lt 00 Proof Let x E F be transcendental over h Then F lt 00 by the previous result We claim that k k S F Once we prove this then we will have proved the proposition Let t1 tn Q h be a linearly independent set over k We claim it is also linearly independent over For if there is an equation tifi 0 with E kz by clearing denorninators we may assume that E Write Z aha with aij E h Then tiaijxj 0 which we may write as Z aijti 0 This would show that z is algebraic over k unless all coef cients are 0 a contradiction since k h is algebraic and z is transcendental over k So aijt 0 for each j Since aij E k and the t are independent over h each aij 0 This shows that each 0 and so the claim that t1tn is independent over Therefore n lt F This forces k k S F as desired D If ay 6 May is an absolutely irreducible polynomial and X Zf is the corre sponding curve over k then h is the eld of constants of We probably will not use this result and we will not prove it A proof can be found in Section 22 of my book Field and Galois Theory That section studies function elds of af ne algebraic varieties not just algebraic curves 31 Discrete Valuation Rings We note that the rational function eld itself is the most simple example of an algebraic function eld One special property for is that since this eld is the quotient eld of the unique factorization dornain km every element x of can be uniquely represented in the form We PM pnz5quot q1f1 qmfm7 where pi pnq1 qm are distinct irreducible polynornials n and m are positive integers and each exponent el and f is nonnegative We note that we may need to use 0 exponents to write an element in this form The element p de nes a function to h whose domain is a subset of k we may talk of zeros and poles of p a zero of p is obviously an element a with a 0 A pole of p is an element b for which the denominator of p is 0 Such an 26 element is a point at which p is not de ned lf b is a pole7 then the denominator of p can be written in the form x 7 bTU for some polynomial 039 for which b is not a root7 and for some integer r 2 1 We then see that7 while x is not de ned at b7 the rational function x 7bfltpz is both de ned and nonzero at b We then de ne the order of the pole b to be r Similarly7 a zero a of p has order 5 if x x 7 a577 where T is a rational function de ned at a and for which 7a 31 0 The problem of Riemann is to determine7 given points P17 7 P777 Q17 7 Qm of a curve X and positive integers 617 7 en7f17 7fm7 which functions p E kX have a zero at P7 of order at least 67 and a pole at Q7 of order at most f7 While this problem is well de ned for curves whose function eld is kz7 it is not well de ned for other curves One point of the rst chapter of Stichtenoth is to de ne what is a zero and a pole of a rational function on a curve For this we need to discuss in some detail the concept of a discrete valuation ring Before giving the de nitions7 we give two examples7 one from number theory and the other from algebraic geometry Example 33 Let p be a prime number We can write every rational number in the form p 7 where n E Z and where a and b are integers neither divisible by p The exponent n is uniquely determined7 this is a consequence of unique factorization We de ne a function 11p Q a Z by vpx n if z p 7 as above We point out two properties of this function Let my 6 Q5 Then vpxy vpz 1y This follows from the laws of exponents and from the de nition of a prime7 that implies that if U71 are not divisible by p7 then neither is M Second7 if z y 31 07 then vpz y 2 min vp7vpy To see why this is true7 write x p and y 10mg For sake of argument7 suppose that n 2 m Then 0 zyp 31 3119 3 7 m pn mad be p bd 39 The denominator be is not divisible by p since neither b nor 0 is divisible by p The numerator may or may not be divisible by p Therefore7 vpz y 2 m min vp7vpy By using the function 1177 we can de ne a subring of Q by W l Opz Qvpz20U0 a7b Z7plb Moreover7 017 has an ideal Mp de ned by Mpx Qvpzgt0U0 a7b Z7pla7plb Furthermore7 every element of 0p M17 is a unit inOp7 as such an element has the form 27 with both ab not divisible by p Then 3 6 Op is the inverse of in Op Thus 017 is a local ring with unique maximal ideal Mp This ring is sometimes called the p adic valuation ring of Q The residue eld OpMp is isomorphic to FF via the map gt gt a 54 Example 34 Let pp be an irreducible polynomial in Every rational function in can be written in the form pz where ap and bx are polynomials neither divisible by We de ne a function vp hx a Z by vpltpp n if Mp as above With the same arguments as in the previous example we see that vpfpgx Z112f96 Up996 and WWW 9 2 min Upf96 p9 for a11f967996 6 WW Also we have a local ring 0pm WW 6 MW ivpltP95 Z 0 U 0 x azbz 6 km pz lbw Q A 2 6 with unique maximal ideal Mp W 6 kW vaa gt 0 u 0 altzgtbltzgt 6 km pa lam ammo 7 7 71 Finally we note that OpM g via the map gt gt This eld is a nite dimensional eld extension of k recall that up to isomorphism is the eld extension of h obtained by adjoining to h a root of the polynomial pp so its dimension over h is equal to the degree of We now give de nitions of a discrete valuation ring and a discrete valuation These de nitions are equivalent in some sense as the previous examples indicate although we make this equivalence more precise below De nition 35 A discrete valuation ring of a eld F is a principal ideal domain R Q F such that for every a E F either a E R or a 1 E R De nition 36 A discrete valuation on a eldF is a nonzero functionv F a Z satisfying the properties vab va vb for all ab E F and ii va b 2 min vavb for allab E F with ab 31 0 A discrete valuation is then a group homomorphism from the multiplicative group of F to the additive group Z Our assumption that v is nonzero means v is not identically equal to 0 Since the image ofv is a nonzero subgroup of Z this image is nZ for some n gt 1 By replacing the function v by iv we assume that v is surjective To connect these de nitions rst let v be a discrete valuation on a eld F Set Ova Fva20U0 28 The de nition of a discrete valuation implies that OH is a subring of F We call On the valuation ring of 1 Furthermore if My is de ned by Mva Foagt0U0 then My is an ideal of OW Moreover every element a of Ow My satis es oa 0 and since this implies that 11a 1 0 the element a is a unit of OE This shows that My is the unique maximal ideal of OW We claim that OH is a discrete valuation ring One property is easy to prove Let a E F Since 1 is a homomorphism 11a 1 711a Therefore either oa 2 0 or 11a 1 2 0 this implies that a E Ow or a 1 E OE To nish the claim we need to show that OH is a principal ideal domain Let I be a nonzero ideal of OE Let n min 1a a E Ia 31 0 and pick a E I with oa n We claim that I aOv The inclusion aOv Q I is clear For the reverse inclusion let x E I Then 11z 2 n by de nition of n Then oza 1 2 0 so za 1 6 OW Therefore z E aOv Therefore I Q aOv and so I aOv as desired This nishes the proof that Ow is a discrete valuation ring We now show that every discrete valuation ring of a eld F is the valuation ring of some discrete valuation on F Let R be a discrete valuation ring of F We rst note that R is a local ring with unique maximal ideal M r E R r Pi the set of nonunits of B To prove this it suf ces to show that M is an ideal of R It is clear that if z E M and r E R then zr E M It is also clear that if z E M then 7z E M So we are left to prove that if zy E M then x y E M It is enough to assume that both z and y are nonzero Since R is a discrete valuation ring either zy l E R or yd l zy 1 1 E R Suppose that 114 6 R Then z yr for some r E R Then z y yr y y1 r This cannot be a unit since it is a multiple of the nonunit y Therefore z y E M Therefore M is an ideal of R and so M is the unique maximal ideal of R We de ne a valuation i on F as follows First for z E R with z 31 0 set 11z n if z E M Mn That is Ultgt n if n is the maximum integer satisfying z E M We view B M0 in this de nition To prove that i is well de ned on R we need to prove that there is such a maximum integer for any nonzero x This amounts to proving that Hill M We do this in the following lemma Lemma 37 Let R be a local principal ideal domain IfM is the mamirnal ideal of R then mil M n 0 Proof Let J f M an ideal of R Since R is a principal ideal domain there is an a E R with J aR Write M tR for some t E R Then M tnR For each ii there is an rn E R with a tnrn In particular trL tnrn for each n Since R is a domain r1 tn lrn This shows that r1 6 J If we write r1 as then a trL tas If a 31 0 then cancelling a gives 1 ts This forces t E M to be a unit which is false Therefore a 0 and so J 0 D We point out that this lemma is an easy special case of the Krull intersection theorem which states that if R is a local Noetherian ring with maximal ideal M then f M 29 As we pointed out before the lemma we now have a well de ned function i R0 a Z de ned by 0a n if z E M Mn Since R is a principal ideal domain there is a t E M with M tR Then M tnR Therefore each nonzero z E R can be written uniquely in the form x t u for some unit u 6 Pi From this fact it is easy to prove that i is a discrete valuation First if Ly E R then x t u and y tmw for some nonnegative integers nm and units uw Then xy tnmuw Since u and w are units the product no is also a unit Then May n m 0a Second suppose that n 2 in Then z y tmt mu w If we write tn mu w tfs for some unit 5 then x y tm s and so oz y m r 2 m minooy We extend i to F by de ning oab ia 7 11b This is well de ned if ab cd then ad be Therefore oad 11bc or ia id 11b oc so ia 7ob 110 7od We have prove that May 0a Mg and 11z y 2 min oxoy for Lg E R A short argument shows that these properties hold for Lg E F which will prove that i is a discrete valuation ring We point out some more properties of a discrete valuation ring of F First we will show that a discrete valuation ring 0 of F is integrally closed in F Recall that if R Q S are commutative rings an element a E S is said to be integral over R if a satis es an equation of the form a rn1a 1 r0 0 with each r E R That is a is integral over R if a satis es a monic polynomial equation with coef cients in R A ring R with quotient eld F is said to be integrally closed if whenever a E F is integral over R then a E R Lemma 38 Let 0 be a discrete valuation ring of a eld F Then 0 is integrally closed Proof Let a E F be integral over 0 Then there are r E O with anrn1a 1 r0 0 Let i be a valuation on F whose valuation ring is 0 Then from the equation 7a rn1a 1 r0 we have iiia i a Q 0512175171 ora 0SI I 151711T1 iia If this minimum occurs at i j then iiia 2 1177 iia 2 iia since r E 0 Then n 7 iia 2 0 which forces ia 2 0 Thus a E O as desired 1 We will use the following lemma to help describe the discrete valuation rings of although it has other uses Lemma 39 Let 0 be a discrete valuation ring of F Then 0 is a mamirnal subring of F That is ifA is a ring with O Q A Q F then A O orA F Proof Suppose that A is a ring with O C A Q F Let a E A 0 Then a 1 E 0 Suppose that t is a uniformizer of 0 Then a 1 6 t0 so a 1 t u for some unit u E 0 So a t u 1 Because 0 Q A we see that tn lua t 1 E A We use this to prove that A F Let x E F be nonzero and set in Then t mx E 0 So z tmc for some 0 E 0 However since t E O and t 1 E A both are in A so t E A Thus z tmc E A which proves that A F D We nish this section by discussing valuation rings of a eld and of a sub eld Let KF be a nite dimensional extension of elds and let 0 be a discrete valuation ring of K Suppose also that 1 is a discrete valuation of K whose valuation ring is 0 It is immediate to see that Ulp is a valuation on F once we see that Ulp is not identically 0 If Ulp is the 0 function then F Q 0 Since K is algebraic over F every element of Kwould be integral over 0 which would force K 0 since 0 is integrally closed This is false thus Ulp is nontrivial The valuation ring of Ulp is clearly F W 0 We will use this in the following way Let Fk be an algebraic function eld in one variable and let x E F be transcendental over k Then is a nite extension lf 0 is a discrete valuation ring of F then 0 is a discrete valuation ring of We will be able use this idea once we know what are the discrete valuation rings of We describe these rings in the next section 32 Discrete Valuation Rings of In this section we determine the discrete valuation rings of Recall that we have shown that if k is algebraically closed then is the function eld of the af ne curve Zy and that there is a 1 1 correspondence between points 10 on this curve and the discrete valuation rings kmm z of We generalize this example by allowing k to be arbitrary We will see more valuation rings which correspond to non rational points of Let 0 be a discrete valuation ring of Then z E O or z 1 E 0 We rst assume that z E 0 Then Q 0 Let P be the maximal ideal of O and set M P klz a prime ideal of Then Mle Q 0 since M Q 0 P and O P 0 is the set of units of 0 Since 0 31 kz the ideal M is nonzero Since nonzero prime ideals of are maximal we now know that M is maximal Therefore M for some irreducible polynomial We have seen that klzkpm is a discrete valuation ring of Moreover since klzlmm Q 0 and since discrete valuation rings are maximal subrings of their quotient eld we have 0 pm This proves that any discrete valuation ring of that contains z is of the form klzlwm Now suppose that z 0 Then z 1 E O furthermore z 1 E P else z is a unit in O and then x E 0 We consider the ring kz 1 which is isomorphic to the polynomial ring via the map fx 1 gt gt Thus we may think of z 1 as the variable of the polynomial ring klz ll By the previous argument M kz 1 P is a maximal ideal of klz ll However z 1 E M so z l Q M However z 1 is clearly irreducible so M z l Continuing to mimic the argument above this forces 0 kz 1171 We have thus shown that the discrete valuation rings of are klzlww pz E is irreducible U klzilkfg To make a geometric correspondence if X A1 K with K an algebraic closure of k and if a E K is a point of X let pz be the minimal polynomial of a over k Then pz is an irreducible polynomial and arguments similar to those we saw earlier show that OpXk pm Since every irreducible polynomial pz has a root in K every valuation ring of 31 the form klxkpm occurs as the local ring of a point of X These account for all but one of the discrete valuation rings of If we consider the projective line P1K by viewing A1 Q P1 via a gt gt a 1 then we have accounted for the local rings of all but one point the point 1 0 If we view A1 inside P1 in a different way via a gt gt 1 a then these two copies have a great deal of overlap If a 31 0 then a 1 1 a l Therefore if we interpret z E as the rational function for which za 1 a then x1 a l a Therefore z 11 a l a Thus z l acts as the variable77 on the second af ne piece This indicates that replacing the rst af ne line by the second should correspond to replacing x by d l Since the valuation ring of 0 1 is klzlm by this symmetry argument the valuation ring of 1 0 is klz lkfi Therefore by considering the projective line instead of the af ne line we account for all discrete valuation rings of 33 Discrete Valuation Rings of Fk Let Fk be an algebraic function eld in one variable A discrete valuation ring of F that contains h is said to be a discrete valuation ring of If h is a nite eld then an exercise shows that any discrete valuation ring of F contains k Therefore in this case any discrete valuation ring of F is also a discrete valuation ring of Lemma 310 Let O be a discrete valuation ring of If k is the eld of constants of Fh then k Q O Proof Let a E F be algebraic over k We wish to show that a E O Since a is algebraic over k there are 04 E h with a an1a 1 040 0 Note that each 04 E k Thus a is integral over O Since O is integrally closed in F we see that a E O as desired D A place of Fk is a maximal ideal of a discrete valuation ring of Let PF be the set of all places of This set takes the place of an algebraic curve if h is algebraically closed then there is a 1 1 correspondence between points on a nonsingular projective curve X and places of To mimic the notation of algebraic curves we typically denote places of Fk by P and we denote the valuation ring corresponding to the place P by Op Since Op is a discrete valuation ring P tOp for some t such an element is called a uniformizer of Op Given a place P the residue eld of P is de ned as kP Op P Furthermore we de ne the degree of P by degP The eld h is a sub eld of hP since h is a subring of Op and h P Thus h g embeds in OpP Proposition 311 Let Fk be an algebraic function eld in one variable and let P 6 PF Ifx E P then degP S F lt 00 Proof First note that any element a E k is a unit in Op since both a and a 1 are in Op Thus if z E P then x is transcendental over k so F lt 00 We are left to prove 32 that degP S F Let 11 an 6 Op such that their images 71 Q in kP are linearly independent over k We claim that 11 an is linearly independent over Note that since z E P Q Op we have Q Op Furthermore P since z E P and is a maximal ideal of Thus Mka Q Op Since Mka is a discrete valuation ring of kz we have Op Irma The residue eld of the valuation ring kmm is k since kmmmkmm E k Now to prove our claim suppose that there are c E with Clo 0 lf 1 is the valuation on corresponding to the valuation ring klxlm by dividing through by the coef cient 0 that has smallest value with respect to 1 we may assume that the c are elements of the valuation ring klzlm and that at least one has value 0 Reducing modulo P we have CT 0 Now 07 lie not just in the residue eld kP OpP but in fact the lie in the residue eld h of Irma Therefore this equation is a linear dependence relation over h of the 17 Since 71 417 is independent over h each 07 0 However one of the c has value 0 so 07 31 0 This contradiction shows that 11 an is linearly independent over kx as desired 1 Let f E F We view f as a function on PF by de ning fP T E OpP kP if f 6 Op If f Op then we do not de ne fP We call P a zero of f if fP 0 and P a pole of f if fP is not de ned Since fP 0 if and only if f E P and fP is not de ned if and only if f Op if Up is the valuation of F corresponding to Op then P is a zero of f if and only if opf gt 0 and P is a pole of f if and only if opf lt 0 If P is a zero of f then the positive integer opf is called the order of the zero P If P is a pole of f then the positive integer 71f is called the order of the pole P Note that P is a zero of f if and only if P is a pole of f l We will prove that every nonconstant f E F has only nitely many zeros and only nitely many poles In fact we will give much more detailed information about the number of zeros and poles and their orders If F kz then these de nitions of zero and pole are the same as we gave earlier as is the de nitions of order For if x g is a rational function having a zero of order r at a then g x 7afglx for some polynomial 91 with 911 31 0 Also we assume that is in reduced form so ha 7E 0 lf 1 is the valuation corresponding to the discrete valuation ring klzkka then oltpz ogx 7ohx r since z 7 a generates the maximal ideal P of this valuation ring and so x E PTPT1 The argument for poles is similar We now show that every f E F has at least one zero and at least one pole provided that f is not a constant We use a theorem from the handout on Dedekind domains if KF is a nite extension if A is a Dedekind domain of F and if B is the integral closure of K then B is a Dedekind domain Recall that a discrete valuation ring is a special example of a Dedekind domain being nothing more than a Dedekind domain that is also a local ring Proposition 312 Let x E F be nonconstant Therz x has at least one zero and at least one pole Proof Let x E F be nonconstant and consider the nite extension Let B be the integral closure of klxlm in F Then B is a Dedekind domain of F If M is a maximal ideal 33 of B then BM is a discrete valuation ring Moreover M Mka is a nonzero prirne ideal of klzlm so M kmm skmm Therefore z E M Thus if we set BM Op with P 6 PF then x E P as M P B Therefore P is a zero of x By using the same argument with z replaced by 4 we see that z 1 has a zero Q Then Q is a pole of x B Let Fk be an algebraic function eld in one variable and let k be the eld of constants For any P 6 PF we have seen that k Q Op The eld k then ernbeds in the residue eld OpP kP Via f gt gt fP Therefore if f E k is nonzero vpf 0 for all P Therefore fP f P E kP is equal to f under the embedding k Q Op Thus by Viewing f as a function on PF it is a constant function This justi es our narning k to be the eld of constants 4 Divisors and the RiemannRoch Theorem ln algebraic number theory the main ring of study is the ring of integers in an algebraic number eld F This is the integral closure A of Z in F The ring A is a Dedekind domain Associated to A is the group of fractional ideals a group under multiplication of ideals The principal ideals form a subgroup and the resulting quotient group is called the ideal class group of A The study of this group has important consequences For example A is a unique factorization domain if and only if the class group is 0 We start by discussing the geometric analogue of this notion which are divisors principal divisors and divisor classes 41 Divisors of a Function Field Let Fh be an algebraic function eld in one variable By replacing h by the eld of constants we will assume that h is the exact constant eld of Fh in this chapter Recall notation we used in the previous chapter Pp is the set of places of If P 6 Pp then Op is the corresponding discrete valuation ring and vp is the discrete valuation associated to Op De nition 41 Let Fh be an algebraic furietiori eld in one variable A divisor is an elemerit of the free Abeliari group on Pp That is a divisor is a riite sum anP where anZ aridPEPp As part of this de nition two divisors 2P in and 2 mpP are equal if and only if np mp for all P In particular 2P in 0 if and only if each np 0 If D 2P in is a divisor we de ne the support of D to be suppD P np 31 0 This is a nite subset of Pp If P 6 Pp and D is a divisor we set vpD to be the coef cient of P in D Then vpD 0 for all but nitely many P There is a partial ordering S on the set Dp of divisors of Fh given by D S E if vpD S vpE for all P 6 Pp A divisor is said to be positive if D 2 0 Note that a divisor need not be positive or negative For example if P Q 6 Pp then the divisor P 7 Q is neither positive or negative The degree of a divisor is de ned as degD 2P vpD degP This is a nite sum since only nitely many vpD are nonzero and since degP is nite for any P If D is a positive divisor then degD 2 0 Furthermore if D S E then degD S degE Lemma 42 Let D arid E be divisors ori Theri degD E degD degE arid deg7D 7degD Thus the degree map is a group homomorphism from Dp to Z Proof Let D 2P in and E 2P mpP Then D E mep mpP Thus degD E Zmp mp degP 2 up degP 2 mp degP P P P degD degE Also deg7D 2P 771p degP 7 2 up degP so deg7D 7 degD D 35 De nition 43 Let f E F The divisor f off is de ned as f 2P vpfP The zero divisor off is f0 2 be vpfP and the pole divisor is ee ZuPUKO 7vpfP A divisor D is said to be principal if D f for some f E F If f is a constant then vpf 0 for all P so f 0 Note that f f0 7 ee In order to see that the de nition above makes sense we need to prove that any f E F has only nitely many zeros and only nitely many poles This result is perhaps the most important theorem leading up to the Riemann Roch theorem The following lemma will allow us to use one of the results in the Dedekind domain handout in proving this theorem Lemma 44 Let Fk be an algebraic function eld in one variable and let x E F be transcendental over h Let O be a discrete valuation ring of Fk with mawirnal ideal P such that z E P Then O kzm Proof Since z E P Q O and since h Q O by hypothesis Q O Let M P km a prime ideal of We have x E M so Q M However z is irreducible in km so is a maximal ideal Thus M Furthermore since Q O P O the group of units of O we have hmm Q O Finally as hmm is a discrete valuation ring hmm O since discrete valuation rings are maximal subrings of their quotient elds D Theorem 45 Let x E F h Then degz0 deg F Therefore x has only nitely many zeros and only nitely many poles Proof Let x E F be nonconstant Suppose that P1B are zeros of x Thus z E P for each i By the previous lemma Op contains hzm Conversely if P is a point with kmm Q Op then x E P since Op hmm zhzm The zeros of z are then precisely those P for which Op contains Irma We proved in the Dedekind domain handout that there are only nitely many discrete valuation rings of F that contain the discrete valuation ring Irma Therefore x has only nitely many zeros Moreover if P is a zero of order ei then vpz ei Let B be the integral closure of hmm in F If M P B then zB Mf1 Mf which was proved in the handout Thus by the handout F 1 eifi where f hzmzhzm The eld kmkm is equal to h and BMl OpiPZ Thus f degP Therefore F edegP and since z0 eiPi we have degz0 F Finally the pole divisor of z is equal to the zero divisor of x l Thus this argument applied to z l shows that degltltzgtmgt degltltr1gtogt F km E Corollary 46 Let f E F Then degf 0 Proof In the previous theorem we proved that degz0 degdoo Since x0 7 moo and the degree function preserves addition degz deg z0 7 degzoo 0 D Corollary 47 Let D be a divisor with degD 0 Then the following conditions are equivalent 1 D is principal 2 dirnD 2 1 3 dirnD 1 Proof We have seen in an example that if D f is principal7 then LD f lk There fore7 dimD 1 This proves 1 gt The proof of 3 gt 2 is obvious Finally7 for 2 gt 17 suppose that dimD 2 1 Take f E LD be nonzero Then D f 2 0 This is then a positive divisor of degree equal to degD 0 by Corollary 46 This forces Df 07 so D7f f l is principal D By using principal divisors7 we can de ne a new group First7 note that f g fg and 7f f l Therefore7 the set of principal divisors is a subgroup of DF The divisor class group CF is de ned to be the quotient group of DF modulo the subgroup of principal divisors If D and E are divisors that have the same divisor class in CF then we say that D and E are equivalent In other words7 D is equivalent to E if there is an f E F with D E We will write D N E if D and E are equivalent De nition 48 Let D be a divisor ori Theri LD f E F D f 2 0 U Example 49 If 0 is the trivial divisor7 then LO k To see this7 if f E LO is nonzero7 then f 0 2 07 so f 2 0 Therefore7 vpf 2 0 for all P 6 PF This forces f to have no poles Since any nonconstant function has a pole7 f must be a constant Conversely7 if f is a constant7 then f 07 so f 0 2 0 This proves the equality LO k Example 410 Let D lt 0 be a negative divisor Then LD 0 For7 if f E LD was nonzero7 then D f 2 0 However7 this would force degD 2 0 Since degD degD degf degD lt 0 by Corollary 467 this is impossible Thus7 f 0 Example 411 Let D f be a principal divisor We claim that LD f lk First7 it is clear that f lk Q LD since if a is any constant7 then af l f f l f 0 Conversely7 if g E LD7 then 9 f 2 0 Then gf is a positive divisor This forces gf to be a constant since any nonconstant function has a pole So7 gf a for some a E k and so 9 af L E f lk Example 412 Let F kz7 and let Po0 correspond to the discrete valuation ring klz lhfi We rst note that if v00 is the valuation that corresponds to this ring7 then voofz idegfx for any f E To see this7 we have vooz 1 1 since z 1 is a generator of P00 Therefore7 voox 71 idegx For f ans a0 a polynomial of degree ii7 we can write f an an1x 1 aox The sec ond term has value 0 since vooan 0 and all other terms have positive value Thus7 7Joof96 WWW in 7 degf9 Let n be a positive integer and consider the divisor nPoo Then LnPoo E nPoo 2 0 This means x E LnPoo if and only if vooltpz 2 7n and opltpz 2 0 for all P 31 Poo If we write x f in reduced form then for any irreducible factor p of 9z we have opmltp lt 0 since this is impossible we see that 9a 1 Therefore x E By our description of 1100 we then see that LnPoo E degfx S This is a k vector space with basis 1 x xn so its dimension is n 1 Note that degPoo 1 so degnPoo n In other words we see that dimkLnPoo 1 degnPoo Example 413 Let F kz and let a1 an E k Consider P1 Pn corresponding respectively to the irreducible polynomials z 7 a1 z 7 an We calculate LD where D e1P1 enPn given positive integers ei Let x E LD Then opltpz 2 7e Write x in reduced form thus f and 9a have no common term Then the poles of a are the zeros of 9z and so a can be a zero of 9a of order at most ei Note that a need not actually be a pole of Since 9a can have no zeros other than the various a we see that 9a x 7a1dl x 7 any with 0 S d S e for each i If we write x 7 a1d1x 7 any then 11p 2 7e as desired Furthermore if P is any other place other than Poo then opltpz 2 0 since the value of the denominator is 0 We only need to consider Poo whose valuation boo satis es voofx 7 degfx for any polynomial Therefore vooltpz di 7 degf Thus a 1 f 96 MD 3 fW E kl17degf S MD WW fz 6 km 0 S di S i7deE ltfgt S M i In fact one can give a slightly simpler description of LD as Therefore LD is isomorphic to the space of all polynomials of degree 3 21 e degD This is a space of dimension 1degD lt is not a coincidence that dimkLD 1degD We will see that this is a general fact for divisors of of positive degree We point out some simple properties about the spaces LD Lemma 414 Let Fk be an algebraic function eld in one variable and let D be a divisor on Z ff 6 F then f E LD if and only ifopf 2 7opD for all P 6 PF 2 The set LD is a k subspaee of F 3 The space LD is nonzero if and only ifD is equivalent to a positive divisor 4 IfE is equivalent to D then LE E LD as h vector spaces Proof 1 We have f E LD if and only if D f 2 0 if and only if vpD vpf 2 0 for all P 6 PF and this occurs if and only if vpf 2 7vpD for all P 6 PF This proves 1 For 2 let fg E LD Then vpf 2 7vpD and vpg 2 7vpD for all P By the de nition of a discrete valuation vpfg 2 min vpfvpg Thus vpfg 2 7vpD for all P so f g E LD Also if f E LD and a E h then of a f Therefore D of D f 2 0 so of E LD Thus LD is a h vector space To prove 3 recall that D is equivalent to D f for any f E F If f E LD is nonzero then D f 2 0 This is a positive divisor equivalent to D Conversely if D is equivalent to a positive divisor E then E D f for some f E F and then f E LD since E 2 0 Thus LD is nonzero This proves Finally for 4 suppose that E is equivalent to D Then E D f for some f E F We de ne a map p LD a LE by ltpx xf Then Eltsoltzgtgt Eltzfgt Eltzgtltfgt Dltzgt Therefore E 2 0 if and only if D 2 0 Therefore z E LD if and only if ltpx E This shows that p does map LD to LE and that p is surjective The map p is h linear since ltpx y x yf zf yf ltpx ltpy for any Ly E LD and ltpax azf azf altpx for any a E h Finally p is injective since if ltpx 0 then xf 0 Since f 31 0 and F is a eld z 0 Therefore LE LD as h vector spaces D If D is a divisor we set dimD dimkLD We will show later that LD is a nite dimensional h vector space Corollary 415 Let Fk be an algebraic function eld in one variable 1 Let D be a divisor with degD lt 0 Then dimD 0 2 IfE and D are equivalent divisors then degE degD and dimE dimD Proof To prove 1 iff E LD is nonzero then Df 2 0 Then degDf degD lt 0 while degD 2 0 since D f 2 0 This is a contradiction Show LD 0 so dimD 0 For 2 we proved in the lemma that if E and D are equivalent then LE g LD Therefore dimE dimD We have already seen that degE degD since the degree of any principal divisor is 0 by Corollary 46 D We now start to investigate the dimensions of divisors The next lemma shows that the quotient space LELD is nite dimensional if D S E This is the rst step toward showing that LE itself is nite dimensional Lemma 416 Let D and E be divisors with D S E Then LD Q LE and the quotient vector space LELD satis es dimkLELD S degE 7 deg 39 Proof We rst suppose that E DP for some P 6 PF Then degE 7degD degP Choose t E F with vpt opE Note that opE opD 1 If f E LE then oQf 2 711QE for all Q In particular opf 2 7opE 7opt Thus optf 2 0 We de ne a map p LE 7 kP by f tfP This is well de ned since tf 6 Op and so tfP H E OpP kP is de ned We show that p is a k linear rnap its kernel is LD and that it is surjective This will prove that LELD is isomorphic to a k subspace of kP as k vector spaces and so dirnkLELD S dirnkkP degP degE 7 degD First if to E LE7 then rf 9 W 9P W t9P ifP t9P7 SO rltf 9 W ltP9 Next if a 6 k then Maf tafP atfP WU The second to last equality holds because a is a constant function so taf P aP tf P atfP Alternatively with the identi cation of k as a sub eld of kP under the map a gt gt 6 we have tafP W 6 H 6 f Thus p is k linear Finally to see that kerltp LD if f E LD then opf 2 7vpD 7opE 1 Thus opf gt 7opE 7opt so optf gt 0 Therefore tfP 0 so f 0 Conversely if f E LE with f 0 then tfP 0 so optf gt 0 This yields opf gt 7opt or opf 2 17 opt 17opE 7opD For any Q 31 P we have UQltDgt UQE so since oQf 2 7UQE we have oQf 2 7UQD All these inequalities say that f E LD This nishes the proof in the case that E D P In the general case we may write E D P1 P with the P not nec essarily distinct Set D D P1 P so E D By the previous para graph dirnkLD1LD S degP for all i By induction if dirnkLD1LD S degP1 degP1 then dimkLDiLD dimkLDiLDi71 f dimkLDi71LD S degP1 degPi71 degPi because LDiLD BodMD and the dimension of a quotient is the difference of the dimensions of the terms So induction shows that dirnkLELD S 2 degPi degE 7 degD 1 HZ LDiLDi71 If D is a divisor we may write D D1 7 D where D1 is the sum of the terms in D with positive coef cients and D is the negative of the sum of the terms with negative coef cients Then D1 and D are both positive divisors The following result shows that the spaces LD are actually nite dimensional over k Proposition 417 Let E be a divisor Then dirnE S degE 1 Therefore LE is a nite dimensional k Ueetor space Proof We have E1 7 E E a positive divisor so E 3 E1 Therefore LE Q LE lt suf ces to prove that dimE S degE 1 Therefore by replacing E by E1 we may assume that E is a positive divisor By the previous lemma applied to E and D 0 we have dimkLEL0 S degE 7 deg0 However L0 k so dim0 1 Also deg0 0 Thus dimkLE dimkLEL0 1 3 degE 1 as desired D This result shows that for a positive divisor E we have dimE 7 degE S 1 In fact if D is any divisor with degD gt 0 the same inequality holds For if LD 0 then dimD 7 degD 7degD lt 0 However if f E LD is nonzero then E D f is positive Since dimE dimD and degE degD the inequality dimE 7degE S 1 says that dimD 7 degD S 1 Written another way 0 3 1 degD 7 dimD This gives a lower bound for this expression We will now show that there is an upper bound for this expression as D ranges over all divisors Lemma 418 There is a constant 7 depending only on Fh such that ifE is any divisor of Fh then degE 7 dimE S y In particular 1 degE 7 dimE is bounded above by y 1 for any divisor E Proof We give a technical looking argument to prove this lemma Let x E F be nonconstant and set E lfn F then Theorem 45 shows that degB ii Let ul un be a h basis for F Let C be any positive divisor containing u1 un To see that such a divisor exists let P1 P be a set of points containing all the poles of all the iii For each j let e 2 max7vpju 1 g i g Then vpju 2 7ej for all i Therefore if C 7 eij then u E LC for each i We next claim that for any positive integer l the divisor lB 0 satis es dimlB C 2 l1n To see this we note that u E LlB C for all j and for all i with 0 S i S l For lB C lB C Now uj 0 2 0 Furthermore izlB i07izooldoo i0 l7i Since x0 and Moe are positive and 0 S l7i the divisor lB is positive and so lB C is positive There are l 1n elements in the set xiuj 0 S i S l 1 S j S n and these elements are k linearly independent since z is transcendental over h and the u are hx linearly independent This proves our claim that dimlB C 2 l 1n On the other hand Proposition 417 and Lemma 416 applied to E lB C and D C shows that dimlB C 7 dimlB S degC These two inequalities show that l1n S dimlB C S dimlB degC Written another way as deglB ldegB ln deglB 7 dimlB S y if y n 7 degC This holds for all positive integers l Let A be any divisor Our nal claim is that there are divisors A1 and D such that A1 2 A7 A1 D7 and D S lB for some l Given this claim7 we have degA 7 dimA S degA1 7 dimA1 degD 7 dimD S deglB 7 dimlB where the rst and third lines follow from Lemma 4167 the second line holds since A1 and D are equivalent7 and the nal line comes from the inequality above The only thing remaining is then to prove the claim Choose A1 any divisor with A1 2 A By Lemma 4167 we have dimlB 7 A1 2 dimlB 7 degA1 2 degUB 7 v 7 degMi nl 7 y 7 degA1 2 0 if l is large enough7 since n 2 1 This nishes the proof of the lemma D This lemma shows that maXdegD 7 dimD 1 D E DF exists We de ne the genus of Fk to be the integer g max degD 7 dimD 1 D E DF By de nition7 we have g 2 deg0 7 dim0 1 Since deg0 0 and dim0 17 we see that g 2 0 The de nition and existence of g is usually stated as Riemann7s theorem Theorem 419 Riemannls Theorem Let Fk be an algebraic function eld in one variable and let g be its genus Z IfD is any divisor of Fh then dimD 2 degD17 g 2 There is a constant c depending only on Fh such that ifdegD 2 c then dimD degD 1 7 g Proof The rst statement is just the de nition of g To see the second7 by the de nition of g7 there is a divisor A with g degA 7 dimA 1 Let c degA g If degD 2 c7 then dimD 7 A 2 degD 7 A 1 7 g 2 degD 7 degA17 g degD17c21 Therefore7 LD 7 A is nonzero Let x E LD 7 A be nonzero Set D D Then D 2 A since D 7 A 2 0 Then degD 7 dimD degD 7 dimD 2 degA 7 dimA g 7 1 Therefore7 dimD S degD 1 7 g Since the reverse inequality holds by de nition of g7 we have dimD degD 1 7 g D 42 The RiemannRoch Theorem Riemann7s theorem is a very useful fact about divisors but because it gives an inequality it does not always allow us to determine the dimension of a divisor The Riemann Roch theorem is an improvement of this result Theorem 420 RiemannRoch Let Fh be an algebraic function eld in one variable IfC is a canonical divisor of Fh then for any divisor D we have dimD degD 1 7 g dimO i D To save time and tedious details we aren7t going to prove this theorem in class However we will de ne what is a canonical divisor To do this we discuss derivations and di feren tials We point out that Section 3 of Chapter 4 of Stichtenoth gives the relation between differentials in the sense we will introduce with the differentials he de nes in Chapter 1 A di erential of Fh is a F linear combination of formal symbols df with f E F These symbols satisfy the properties da 0 if a E h df g df dg and dfg gdf fdg The F vector space of differentials is denoted Sipk We formally de ne Sipk as the quo tient of the F vector space with basis all symbols of the form df for f E F modulo the submodule generated by all elements of the form da for a E h and all elements of the form df g 7 df 7 dg and all elements of the form dfg 7 gdf 7 fdg For the rest of this discussion we assume that F contains an element x E F transcendental over h for which is a separable extension We state without proof that if h is a perfect eld then such an element exists In particular if h is a nite eld such an element exists We need to know that Sipk is a 1 dimensional F vector space It is fairly easy to prove that its dimension is at most 1 and we do so below to see that it is nonzero we need a little more information about Sipk First we de ne a related notion If M is an F vector space then a h derivation is a function D F 7 M is a h linear function such that Dab DabaDb for all ab E F We point out the universal mapping property for Sipk if D F 7 M is a derivation then there is a unique F linear map p QFk 7 M such that Dz dx for all z E F The proof of this fact is not hard although we do not give it Proposition 421 With x as above the F vector space Sipk is one dimensional with basis dz Proof By assumption is separable Let t E F If pT E T is the minimal polynomial of t over hz then pT has no repeated roots Let pT 20 aixTi Then 0 20 aiti Applying the de ning rules for Sipk we see that dax a zd where a is the derivative of ax Thus we have 0 Z aztid aiziti1gt dt Zaxd idx ptdt 39 i0 i0 n i0 t so vpdxdt vpdxdt For a general differential in fdz we set w f Therefore any two differentials have equivalent divisors so the class of the divisor of a differential is uniquely determined Any such divisor is called a canonical divisor Example 422 Let F We compute a canonical divisor of We can use z since F is obviously separable over First consider a point P that is the maximal ideal of klxkpm for some irreducible polynomial Then p is a generator of the maximal ideal We have dpx p xd Now since p is not divisible by px we see that vpmp 0 Thus vpmd 0 This handles all but one point The remaining point is Poo the point at in nity The element z 1 is a generator of this maximal ideal Since dx l ix zdz and 0000 7 degf we see that 000dz 7000z 2 deg 2 72 Therefore the canonical divisor dm is 72PM Since we see that deg72Poo 72 and its degree is 29 7 2 as we will see in the next corollary this yields 9 0 We will give another proof of this in the next section We give some elementary consequences of the Riemann Roch theorem Corollary 423 Let C be a canonical divisor of Then degC 2g 7 2 and dimC 9 Proof The Riemann Roch theorem applied to D 0 yields 1 1 7 g dimC Therefore7 dimC g Next7 applying the theorem to D 0 yields dimC degC17gdim07 or g degC17 g dim07 which then gives degC 2g 7 2 B These formulas for the degree and the dimension uniquely determine the divisor class of C as the next corollary shows Corollary 424 A divisor D is a canonical divisor if and only if degD 2g 7 2 and dimD 2 9 Proof Let D be a divisor with degD 2g 7 2 and dimD 2 g If C is a canonical divisor7 we need to show that D N C By the Riemann Roch theorem7 dimD degD 1 7 g dimC7D Since dimD 2 g and degD 29727 we get 9 S 297217gdimC7D7 or 1 S dimC 7 D Since degC 7 D degC 7 degD 07 Corollary 47 shows that C 7 D is principal Thus7 D N 0 Note that this forces dimD g D We can improve on the second statement of Riemann7s theorem Corollary 425 IfD is a divisor with degD gt 29 7 2 then dimD degD 1 7 9 Proof Suppose that degD gt 29 7 2 Then degC 7 D lt 07 so dimC 7 D 0 The corollary then follows from the Riemann Roch theorem 1 We will not say very much about how one can nd the genus of a function eld in fact7 it is dif cult in general to calculate the genus We do state without proof one fact lf Fk is the function eld of the af ne curve Zfxy for some polynomial ay 6 May of degree d Then the genus g of Fk satis es 9 S d 7 1d 7 27 and equality holds if the projective version of this af ne curve is nonsingular We will use this below to show that the function eld of an elliptic curve has genus 1 43 Fields of Genus 0 and 1 In this section we classify elds of genus 0 and 1 We rst consider genus 0 To start7 let us see a second argument for why the genus of the rational function eld is 0 Let P Pgt07 a point of degree 1 The valuation vp corresponding to Op is given by vpltpz 7 degltpx If D nP7 then by Riemann7s theorem7 if n is large enough7 then dimD degD 1 7 9 Since degD n7 it is enough to calculate LD As we saw in an earlier example7 the space LD is the set of all f E with degfz S n7 which has dimension n 1 Then 9 degD 7 dimD 1 0 Note that Po0 is a rational point of For a partial converse suppose that Fk has genus 0 Recall that we are assuming that k is the exact eld of constants Moreover suppose that there is a rational point P 6 PF That is suppose that degP 1 By Riernann7s theorern dirnP 2 degP 1 2 Let x E LP be nonconstant Then z exists since the set of constants has dimension 1 and dirnLP 2 2 Because z E LP we have vpz 2 7vpP 71 and vQz 2 0 for all Q 31 P As the nonconstant z must have a pole we see that vpz 71 Then z 1 E P By Theorem 45 F degxoo However since P is the only pole of x and since vpz 71 the pole divisor moo P So degzoo degP 1 which shows that F 1 Therefore F kz so F is a rational function eld in one variable over k The assumption that F has a rational point is necessary Let k R and let F be the function eld of the curve 2 y2 1 0 over k Then F It can be proven that Fk has genus 0 but that F has no rational point Thus F is not isomorphic to a rational function eld in one variable over k However if k C then the genus of Fk is also 0 but any point is a rational point so k171 7 x2 is isomorphic to a rational function eld in one variable In particular an exercise will show that F is generated over k by 239 7 x71 V717 2 We now consider elds of genus 1 We restrict to the case chark 31 2 We will show that a function eld Fk with a rational point has genus 1 if and only if F for some y E F with y2 f for some cubic polynornial fx with no repeated roots in k This means that such a eld is precisely the function eld of an elliptic curve First suppose that F kzy where z is transcendental over k and with y2 fx for some cubic f with no repeated roots We rst show that Fk has a rational point P Consider the extension a quadratic extension Let P be any point for which Op extends the valuation ring k 1f1 of Let 6 be the rarni cation index and f the residue degree Since the residue eld of kx 1f1 is k we see that f degP Thus to show that P is a rational point we need to show that f 1 Recall from the Dedekind dornain handout that 6f 3 F 2 Thus if we prove that e 2 then f 1 is forced upon us To see this let Up be the valuation corresponding to Op Then 1mm woo where UGO is the valuation on corresponding to kx 1fi Since vooltpz 7degltpz and y2 fx we have 1100fz 73 So 2vy 0y2 736 For My 6 Z this forces 6 2 as desired To see that he genus is 1 let gy y2 7 Then 9 is a polynomial of degree 3 Its hornogenization is 122 7 23fzz We have stated that the corresponding projective variety is nonsingular since f has no repeated roots Therefore by the genus formula given earlier 9 3 713 7 2 1 Alternatively we could compute the genus by computing the degree of a canonical divisor If we do so we would see that degdz 0 Since this degree is 29 7 2 we obtain 9 1 Now suppose that Fk has genus 1 and that Fk has a rational point P We produce Ly E F such that z is transcendental over k and y2 f for some cubic fx with no repeated roots in k By the Riemann Roch theorern if degD gt 29 7 2 0 then dirnD degD 1 7 g degD Therefore dirnnP n if n gt 0 Furthermore LnP Q Ln 1P since 71 1P 2 nP There are then elernents z E L2P LP and 46 y E L3P L2P So for all Q 31 P we have vQx 2 0 and vQy 2 0 while at P we have vpz 72 and 11py 73 The element x is transcendental over k since it is not constant as vpz 31 0 Since P is the only pole of z or y we see that moo 2P and moo 3P From Theorem 45 this implies that F 2 and F 3 So and have no intermediate elds since these degrees are prime Therefore F We wish to show that y satis es a cubic over To see this note that the seven elements 1z23yzyy2 all lie in L6P Since dim6P 6 these elements are linearly dependent over k There are constants with ayz bzy By dx3 6x2 bx j 0 At least one of abc is nonzero since z is transcendental over k From this it follows that a 31 0 else y E kx which is not true By dividing by a we may assume that a 1 Similarly d 31 0 else z satis es a quadratic over My which is impossible since F and F 3 By completing the square we have y bz 02 7 f 0 where f is a cubic in x Finally if we replace y by y bz c then F is still equal to kxy and y2 It remains to show that f has no repeated roots lf instead we can write f 04x 7 2z 7 y for some 0467 6 k then i 52 04 7 y Replacing y by 7 B we still have that my generate F over k However since yz az 7 y we see that z E Therefore F My and so F is a rational function eld in one variable over k However this would yield 9 0 while we are assuming that g 1 Thus f has no repeated roots in k 5 Goppa Codes We now put the Riemann Roch theorem to work to de ne a class of codes These codes were discovered by Goppa in 1981 Let Fqu be an algebraic function eld in one variable We assume that E is the exact constant eld We denote the genus of Fqu by 9 Let P1 713 be rational points in PF Recall that P is a rational point if degP 1 Thus7 the residue eld of a point is E if and only if the point is a rational point Consider the divisor D P1 Pn7 and let G be a divisor whose support is disjoint from 1317 Pn De ne an evaluation function evD LG 7 F by evDf 1317 7fPn To see that this is well de ned7 if f E LG7 then Upif 2 0 since P1 is not in the support of G Therefore7 fPZ is de ned and is an element of E1 FAB for each i It is easy to see that evD is Fq linear For addition7 we have eVDf g f 9P17 7 f 9Pn 90 fPn 9Pn fPn 9P1 79Pn eVDf eVD9 If a E E1 then eVDaf afP17 7 afPn 0P1fP17 70PnfPn afP17 afPn aeVD ltMf since a is a constant function The image of evD is then a Fq subspace of F2 De nition 51 With notation as above the Goppa code CLDG is the image of evD That is OLDG fP1fPn f E LG From basic facts about divisors7 we can determine the parameters of a Goppa code Theorem 52 The parameters of the Goppa code CLD7 G has length n degD dimen sion h dimG 7 dimG 7 D and distance d 2 n 7 degG Proof The length of the code is the number n of rational points P17Pn Since D 21 P1 and each R has degree 17 we see that n degD To determine the dimension7 since evD is linear7 we see that h dimLG 7 dimkerevD We show that kerevD LG 7 D Let f E kerevD Then 0 for all i Then Upif 2 17 so Upif 2 7UPZG 7 D 1 If Q is any other point7 then oQf 2 7UQG 7UQG 7 D since f E LG Therefore7 we see that f E LG7D Conversely7 iff E LG7D7 then f E LG since G7D S G and fPZ 0 for allisince Upif 2 7opiG7D 1 Thus7 f E kerevD We have then shown that kerevD LG 7 D7 and so h dimLG 7 dimLG 7 D dimG 7 dimG 7 D Finally7 let evDf E CLDG have weight d Then n 7 d of the 48 coordinates of evDf are zero If these zeros are at the points Pil Paid then we see that f E LG7 Pi177 inid This space is then nonzero so the degree of G7 Pi 7 7 Find is nonnegative In other words degG 7 n 7 d 2 0 so d 2 n 7 degG as desired D 7 The Riemann Roch theorem will give us more information about the parameters of this code Corollary 53 Suppose that degG lt n Then evD is injectiue and so h dimG 2 degG 1 7 g and d 2 n 7 degG Furthermore if n gt degG gt 2g 7 2 then h degG 1 7 9 Proof We are claiming nothing new about d If degG lt n then degG 7 D lt 0 so LG7 D 0 Thus kerevD 0 so evD is injective This implies that h dimG From Riemann7s theorem we have dimG 2 degG 1 7 g which yields h 2 degG 1 7 g Finally if n gt degG gt 2g 7 2 then the Riemann Roch theorem yields dimG degG 17gsodegG17g D As an immediate consequence if 2g 7 2 lt degG lt n then kd 2 n17g Therefore if F Fqz is a rational function eld over 19 then g 0 and so h d 2 n 1 However the Singleton bound says that h d S n 1 Therefore h d n 1 In other words Goppa codes associated to FAQE are MDS codes that is codes that attain the Singleton bound The only problem with this is that there are only q 1 rational points of RAME these are the points corresponding to the discrete valuation rings kmw z for a E E1 along with k71m71 So if q 2 we can only build Goppa codes of length at most 3 if we use F Proposition 54 Suppose that degG lt ii If p1 zk is a basis for LG then 1031 1032 1Pn 2031 2032 quot39 Mann an ma ms is a generator matrip for CLD G Proof Recall that a generator matrix for a code is a h gtlt n matrix whose rows form a basis for the code Since degG lt n the map evD LG 7 F is injective Therefore if x1 xk is a basis for LG then evDp1 evDzn is a basis for the image of evD which is CLD G The i th row of the matrix above is simply the vector evDx so the matrix is indeed a generator matrix for the Goppa code D De nition 55 The integer d n 7 degG is called the designated distance of the code CLD G It is a lower bound for the actual distance of the code 51 Goppa Codes coming from FqxFq In this section we will relate BCH codes and Reed Solomon codes to Goppa codes We will see that such codes arise from Goppa codes associated to the function eld Fqz First note that there are q 1 rational points of FqxFq these correspond to the discrete valuation rings Fqzma for a 6 RI and to Fqld lkfi Therefore any Goppa code constructed from this function eld is limited to have its length n satisfy n S q 1 In particular if we work with F2 we can have codes of length at most 3 To avoid repeatedly making reference to the function eld PAME we call a Goppa code a rational Goppa code if it is associated to PAMEr Recall our standard notation for the parameters of a code n is the length of the code h is the dimension of the code and d is the distance of the code We note what are the parameters of a rational Goppa code in terms of the degree of G in the next proposition Proposition 56 Let GLD G be a rational Goppa code with parameters n k and d Then h 0 if and only if degG lt 0 and h n if and only if degG gt n 7 2 Furthermore if 0 S degG S n 7 2 then h 1 degG and d n 7 degG In particular GLDG is an MDS code Proof If degG lt 0 then LG 0 so GLD G 0 If degG gt n72 then degG7D gt 72 2g 7 2 where g 0 is the genus of PAMET Thus by the Riemann Roch theorem dimG7DdegG7D17gdegG7D1degG7n1 Since h dimG 7 dimG 7 D and dimG degG 1 7 9 also by Riemann Roch we see that h degG 1 7 degG 771 1 n Finally if 0 S degG S n 7 2 then dimG 7 D 0 since degG 7 D S 72 lt 0 Thus h dimG degG 1 by Riemann Roch Also d 2 n 7 degG Since h d S n 1 by the Singleton bound this forces d n 7 degG The code is then an MDS code since the Singleton bound is met D Recall the de nition of a Reed Solomon code over Fq Let n q 7 1 and let 04 be a primitive element of Fq We identify the vector space l9 with the vector space of polynomials of degree less than n under the association a0 an1 gt gt a0 an1z 1 The code consisting of all polynomials p of degree less than n with pa pa2 1001 1 0 is a Reed Solomon code of dimension n 1 7 d As we will see this is almost a special example of a Goppa code In fact we look at a larger class of codes which are called generalized Reed Solomon codes To motivate the de nition we look at the original de nition of Reed and Solomon which is different than the de nition we gave Set n q and let 04 be a primitive element of Fq Set 04 0 for 0 S i S q 7 2 and set 041 0 Pick an integer k lt n and let L be the vector space of polynomials in Elm of degree less than h Then the code C fao77faq71 t f E L is the type originally de ned by Reed and Solomon This is a code of length n q and dimension k we note that the linear transformation f gt gt fozofozq1 is injective since k lt n and a nonzero polynomial of degree lt k cannot have n roots Thus dimC dimL k Moreover given any nonzero f since f has at most k71 roots at least n7k1 of the components of the vector f0 0 fozq1 are nonzero This says d 2 n 7 k 1 The Singleton bound shows that d n7k1 and so this code is an MDS code To generalize this de nition let 1 11 0 6 IF have all nonzero entries let Oz 041 Ozn have distinct components in Fq and consider the code GRSkOL1U1f0411nfoznf E L This is called a generalized Reed Solomon code If 1 1 1 and Oz 040 ozq1 as above then this Reed Solomon code is the code originally de ned by Reed and Solomon To see the connection between these codes and the Reed Solomon codes we de ned earlier we rst de ne the amended code 5 of a code C This is the code n1 6 clcn1 619 01cn 6 020 0 i1 If H is a parity check matrix for C then the matrix whose top left part is H whose bottom row has every entry 1 and whose last column has all 0 entries except for the bottom is a parity check matrix for 5 The code 5 has length 1 more than the length of C but whose dimension is the same as the dimension of C The distance of is either equal to the distance d of C or to d 1 depending on whether or not every word in C of weight d has the sum of its coef cients di ferent than 0 We will now show that an extended Reed Solomon code is a generalized Reed Solomon code Let Oz 040 ozq1 with the 04 de ned as above First note that since every element of E1 is a root of 7 z every nonzero element is a root of 74 71 z 71z 7 2 z 1 Thus if c E E1 then 2501 0 if c 31 01 Now let fz 23 ajzj and set c foz og for 0 S 239 S q7 2 lf1 1 q7 k 7 1 then since 25oHquot 0 by the comment above since 1 S lj S q 7 2 Therefore tak ing the Reed Solomon code for d q 7 k we see that 225 cizi lies in this code Un der the correspondence between polynomials and n tuples this polynomial corresponds to co cq2 f0 0 fozq2 So given 00 cq1 E GRSk1 04 there is a unique fz of degree lt n such that og Ci and then 00cq2 lies in the Reed Solomon code of dimension k and distance q 7 k A similar calculation to that above shows that 1 1 20 c 0 since 21 q1 22 22 111 1 fa1 W Z fW amt 10 10 10 10 10 h1 12 12 a0 a 01 l a0a0lt 1 0 i0 i0 Thus GRSk1 01 is the extended code corresponding to the Reed Solomon code for d qk We now show that any generalized Reed Solomon code is a rational Goppa code Proposition 57 Every generalized Reed Solomon code is a rational Goppa code Proof Let G nPoo and let P be the point corresponding to Fqlzlwwi Set D P1 P Then LG f 6 Elm degf lt h and so CAGED fP111fP11 i f E LG fa111fan i f E L is the generalized Reed Solomon code GRSk101 To deal with a general 1 o1 on nd 9 E with gP i for each i This is possible to do with a polynomial of degree n 1 nding such a polynomial amounts to solving an appropriate system of n equation and n unknowns Consider now G nPoo The support of G is disjoint from that of D since each i is nonzero so no P is a zero or pole of 9 Moreover LG gLnPoo Thus if f E LnPoo then gf E LG and so eVDgf gfP111gfP1 9P1fP11 19P1fP1 111C031 1fP which shows that GLGD GRSkooz 1 We now look at the connection between BCH codes and rational Goppa codes Recall the de nition of a BCH code We set n qm 1 and let 01 be a primitive element of y We choose a positive integer e S n and we consider the code of all polynornials pz 6 Elm of degree lt n with p01 10015 1 0 The resulting code has designated distance e which is a lower bound for the actual distance d A Reed Solomon code is nothing but a BCH code when m 1 To connect these codes to Goppa codes we need to relate codes over E to codes over Elm De nition 58 Let G be a code of length n over qu Then Olyq G l9 is called a sub eld subcode of G or the restriction ofG to Eq 52 In other words Olyq consists of all codewords 01 0 E C such that each c 6 Pg Assuming that Olyq is not the trivial code its minimum distance is at least that of C and its length is 71 Moreover dim3qClyq S dimqu C to see this suppose that 041 am is an Fq basis for qu Take 111 1 E Olyq that are linearly independent over E If there are a 6 PW with air 0 write a Z xijozj for some 7 6 Pg We then have 0 Z l ll39 Z 170 0139 Z 04739 1745 z r J J 1 By looking at each component of 11 which is an element of Fq we see that since a1 am is an Fq basis of qun we have xiio 0 for each j However the Fq independence of the 1 implies that each xij 0 and so each a 0 Thus an Fq basis of Olyq is an qu independent set in C which yields the result about dimensions It is possible that dim3qClgq lt dimqu C for example if C is a 1 dimensional code generated by a vector 01cn where 0102 Fq then for no 04 6 PW do we have 0401 E E1 and 0402 6 Pg Therefore Olyq 0 Proposition 59 Every emterzded BCH code is a sub eld subcode for some rational Goppa code Proof Let n gm 7 1 and let 04 be a primitive element of qu Let C pm 6 Fql cl idegp96 lt 71 Ma pW l 0 be a BCH code As usual we associate the n tuple a0an1 with the polynomial a0 an1x 1 lf 0 pm 6 mm M degmz lt n pm pm 0 then C is a Reed Solomon code and C O yq We claim that 5 El First if 01 76n1 E 5 then 01 0 E C O yq and on 7 21 Ci Thus 01 70n1 E a and so it lies in my since on 6 Pg Conversely if 01cn1 6 Wm then 01cn E O yq 0 since all c 6 Pg and as on 7210 we see that on E E1 and 01 0 E 6 Thus 6 Em as claimed By the previous result 3 is a rational Goppa code Therefore 5 is a sub eld subcode of a rational Goppa code E q39 In order to get Goppa codes of arbitrary large length over a xed nite eld E we need to work with function elds other than Fqx qu We will look in detail at some other examples of function elds in the next chapter Later we will consider the problem of determining how many rational points there is on a curve We will prove the Hasse Weil theorem which gives a bound in terms of q and of the genus of the curve for the number of rational points 6 Examples of Function Fields In this chapter we work in detail with two examples of curves de ned over a nite eld To help us make computations we describe without proof the connection between points on the curve and discrete valuation rings of its function eld This connection is a bit more complicated than the situation for algebraically closed elds where there is a 1 1 correspondence between the points on a nonsingular curve and the discrete valuation rings of its function eld 61 The Connection Between Points and Places To describe the correspondence let k E be a nite eld let G be a nonsingular projective curve de ned over k and let F MC be its function eld Recall that for C to be de ned over k means 0 Zf for some homogeneous polynomial f E kxyz Let K be an algebraic closure of k An exercise using a small amount of Galois theory will show that we may take K to be the union of all nite elds containing k We view our curve in projective space P2K over K Let P be a point on the curve The local ring of P is OpCk ltp E F P is de ned Its unique maximal ideal is MPCk W E F i ltMP 0 We will write Op for OpCk and Mp for MpCk Let kP be the residue eld of this valuation ring that is kP OpMp Then kP is a nite extension of k In fact if eVp is the evaluation map Op a K given by eVpf fP then eVp is a ring homomorphism from Op to K with kernel Mp Thus eVp induces an injective homomorphism kP a K and this sends the coset ltp Mp of a function p to the evaluation fP If P a b c E C then P abc Recall that we may represent elements of F MC as quotients of homogeneous polynomials over k of the same degree Thus we see that a bc E ka bc the eld generated over k by the coordinates of P From this we see that kP kabc In particular kP k if and only if abc E k That is kP k if and only if P is a k rational point For any point P we set degP This is the same number as degMp since degMp was de ned to be OpMp k So if P E Ck is a k rational point of C then degMp 1 so Mp is a rational point of Pp Furthermore it can be shown that given any place M of Fk with degM 1 then there is a unique k rational point P E C with M Mp Thus there is a 1 1 correspondence between rational points of C and rational points of Pp For points of degree greater than 1 the correspondence is somewhat more complicated Suppose that P a b c E 0 Any automorphism 039 of Kk yields a new point 0P Uaabac E P2K If C is the curve Zf with f E kxyz then the 54 coef cients of f are xed by 0 Thus foP fP 0 so oP E C We state without proof that the set of points 7P 739 E GalKk satis es lTP 739 E GalKhl degP and that this set is the set of all points whose local ring is equal to Op Moreover we have a 1 1 correspondence between places of PF and sets of points on C where the correspondence sends Mp to TP 739 E GalKk Moreover if P a b c has its coordinates in the nite eld Fqi then Galquil q is generated by the automorphism o where ot tq Then oiP 1 S i S r is the set of points who share the same local ring lf 739 E GalKk and p 6 MC by representing p as a quotient of homogeneous polynomials with coef cients in k we see that pTP TltpP pP Therefore p is de ned at P if and only if p is de ned at 7P This is why P and 7P share the same local ring Finally we point out that fP f Mp for any P E 0 although the proof is not too dif cult Since we have de ned evaluation off 6 F at a place Mp 6 Pp by fMp fMp we see that our two de nition of evaluation agree that is fP fMp To summarize this connection we state it as a theorem Recall that an extension EliE of nite elds is a cyclic Galois extension with Galois group generated by the P robenius automorphism o where o is given by the formula 0a aq for all a 6 Eli Theorem 61 Let F be the function eld of a nonsingular projective curve 0 de ned over the base eld h Pg 1 There is a 1 1 correspondence between rational points on the curve and places of degree 1 of Fh where a rational point P corresponds to the place Mp 2 There is a 1 1 correspondence between places of Fk of degree n and sets of points on the curve with coe cients in Fqn where a place corresponds to the set of points all having the same local ring Alternatively ifK is the algebraic closure of k and if P E O the place Mp corresponds to the set 7P 739 E GalKk If hP Fqn then P E CFqn and this set of points is equal to 0P o E GalqunFq 3 IfP E C and if Mp is the corresponding place then kP hMp Moreover if f E F then fP is de ned if and only if fMp is de ned and fP fMp 62 Connections Between Divisors To help to compute divisors of functions we de ne the intersection divisor of a curve Recall that we have de ned divisors of Fk we will refer to the group of such divisors as DivFk These divisors are integral linear combinations of points in Pp The divisors we are about to de ne will be combinations of points on a curve C We will refer to the group of these divisors as DivC Let C Zg be a curve If f is a homogeneous polynomial then the intersection divisor Zg Zf E DivC is the divisor 2P in where P is summed over 55 all points in both Zg and Zf7 and where 71 is the multiplicity of P in this intersection If D niPi E DivC7 then the degree of D is de ned to be 711 The degree of the intersection divisor Zf Zg is degf degg7 by the following theorem of Bezout Theorem 62 B zout Let Zf arid Zg be projective curves in P2K where K is an algebraically closed eld Theri the number ofpoirits courited with multiplicity on both Zf arid Zg is equal to degf degg We will not give the de nition of rnultiplicity7 but we indicate its meaning in examples below If fg E hxyz and if P E Zf Zg7 then any point of the form 7P for 739 E GalKk is also in Zf Zg The difference and connections between a divisor of Fk and an intersection divisor of C will be addressed in examples below As means of using intersection divisors in DivC to compute divisors in DivFh7 we describe the connection between DivFk and DivC If P E PF7 set TP Pi where Pl 6 C are the points whose local ring is Op That is7 TP is the sum of the points on 0 whose local ring is the discrete valuation ring corresponding to P The map KI DivFk 7 DivC is a group homomorphism7 and it is degree preserving that is7 degID degD This follows since if M is a place7 then there are degM points on 0 whose corresponding place is M The map KI is injective since distinct places correspond to distinct sets of points The connection between the divisor of a function and an intersection divisor is the following If f E MC7 then we may write f gh for some homogeneous polynomials 97h Then f I 1D 7 E7 where D is the intersection divisor 0 Zg and E is the intersection divisor 0 Zh ln examples below7 we will compute principal divisors using this fact 63 Elliptic Curves An elliptic curve is a curve given by an af ne equation of the form yz x where u is a cubic with no repeated roots We look at a speci c example Let h F3 and let G be the projective curve over h given by the equation yzz 3 7 22 7 23 This is an elliptic curve since the dehomogenized version of this equation is y2 x3 7 z 7 17 and x3 7 z 7 1 can be seen to have no repeated roots over F3 Let s dz and t yz7 elements of the function eld F Note that t2 53 7 s 7 1 and that F kst Moreover7 we claim that h is the exact constant eld of To see this7 we could give a direct argument7 although instead we quote a theorem that says h is the exact constant eld of Fk provided that the de ning polynomial or the dehornogenized polynomial is absolutely irreducible This latter condition means that the polynomial is irreducible over Ls7 t for any eld extension L of h To see this is true in our example7 let f y2 7 x3 z 1 If f factors in Lley as f gh7 then either the degrees in y of both 9 and h are 17 or one has degree 0 and the other degree 2 This latter case cannot happen since one of g and h is then a polynomial in x which would divide every coef cient of f7 which is clearly false If g and h are both linear in y7 then we may write 9 ay b and h cy d with ab7 ed 6 Then f acy2 adbcy bd 56 Then ac 1 so 10 are units in They are then constants and by multiplying and dividing accordingly we may assume that a c 1 Then b d 0 and bd 7x3 z 1 This yields b2 3 7 z 7 1 which is false since x3 7 z 7 1 has no repeated roots and is square free This contradiction shows that f is irreducible in Lxy for any eld L 2 k Since G Zf is nonsingular the genus of Fk is given by the formula 9 7 ltdegltfgt 71gtltdegltfgt 7 2gt 1 lt is easy to check that C is a nonsingular curve essentially this follows from the fact that x3 7 z 7 1 has no repeated roots over F3 Therefore each local ring Op for P E C is a discrete valuation ring of F This curve has a unique rational point to see this note that if a b c E Cng and if c 31 0 then we may assume that c 1 Then b2 13 7a71 Checking the three possibilities of a 6 F3 we see that there are no solutions to this equation However if c 0 then 0 13 so a 0 Finally since b 31 0 else a b c is not a valid point in P2 we see that by dividing by b we may assume that b 1 Therefore Po0 0 1 0 is the unique rational point of this curve and it is the unique point at in nity recall that we can view the af ne curve yz 3 7 z 71 as sitting inside 0 by sending a point a b to a b 1 Thus points at in nity are points whose third component is 0 Let 00 be the af ne curve yz x3 7 z 7 1 viewed inside 0 So 0 00 U Poo We now consider points of larger degree To do this we need to consider eld extensions of F3 The polynomial T2 1 is irreducible over F3 Therefore FnglTz 1 is a eld and since it has dimension 2 over F3 it is isomorphic to F9 If 04 is a root of T2 1 then F9 F3a A computation will show that Cng0a107a1104117a171a1717oz1P00 In fact from the equation yz z 71z1 71 we see that if z 0 1 71 then yz 71 so y id All of these points have degree 2 except for Poo since their coef cients generate F9 By a Maple calculation one can show that CF27 has 28 points All points in Cle but Po0 is a point of degree 3 since each has at least one coordinate outside F3 so the coordinates generate F27 These calculations can be done with the Maple worksheet POINTSMWS which de nes procedures for computing points on an af ne or a projective plane curve We now look at examples of computing the intersection divisors To give some notation to connect divisors of Fk and divisors of C if a b 1 E C we write the corresponding place in PF as Pub or as PM We use the notation Po0 both for the point 0 1 0 and the corresponding place of Example 63 Consider the intersection divisor 0 If a b c E C then a b c E Zz if and only if c 0 Thus the only point on both curves is Poo Since G is the zero set of a cubic and z is linear their degrees multiply to 3 Therefore this intersection divisor is 3PM Similarly if we consider 0 Zx then P a b c is on both curves if a 0 57 and bzc a3 7 102 7 03 703 Therefore if c 0 then b 1 and we get Poo lf 0 31 0 we may assume that c 1 and then b2 71 so b id We then have two points 0 Oz 1 and 0 7o 1 There are three points in total in this intersection so the multiplicity of each is 1 Thus the intersection divisor is Po0 0 Oz 1 0 7o 1 This is a divisor in DivC Under the map 1 de ned above this divisor is 1Pc0 POW Example 64 Let us consider the divisor of s sz in DivFk As we calculated earlier the intersection divisors C Zx and C Zz are Poo0 Oz 10 7o 1 1P0aPoo and 13P00 respectively Thus by the relation between principal divisors and intersection divisors s POD 7 2PM Example 65 We calculate the divisor oft We have already calculated the inter section divisor 0 Zz getting 3PM We next must see what is the intersection divisor 0 A point a b c is on Zy if and only if b 0 Since bzc a3 7 acz 7 03 we have a3 102 03 lf 0 0 then a 0 thus 0 31 0 By dividing by c we may assume that c 1 Then a3 7 a 7 1 0 There are three roots in K to this equa tion If we label them 11 a2 and 13 then the points on this intersection divisor are 11 0 1 a2 0 1 and a3 0 1 Therefore the multiplicity of each point is 1 and so 0 Zy a1 0 1 a2 0 1 a3 0 1 Each 1 generates the eld 193ai F27 since 3 7 z 7 1 is easily seen to be irreducible over F3 because it has no roots in F3 These three points are of the form 7a1 0 1 739 E GallFZ7lF3 There fore the three correspond to the one place Pale This is then a place of degree 3 and so C Zy 1Pa0 Since 9 is the difference of the preimages of these intersection divisors we have t Pale 7 3PM Example 66 We calculate the space LnPoo If n gt 0 2g 7 2 then the Riemann Roch theorem yields dimnPoo degnPoo 1 7 g n We claim that W 0 320 j12 3j n is a basis for LnPoo These elements clearly are linearly independent over k since 1t is a basis for F over ks and since 5 is transcendental over k Moreover it is not hard to see that there are n elements in the set so we only need to show that all are in LnPoo We see that this is true since gigti 25 jt POD 7 2PM 30310 7 3PM 21300 jPalO i 2239 3jPoo Therefore if 2239 3739 S n then sitj 71PM 2 0 We nish this section by giving a connection between divisors of Fk of degree 0 and rational points on 0 While we will not need this for our coding theory purposes it will help to illustrate one of the rst notions in the theory of elliptic curves First recall that P00 is a 58 rational point If Q is another rational point7 then Q 7 P00 is a divisor of degree 0 We claim that every divisor of degree 0 is equivalent to a divisor of the form Q 7 P00 for some rational point Q To prove this7 let D be a divisor of degree 0 Then D P00 has degree 1 Since the genus of C is 17 the Riemann Roch theorem yields dimD Poo degD Poo 1 Thus7 there is a nonzero f E LD Poo Then D P00 f 2 0 is a positive divisor of degree 1 The only way for this to happen is if D P00 f Q for some rational point Q7 which shows that D is equivalent to Q 7 P00 We have thus proven that every divisor D of degree 1 is equivalent to a divisor of the form Q 7 P00 If D is principal7 then D is equivalent to 0 P00 7 P00 On the other hand7 if Q 31 Pgt07 we claim that Q 7 P00 is not principal For7 if Q 7 P00 f for some f7 then f0 Q and ee Poo However7 degfoo F kf7 which would force F 17 or F Since a rational function eld has genus 07 this cannot happen Thus7 Q 7 P00 is not principal Similarly7 if Q 31 Q are both rational points7 then Q 7 P00 is not equivalent to Q 7 P00 As a consequence7 the subgroup of divisor classes of degree 0 is Q 7 P00 Q E Example 67 To give some motivation for looking at the group of divisor classes of degree 07 a classical result about elliptic curves is that one can de ne a group structure on the set of rational points such that P00 is the identity of the group Given rational points P and Q7 to de ne P Q7 consider the line connecting P and Q This line will intersect the elliptic curve in three points Two of these points are P and Q let R be the third point Next7 the line connecting P00 and R hits the curve at a third point This third point is the point P Q It is tedious to show that we do get a group structure from this In the picture 0 Pgt07 T P gtk Q7 and R P Q However7 The function that sends Q to the divisor class of Q 7 P00 is a group isomorphism from this group of rational points to the group of divisor classes of degree 0 Thus7 a non geometric way to de ne P Q is that P Q is the unique point B that satis es R 7 P00 Q 7 P00 P 7 P00 To see why this is true7 say that the line L passing through P and Q hits 0 at T7 and the line L passing through P00 and T hits 0 at R There are linear polynomials f and g with L Zf and L Zg The divisor fg is the di erence between the intersection divisor C Zf and the intersection divisor 0 Zg These divisors are P Q T and P00 T R7 respectively Thus7 in the divisor class group7 we have 0 fg P Q 7 P00 7 R Thus7 RPQ7P007andsoR7P00Q7PooP7Poo 59 64 Hermitian Curves In this section we work with a curve that has many rational points This curve will allow us to construct Goppa codes of large length Let q be a power of a prime and let G be the Hermitian curve given by the equation 11 yzq xq over the eld k qu2 We rst note that C is a nonsingular curve since the three partial derivatives of 11 yzq 7 simultaneously vanish only at z y z 0 Let F be the function eld F MC of C and let 5 mz and t Then F kst and 7 t 5 7 The base eld k is the exact eld of constants because 7 yq 7 y is absolutely irreducible To see that this is true let L be any eld extension of k The Eisenstein criterion applied to the principal ideal domain Ly shows that 7 yq 7 y is irreducible over Ly since yq y is a square free polynomial since its derivative is relatively prime to yq y As a consequence of this we see that F q and F q 1 For the rst equation we have F ks t so F is equal to the degree of the minimal polynomial of t over This polynomial clearly is Tq T 7 5 74r1 E ksT which has degree q A similar argument holds for the second equation We rst consider what are the k rational points of C We claim that there are 1 qg such points First let 11 0 E C lf 0 0 then a 0 and so b 31 0 By dividing by b we see that the only such point is 0 1 0 This is the only point at in nity that is it is the only point not on the af ne curve yq y xq consisting of all points whose third component is 1 This point is clearly a k rational point Now consider rational points a b 1 If a E k is any element then we note that 1 74r1 6 Pg For since a E qu2 we have 1 72 1 Then q 2 2 aq at M at at aaq aq17 so 1 74r1 is a root of 7 x Since the elements of qu are precisely the roots of this polynomial we see that aq E E as claimed Next if 04 6 Pg we claim that there are q elements of qu2 satisfying yq y a To see this recall that qu2 is a Galois extension of qu and the Galois group is the cyclic group generated by 039 where 0t 7 for all t E qu2 Moreover 039 has order 2 since qu2 Pg 2 An element y satisfying yqy 04 is then an element y satisfying 0y y a Consider the polynomial t2 7 at B E qut where B E qu is chosen so that this polynomial is irreducible over qu This is possible since qu2 can be generated over qu by an element whose minimal polynomial has degree 2 over qu This minimal polynomial then has two roots in quz and these roots are not in qu Say the roots are y and y Since 0y is also a root we have y 0y Then t2 7ozt t7y t 7 Uy shows that y Uy 04 or yq y a Thus qu2 contains a root of yq y a This polynomial then splits over qu2 since qu2 is Galois over qu Furthermore since the derivative of yq y 7 04 is 1 which is not zero there are no repeated roots So there are exactly q elements of quz satisfying yq y 7 a Since this is true for every 04 6 Pg and since 6 qu for all z E qu2 we see that the number of pairs 11 satisfying bq b 1 74r1 is q3 as there are 12 choices for a and for each a there are q solutions to bq b aq Finally adding our point 0 1 0 at in nity we have 1 q3 points over E12 The Hermitian curve yqzyzq is an example of a curve that meets the Hasse Wail bound We will see that this bound is a consequence of the Hasse Weil theorem which proves the Riemann Hypothesis for algebraic function elds over nite elds This bound says that the number of k rational points N of a curve of genus 9 satis es lNLHMM 2m H If k E12 then 12 and so this bound is 1N 7 1 q2l S 2qg Because our curve is nonsingular the genus is given by the formula 9 degf 7 1degf 7 2 where f is the polynomial de ning the curve This degree is q 1 so 9 qq 7 1 The Serre bound then simpli es to N711q53q q7lt or f72f715JV31f Therefore N is as large as possible for a curve of genus 9 that de ned over E12 We now look into calculating some divisors on C We mimic our notation in the previous section by writing Pub for the place corresponding to the point a b 1 and Po0 for the place corresponding to 0 1 0 Example 68 First let us consider the divisor of We calculate this as we calculated principal divisors for the elliptic curve example So consider rst the intersection divisor C Zx Since points on the curve satisfy the equation yqzyzq zq if a point a b c is in Zx then a 0 and 19 bcq 0 lf 0 0 then b 31 0 to have a legitimate point of projective space so the point is 0 1 0 lf 0 31 0 we may divide by c to assume that c 1 Then bqb 0 There are exactly q solutions in k to this equation see the argument above for rational points of 0 say they are b1 bq Then the points in 0 Z are 0 1 00 b1 10 bq 1 Thus the multiplicities of each must be 1 and so C Zz 0 1 00 b1 10 bq 1 As forC Zz we see that if 11 0 E C has 0 0 then a 0 and so the only point we get is 0 1 0 This point must then have multiplicity q 1 so 0 Zz q 10 1 0 Therefore 1xz0b110bq17q010 The divisor is then equal to PO71 Pebq 7 11300 In particular this says that Po0 is a pole of It is not at rst apparent that this is so because Po0 is a zero of both z and of 2 However we can manipulate the de ning equation of the curve to see why this is so From 11 yzq zq dividing by xqz gives 7WWH 27 x 39 61 The numerator of the right hand quotient is de ned at P00 and is not 0 at P00 while the denominator vanishes at P00 Thus P00 is indeed a pole of Moreover we see that DpwZ 7qvooz which shows that the order of the pole P00 is at least q since 000z 2 1 By realizing that P00 is the only pole of zz we see that qPoo since F q which we pointed out earlier Example 69 For another example of calculating divisors consider f Since we have already calculated C Zz q1Poo we only need to calculate C Zy If a b c E 0 lies on Zy then b 0 Therefore we get a 0 since 19 bcq aq Then 0 31 0 and o a b c 0 0 1 This point then must have multiplicity q 1 Our divisor is then equal to 91P00Q1Poo Q1P00Poo Next we calculate the space LnPoo where n is an arbitrary positive integer Proposition 610 The space LnPoo has basissitj 0 3 20 Sj S q 7121 jq 1 S Proof We have shown that s P0171 Pebq 7 qPoo and t q 1P00 7 P00 Therefore Sitj 5jt P0b1quot39 P0124 qPoojQ1P00 Poo Thus sitj E LnPoo if and only if it jq 1 S 71 Moreover since F q the 1 span F as a ks vector space We then see that the set of elements given above is k linearly dependent since 5 is transcendental over It remains to show that they span LnPoo To see this let f E LnPoo and write elements 1t tq 7 f0 flt 39 39 39 fq71tq71 g 7 f where fig 6 Ms and g has no factor in common with every fi Since f nPoo 2 0 the only possible pole of f is Poo Thus the only zero of 9 can be Poo However since 9 is a polynomial P00 is not a zero of g and that g has a zero unless it is constant Suppose that g is not a constant and let a be a zero of 9 Since there is no factor of 9 common to each f we see that fa 31 0 for some j The polynomial fiaTi is nontrivial so there is some nonzero value b in an algebraic closure of E12 for which the polynomial evaluated at b is nonzero Then fabi 31 0 Consider a point a b c where c 31 0 is any element satisfying a b c E 0 Finding a c is possible since 0 can be any root of the polynomial qu qu 7 aq and this polynomial has a root in an algebraic closure of 1912 since b 31 0 If P is a place corresponding to the point a b c then P is a pole of f which is a contradiction since the only pole of f is P00 and P 31 P00 as c 31 0 This forces 9 to be a constant so f is a linear combination of sitj 0 S j S q 7 10 3 239 as desired D 62 With this example we can nicely illustrate the dependence of dimD on degD The Maple worksheet HERMITIANMWS will calculate the size of the basis of LnPoo using it7 we obtain the following table in the case q 8 Recall that since the genus is qq 7 12 287 if degD gt 29 7 2 547 then dimD degD17 97 while dimD 2 degD17 g in any case n degnPoo 1 7 g dimnPoo 45 18 21 46 19 21 47 20 21 48 21 22 49 22 23 50 23 24 51 24 25 52 25 26 53 26 27 54 27 28 55 28 28 56 29 29 57 30 30 7 The HasseWeil Theorem In this nal chapter we will introduce the Riemann zeta function of a curve de ned over a nite eld and see how knowledge of its zeros yield bounds on the number of rational points of the curve We start off by describing the classical Riemann zeta function including writing it in a way that will allow us to de ne a zeta function of an arbitrary algebraic number eld and then to de ne a zeta function for an algebraic function eld F in one variable over a nite eld 71 The Riemann Zeta Function The classical Riemann zeta function 5 is de ned by the formula 5 27175 From the integral test we see that this series converges if s gt 1 By making use of complex function theory we can view 5 as a function of a complex variable 5 and by doing so we get a complex analytic function that converges for Res gt 1 This function encodes a lot of information about primes one indication of this is the product formula for 5 which says that as Hu 7 psrl p where the product is over all prime numbers The proof of this fact can be found in standard complex analysis texts To give an intuitive idea of why it is true rst consider 51 7 25 We have 00 00 ltltsgtlt17 25 Z711 Zenrs Zm n1 where m runs over all odd numbers Next ltltsgtlt17 29gtlt17 35 2 me 7 Zamrs Z where 7 runs over all integers not divisible by 2 or 3 Continuing this reasoning and being careful with the limits will yield the result The Riemann hypothesis probably the most important unsolved problem in mathe matics has to do with the zeros of 5 The functional equation for C says that 5 257T5 1 sin7Ts2F17 5C1 7 s where Us is the Gamma function a generalization of the factorial function in that Fn n if n is a nonnegative integer From this equation we see that there are trivial77 zeros of 5 occurring when sin7Ts2 0 Other zeros of 5 are then called nontrivial zeros Using the product formula for 5 it follows that 5 has no zeros in the half plane Res gt 1 From the functional equation one sees that the only non 64 trivial zeros occur in the strip 0 S Res S 1 The Riemann hypothesis conjectures that all of the nontrivial zeros of 5 are on the line Res 12 While this is still unproven Weil proved in 1941 that the analogue of the Riemann hypothesis for zeta functions associated to function elds of curves over nite elds is true It is this fact that we will see in this chapter and how this fact leads to a bound on the number of rational points of a curve 72 Riemann Zeta Functions of Number Fields We will de ne an analogue of the Riemann zeta function which will be associated to a curve de ned over a nite eld However to motivate its de nition we rst discuss the Riemann zeta function of an algebraic number eld To motivate the de nition of these functions we view 5 in another light To do this we recall some of the theory of Dedekind domains Let A be a Dedekind domain with quotient eld K A fractional ideal of A is a nonzero A submodule I of K such that aI Q A for some nonzero a E A For example if r ab E K then r TA is a fractional ideal of A since TA is an A submodule of K and brA aA Q A We call 7 a principal fractional ideal Furthermore as this example indicates if I is a fractional ideal and aI Q A for some a then aI is an ordinary ideal of A We extend the de nition of product of two ideals to the case of fractional ideals by setting for fractional ideals I and J V L IJ Zab a 6 Lb e Jn 21 i1 the usual formula for the product of two ideals It is easy to show that IJ is a fractional ideal Let G be the set of all fractional ideals of A Then this multiplication is an operation on A and the fractional ideal A is the identity for this operation Multiplication of fractional ideals is an associative operation so we have a semigroup with identity However one very important fact about Dedekind domains is that any fractional ideal has an inverse If I is a nonzero fractional ideal then AIzEKzIQA is another fractional ideal and it is the inverse of I in G Therefore G is an Abelian group The set of principal fractional ideals is a subgroup of G and the resulting quotient group is called the ideal class group of A The ideal class group of A is often denoted by ClA If A Z then ClA 0 More generally a Math 582 exercise shows that ClA 0 if and only if A is a unique factorization domain Therefore ClA measures the factorization properties of A in some sense For some terminology we call a fractional ideal positive if it is an ordinary ideal and we write I 2 0 when this occurs We now rephrase the de nition of the Riemann zeta function Since Z is a principal ideal domain every nonzero ideal of Z can be written uniquely in the form for some n 2 1 If I n then we recover n as n lZIl We call this number the norm of I and write NI for this In other words NU lZIl Therefore 8 Z NUquot 120 In this formulation we can extend the de nition easily to other number elds Recall that an algebraic number eld K is a nite extension of Q If A is the integral closure of Z in K that is A is the set of all elements of K that satisfy a monic polynomial equation with coef cients in Z then A is a Dedekind domain If I is an ideal of A we set NI lAIl We de ne the Riemann zeta function CK5 by ltKltsgt 2 MW 120 where the sum is over all nonzero ideals of A To view this in a slightly different way recall that since A is a Dedekind domain any nonzero ideal I can be written uniquely in the form I Pf1 PfT where the P are prime ideals and e 2 1 By the Chinese remainder theorem it follows that AI APf1 APfh A calculation which can be found in the handout on Dedekind domains shows that lAPel lAPle for any prime ideal P and positive integer 6 Therefore lAIl H iAPm Thus NI H NP5i 73 Riemann Zeta Functions of Curves To de ne zeta functions for curves we point out that the group of fractional ideals of A is the free Abelian group on the set of nonzero prime ideals of A This is just a fancy way to say that every ideal can be written uniquely as a product of prime ideals Every fractional ideal is in multiplicative notation an integer linear combination of prime ideals This indicates that the group of fractional ideals of a number ring is the analogue of the group of divisors of a function eld Fqu Let PF be the set of places of Fqu this set is the analogue of the set of nonzero prime ideals of a Dedekind domain For P 6 PF we de ne the norm NP of P by NP lquPl which is equal to qdegP since degP FqP E1 If E 2171 eiP is a positive divisor of Fqu we de ne the norm of E by H H qdegPiei 12515239 degPi i1 i1 qdegE39 Finally we de ne the Riemann zeta function 195 by ltFltsgt Z NltEgt E20 where the sum is over all positive divisors of Fqu We can rewrite this a little which will lead us to a related function We have ltFltsgt 2 MW Ewes Erodes E20 E20 E20 Let us write t 1 9 for the moment Then 195 Erie 0 Z t ism n1 E20 n1 degEn where An E E DivFqu E 2 0degE nl Note that A1 is the number of places of degree 1 this follows since a positive divisor of degree 1 must be a single place and then the place is necessarily of degree 1 We need to investigate the sets E E 2 0degE n in order to know that this for mulation of 195 is well de ned However before we do this we set 2F t 0 Ant n1 and call this the Z function associated to the function eld Fqu Lemma 71 Ifn is a positive integer then there are only nitely many positive divisors of degree n Proof If E eiP is a positive divisor of degree n then n e degP so 0 S e S n and degP S n It is then enough to prove that there are only nitely many places of degree less than or equal to ii To do this suppose that F is the function eld of a curve X de ned over Fq We have seen that the places of degree r correspond to sets of r conjugate points with coef cients in EXT Since there are only nitely many such points since EXT is 67 nite there can be only nitely many places of degree r Since this is true for any r there are only nitely many places of degree 3 n D Let CF be the divisor class group of Fqu We set 0 to be the subgroup of divisor classes of degree 0 Lemma 72 The group 0 is a nite group Proof Let E be a divisor of degree n 2 g where g is the genus of Fqu We set C D degD It is elementary to check that C is the coset 0 Therefore ngl It then suf ces to show that C is a nite set By Riemann7s theorem dimE 2 degE 1 7 g 2 1 by the choice of E Therefore there is a nonzero f E LE the divisor E f is then a positive divisor similar to E Thus every divisor of degree n is equivalent to a positive divisor of degree This yields C C 2 0degC By the previous lemma there are only nitely many positive divisors of a given degree so lC l lt 00 and so ngl lt 00 as desired D Corollary 73 If An E DivFqu E 2 0degE nl then An is a nonnegatiue integer The degree map deg DivFqu 7 Z is a group homomorphism The image is a subgroup of Z therefore there is a positive integer 3 such that the image is 3 The integer 3 then can be characterized as the smallest degree of a nonzero divisor or the greatest common divisor of the degrees of places Furthermore every divisor7s degree is a multiple of 3 We will see later that 3 1 De nition 74 The class number of Fqu is the integer h ngl Note that from the proof of the lemma above if C is the set of divisor classes of degree n and if C is nonempty then C is a coset of 00 so lC l h We now investigate the function Zt We need further information about the numbers An Lemma 75 Let C E OF Then HA 6 a A 2 0H 1171qdmc 1 Furthermore ifn gt 2g 7 2 and 3 l n then h 177 11771 g 1 q Proof Let C be a divisor and set n dimC Then any divisor A E C is of the form C f for some f E F and if A 2 0 then f E LC by de nition of LC Now 0 f C g if and only if f 7 g 0 or fg l 0 Therefore as a principal divisor is 0 only when the function is a constant we see that C f C g if and 68 only if g of for some 04 6 Pg Therefore for every A E C with A 2 0 there are q 7 1 elements f E LC with A C Furthermore the number of nonzero f E LC is lLCl 71 qdjm0 7 1 Thus the formula 1 1 HA 6 Cl i A 2 0H 71qd mw 1 q 7 is true For the second statement suppose that n gt 29 7 2 and 3 l n The divisibility condition implies that there are divisors of degree 71 Moreover if degC n then dimC degC 1 7 g by the Riemann Roch theorem By the rst part of the lemma and the de nitions of An and h we have A HlOl deg0nl1A 101A 2 0H 7 dime 7L M179 7h 11q 17 q 1 since h 101091 10 which was shown in the proof of the previous lemma D Corollary 76 The power series Zt 20 Ant converges for ltl lt 1 Proof We have Zt 2200 Amatma A little algebra and a basic fact about power series shows that if An1 An then the power series converges for ltl lt 0 1 From the previous lemma we have for 71 large enough lim 7 naoo 7 i An1 1 1 qn w 7 1 i qn g 1 l1m l1m h 1m 7700 A 7700 7H gram 7 1 700 qnfl g 7 1 q 7 9 117 W 1 Thus the power series converges for ltl lt 1 D Lemma 77 Let Fqu have germs 0 Then h 1 Proof We need to prove that every divisor class of degree 0 is principal Let C be a divisor with degC 0 By Riemann7s theorem we have dimC 2 degC 1 7 g 1 Thus there is a nonzero f E LC Then f C 2 0 and f C also has degree 0 This forces f C 0 so 0 7f f l is principal 1 To help with the proof of the following proposition recall that the power series represen tation of 11 7 z in the range lt 1 is 20 z Moreover 22 z zT17 z on this range We now see that Zt is a rational function This indicates how much simpler zeta functions for curves over a nite eld are than zeta functions of number elds 69 Proposition 78 Let g be the genus of Fqu Consider Zt 0n the interval of convergence M lt 14 1 Ifg 0 then 7 L7 Zt qil1eltqtgta Hal 2 Ifg 21 then Zt Ft Gt where F Z qdi1n0tdeg07 7 1 OSdegC 2972 where the sum is over all divisor classes 0 with 0 S degC 3 2g 7 2 and h qligqt29723 1 Glttgt W Proof First suppose that g 0 By the previous lemmas and since An 0 if 3 does not divide n 171 i0 Aamtam 0 L q3m1 7 tam n0 m0 m0 i ltq2ltltqtgt3gtm 7 29 m0 m0 i 1 q 1 7q71 17qt3 lita 39 The nal equality holds for ltl lt 1 since both power series converge on this range to the respective rational functions For the second part suppose that g 2 1 We have by the de nition of An and the lemmas 00 d c qdjmw 1 d 0 2M 2 1A6101A201teg 1lt7H n0 CdegCZO CdegCZO q 7 1 1 qdimCtdegC 7 1 1 tdegC q 7 CdegCZO q 7 CdegCZO 1 t 1 l 1 Z qdimCtdegC 1 qd1mCtdegC q 7 OSdegCS2972 q 7 degCgt2972 1 Z tdegltcgt 7 1 1 degC20 The rst term by de nition is To analyze the second and third terms recall that if n is divisible by 3 then there are exactly h divisor classes of degree n the set of these divisor classes is the coset E C degE n C E 0 of 0 and h ngl Therefore the third term simpli es as 71 d 7h 7h 1 7 Z teglt0gt77ztr777 7 7 7 a 1 1deglt101gt20 1 1 770 q 1 1 t For the second term we have by Riemann Roch L qd1mltcgttdegltcgt L qdegltcgt17gtdegltcgt q 7 1 d 7 q 7 1 7 egltlClgt29 2 degClgt29 2 5 Z wereg Z aqua degClgt2972 3mgt2972 W 00 a m hq1 9qt29 23 a m 7 qt qt H 22g H gr ga H m a i hq17gqt29723 1 7 171 17W because 3 divides 2g 7 2 since 29 7 2 is the degree of a divisor namely the canonical divisor The sum of the second and third terms is Gt Since the sums for Gt converge on ltl lt 1 the series for Zt converges on the same interval D Corollary 79 The function Zt is a rational function with a simple pole at t 1 Proof The formula for Ft shows that Ft is a polynomial since there are only nitely many divisor classes 0 with 0 S degC 3 2g 7 2 The formula for Gt shows that it is a rational function with a simple pole at t 1 Therefore Zt Ft Gt is a rational function with a simple pole at t 1 D We now give a product representation for Zt Proposition 710 The function Zt can be represented on ltl lt 1 1 as Zt H 1 7 idegltPgt1 PEPF Therefore Zt 31 0 for ltl lt 1 Proof We need a fact about convergence of in nite products the product H11 an converges absolutely if and only if 20 an converges absolutely This fact can be found in most complex analysis books Therefore the product above converges absolutely on ltl lt 1 1 since Z 15 iAnt Zt rLO PEPF converges absolutely on ltl lt 1 By writing each term of the product as a geometric series and doing lots of rearranging of terms we have 1 CO H H 2W5 PePF PePF n0 0 2w E20 n0 2o D To prove that 3 1 we will need to consider how function elds behave under base extension Let X Zf be a nonsingular irreducible projective curve de ned over Fq and let X be its function eld Then F Fqst where s mz and t yz viewed as rational functions on X We assume that E is the exact constant eld of Fqu Let r 2 1 and consider the extension Fi E1 We can view X as a curve over Fly and the function eld of X over Ff is Fqlst We write F for this function eld What we need to know is information about places of F in relation to places of F Note that in any case our curve X is the set of all points a b c E FRET such that fab c 0 where E is an algebraic closure of E1 or of Ell If P is the place of F that corresponds to a b c then P ltp E F abc 0 Also if P is the place of F that corresponds to a b c then P ltp E F abc 0 Thus P F P In the proof of the following lemma we need to know some facts about nite elds First recall that if q is any prime power then the elds containing E are exactly the elds of the form Ell for some r 2 1 Conversely if T is a sub eld of Ell containing Fq then T Fqs for some integer s that divides r since if t Ell T then q ll qll lTlt q Moreover the eld Fqs is characterized as the sub eld of Ell consisting of the solutions to the equation zqs x Lemma 711 Let P be a place of F and let P be a place of F for which P F P If m degP therz degP m gcdmr Moreover there are gcdmr many such P Proof Let X be a curve whose function eld over E is F Choose a point a b c E X 72 whose corresponding place in F is P and whose place in F is P The residue eld of P is Fqabc and the residue eld of P is Fqiabc Then by de nition degP Eli a bc quT and degP Fqa bc E1 Set L Fqab 0 Then the residue eld of P is the composite eld LquT By the theorem of natural irrationalities frorn Galois theory LquT Eli L L Eli Note that since m degP we have L qu We claim that Eli f7qu qu where d gcdmr If this is true then i i L will qu 3Fql i m L 39 LN Low rq qu 19 E as desired To prove the claim note that qu g Eli qu since d divides both 7 and m For the reverse inclusion let a 6 Eli qu Then aqf a and aqm 1 Since d gcdmr we may write d mm r for some integers 046 Since dmr are all positive one of 046 is positive and the other is negative Suppose that 04 gt 0 Then d 7 Br am A simple induction shows that aqm a and aqim 1 Therefore d 9 dim d 7M 7m q d aaq aq aqq aq gt aq This shows that a E qu as desired Note that as a consequence of this we see that the composite ququT E11 where l mrd lcrnm 7 To prove the second staternent let P be a place of F of degree m and let S be the set of points of X whose corresponding place is P This set has m elements The places of F lying over P are exactly the places corresponding to points in S However each such place of F corresponds to lr points this is because if p is a point in S then its residue eld E1410 with respect to F has degree Fqip Eli qu Eli lr as mentioned above Now l mrd so lr md Therefore if PPt are the places of F lying over P each corresponds to md points of S and every point of S corresponds to exactly one of the Pi Therefore there are d such P This nishes the proof D To help prove the following proposition we need a result about roots of unity Let r and m be positive integers and set d gcdr Then We 1V Hwyquot 6 where the product runs over the r th roots of unity in C That is 6 ranges over the elements exp2m r 0 S 239 lt 7 To see why this equation is true we note that both sides are rnonic polynomials of degree 7 It is enough to see that they have the same roots The roots of the left hand side are precisely the rd th roots of unity lf 6 is an r th root of unity then WWd Wmd 1 Conversely any rd th root of unity is of the form 6 quot for some r th root of unity 6 as a group theory exercise shows If we substitute z t and then multiply both sides by tW7 a little algebra will show that 17tWdd 7 H1 7 16w 6 We will use this in the following proposition Proposition 712 Let Zt be the Z furzetz39orz associated to F Then ZTtT H6 Z6t where 6 rurzs over all r th roots of unity in C Proof Since two complex analytic functions are equal if they agree on a nontrivial open disk of the complex plane7 and because both sides of the equation above are rational functions in t7 it is enough to prove the equality for ltl lt 1 We will write P l P if P F P In the region ltl lt fl7 the product representation yields Zrtr H 7 trdegP 71 H H 7 trdegP 7139 P EPFT PePF PP For a xed P E PF7 let m degP and d gcdr7 We then have7 by the lernrna7 H lt17trdegP gt 1 7 trot111 PP HG 50m 6 H176tdegltPgt 6 Therefore7 27017 H Hlt17lt6tdegP 1 PEPF 6 H H 1761degltPgt 17HZ61 6 PEPF 6 Corollary 713 fa gcd degE E E Dp then 3 1 Proof Set r 3 Then 1 7 6tdegP 1 7 tdegP since degP is a multiple of 3 so 6degP 1 Therefore7 Z6t 7 H 1761degltPgt 1 7 H 17tdegltPgt 1 7 21 PEPF PEPF Therefore7 by the proposition7 ZTtT Zt By the description of Zt in Proposition 7107 applied to ZTt7 the Z function associated to FTquT7 we see that Zt has a simple 74 pole at t 1 However7 Zt has a pole of order 7 at t 1 Thus7 as ZTtT Ztf7 we see that r 1 In other words7 3 1 D This corollary is not true if the base eld is in nite For exarnple7 if F is the function eld of the curve 2 y2 22 0 over R then the curve has no R rational point Since the residue eld of any point is a nite extension of R each residue eld must then be C Therefore7 the degree of every place is 27 and so 3 2 for this function eld Corollary 714 Any function eld Fqu of genus 0 is a rationalfunetz39on eld and lt17tgtlt17qtgt39 Proof Since 3 17 there is a divisor E of degree 1 By Riemann Roch7 dirnE 27 so LE is nonzero Taking a nonzero f E LE7 we have f E 2 07 a positive divisor of degree 1 Then f E P for some single place P7 which is then necessarily of degree 1 Therefore7 Fqu has a place of degree 1 We have proven that a function eld of genus 0 with a rational point is a rational function eld7 which says that F is isomorphic to Fqz Finally7 frorn Proposition 7107 we have 1 q 1 1 q 1 Zt7 777 7 777 q7117qt3 17t3 q7117qt 17t lt17qtgtlt17tgt as a little algebra shows D Zt Corollary 715 If Fqu has genus g 2 1 then Zt Ft Gt where Fa L 2 qdimCtdegC q T 1 OSdeglClS2972 h 9t29 1 1 on 7 q 7 7 q 7 1 1 7 qt 1 7 t The following result is an analogue of the functional equation mentioned at the beginning of this chapter We will use the functional equation to get information about the zeros of Zt and Proposition 716 If Zt is the Z funetz39on of Fqu then Zt qg ltzg ZZuqt First suppose that g 0 Then Zt 11 7 t1 7 qt Therefore7 q7lt72 wwwmum75ig mfm i 1 1 qtilwil 1ft1qt39 Next7 suppose that g 2 1 We write Zt Ft Gt as earlier Let C be a canonical divisor Recall that degC 2g 7 2 Then7 by the Riemann Roch theorern7 m7umwi WWMW OSdeglt1E1gt32972 qdegE 17gdjmC7E tdegE 0Sd6glEl32972 qg71t2972 Z qdegE7Zg72dimC7E tdegE7Zg72 OgdeglEl32972 If E is a divisor with 0 S degC 3 2g 7 27 then degC 7 E degC 7 degE 2g 7 2 7 degE7 and so 0 S degC 7 E 3 2g 7 2 Furtherrnore7 by Riemann Roch7 dirnC 7 E degC 7 E 1 7 g dirnE Moreover7 as ranges over divisors of degree between 0 and 29 7 27 so does 0 7 Therefore7 setting D C 7 E7 we have q 7 1 Fa qgi 1129 Z qdimltDgt7degltDgtt7 degltDgt OSdeglDlS2972 de D g7lt2972 Z djmD glt q q t OsdeglDl32972 1 f mwwm Next7 for Gt7 we have7 by the corollary7 qg71t2972g1qt h q971t2972 qg i2971 1 7 1 q 71 qt 17 Qqt 1 1 7 910 h 1 1 91941294 i 1 1 E1t 171qt 1 h 1 91294 7 7 7 q Gt q 7 1 t7 1 qt 7 1 Therefore7 q9 1t29 2Z1qt Zt We can phrase the functional equation in terms of the zeta function 5 Recall that 5 Zq 5 Substituting t 1 9 in the functional equation yields 8 q9 1q 929 22q9 1 q9 1q 929 2lt 1 7 8 q91129lt 1 7 5 In this case unlike the classical case we have no trivial zeros7 Also if s is a zero of 5 then so is 17 s The function Zt is a rational function with denominator 1 7 t1 7 qt We work with the numerator of Zt De nition 717 The polynomial Lt 1 7 t1 7 qtZt is called the L polynomial of Fqu By the previous two corollaries we see that Lt 1 if the genus of Fqu is 0 and that in any case degLt 3 29 this comes from the fact that Ft is a polynomial of degree at most 29 7 2 and Gt is a rational function of degree 29 7 2 Thus the degree of Zt is at most 29 7 2 We will see that much information is encoded in this polynomial Note that Zt is determined by Lt as Llttgt lt17tgtlt1eqw Therefore Lt encodes all the information about the numbers An that Zt encodes Zt Theorem 718 Let Lt be the L polynomial of the function eld Fqu Z Lt has coe cients in Z and degLt 2g 2 Lt q9t29L1qt 3 L1 h the class number of Fqu 4 If Lt a0 a1t a29t29 then a0 1 129 Q9 and a1 N 7 q 1 where N is the number of places of PF of degree 1 Proof If the genus g 0 then Lt 1 and the result is trivial in this case We thus assume that g 2 1 Since Lt 1 7 t1 7 qt 20 Ant the polynomial Lt is a power series in t with integer coef cients Thus its coef cients as a polynomial are the same integer coef cients We have already remarked that degLt 3 29 When we prove that the coef cient 129 qg we will have degLt 29 For the functional equation we have Zt q9 1t29 2Z1qt Therefore Mt 1 i 01 7 10200 1 i 01 7 a qg ltzg ZZUqt i 7 7 9712972 1 t 1th t 17qt 11Qqt71 971 2972 I mqt Ui imq t HTE 3HfF i mum 717t1qtqt m q9t29L1qt This proves the functional equation To prove 3 since Zt Ft Gt we may write M007w 7mF g f 7w7 7m by our description of Gt Since Ft is a polynomial F1 is de ned and this formula then yields h L 1 7 7 1 7 h m gig lt m Finally write Lt a0 a1t aggtzg From Lt 1 7 t1 7 1023 Ant if we multiply this out we see that 10 A0 and a1 A1 7 q 1 We have noted earlier that A1 is the number of places of degree 1 in other words A1 N Therefore 11 N 7 q 1 Also A0 1 since the only positive divisor of degree 0 is 0 Thus 10 1 Finally the functional equation yields Lt 9t29L1 t 9W E 029 o q ltmgtq m WW a a 3 29 113 alqgil gil aoqthg39 19 19 1 Equating the constant terms gives 10 aggqg so 129 aoqg qg This completes the proof D The polynomial Lt has degree 29 If we work over C then the fundamental theorem of algebra says that we can factor Lt into linear factors It is more convenient to work with the reciprocals 041 0429 of the roots of Lt Since the leading coef cient of Lt is qg we write 29 29 W i qg Ho 7 a qg H 7a117 an i1 11 19 29 1 7 out gt Now since the constant term of Lt is 1 we see that the coef cient in front of the product is 1 Therefore 29 Lt HQ 7 am i1 Corollary 719 IfN is the number of places of degree 1 in F then N q 1 7 21 04 Proof This follows immediately from Lt H17 Edit a0 a1t aggtzg and a1 N7 q 1 once we multiply out the product and gather together the linear terms D From this corollary in order to determine the number of rational points of Fqu we need to know about the reciprocals of the roots of Lt The Riemann hypothesis for function elds of curves over nite elds proved by Andre Weil in 1941 gives the needed information about the 04 Theorem 720 Hasse Weil The roots of the Riemann zeta function 195 lie on the line Res 12 Therefore ifa10429 are the reciprocals of the roots ofLt then 1041 Because of time constraints we will not prove this theorem However to relate the two statements of the theorem recall that 5 and Zt are related by the equation 195 Zq 9 Furthermore since L t zlttgt ltgt 1 i 01 qt the roots of Lt are the zeros of Zt So if s is a zero of 105 then Zq 9 0 If we write 5 12 yi then 179 671nqs 67lnq12yi 67121nq67ylnqi39 712h1q Therefore q si le fl2 recall that e cos0 isin0 so lewl 1 for any 0 Thus a root of Lt has absolute value 1 so any a has absolute value Corollary 721 HasseWeil Bound IfN is the number of places of Fqu of degree 1 then 1Nq11 S 29 Proof We have N q 1 7 21 04 Therefore 29 2m i1 29 As we have seen Herrnitian curves attain this bound so this is in general the best one can do D iN Q1i 29 3 Z W i1 Dedekind Domains Mathematics 601 In this note we prove several facts about Dedekind domains that we will use in the course of proving the Riemann Roch theorem The main theorem shows that if KF is a nite extension and A is a Dedekind domain with quotient eld F7 then the integral closure of A in K is also a Dedekind domain As we will see in the proof7 we need various results from ring theory and eld theory We rst recall some basic de nitions and facts A Dedekind domain is an integral domain B for which every nonzero ideal can be written uniquely as a product of prime ideals Perhaps the main theorem about Dedekind domains is that a domain B is a Dedekind domain if and only if B is Noetherian7 integrally closed7 and dimB 1 Without fully de ning dimension7 to say that a ring has dimension 1 says nothing more than nonzero prime ideals are maximal Moreover7 a Noetherian ring B is a Dedekind domain if and only if BM is a discrete valuation ring for every maximal ideal M of B In particular7 a Dedekind domain that is a local ring is a discrete valuation ring7 and vice versa We start by mentioning two examples of Dedekind domains Example 1 The ring of integers Z is a Dedekind domain In fact7 any principal ideal domain is a Dedekind domain since a principal ideal domain is Noetherian integrally closed7 and nonzero prime ideals are maximal Alternatively7 it is easy to prove that in a principal ideal domain7 every nonzero ideal factors uniquely into prime ideals Let F be an algebraic number eld That is7 F a nite dimensional eld extension of Q If A is the integral closure of Z in F7 then Theorem 4 below shows that A is a Dedekind domain The ring A is called the ring of integers in the number eld F The study of the arithmetic of A is the heart of algebraic number theory Example 2 Let X be a nonsingular af ne curve over a eld k The coordinate ring PXh is Noetherian7 since it is a quotient of the Noetherian ring It can be shown that for any maximal ideal M of l Xh7 there is a point P E X with M f E PXh fP 0 Moreover7 the local ring OpXh is equal to the localization of PXh at M Since each local ring OpXh is a discrete valuation ring7 PXh is a Dedekind domain Lemma 3 Let A be an integrally closed domain with quotient eld F and let K be an eatension eld of F Ifb E K is algebraic over F then b is integral oyerA if and only if the minimal polynomial ofb over F is in Proof One direction is easy Let p be the monic minimal polynomial of b over F If p 6 AM then b is integral over A Conversely7 suppose that b is integral over A Let L be the normal closure of FbF7 and let G be the integral closure of A in L Since b is integral over A7 there is a monic g 6 AM with gb 0 Then p divides g and since LF is normal7 p splits in L recall that if MF is a normal extension and a 6 M7 then the minimal polynomial of a over F splits in Any root of g is integral over A7 and any root ofp is a root of g by the divisibility relation Therefore7 all the roots ofp lie in 0 Say p x 7 cl x 7 on E Since each 01 E C the coef cients of p also lie in 0 since they are sums of products of the 01 Thus7 the coef cients lie in 0 F This ring contains A7 is contained in F7 and is integral over A Since A is integrally closed7 we see that 0 F A7 so the coef cients ofp lie in A Thus7 p E D Theorem 4 Let A be a Dedekind domain with quotient eld F let K be a nite extension of F and let B be the integral closure ofA in K Then B is a Dedekind domain Proof We rst assume that KF is separable Recall that a Dedekind domain is a Noethe rian7 integrally closed domain of Krull dimension 1 The ring B is integrally closed because if z E K is integral over B7 then x is integral over A since integrality is a transitive property Therefore7 z E B Moreover7 since dimA 1 and BA is integral7 dimB 1 by the incomparability theorem lf Q1 Q Q2 are prime ideals of B with Q1 A Q2 A7 then Q1 Q2 Therefore7 a chain of prime ideals of length more than 1 cannot contract to a chain in A of length at most 17 so dimB S 1 Moreover7 if dimB 07 then 0 is a maximal ideal of B7 so B is a eld But then A would be a eld7 which is false Thus7 dimB 1 The last step in the case that KF is separable is to show that B is Noetherian We prove more by showing that B is a Noetherian A module Since ideals of B are A submodules of B7 ascending chain condition on A submodules is a stronger condition than ascending chain condition on ideals7 so B is then a Noetherian ring We will assume that KF is Galois7 since if N is the normal closure of KF and if C is the integral closure of A in N7 then if C is a Noetherian A module7 then B is also a Noetherian A module since B is a submodule of C So7 assume KF is Galois7 and let b1 bn E B be an F basis for K We can choose the bl E B for the following reason Given any basis 1 an we can write cidi with each 01 dl E B By using common denominators7 we can then write each m in the form bid with bi d E B The bl then form a basis for K Now7 recall that the trace map T K a F is nonzero Therefore7 the bilinear map K gtlt K a F given by a7b a Tab is nondegenerate Given the basis b1 bn7 there is a dual basis y17yn with Tbiyj 1 ifi j and 0 otherwise Let b 6 B7 and write b oziyi with each ai E F Then bl aiyibj so Tbb 047 However7 Tbb 06G31KFobbj and 0B B for each 0 Therefore7 Tbb E B F A7 so each oo 6 A This shows that B Q Ayi7 so B lies in a nitely generated A module This module is Noetherian since A is a Noetherian ring7 so B is also a Noetherian A module7 as we wanted to show We now remove the assumption that KF is separable We may assume that charF p gt 0 Let S be the separable closure of F in K7 and let B0 be the integral closure of 2 A in S By what we have just proven B0 is a Dedekind domain and note that B is the integral closure of B0 in K Thus by replacing F by S it suf ces to assume that KF purely inseparable Recall that since K F lt 00 there is an n with K F p and that a E F for all a E K For simplicity write q K Let M be the splitting eld over F of all polynomials of the form pg 7 04 with 04 E F Then K Q M Let C be the integral closure of A in M Note that the map p M a F de ned by Mp is injective since charF p and q is a power of p and p is surjective by the de nition of M Thus p is a eld isomorphism Moreover ltpC A The inclusion A Q 0 follows from the de nition of integrality and the other inclusion is true by the lemma Therefore A and C are isomorphic so 0 is a Dedekind domain To nish the proof we recall another criterion for a ring to be a Dedekind domain B is a Dedekind domain if and only if every ideal of B is invertible Let I be an ideal of B Then IC is an ideal of C so IC is invertible Thus there are b E I and d E C IO with bid 1 From this we get bid 1 Moreover d E F by the de nition of M Let e bfildf E K Then Zibiei 1 The elements d satisfy diI Q C by de nition so eiI Q C lntersecting with K gives eiI Q 0 K B so e E B I Thus we have proven that IB I B so I is invertible and so B is a Dedekind domain D An important fact we need is to determine the discrete valuation rings of a nite extension F of the rational function eld that contain Irma Proposition 5 Let KF be a nite eld ecctension let 0 be a discrete valuation ring of F and let B be the integral closure ofO in K Then the discrete valuation rings ofF that contain 0 are the localizations BM at mapimal ideals of B Proof Let B be the integral closure of O in K Then B is a Dedekind domain by Theorem 4 Thus if M is a maximal ideal of B then BM is a discrete valuation ring Since 0 Q B because 0 is integrally closed 0 Q BM Conversely let V be a valuation ring of K that contains 0 If b E B then b is integral over 0 Thus b is also integral over V Therefore b E V since V is integrally closed Therefore B Q V Let P be the maximal ideal of V Then P B M is a prime ideal of B We have BM Q Vp V and since V 31 K the ideal M is nonzero Thus since every nonzero prime ideal of B is maximal M is maximal and since BM is a discrete valuation ring BM V since discrete valuation rings are maximal subrings of their quotient elds Thus the valuation rings of K containing 0 are precisely the localizations BM at maximal ideals of M D This proposition tells us how to obtain the discrete valuation rings of K that contain 0 We do not yet know how many there are In particular we have not yet shown that there are only nitely many such valuation rings We do this now Proposition 6 With notation in the previous proposition there are only nitely many mapimal ideals of B and so there are only nitely many discrete valuation rings ofK that contain 0 Proof Let t be a uniformizer of 0 Then tO is the maximal ideal of 0 Consider the ideal tB of B Then this ideal factors into a product of prime ideals since B is a Dedekind domain Thus we may write tB M181 M959 for some integers e 2 1 and distinct maximal ideals M1 Mg We claim that M1 M9 is exactly the set of maximal ideals of B These are maximal ideals since they are nonzero prime ideals Conversely let M be a maximal ideal of B Then BM is a discrete valuation ring of K and BM contains 0 since B contains 0 Thus MBM O tO because MBM is the maximal ideal of BM Since MBM B M we see that t E M Then tB Q M So Mf1Mgeg Q M Because M is prime M Q M for some i However M is maximal which forces M M We have thus proved that M1 M9 is the set of maximal ideals of B and so BMI BMg are all the discrete valuation rings of K that contain 0 D We nish this note with a technical looking result which will be the heart of the proof that principal divisors of an algebraic function eld have degree 0 We rst need two lemmas which are useful facts about Dedekind domains or discrete valuation rings Lemma 7 Let B be a Dedekind domain with only nitely many mamimal ideals Then B is a principal ideal domain Proof Let M1 Mg be the maximal ideals of B By the prime avoidance theorem M1 is not contained in M12 U M2 U U My Therefore there is a t 6 M1 with t M12 and t M for each i 2 2 Consider the ideal tB This factors as a product of maximal ideals Thus there are integers e 2 0 with tB M181 Mgeg For i 2 2 if e 2 1 then tB Q Mi Q Mi which says t 6 Mi Since this is false e 0 for i 2 2 Therefore tB M151 Since t M12 we have e Z 2 So e1 1 and so tB M1 This proves that M1 is principal Similarly each M is principal Finally since every nonzero ideal of B is a product of maximal ideals each ideal is principal B Let B be a Dedekind domain with nitely many maximal ideals M1 Mg Let i be a valuation on the quotient eld F of B whose valuation ring is BM If x E B and zB factors as zB M181 Mg then xBMZ MfiBMi and so ei Thus we can recover the values of z in terms of the factorization into prime ideals of the ideal zB Lemma 8 Let 0 be a discrete valuation ring Then any nitely generated torsion free O module is free Proof Let T be a nitely generated torsion free O module Pick n minimal such that T is generated by n elements and let t1 tn be a generating set of T We will prove that t1tn is a basis Since this set generates T we only need to prove that it is linearly independent over 0 Suppose that zit 0 with x E 0 Let i be a valuation on the quotient eld F of 0 whose valuation ring is 0 If some x 7 0 then let 11zj min 1 S i S Then i gl E O for eachisinceyxiz1 2 0 Then xglditi 0 and so tj 7 Zijx1xiti is an O linear combination of ti i 31 j Therefore we have a generating set for T with n 7 1 elements This contradiction shows that each m 07 and so t1L7 7tn is independent7 and so is a basis for T as an O module Therefore7 T is a free module D Proposition 9 Let KF be a nite separable extension of elds let 0 be a discrete valuation ring of F with mamimal ideal P t0 and let B be the integral closure of O in K Let M17Mg be the mamimal ideals of B and let fl OP IftB Mf1Mgeg then K F 191 eifi Proof We rst remark that the proof of Theorem 4 shows that B is contained in a nitely generated O module Since 0 is a discrete valuation ring7 any nitely generated torsion free module is free Since 0 is a Noetherian ring7 any nitely generated O module is a Noetherian 9 module7 and so any submodule is nitely generated Thus7 B is nitely generated as an 9 module7 and so B is free by Lemma 8 Its rank as an O module is n K F since B 80 F K this fact is almost really a restatement that the quotient eld of K is B Let h OP7 a eld Then dimk BtB n because BtB g B O k By the Chinese remainder theorem7 BtB BMf1M59 69771BMfi It then suf ces to prove that dimkBMfi eifi Set M Mi e ei and f fi We will prove this by considering the chain B 2 M 2 M2 Q 2 M5 Each quotient MTMTH is a k vector space because each is an O module with tMT Q MT 7 since tB Q M5 We then have a sequence of k vector spaces BMMM2M5 1M5 ln fact7 BMe is a k vector space for the same argument that each quotient above is a k vector space We prove that dimkMTMT1 f for each r Since dimkVW dimkV 7dimkW for any k vector spaces W Q V7 induction shows that dimkBMT rf for each r7 and so dimkBM5 ef7 as desired Now7 to prove this7 we produce a k vector space isomorphism MTMTH E BM The ideal M is principal by the previous lemma say M 5B Then MT STE De ne p B a MTMTH by b sTbMT1 Then p is an O module homomorphism with M 0 Thus7 we have an induced map7 which we also denote by p from BM to MTMTH7 given by ltpb M s b MT This is a k vector space map since it is an O module map It is clear that p is surjective To prove that it is injective7 suppose that ltpb M 0 Then 57b 6 MTH sTHB So7 there is a c E B with s b ST1C Then b so 6 M7 so b M 07 as desired This nishes the argument D Characterizing Discrete Valuation Rings Matematics 601 If F is a eld and i is a discrete valuation on F then Ow a E F l oa 2 0 U 0 is a ring with quotient eld F This ring is called a discrete valuation ring DVR In other words a ring R is a DVR if there is a discrete valuation i on the quotient eld F of R such that R a E F l oa 2 0 U The following theorem gives the main ring theoretic information about DVR7s Theorem 1 Let R be a local Noetherian integral domain with dimR 1 Ifm is the unique mamimal ideal ofR and h Rm then the following conditions are equivalent 1 R is a DVR 2 R is integrally closed 3 The mamimal ideal m ofR is principal 4 dimkmm2 1 5 Every proper nonzero ideal ofR is a power ofm Proof Before we get into the proof of the equivalences of these conditions we point out two things First run 31 m for each n For if m 1 m then the nitely generated R module M 111 would satisfy 111M M so M 0 by Nakayama This is false so run 31 111 Second if I is any nonzero ideal of R then 111 since I is the intersection of all prime ideals containing I and since R is local and 1 dimensional the only nonzero prime ideal of R is m From this we can show that 111 Q I for some ii To see why suppose m 1 zr Since 111 there are integers n with E I If n 271 then m is generated by the elements xii w r with Z s n For each such generator there is some i with s 2 71 so E I hence the product xii a E I This means 111 Q I We now begin the series of implications to prove this theorem 1 a 2 Let F be the quotient eld of R Suppose oz 6 F is integral over R Then 04 Lilo 1 a0 0 for some a E R If R Ow for some valuation i then Mai 2 0 for each i Suppose 04 R Then Ma lt 0 From the polynomial equation above we get Oz 7an1 awgofl 304 The right hand side is in R since 1104 1 gt 0 so of1 E B This shows 04 E B so R is integrally closed 2 7 3 Let a E m with a 31 0 By the second comment above we have 111 Q a for some 71 Choose 71 minimal so that m 1 gm Take b 6 m 1 7 a and let x ab Then z 1 R since I 1 so z 1 is not integral over R The conditions 111 Q a and b 6 mn 1 imply that x lm Q R If x lm R then m is principal and we would then be done The only other possibility is for x lm Q 111 since x lm is an ideal of R Suppose this happens Let u1 u be generators for m recall R is Noetherian so 111 is nitely generated Then z lui Zrijuj for some n 6 B This means 20 7 Fijs lmi 0 for each 239 hence det n7 7 Fig4 annihilates each uj Since we are in an integral domain this forces det n7 7 Fig4 0 Expanding this out we see that z 1 is integral over R a contradiction Thus x lm Q m is false hence m from the previous case 3 7 4 Suppose m Then as an R module m is generated by x hence 1111112 is generated by z 1112 But then 1111112 is generated as an Rm module by z 1112 so dimkmm2 S 1 But 1111112 31 0 by the rst comment above so dimkmm2 1 4 7 5 Let x m2 be a generator for the Rm vector space 1111112 Then m Pm m2 Pm m m Therefore by Nakayama7s lemma with M m we see 111 Let I be any proper nonzero ideal of R Then I Q 111 since 111 is the unique maximal ideal of B Let r be maximal with I Q 111 so I Q 1117 There is such an 7 since if I Q m7 for every 7 then by the second comment above 111 Q I for some n which forces In run for each s 2 0 This is false by the rst comment above Take y E I with y 111 Then y ax for some a E R since 111 Moreover since y 111 the element 1 111 hence a is a unit in R Thus m7 and since Q I Q 1117 this forces I m7 to be a power of m 5 7 1 Let F be the quotient eld of R and suppose every ideal of R is a power of 111 To de ne a valuation on F for a E R 7 0 de ne 0a n if a m We set 1110 R to make this de nition apply to every element of R From this de nition it follows that 0ab 0a 11b and va b 2 min i11b for ab E R The rst statement holds since a b ab and the second holds since a b Q a bTo extend 1 to F de ne 0ab a 711b for ab E R7 This is well de ned since if ab cd then ad be so 0a vd Mb vc hence 0a 7vb 00 7vd An elementary calculation shows that 1 has the properties of a discrete valuation Moreover it is clear from the de nition of 1 that R Q OH and 111 Q my For the reverse inclusion suppose ab E F with ab E R Then 0 S 0ab 0a 7 Mb so 0a 2 11b We wish to show a bx for some z E R which will show ab E B Let 7T 6 m 7 1112 Then 7T 111 since all ideals of R are powers of m and 7T Q 1112 Thus m is principal so all powers of m are principal so all ideals of R are principal If a m n and b mm Wm then 71 2 m so we can write a 7T x and b me with my units in R Then ab Wn mzy E R since 71 7 m 2 0 and my are units in B so zy E B This proves that On R and since ray is an ideal of R containing 111 we must have m my D

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#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.