Lectures on the Theory of Algebraic Functions of One Variable

Home , Algebraic function, Rational function

Lectures on The Theory of Algebraic Functions of One Variable

by M. Deuring

Notes by C.P. Ramanujam

No part of this book may be reproduced in any form by print, microﬁlm or any other means without written permission from the Tata Institute of Fundamental Research, Apollo Pier Road, Bom- bay - 1

Tata Institute of Fundamental Research Bombay 1959

Contents

1 Lecture 1 1 1 Introduction...... 1 2 OrderedGroups...... 2 3 Valuations, Places and Valuation Rings ...... 3

2 Lecture 2 7 3 (Contd.)...... 7

3 Lecture 3 9 4 The Valuations of Rational Function Field ...... 9 5 Extensions of Places ...... 10

4 Lecture 4 15 6 Valuations of Algebraic Function Fields ...... 15 7 TheDegreeofaPlace...... 17 8 Independence of Valuations ...... 18

5 Lecture 5 23 9 Divisors...... 23

6 Lecture 6 27 10 The Space L(U ) ...... 27 11 The Principal Divisors ...... 28

7 Lecture 7 33 12 The Riemann Theorem ...... 33

iii iv Contents

13 Repartitions ...... 34

8 Lecture 8 37 14 Diﬀerentials...... 37 15 The Riemann-Roch theorem ...... 41

9 Lecture 9 45 16 Rational Function Fields ...... 45 17 Function Fields of Degree Two...... 46 18 FieldsofGenusZero ...... 50 19 FieldsofGenusOne ...... 51

10 Lecture 10 55 20 The Greatest Common Divisor of a Class ...... 55 21 The Zeta Function of Algebraic...... 56

11 Lecture 11 61 22 The Inﬁnite Product for ζ(s, K)...... 61 23 The Functional Equation ...... 62 24 L-series ...... 64

12 Lecture 12 67 25 The Functional Equation for the L-functions ...... 67

13 Lecture 13 71 26 The Components of a Repartition ...... 71

14 Lecture 14 75 27 A Consequence of the Riemann-Roch Theorem . . . . . 75

15 Lecture 15 81 28 Classes Modulo F ...... 81

16 Lecture 16 85 29 Characters Modulo F ...... 85 Contents v

17 Lecture 17 89 30 L-functions Modulo F ...... 89 31 The Functional Equations of the L-functions...... 89

18 Lecture 18 97 32 Extensions of Algebraic Function Fields ...... 97

19 Lecture 19 103 33 Application of Galois Theory ...... 103

20 Lecture 20 109 34 Divisors in an Extension ...... 109 35 Ramiﬁcation...... 114

21 Lecture 21 117 36 Constant Field Extensions ...... 117

22 Lecture 22 123 37 Constant Field Extensions ...... 123

23 Lecture 23 133 38 Genus of a Constant Field Extension ...... 133 39 The Zeta Function of an Extension ...... 144

Lecture 1

1 Introduction

We shall be dealing in these lectures with the algebraic aspects of the 1 theory of algebraic functions of one variable. Since an algebraic function w(z) is defined implicitly by an equation of the form f (z, w) = 0, where f is a polynomial, it is understandable that the study of such functions should be possible by algebraic methods. Such methods also have the advantage that the theory can be developed in the most general setting, viz. over an arbitrary field, and not only over field of complex numbers (the classical case).

Definition . Let k be a field. An algebraic function field K over k is a finitely generated extension over k of transcendence degree at least equal to one. If the transcendence degree of K/k is r, we say that it is a function field in r variables.

We shall confine ourselves in these lectures to algebraic function fields of one variable, and shall refer to them shortly as ‘function fields’. If K/k is a function field, it follows from our definition that there exists an X in K transcendental over k, such that K/k(X) is a finite algebraic extension. If Y is another transcendental element of k, it should satisfy a relation F(X, Y) = 0, where F is a polynomial over K which does not vanish identically. Since Y is transcendental by assumption, 2 the polynomial cannot be independent of X. Rearranging in powers of X, we see that X is algebraic over k(Y). Moreover,

1 2 1. Lecture 1

[K : k(Y)] = [K : k(X, Y)].[k(X, Y) : k(Y)] ≤ [K : k(X)].[k(X, Y) : k(Y)] < ∞

and thus Y also satisfies the same conditions as X. Thus, any transcendental element of K may be used as a variable in the place of X. The set of all elements of K algebraic over k forms a subfield k′ of K, which is called the field of constants of K. Hence forward, we shall always assume, unless otherwise stated, that k = k′, i.e., that k is algebraically closed in K.

2 Ordered Groups

Deﬁnition . A multiplicative(additive) Abelian group W with a binary relation < (>) between its elements is said to be an ordered group if

0(1) for α, β ∈ W, one and only one of the relations α < β,α = β,β < α (α>β,α = β>α) holds.

0(2) α<β,β<γ ⇒ α < γ (α>β,β>γ ⇒ α > γ)

0(3) α < β, δ ∈ W ⇒ αδ < βδ(α > β, δ ∈ W ⇒ α + δ > β + δ)

We shall denote the identity (zero) element by 1(0). In this and the following section, we shall express all our results in multiplicative notation. α > β shall mean the same thing as β<α. Let W0 be the set {α : α ∈ Wα< 1}. W0 is seen to be a semi group by = −1 3 0(2) and 0(3). Moreover, W W0 ∪{1}∪W0 is a disjoint partitioning of −1 W (where W0 means the set of inverses of elements of W0). Conversely, −1 if an Abelian group W can be partitioned as W0 ∪{1} ∪ W0 , where W0 is a semi-group, we can introduce an order in W by defining α < β to −1 mean αβ ∈ W0; it is immediately verified that 0(1), 0(2) and 0(3) are fulfilled and that W0 is precisely the set of elements < 1 in this order. For an Abelian group W, the map α → αn (n any positive integer) is in general only an endomorphism. But if W is ordered, the map is a monomorphism; for if α is greater than or less than 1, αn also satisfies the same inequality. 3. Valuations, Places and Valuation Rings 3

3 Valuations, Places and Valuation Rings

We shall denote the non-zero elements of a field K by K∗. Definition. A Valuation of a field K is a mapping v of K∗ onto an ordered multiplicative (additive) group W (called the group of the valuation or the valuation group) satisfying the following conditions: V(1) For a, b ∈ K∗, v(ab) = v(a)v(b) (v(ab) = v(a) + v(b)); i.e. v is homomorphism of the multiplicative group K∗ onto W.

V(2) For a, b, a+b ∈ K∗, v(a+b) ≤ max(v(a), v(b)) (v(a+b)) ≥ min(v(a) v(b)))

V(3) v is non-trivial; i.e., there exists an a ∈ K∗ with v(a) , 1(v(a) , 0) Let us add an element 0(∞) to W satisfying the following (1) 0.0 = α.0 = 0.α = 0 for every α ∈ W(∞+∞ = α+∞ = ∞+α = ∞),

(2) α > 0 for every α ∈ W(α < ∞). If we extend a valuation v to the 4 whole of K by defining v(0) = 0 (v(0) = ∞, the new mapping also satisfies V(1), V(2) and V(3). The following are simple consequences of our definition. (a) For a ∈ K, v(a) = v(−a). To prove this, it is enough by V(1) to prove that v(−1) = 1. But v(−1). v(−1) = v(1) = 1 by V(1), and hence v(−1) = 1 by the remark at the end of §2.

(b) If v(a) , v(b), v(a + b) = max(v(a), v(b)). For let v(a) < v(b). Then, v(a + b) ≤ max(v(a), v(b)) = v(b) = v(a + b − a) ≤ max(v(a + b), v(a)) = v(a + b)

(c) Let ai ∈ K, (i = 1,... n). Then an obvious induction on V(2) gives n n v( ai) ≤ max v(ai), and equality holds if v(ai) , v(a j) for i , j. 1 i=1 P n (d) If ai ∈ K, (i = 1,... n) such that ai = 0, then v(ai) = v(a j) for 1 at least one pair of unequal indicesPi and j. For let ai be such that 4 1. Lecture 1

n n v(ai) ≥ v(ak) for k , i. Then v(ai) = v( ak) ≤ max(v(ak)) = v(a j) k=1 k=1 kP,i k,i for some j , i, which proves that v(ai) = v(a j).

Let be a ﬁeld. By (∞), we shall mean the set of elements of togetherP with an abstractP element ∞ with the following properties. P α + ∞ = ∞ + α = ∞ for every α ∈ . X α. ∞ = ∞.α = ∞ for every α ∈ ,α , 0. ∞ + ∞ and 0.∞ are not deﬁned. X

5 Definition. A place of a field K is a mapping ϕ of K into U(∞) (where may be any field ) such that P P P(1) ϕ(a + b) = ϕ(a) + ϕ(b).

P(2) ϕ(ab) = ϕ(a).ϕ(b).

P(3) There exist a, b ∈ K such that ϕ(a) = ∞ and ϕ(b) , 0 or ∞. P(1)andP(2) are to hold whenever the right sides have a meaning.

From this it follows, taking the b of P(3), that ϕ(1)ϕ(b) = ϕ(b), so that ϕ(1) = 1, and similarly ϕ(0) = 0. Consider the set Oϕ of elements a ∈ K such that ϕ(a) , ∞. Then by P(1), P(2) and P(3), Oϕ is a ring which is neither zero nor the whole of K, and ϕ is a homomorphism of this ring into . Since is a ﬁeld, the kernel of this homomorphism is a prime ideal YP of Oϕ P 1 Let b be an element in K which is not in O . We contend that ϕ( ) = ϕ b 1 0. For if this mere not true, we would get 1 = ϕ(1) = ϕ(b). ϕ( ) = ∞, b 1 by P(2). Thus ∈ Y , and thus Y is precisely the set of non-units of b Oϕ. Since any ideal strictly containing Y should contain a unit, we see that Y is a maximal ideal and hence the image of Oϕ in is again a ﬁeld. We shall therefore always assume that is preciselyP the image of Oϕ by ϕ, or that ϕ is a mapping onto U(∞).P The above considerations motivateP the 3. Valuations, Places and Valuation Rings 5

Deﬁnition. Let K be a ﬁeld. A valuation ring of K is a proper subring 1 O of K such that if a ∈ K∗, at least one of the elements a is in O. 6 a In particular, we deduce that O contains the unity element. Let Y be the set of non-units in O. Then Y is a maximal ideal. For, let a ∈ O, b ∈ 1 Y . If ab < Y , ab would be a unit of O, and hence ∈ O. This implies ab 1 1 that a = ∈ O, contradicting our assumption that b is a non-unit of ab b O. Suppose that c is another element of Y . To show that b − c ∈ Y , we may assume that neither of them is zero. Since O is a valuation ring, b c b b b − c at least one of or , say , is in O. Hence, − 1 = ∈ O. If c b c c c 1 1 b − c 1 b − c were not in Y , ∈ O, and hence · = ∈ O, b − c b − c c c contradicting our assumption that c ∈ Y . Finally, since every element outside Y is a unit of O, Y is a maximal ideal in O.

Lecture 2

3 (Contd.)

In this lecture, we shall establish the equivalence of the concepts of val- 7 uation, place and valuation ring. Two places ϕ1 : K → 1 ∪(∞) and ϕ2 : K → 2 ∪(∞) are said to be equivalent if there existsP an isomorphism λ of P1, onto 2 such that = = ϕ2(a) λ ◦ ϕ1(a) for every a, with the understandingP that λP(∞) ∞. This is clearly an equivalence relation, and thus, we can put the set of places of K into equivalence classes. Moreover, equivalent places ϕ1 O O and ϕ2 obviously define the same valuation rings ϕ1 and ϕ2 . Thus, to every equivalence class of places is associated a unique valuation ring. Conversely, let O be any valuation ring and Y its maximal ideal. Let be the quotient field O/Y and η the natural homomorphism of O ontoP . It is an easy matter to verify that the map ϕ : K → U{∞} definedP by P η(a) if a ∈ O ϕ(a) = ∞ if a < O  is a place, whose equivalence class corresponds to the given valuation ring O. Let v1 and v2 be two valuations of a field K in the ordered group W1 and W2. We shall denote the unit elements of both the groups by 1, 8 since it is not likely to cause confusion. We shall say that v1 and v2 are equivalent if v1(a) > 1 if and only if v2(a) > 1. Let v1 and v2 be two equivalent valuations. From the definition, it follows, by taking reciprocals, that v1(a) < 1 if and only if v2(a) < 1,

7 8 2. Lecture 2

and hence (the only case left) v1(a) = 1 if and only if v2(a) = 1. Let α be any element of W1. Choose a ∈ K such that v1(a) = α (this is possible since v1 is onto W1). Define σ(α) = v2(a). The definition is independent of the choice of a since if b were another element with −1 −1 v1(b) = α, then v1(ab ) = 1 so that v2(ab ) = 1, i.e. v2(a) = v2(b). Thus, σ is a mapping from W1 onto W2 (since v2 is onto W2). It is easy to see that σ is an order preserving isomorphism of W1 onto W2 and we ∗ have v2(a) = (σ.v1)(a) for every a ∈ K . Thus, we see that the definition of equivalence of valuations can also be cast into a form similar to that for places. Again, equivalence of valuations is an equivalence relation, and we shall that equivalence classes of valuations of a field K correspond canonically and biunivocally to valuation rings of the field K. Let v be a valuation and O be the set of elements a in K such that v(a) ≤ 1. It is an immediate consequence of the definition that O is a 1 1 ring. Also, if a ∈ K, v(a) > 1, then v < 1 and hence ∈ O. Thus, O a! a 9 is a valuation ring. Also, if v1 and v2 are equivalent, the corresponding rings are the same. Suppose conversely that O is a valuation ring in K and Y its maximal ideal. The set difference O − Y is the set of units of O and hence a subgroup of the multiplicative group K∗. Let η : K∗ → K∗/O − Y be the natural group homomorphism. Then η(O∗) is obviously a semi- group and the decomposition K∗/O − Y = η(O∗) ∪{1}U ∪ η(O∗)−1 is disjoint, Hence, we can introduce an order in the group K∗/O − Y and it is easy to verify that η is a valuation on K whose valuation ring is precisely O. Summarising, we have

Theorem. The valuations and places of a ﬁeld K are, upto equivalence, in canonical correspondence with the valuation rings of the ﬁeld. Lecture 3

4 The Valuations of Rational Function Field

Let K = k(X) be a rational function field over k ; i.e., K is got by ad- 10 joining to k a single transcendental element X over k. We seek for all the valuations v of K which are trivial on k, that is, v(a) = 1 for every a ∈ K∗. It is easily seen that these are the valuations which correspond to places whose restrictions to k are monomorphic. We shall henceforward write all our ordered groups additively. Let ϕ be a place of K = k(X) onto ∪(∞). We consider two cases Case 1. Let ϕ(X) = ξ , ∞. Then, the polynomialP ring k[X] is contained in Oϕ, and Y ∩ k[X] is a prime ideal in k[X]. Hence, it should be of the form (p(X)), where p(X) is an irreducible polynomial in X. Now, if g(X) r(X) ∈ K, it can be written in the form r(X) = (p(X))ρ , where g(X) h(X) and h(X) are coprime and prime to p(X). Let us agree to denote the image in of an element c in k by c,¯ and that of a polynomial f over k by f¯ . Then,weP clearly have 0 if ρ> o ϕ(r(x)) =  g(ξ) if ρ = 0  h(ξ) ∞ if ρ< 0   Conversely, suppose p(X) is an irreducible polynomial in k[X] and 11 ξ a root of p(X). The above equations then define a mapping of k(X) onto k(ξ) ∪ {∞}, which is a place, as is verified easily. We have thus determined all places of k(X) under case 1 (upto equivalence).

9 10 3. Lecture 3

If Z is the additive group of integers with the natural order, the valuation v associated with the place ϕ above is given by v(r(X)) = ρ. 1 Case 2. Suppose now that ϕ(X) = ∞. Then ϕ( ) = 0. Then since K = X 1 k(X) = k( ), we see that ϕ is determined by an irreducible polynomial X 1 1 p( ), and since ϕ ) = 0, p(Y) should divide Y. Thus, p(Y) must be Y X X (except for a constant in k), and if

a + a x + − + a xn = 0 1 n , r(X) m , an, bm 0, b0 + b1 x + − + bm x o ifm > n a0 + a1 + − + a ( 1 )m−n xn xn−1 n a ϕ(r(X)) = ϕ X =  n if m = n  b0 + b1 + − + b   bm  xm xm−1 m  ∞ if m < n       The corresponding valuation with values in Z is given by v(r(X)) = m − n = − deg r(X), where the degree of a rational function is defined in the degree of the numerator-the degree of the denominator. We shall say that a valuation is discrete if the valuation group may be taken to be Z. We have in particular proved that all valuations of a rational function field trivial over the constant field are discrete. We shall extend this result later to all algebraic function fields of one variable.

5 Extensions of Places

12 Given a field K, a subfield L and a place ϕL of L into , we wish to 1 1 prove in this section that there exists a place ϕK of K intoP , where is a field containing and the restriction of ϕK to L is ϕLP. Such a ϕKPis called an extension ofP the place ϕL to K. For the proof of this theorem, we require the following

Lemma (Chevalley). Let K be a ﬁeld, O a subring and ϕ a homomorphism of O into a ﬁeld ∆ which we assume to be algebraically closed. Let q be any element of K∗, and O[q] the ring generated by O and q in 5. Extensions of Places 11

K. Then ϕ can be extended into a homomorphism Φ of at least one of 1 the rings O[q], O[ ], such that Φ restricted to O coincides with ϕ. q Proof. We may assume that ϕ is not identically zero. Since the image of O is contained in a field, the kernel of ϕ is a prime ideal Y which is a not the whole ring O. Let O1 = O ∪ a, b ∈ O, b < Y . Clearly, O1 b is a ring with unit, and ϕ has a unique extensionϕ ˜ to O1 as a homomor- a ϕ(a) phism, give byϕ ˜ = . The image byϕ ˜ is then the quotient field b ϕ(b) of ϕ(O). We shall denoteϕ ˜(a) bya ¯ for a ∈ O1. P Let X and X¯ be indeterminates over O1 and respectively.ϕ ˜ can 1 be extended uniquely to a homomorphismϕ ¯ of OP[X] onto [X¯] which takes X to X¯ by defining P n n ϕ¯(a0 + a1X + − + anX ) = a¯0 + a¯1X¯ + − + a¯nX¯ .

Let U be the ideal O1[X] consisting of all polynomials which van- 13 ish for X = q, and let U¯ be the idealϕ ¯(U ) in [X¯]. We consider three cases. P Case 1. Let U¯ = (0). In this case, we define Φ(q) to be any fixed element of ∆. Φ is uniquely determined on all other elements of O1[q] by the requirement that it be a ring homomorphism which is an extension of ϕ˜. In order that it be well defined, it is enough to verify that if any polynomial over O1 vanishes for q, its image by ϕ¯ vanishes for Φ(q). But this is implied by our assumption. Case 2. Let U¯ , (0), , [X¯]. Then U¯ = ( f (X¯)), where f is a non- constant polynomial overP . Let α be any root f (X¯) in ∆ (there is a root in ∆ since ∆ is algebraicallyP closed). Define Φ(q) = α. This can be extended uniquely to a homomorphism of O1[q], since the image by ϕ¯ of any polynomial vanishing for q is of the form f (X¯) f (X¯), and therefore vanishes for X¯ = α. Case 3. Suppose U¯ = [X¯]. Then the homomorphism clearly cannot 1 be extended to O1[q]. SupposeP now that it cannot be extended to O1[ ] q 12 3. Lecture 3

either. Then if δ denotes the ideal of all polynomials in O1[X] which 1 vanish for , and if δ¯ is the ideal ϕ¯(δ)in [X¯], we should have δ¯ = q P n [X¯]. Hence, there exist polynomials f (X) = a0 + a1X + − + anX and P m 1 b0 +b1X +−+bmX such that ϕ¯( f (X)) = ϕ¯(g(X)) = 1, f (q) = g = 0. q! We may assume that f and g are of minimal degree n and m satisfying 14 the required conditions. Let us assume that m ≤ n. Then, we have m a¯0 = b¯0 = 1, a¯i = b¯ j = 1 for i, j > 0. Let g0(X) = b0X + ··· + bm. n Applying the division algorithm to the polynomials b0 f (X) and g0(X), we obtain

n = + O1 b0 f (X) g0(X)Q(X) R(X), Q(X), R(X) ∈ [X], deg R < m. Substituting X = q, we obtain R(q) = 0. Also, acting withϕ ¯, we have

= ¯n ¯ ¯ = ¯ ¯ ¯ + ¯ ¯ = ¯ ¯ ¯ m + ¯ ¯ 1 b0 f (X) g¯0(X)Q(X) R(X) Q(X)X R(X), and hence, we deduce that Q¯(X¯) = 0, R¯(X¯) = 1. Thus, R(X) is a polynomial with R(q) = 0, R¯(X¯) = 1, and deg F(X) < m ≤ n, which contradicts our assumption on the minimality of the degree of f (X). Our lemma is thus prove. We can now prove the

Theorem. Let K be a field and O a subring of K. Let ϕ be a homomorphism of O in an algebraically closed field ∆. Then it can be extended either to a homomorphism of K in ∆ or to a place of K in ∆ ∪ (∞). In particular, any place of a subfield of K can be extended to a place of K.

Proof. Consider the family of pairs {ϕα, Oα}, where Oα is a subring if K containing O and ϕα a homomorphism of Oα in ∆ extending ϕ on O. The family is non-empty, since it contains (ϕ, O). We introduce a partial order in this family by deﬁning (ϕα, Oα) > (ϕβ, Oβ) if Oα ⊃ Oβ and ϕα is an extension of ϕβ.

15 The family clearly being inductive, it has a maximal element by Zorn’s lemma. Let us denote it by (Φ, O). O is either the whole of K or 5. Extensions of Places 13 a valuation ring of K. For if not, there exists a q ∈ K such that neither 1 q nor belongs to O. we may then extend Φ to a homomorphism of q 1 at least one of O[q] or O in ∆. Since both these rings contain O "q# strictly, this contradicts the maximality of (Φ, O). If O were not the whole of K, Φ must vanish on every non - unit of 1 1 O; for if q were a non-unit and Φ(q) , 0, we may define Φ = q! Φ(q) 1 and extend this to a homomorphism of O , which again contradicts "q# the maximality of O. This proves that Φ can be extended to a place of K by defining it to be ∞ outside O. In particular, a place ϕ of a subfield L of K , when considered as a homomorphism of its valuation ring Oϕ and extended to K, gives a place on K; for if ϕ where a homomorphism of the whole of K in ∆, it should be an isomorphism (since the kernel, being a proper ideal in K, should be the zero ideal). But Φ being an extension of ϕ, the kernel contains at least one non-zero element.

Corollary . If K/k is an algebraic function ﬁeld and X any element of K transcendental over k, there exists at least one valuation v for which v(X) > 0.

Proof. We have already shown in the previous section that there exists Y Y a place 1 in k(X) such that vY1 (X) > 0. If we extend this place 1 to a place Y of K, we clearly have vY (X) > 0.

Lecture 4

6 Valuations of Algebraic Function Fields

It is our purpose in this paragraph to prove that all valuations of an al- 16 gebraic function field K which are trivial on the constant field k are discrete. Henceforward, when we talk of valuations or places of an algebraic function field K, we shall only mean those which are trivial on k. Valuations will always be written additively. We require some lemmas.

Lemma 1. Let K/L be a ﬁnite algebraic extension of degree [K : L] = n and let v be a valuation on K with valuation group V. If V◦ denotes the ∗ subgroup of V which is the image of L under v and m the index of V◦ in V, we have m ≤ n.

Proof. It is enough to prove that of any n + 1 elements α1,...αn+1 of V, at least two lie in the same coset modulo V◦.

Choose ai ∈ K such that v(ai) = αi (i = 1,... n + 1). Since there can be at most n linearly independent elements of K over L, we should have

n+1 liai = 0, li ∈ L, not all li being zero . Xi=1

This implies that v(liai) = y(l ja j) for some i and j, i , j (see Lecture 1, § 3). Hence, we deduce that

v(li) + v(ai) = v(liai) = v(l ja j) = v(l j) + v(a j),

αi − α j = v(ai) − v(a j) = v(l j) − v(li) ∈ V◦

15 16 4. Lecture 4

and αi and α j are in the same coset modulo V◦. Our lemma is proved. 17 Let V be an ordered abelian group. We shall say that V is archimedean if for any pair of elements α, β in V with α> 0, there corresponds an integer n such that nα > β. We shall call any valuation with value group archimedean an archimedean valuation.

Lemma 2. An ordered group V is isomorphic to Z if and only if (i) it is archimedean and (ii) there exists an element ξ > 0 in V such that it is the least positive element; i.e., α> 0 ⇒ α ≥ ξ.

Proof. The necessity is evident. Now, let α be any element of V.

Then by assumption, there exists a smallest integer n such that nξ ≤ α< (n + 1)ξ. Thus,

0 ≤ α − nξ < (n + 1)ξ − nξ = ξ,

and since ξ is the least positive element, we have α = nξ. The mapping α ∈ V → n ∈ Z is clearly an order preserving isomorphism, and the lemma is proved.

Lemma 3. If a subgroup V◦ of ﬁnite index of an ordered group V is isomorphic to Z, V is itself isomorphic to Z.

Proof. Let the index be V : V◦ = n. Let α, β be any two elements of V, with α> 0. Then nα and nβ are in V◦, n α> 0.

Since V◦ is archimedean, there exists an integer m such that mnα > nβ, from which it follows that mα > β. 18 Again, consider the set of positive elements α in V. Then nα are in V◦ and are positive, and hence contain a least element nξ (since there is an order preserving isomorphism between V◦ and Z). Clearly, ξ is then the least positive element of V. V is therefore isomorphic to Z, by Lemma 2. We ﬁnally have the

Theorem. All valuations of an algebraic function ﬁeld K are discrete. 7. The Degree of a Place 17

Proof. If X is any transcendental element of K/k, the degree K : k(X) < ∞. Since we know that all valuations of k(X) are discrete, our result by applying Lemma 1 and Lemma 3.

7 The Degree of a Place

Let Y be a place of an algebraic function field K with constant field k onto the field kY ∪∞. Since Y is an isomorphism when restricted to k, we may assume that kY is an extension of k. We shall moreover assume that kY is the quotient OY /MY where OY is the ring of the place Y and MY the maximal ideal. We now prove the

Theorem. Let Y be a place of an algebraic function ﬁeld. Then kY /k is an algebraic extension of ﬁnite degree.

Proof. Choose an element X , 0 in K such that Y (X) = 0. Then X should be transcendental, since Y is trivial on the ﬁeld of constants.

Let K : k(X) = n < ∞. Let α1,...αn+1 be any (n + 1) elements of 19 kY . Then, we should have αi = Y (ai) for some ai ∈ K(i = 1,... n + 1). There therefore exist polynomials fi(X) in k[X] such that n+1 fi(X)ai = 0, not all fi(X) having constant term zero. i=1 P Writing fi(X) = li + Xgi(X), we have

n+1 n+1 liai = −X aigi(X), and taking Xi=1 Xi=1

n+1 n+1 the Y -image, liαi = −Y (X) aigi(Y X) = 0, li ∈ k, not all li being i=1 i=1 zero. P P Thus, we deduce that the degree of kY /k is at most n. The degree fY of kY over k is called the degree of the place Y . Note that fY is always ≥ 1. If the constant ﬁeld is algebraically closed (e.g. in the case of the complex number ﬁeld), fY = 1, since kY , being an algebraic extension of k, should coincide with k. 18 4. Lecture 4

Finally, we shall make a few remarks concerning notation. If Y is a place of an algebraic function ﬁeld, we shall denote the corresponding valuation with values in Z by vY (vY is said to be a normed valuation at the place Y ). The ring of the place shall be denoted by OY and its unique maximal ideal by Y . (This is not likely to cause any confusion).

8 Independence of Valuations

20 In this section, we shall prove certain extremely useful result on valuations of an arbitrary ﬁeld K.

Theorem. Let K be an arbitrary ﬁeld and vi(i = 1,... n) a set of valuations on K with valuation rings Oi such that Oi 1 O j if i , j. There is then an element X ∈ K such that v1(X) ≥ 0, vi(X) < 0 (i = 1,... n).

Proof. We shall use induction. If n = 2, since O1 1 O2, there is an X ∈ O1, X < O2, and this X satisﬁes the required conditions. Suppose now that the theorem is true for n − 1 instead of n. Then there exists a Y ∈ K such that

v1(Y) ≥ 0, vi(Y) < 0 (i = 2,... n − 1).

Since O1nsubsetOn, we can ﬁnd aZ ∈ K such that

v1(Z) ≥ 0, vn(Z) < 0

Let m be a positive integer. Put X = Y + Zm. Then

m v1(Y + Z ) ≥ min(v1(Y), mv1(Z)) ≥ 0

Now suppose r is one of the integers 2, 3,... n. If vr(Z) ≥ 0, r cannot be n, and since vr(Y) < 0, we have

m vr(Y + Z ) = vr(Y) < 0. 8. Independence of Valuations 19

m 21 If vr(Z) < 0 and vr(Y + Z r ) ≥ 0 for some mr, for m > mr we have

m mr m mr mr m mr vr(Y + Z ) = vr(Y + Z + Z − Z ) = min(vr(Y + Z ), vr(Z − Z )),

m m m m−m and vr(Z − Z r ) = vr(Z r ) + vr(1 − Z r ) = mrvr(Z) < 0, m−mr Since vr(1 − Z ) = vr(1) = 0 m Thus, vr(Y + Z ) < 0 for large enough m in any case. Hence X satisﬁes the required conditions. If we assume that the valuations vi of the theorem are archimedean, then the hypothesis that Oi 1 O j for i , j can be replaced by the weaker one that the valuations are inequivalent (which simply states that Oi , O j for i , j). To prove this, we have only to show that if v and v1 are two archimedean valuations such that the corresponding valuation rings O and O1 satisfy O ⊃ O1, then v and v1 are equivalent. For, consider an element 1 1 a ∈ K∗ such that v(a) > 0. Then, v < 0, and consequently is not a! a 1 in O, and hence not in O1. Thus, v1 < 0, v1(a) > 0. Conversely, a! suppose a ∈ K∗ and v1(a) > 0. Then by assumption, v(a) ≥ 0. Suppose now that v(a) = 0. Find b ∈ K∗ such that v(b) < 0. If n is any positive n integer, we have v(a b) = nv(a) + v(b) < 0, anb < O. But since v1 is archimedean and v1(a) > 0, for large enough n we have v1(anb) = nv1(a) + v1(b) > 0 , anb ∈ O1.

22 This contradicts our assumption that O1 ⊂ O, and thus, v(a) > 1 0. Hence v and v are equivalent. Under the assumption that the vi archimedean, we can replace in the theorem above the ﬁrst inequality V1(X) ≥ 0 even by the strict inequality v1(X) > 0. To prove this let ∗ X1 ∈ K satisfy v1(X1) ≥ 0, vi(X1) < 0, i > 1. Let Y be an element in ∗ m K with v1(Y) > 0. Then, if X = X1 Y, where m is a suﬃciently large positive integer, we have = m = + v1(X) v1(X1 Y) mv1(X1) v1(Y) > 0, = m = + = vi(X) vi(X1 Y) mvi(X1) vi(Y) < 0, i 2,..., n. 20 4. Lecture 4

We shall hence forward assume that all valuations considered are archimedean. To get the strongest form of our theorem, we need two lemmas.

Lemma 1. If vi (i = 1,... n) are inequivalent archimedean valuations, and ρi are elements of the corresponding valuations group, we can ﬁnd Xi(i = 1,... n) in K such that vi(Xi − 1) > ρi, v j(Xi) > ρ j, i , j

Proof. Choose Yi ∈ K such that

vi(Yi) > 0, v j(Yi) < 0 for j , i.

= 1 Put Xi + m . Then, if m is chosen large enough, we have (since the 1 Yi valuation are archimedean) = + m = v j(Xi) −v j(1 Yi ) −mv j(Yi) > ρ j, i , j m −Yi 23 and v (X − 1) = v ( ) = mv (Y ) − v (1 + Ym) = mv (Y ) > ρ . i i i + m i i i i i i i 1 Yi + m = since v − i(1 Yi ) 0.

A set of valuations vi(i = 1,... n) are said to be independent if given any set of elements ai ∈ K and any set of elements ρi in the respective valuation groups of vi, we can ﬁnd an X ∈ K such that

vi(X − ai) > ρi. We then have the following Lemma 2. Any ﬁnite set of inequivalent archimedean valuations are independent.

Proof. Suppose vi(i = 1,... n) is a given set of inequivalent archimedean valuations. If ai ∈ K and ρi are elements of the valuation group of the n vi, put σi = ρi − min vi(a j). Choose Xi as in Lemma 1 for the vi and σi. j=1 n Put X = aiXi. Then 1 P n vi(X − ai) = vi a jX j + ai(Xi − 1) >σi + min vi(a j) = ρi,   j=1 Xj,i      8. Independence of Valuations 21 and our lemma is proved.

Finally, we have the following theorem, which we shall refer to in future as the theorem of independence of valuations.

Theorem . If vi(i = 1,... n) are inequivalent archimedean valuations, ρi is an element of the value group of vi for every i, and ai are given 24 elements of the ﬁeld, there exists an element X of the ﬁeld such that

vi(X − ai) = ρi

Proof. Choose Y by lemma 2 such that vi(Y − ai) > ρi. Find bi ∈ K such that vi(bi) = ρi and an Z ∈ K such that vi(Z − bi) > ρi. Then it follows that vi(Z) = min(vi(Z − bi), vi(bi)) = ρi. Put X = Y + Z. Then,

vi(X − ai) = vi(Z + Y − ai) = vi(Z) = ρi, and X satisﬁes the conditions of the theorem.

Corollary. There are an inﬁnity of places of any algebraic function ﬁeld.

Proof. Suppose there are only a ﬁnite number of places Y1,..., Yn. = + Choose an X such that vYi (X) > 0 (i 1,... n). Then, vYi (X = = Y 1) vYi (1) 0 for all the places , i, which is impossible since X an consequently X + 1 is a transcendental element over k.

Lecture 5

9 Divisors

Let K be an algebraic function ﬁeld with constant ﬁeld k. We make the 25

Deﬁnition. A divisor of K is an element of the free abelian group generated by the set of places of K. The places themselves are called prime divisors.

The group ϑ of divisors shall be written multiplicatively. Any element U of the group ϑ of divisors can be written in the form

U = Y vY (U ) YY where the product is taken over all prime divisors Y of K, and the vY (U ) are integers, all except a ﬁnite number of which are zero. The divisor is denoted by n. We say that a divisor U is integral if vY (U ) ≥ 0, for every Y , and that U divides δ if δU −1 is integral. Thus, U divides δ if and only if vY (δ) ≥ vY (U ) for every Y . Two divisors U and δ are said to be coprime if vY (U , 0 implies that vY (δ) = 0. The degree d(U ) of a divisor U is the integer

d(U ) = fY vY (U ), XY where fY is the degree of the place Y . 26

23 24 5. Lecture 5

The map U → d(U ) is a homomorphism of the group of divisors ϑ into the additive group of integers. The kernel ϑ◦ of this homomorphism is the subgroup of divisors of degree zero. An element X of K is said to be divisible by the divisor U if vY (X) ≥ vY (U ) for every Y . Two elements X, Y ∈ K are said to be congruent modulo a divisor U (written X ≡ ( mod U )) if X − Y is divisible by U . Let S be a set prime divisors of K. Then we shall denote by Γ(U /S ) the set of elements X ∈ K such that vY (X) ≥ vY (U ) for every Y in S . It is clear that Γ(U /S ) is a vector space over the constant field k, and also that if U divides δ, Γ(δ/S ) ⊂ Γ(U /S ). Also, if S and S 1 are two sets of prime divisors such that S ⊂ S 1, then Γ(U /S 1) ⊂ Γ(U /S )). Finally, Γ(U /S ) =Γ(δ/S ) if U δ−1 contains no Y belonging to S with a non-zero exponent. If S is a set of prime divisor which is fixed in a discussion and U a divisor, we shall denote by U◦ the new divisor got from U by omitting vY (U ) all Y which do not occur in S : U◦ = Y Y ∈S Q Theorem. Let S be a finite set of prime divisors and U , δ two divisors such that U divides δ. Then, Γ U ( /S ) −1 dim = d(δ◦) − d(U◦) = d(δ◦U ). k Γ(δ/S ) ◦

27 Proof. By our remark above, we may assume that U = U◦ and δ = δ◦. Moreover, it is clearly suﬃcient to prove the theorem when δ = UY , where Y is a prime divisor belonging to S . For, if δ = UY1..Yn we have Γ(U /S ) Γ(U /S ) dimk = dimk + Γ(δ/S ) Γ(UY1/S ) Γ UY Γ U ( 1/S ) ( Y1···Yn−1 /S dimk + ··· + dimk Γ(UY1Y2/S ) Γ(δ/S )

and d(δ) − d(U ) = d(Y1) + d(Y2) + ··· + d(Yn). Hence, we have to prove that if U is a divisor such that all the prime divisors occurring in it with non-zero exponents are in S , and Y 9. Divisors 25 any prime divisor in S , we have

Γ(U /S ) dim = fY = f. k Γ(UY /S ) By the theorem on independence of valuations, we may choose an element u ∈ K such that 28

vU (u) = vU (U ) for all U in S.

If X1, X2,... X f +1 are any f + 1 elements of Γ(U /S ), the elements −1 −1 X1u ,... X f +1u are all in UY . But since the degree of kU = OY /Y over k is f , we have

f +1 −1 aiXiu ∈ Y , ai ∈ k, not all ai being zero . Xi=1

f +1 Hence, aiXi ∈ Γ(UY /S ), ai ∈ k, not all ai being zero, thus i=1 P Γ(U /S ) proving that the dimension over k of the quotient is ≤ f . Γ(UY /S ) Now, suppose Y1,... Y f are f elements of OY such that they are lin- Y 1 early independent over k modulo . ChooseYi ∈ K such that 1 1 U Y U = vY (Yi − Yi) > 0, vU (Yi ) ≥ 0 for , , ∈ S.(i 1, f ). 1 Y 1 By the ﬁrst condition, Yi ≡ Yi( mod ), and hence Yi and Yi de- termine the same element in kY . But by the second condition, the ele- 1 Γ U ments uYi belong to ( /S ), and since Y1,... Y f are linearly indepen- Y 1 1 ﬃ dent mod , no linear combination of uY1 ,... uY f with coe cients in k- at least one of which is non-zero-can lie in Γ(UY /S ). Thus, Γ(U /S ) dim ≥ f . k Γ(UY /S ) This proves our theorem.

Lecture 6

10 The Space L(U )

Let U be any divisor of an algebraic function ﬁeld K. 29 We shall denote by L(U ) the set of all elements of K which are divisible by U . Clearly L(U ) = Γ(U /S ) if S is the set of all prime divisors of K, and thus we deduce that L(U ) is a vector space over k and if U divides δ, L(U ) ⊃ L(δ). We now prove the following important Theorem . For any divisor U , the vector space L(U ) is ﬁnite dimensional over k. If we denote its dimension by l(U ), and if U divides δ, we have l(U ) + d(U ) ≤ l(δ) + d(δ). Proof. Let S be the set of prime divisors occurring in U or δ. Then, it easy to see that

L(δ) = L(U ) ∩ Γ(δ/S )

Hence, by Noether’s isomorphism theorem, L(U ) L(U ) L(U ) +Γ(δ/S ) L(U /S ) = ≃ ⊂ , L(δ) L(U ) ∩ Γ(δ/S ) L(δ/S ) Γ(δ/S ) L(U ) Γ(U /S ) and therefore, dim ≤ dim = d(δ) − d(U ) k L(δ) k Γ(δ/S ) Now, choose for δ any integral divisor which is a multiple of U and is not the unit divisor n. Then, L(δ) = (0); for if X were a non-zero element of L( ), it cannot be a constant since vY (X) ≥ vY (δ) > 0 for

27 28 6. Lecture 6

at least one Y , and it cannot be transcendental over k since vY (X) ≥ 30 vY (δ) ≥ 0 for all Y .

This is not possible. This together with the above inequality proves that dimk L(U ) = l(U ) < ∞, and our theorem is completely proved. Since L(N ) clearly contains only the constants, l(N ) = 1.

11 The Principal Divisors

We shall now associate to every non-zero element of K a divisor. For this, we need the

Theorem . Let X ∈ K∗. Then there are only a ﬁnite number of prime divisors Y with vY (X) , 0.

Proof. If X ∈ k, vY (X) = 0 for all Y and the theorem is valid. Hence assume that X is transcendental over k. Let [K : k(X)] = N. Y Y = Suppose 1,... n are prime divisors for which vYi (X) > 0. Let δ n Y = Y Y ivYi (X), and S { 1,... n}. i=1 Q n Γ(N /S ) = = Then, dimk Γ(δ/S ) (δ) fYi vYi (X). We shall show that this is i=1 at most equal to N. P Let in fact Y1,..., YN+1 be any (N + 1) elements of Γ(N /S ). Since N+1 [K : k(X)] = N, we should have f j(X)Y j = 0, f j(X) ∈ k[X], with j=1 P at least one f j having a non-zero constant term. Writing f j(X) = a j + N+1 N+1 Xg j(X), the above relation may be rewritten as a jY j = −X g j 1 1 (X)Y j, not all a j being zero, and hence P P

N+1 N+1 vY a Y = vY (X) + vY g (X)Y ≥ vY (X) ν  j j ν ν  j j ν X1  X1      31     11. The Principal Divisors 29

N+1 This proves that a jY j ∈ Γ(δ/S ), and therefore 1 P n Γ(N /S ) n ≤ fY vY (X) = d(δ) = dim ≤ N, i I k Γ(δ/S ) Xi=1 1 By considering instead of X, we deduce that the number of prime X divisors Y for which vY (X) < 0 is also finite, and our theorem follows. The method of defining the divisor (X) corresponding to an element ∗ X ∈ K is now dear. We define the numerator zX of X to be the divisor Y vY (X) (the product being taken over all Y for which vY (X) > vY (X)>0 Q −vY 0), the denominator NX of X to be the divisor Y (X) , and the vY (X)<0 Qz principal divisor (X) of X to be Y vY (X) = X . v (X),0 NX y Q If X, Y ∈ K∗, clearly (XY) = (X)(Y) and (X−1) = (X)−1. Thus, the principal divisors form a subgroup Z of the group of divisors ϑ. ϑ The quotient group R = is called the group of divisor classes. The Z following sequence of homomorphisms is easily seen to be exact (i.e., the image of a homomorphism is equal to the kernel of the next).

1 → k∗ → K∗ → ϑ → R → 1.

In the course of the proof of the above theorem, we proved the inequalities d(Nx) ≤ N, d(zx) ≤ N, for a transcendental X, where [K : k(X)] = N. We will now show that equality holds 32 Theorem . Let X be a transcendental element of K, and put N = [K : k(X)]. Then, d(zX) = d(NX) = N. In order to prove the theorem, we shall ﬁrst prove a lemma. We shall say that Y ∈ K is an integral algebraic function of X, if it satisﬁes a relation

m m−1 Y + fm−1(X)Y + − + fo(X) = o, fi(X) ∈ k[X]. 30 6. Lecture 6

We then have the Lemma. If Y is an integral algebraic function of X, and a prime divisor Y does not divide NX, it does not divide NY .

Proof. Since Y does not divide NX, vY (X) ≥ 0, and hence

m m−1 mvY (Y) = vY (Y ) = vY ( fm−1(X)Y + m−1 ··· + f0(X)) ≥ min(γvY (Y)) = ν0vY (Y), ν=0

for some ν0 such that 0 ≤ ν0 ≤ m − 1. This proves that (m − ν0)vY (Y) ≥ 0, vY (Y) ≥ 0, and therefore Y does not divide NY . Now, let Y be any element of K satisfying the equation m fm(X)Y + ··· + f0(X) = 0

Then, the element Z = fm(X)Y satisfying the equation m m−1 Z + gm−1(X)Z + ··· + go(X) = 0, = m−k−1 33 where gk(X) fk(X) fm (x), and hence Z is an integral function of X

Suppose then that Y1,... YN is a basis of K/k(X). By the above remark, we any assume that the Yi are integral functions of X. The i elements X Y j(i = 0,... t; j = 1,... N) are then linearly independent over k, for any non-negative integer t. By the above lemma, we can ﬁnd N s N s+t i integer s such that X (Y j) are integral divisors. Hence, X (X )(Y j) is an integral divisor for (i = 0,... t, j = 1,... N), which implies that i N −s−t X Y j are elements of L( X ). Since these are linearly independent and N(t + 1) in number, we obtain + N −s−t N + N N −s−t = + + N N(t 1) ≤ 1( X ) ≤ l( ) d( ) − d( X ) 1 (s t)d( x), N −s−t N the latter inequality holding because X divides . Nt + N − 1 Thus, d(N ) ≥ → N as t → ∞, which taken together X s + t with the opposite inequality we proved earlier shows that d(NX) = N. It is clearly suﬃcient to show that d(NX) = N, since the other fol- 1 lows on replacing X by and observing that k(X) = k(1/X). X 11. The Principal Divisors 31

Corollary 1. If X ∈ K∗, d((X)) = 0. This is clear when X is a constant. If X be a variable, d((X)) = d(zX) − d(NX) = N − N = 0. Hence we get the exact sequence

∗ ∗ 1 → k → K → ϑ → R0 → 1, ϑ where ϑ is the group of divisor of degree zero and R = 0 is the group 0 0 Z of divisor classes of degree zero.

Corollary 2. Suppose C ∈ R is a class of divisors. IF U , δ are two divisors of this class, there exists an X ∈ K∗ such that U = (X)δ. Hence, d(U ) = d((X)) + d(δ) = d(δ), and therefore we may deﬁne the 34 degree d(C) of the class C to be the degree of any one of its divisors.

Corollary 3. If X is any transcendental element, there exists an integer Q dependent only on X such that for all integral m, we have

N −m + N −m l( X ) d( X ) ≥−Q.

Proof. We saw in the course of the proof of the theorem that for t ≥ 0,

N −s−t + = N + l( X ) ≥ N(t 1) d( X)(t 1), and writing m = s + t, we obtain for m ≥ s,

N −m + N −m N = l( X ) d( X ) ≥ (1 − s)d( X) −Q. N −s N −m For m < s, since X divides X , we have N −m + N −m N −s + N −s l( X ) d( X ) ≥ l( X ) d( X ) ≥−Q.

Lecture 7

12 The Riemann Theorem

In the last lecture, we saw that if X is any element of K, the integer 35 N −m + N −m l( ) d( X ) remains bounded below as m runs through all integral values. We now prove the following stronger result, known as Riemann’s theorem. Theorem . Let X be any transcendental element of K and (1 − g) the N −m + N −m U lower bound of l( X ) d( X ). Then, for any divisor , l(U ) + d(U ) ≥ 1 − g. U = U U −1 U U Proof. Let 1 2 , where 1 and 2 are integral divisors. Then U −1 U clearly 2 divides , and we have U + U U −1 + U −1 l( ) D( ) ≥ l( 2 ) d( 2 ). U −1 It is therefore enough to prove the inequality with 2 in the place of U .

The key to the proof lies in the statement that l(δ) + d(δ) is unaltered when we replace δ by δ(Z), where (Z) is a principal divisor. To prove this, consider the map deﬁned on L(δ) by Y ∈ L(δ) → YZ This is clearly a k-isomorphism of the vector space L(δ) onto the vector space L(δ(Z)). This proves that l(δ) = l(δ(Z)), and since we already know that d(δ) = d(δ(Z)), our statement follows.

33 34 7. Lecture 7

N −m Observe now that for any non-negative integer m, we have l( X 36 U + N −mU N −m + N −m 2) d( X 2) ≥ l( X ) d( X ) ≥ l − g. Since X is transcendental, d(NX) > 0 and it follows that for large enough m, N −mU N U + l( X 2) ≥ md( X − d( 2) l − g > 0 N −m For such an m therefore, there exists a non-zero element Z in L( X U N m U −1 ). This clearly means that the divisor (Z) X 2 is integral, or that N −m U −1 X divides (Z) 2 . Hence, we deduce that U −1 + U −1 = U −1 + U −1 l( 2 ) d( 2 ) l((Z) 2 ) d((Z) 2 ) N −m + N −m ≥ 1( X ) d( X ) ≥ l − g which proves our theorem. The integer g is called the genus of the field. Since 1 + 0 = l(N ) + d(N ) ≥ 1 − g, it follows that g is always non-negative. The integer δ(U −1) = l(U ) + d(U ) + g − 1, which is non-negative by the above theorem, is called the degree of speciality of the divisor U . We say that U is a non-special or special divisor according as δ(U −1) is or is not equal to zero. We shall interpret δ(U −1) later. Incidentally, we have proved that if U is any divisor and X ∈ K∗, the dimensions of the spaces L(U (X)) and L(U ) are the same. This enables us to define the dimension of a divisor classC. Choose any element U −1 in C and define the dimension N(C) of C to be l(U ). By the remark, this is independent of the choice of U −1 in C.

13 Repartitions

37 We now consider the following question, to which we are led naturally by the theorem of §7. If for every place Y of K, we are given an element XY of K, can we ﬁnd an X in K such that vY (X−XY ) ≥ 0 holds for every Y ? A necessary condition for such an X to exist is that vY (XY ) ≥ 0 for all but a ﬁnite number of Y . For, suppose vY (XY ) < 0 for some Y . Then, since vY (X − XY ) ≥ 0,

vY (X) = vY (X − XY + XY ) = min(vY (X − XY ), vY (XY )) = vY (XY ) < O, 13. Repartitions 35 and this can hold for at most a ﬁnite number of Y . We now make the following

Definition. A repartition C is a mapping Y → GY of the set of prime divisors Y of K into the field K such that vY (CY ) ≥ o for all but a finite number of Y . We can define the operations of addition and multiplication in the space X of repartitions in an obvious manner. If C and G are two repartitions, and a an element of the constant field k,

(C + G )Y = CY + GY , (C G )Y = CY GY , (aC )Y = aCY . The newly defined mappings are immediately verified to be repartitions. Thus, X becomes an algebra over the field k. We can imbed the field K in X by defining for every X ∈ K the repartition CX by the equations (CX)Y = X for every Y . The condition for this to be a repartition clearly holds, and one can easily verify that this is an isomorphic 38 imbedding of K in the algebra X. We can now extend to repartitions the valuations of the field K by defining for every place Y ,

vY (C ) = vY (CY ). Clearly, we have the following relations

vY (C G ) = vY (C ) + vY (G ) vY (C + G ) ≥ min(vY (C ), vY (G )), and vY (CX) = vY (X) This leads to the notion of the divisibility of a repartition C by a divisor U . We shall say that C is divisible by U if vY (C ) ≥ vY (U ) for every Y , and that C and G are congruent modulo U (C ≡ G (U ) in symbols) if C − G is divisible by U . The problem posed at the beginning of this article may be restated in the following generalised form. Given a repartition C and a divisor U , to ﬁnd an element X of the ﬁeld such that X ≡ C (U ). (The original problem is the case U = N ). If U is a divisor, let us denote by ∧(U ) the vector space (over k) of all repartitions divisible by U . Then we have the following 36 7. Lecture 7

Theorem . If U and δ are two divisors such that U divides δ, then ∧(U ) ⊃ ∧(δ) and ∧(U ) dim = d(δ) − d(U ). k ∧(δ)

39 Proof. Let S denote the set of prime divisors occurring in either U or Γ(U /S ) δ with a non-zero exponent. Since dim = d(δ) − d(U ), it is k Γ(δ/S ) Γ(U /S ) ∧(U ) enough to set up an isomorphism of onto the space . Γ(δ/S ) ∧(δ)

If x ∈ Γ(U , /S ), deﬁne a repartition Gx as follows:

X if Y ∈ S (GX)Y = 0 if Y < S   Clearly, GX ∈ ∧(U ), and X → GX is a k-homomorphism of Γ(U /S ) into ∧(U ). The image of an element X ∈ Γ(U /S ) lies in ∧(δ) if and only if vY (X) ≥ vY (δ) for every Y ∈ S , i.e., if and only if X ∈ Γ(δ/S ). Γ(U /S ) ∧(U ) Thus, we have an isomorphism of into . We shall show Γ(δ/S ) ∧(δ) that this is onto. Given any repartition C ∈ ∧(U ), ﬁnd X ∈ K such that

vY (X − C ) ≥ vY (δ) for every Y ∈ S.

This means that the repartition GX − C is an element of ∧(δ). Also, the above condition implies that for Y ∈ S, vY (X) ≥ min(vY (C ), vY (δ)) ≥ vY (U ). Thus, X is an element of Γ(U /S ) and its image in ∧(U ) is the coset C + ∧(δ). Our theorem is thus proved. ∧(δ) Lecture 8

14 Diﬀerentials

In this article, we wish to introduce the important notion of a differential 40 of an algebraic function field. As a preparation, we prove the Theorem. If U and δ are two divisors and U divides δ, then ∧(U ) + K dim = (l(δ) + d(δ)) − (l(U ) + d(U )) k ∧(δ) + K X and dim = δ(U −1) = l(U ) + d(U ) + g − 1. k ∧(U ) + K Proof. We have ∧(U ) + K ∧(U ) + (∧(δ) + K) ∧(U ) = ≃ ∧(δ) + K ∧(δ) + K (∧(δ) + K) ∩ ∧(U ) But it is easily verified that (∧(δ) + K) ∩ ∧(U ) = ∧(δ) + L(U ). Hence, we obtain

∧(U ) + K ∧(U ) ∧(U )/ ∧ (δ) ≃ ∧(δ) + K ∧(δ) + L(U ) ∧(δ) + L(U )/ ∧ (δ) ∧ U ∧ ∧ U ∧ ( )/ (δ) = ( )/ (δ) L(U )/L(U ) ∩ ∧(δ) L(U )/L(δ) Thus,

∧(U ) + K ∧(U ) L(U ) dim = dim − dim k ∧(δ) + K k ∧(δ) k L(δ)

37 38 8. Lecture 8

= (d(δ) − d(U )) − (l(U ) − l(δ)) = (l(δ) + d(δ)) − (l(U ) + d(U )),

41 which is the ﬁrst part of the theorem.

Now, choose a divisor L such that

l(L) + d(L) = 1 − g.

µY Putting µY = min(vY (δ), vY ), and U = Y , we see that U Y divides both δ and L. Hence Q

1 − g ≤ l(U ) + d(U ) ≤ l(L) + d(L) = 1 − g, and hence l(U ) + d(U ) = 1 − g.

Moreover, we have X ∧(U ) + K dim ≥ dim = l(δ) + d(δ) − 1 + g = δ(δ−1). k ∧(δ) + K k ∧(δ) + K

To prove the opposite inequality, suppose C1,... Cm are m linearly independent elements of X over k module ∧(δ) + K. If we put νY = γY min(vY (Ci), vY (δ)) and U = Y , clearly all the Ci lie in ∧(U ). i Y We deduce that Q ∧(U ) + K m ≤ dim = (l(δ)+d(δ))−(l(U )+d(U )) ≤ l(δ)+d(δ)−1+g, k ∧(δ) + K X which proves that dim is ﬁnite and ≤ δ(δ−1). The second part k ∧(δ) + K of the theorem is therefore proved.

Deﬁnition. A diﬀerential ω is a linear mapping of X into k which vanishes on some sub space of the form ∧(U + K).

In this case, ω is said to be divisible by U −1.ω is said to be of the first kind if it is divisible by N 42 If δ divides U , clearly ∧(δ−1) + K ⊂ ∧(U −1) + K, and therefore every differential divisible by U is also divisible by δ. 14. Differentials 39

Consider now the set D(U ) of diﬀerentials ω of K which are divisible by U . This is the dual of the ﬁnite dimensional vector space X/∧(U −1)+K, and therefore becomes a vector space over k of dimension X dim D(U ) = dim = δ(U ). k k ∧(U −1) + K

If ω1 and ω2 are two differentials divisible by U1 and U2 respectively, their sum is a linear function on X which clearly vanishes on −1 ∧(U )+K, where U is (U1, U2). (The greatest common divisor (g.c.d.) min(vY (U ),vY (U )) of two divisors U1 and U2 is the divisor U = Y 1 2 .) Y Q Thus, ω1 + ω2 is a differential divisible by U . Similarly, for a differential ω divisible by U and an element X ∈ K, we define the differential Xω by Xω(C ) = ω(XC ). Xω is seen to be divisible by (X)U . We then obtain (XY)ω = X(Yω) (X + Y)ω = Xω + Yω

X(ω1+ω2) = Xω1 + Xω2 It follows that the diﬀerentials form a vector space over K. We now prove the

Theorem. If ω0 is a non-zero differential, every differential can be writ- 43 ten uniquely in the form ω = Xω0 for some X ∈ K. In other words, the dimension over K of the space of differentials of K is one. −1 −1 U Proof. Let ω0 be divisible by δ0 and ω by δ . Let be an integral divisor, to be chosen suitably later. The two mappings

−1 −1 X0 ∈ L(U δ0) → X0ω0 ∈ D(U ) and X ∈ L(U −1δ) → Xω ∈ D(U −1)

−1 −1 are clearly k-isomorphisms of L(U δ0) and L(U δ) respectively into D(U −1). Hence, the sum of the dimensions of the images in D(U −1). is −1 −1 l(U δ0) + l(U δ) ≥ 2d(U ) − d(δ) − d(δ0) + 2 − 2g, 40 8. Lecture 8

and is therefore > dim D(U −1) = δ(U −1) = d(U )+g−1 if U is chosen so that d(U ) is suﬃciently large. With such a choice of U , therefore, we see that the images must have a non-zero intersection in D(U −1). Hence, for some X0 and X diﬀerent from zero, we must have

−1 X0ω0 = Xω,ω = X0X ω0 = Yω0, Y ∈ K.

The uniqueness is trivial.

We shall now associate with every diﬀerential ω a divisor. We require a preliminary

Lemma. If a diﬀerential ω is divisible by two divisors U and δ, it is 44 also divisible by the least common multiple L of U and δ (Deﬁnition of l.c.m. obvious).

Proof. Suppose C ∈ ∧(L−1). Then,

vY (C ) ≥−vY (L) = − max(vY (U ), vY (z)).

Deﬁne two repartitions C 1 and C ′′ by the equations

1 ′′ C = CY , C = 0 for all C such that vY (U ) ≥ vY (δ) Y1 Y 1 ′′ CY = 0, CY = CY for all Y such that vY (U ) < vY (δ).

C 1 and C ′′ are by the above deﬁnition divisible by U −1 and Y −1 respectively, and C = C 1 + C ′′. Hence,

ω(C ) = ω(C 1) + ω(C ′′) = 0.

Since ω must vanish on K,ω must vanish on ∧(L−1)+K. Our lemma follows

Theorem. To any diﬀerential ω , 0, there corresponds a unique divisor (ω) such that ω is divisible by U if and only if (ω) is divisible by U . 15. The Riemann-Roch theorem 41

Proof. Suppose ω is divisible by a divisor U . Then, the mapping X ∈ L(U −1) → Xω ∈ D(N ) is clearly a k-isomorphism of L(U −1) into D(N ). Hence, we deduce that

l(U −1) ≤ dim D(N ) = δ(N ) = l(N ) + d(N ) + g − 1 = g.

On other hand, we have

l(U −1) + d(U −1) = 1 − g + δ(U ) ≥ 2 − g, since δ(U ) ≥ 1. Combining these two inequalities, we deduce that

d(U ) ≤ 2g − 2.

45 This proves that the degrees of all divisors dividing a certain diﬀer- ential ω are bounded by 2g − 2.

Now, choose a divisor (ω) dividing ω and of maximal degree. If U were any another divisor of ω, the least common multiple δ of U and (ω) would have degree at least that of (ω), and would divide ω by the above lemma. Hence, we deduce that δ = (ω) or that U divides (ω). The uniqueness of (ω) also follows from this. Our theorem is proved.

Corollary 1. If X ∈ K∗, (Xω) = (X)(ω). This follows from the easily veriﬁed fact that U divides ω if and only if (X)U divides Xω.

This corollary, together with the theorem that the space of diﬀeren- tials is one dimensional over K, proves that the divisors of all diﬀeren- tials form a class W. This class is called the canonical class

15 The Riemann-Roch theorem

Let C be a class and U any divisor of C. If Ui(i = 1,..., n) are elements −1 of C, UiU = (Xi) are principal divisors. We shall say that the divisors Ui, are linearly independent if Xi(i = 1,..., n) are linearly independent over k. This does not depend on the choice of U or of the respectively −1 Xi of UiU , as is easy to verify. We now prove the 42 8. Lecture 8

Lemma. The dimension N(C) of a class C is the maximum number of linearly independent integral divisors of C. 46

Proof. Let U be any divisor of C. Then, the divisors (X1)U , (X2)U ,..., (Xn)U are linearly independent integral divisors of C if and only if −1 X1,..., Xn are linearly independent elements of L(U ). Our lemma follows.

We now prove the celebrated theorem of Riemann-Roch.

Theorem. If C is any divisor class,

N(C) = d(C) − g + 1 + N(WC−1).

Proof. Let U ∈ C. Then,

N(C) = l(U −1) = d(U ) − g + 1 + δ(U ) = d(C) − g + 1 + δ(U ).

But δ(U ) being the dimension of D(U ) is the maximum number of linearly independent diﬀerentials divisible by U . Hence, it is the maximum number of linearly independent diﬀerentials ω1,...,ωn such that −1 −1 −1 (ω1)U , (ω2)U ,..., (ωn)U are integral. By the above lemma, we conclude that δ(U ) = N(WC−1), and our theorem is proved. Corollaries E and W shall denote principal and canonical classes respectively.

(a) N(E) = l(n) = 1, d(E) = d(N) = 0. 1 = N(E) = d(E) − g + 1 + N(W) =⇒ N(W) = g. g = N(W) = d(W) − g + 1 + N(E) =⇒ d(W) = 2g − 2.

(b) If d(C) < 0, or if d(C) = 0 and C , E, N(C) = 0. 47 If d(C) > 2g − 2 or if d(C) = 2g − 2 and C , W,

N(C) = d(C) − g + 1 15. The Riemann-Roch theorem 43

Proof. Suppose N(C) > 0. Then there exists an integral divisor U in C, and hence d(C) = d(U ) ≥ 0, equality holding if and only if U = N or C = E.

The second part follows immediately on applying that ﬁrst part to the divisor WC−1.

N(C) = d(C) − g1 + 1 + N(W1C−1),

We must have W = W1 and g = g1.

Proof. Exactly as in (a), we deduce that N(W1) = g1, d(W1) = 2g1 − 2. Again as in the second part of (b), we deduce that if d(C) > 2g1 − 2, N(C) = d(C) − g1 + 1. Hence, for d(C) > max(2g − 2, 2g1 − 2),

N(C) = d(C) − g + 1 = d(C) − g1 + 1, g = g1.

Hence N(W1) = g and d(W1) = 2g − 2, and it follows from the second part of (b) that W = W1.

This shows that the class W and integer g are uniquely determined by the Riemann-Roch theorem. Let us give another application of the Riemann-Roch theorem. We shall say that a divisor U divides a class C if it divides every integral divisor of C. We then have the following

Theorem. If C is any class and U an integral divisor,

N(C) ≥ N(CU ) ≤ N(C) + d(U )

The ﬁrst inequality becomes an equality if and only if U divides the class CU , and the second if and only if U divides the class WC−1.

Proof. Since the maximum number of linearly independent integral di- 48 visors in CU is clearly greater than or equal to the number of such divisors in C, the ﬁrst part of the inequality follows. Suppose now that 44 8. Lecture 8

equality prevails. Then there exists a maximal set δ1,...δn of linearly independent integral divisors in C, such that U δ1,... U δn forms such a set in CU . But since every integral divisor in CU is ‘linear combination of divisors of such a set’ with coeﬃcients in k (in an obvious sense), every integral divisor of CU is divisible by U .

Now, by the theorem of Riemann-Roch,

N(CU ) = d(C) + d(U ) + 1 − g + N(WC−1U −1), and the second inequality together with the condition of equality follows by applying the ﬁrst to WC−1U −1 instead of C.

Corollary. For any class C, N(C) ≥ max(0, d(C) + 1).

Proof. If N(C) = 0, there is nothing to prove, if N(C) > 0, there exists an integral divisor U in C. Hence we obtain

N(C) = N(EU ) ≤ N(E) + d(U ) = d(U ) + 1 = d(C) + 1, and our corollay is proved. Lecture 9

16 Rational Function Fields

In this lecture, we shall consider some particular function fields and find 49 their canonical class, genus, etc. as illustrations of the general theory. Let us first consider the rational function fields. Let K = k(X) be a rational function field in one variable over k. We shall first show that k is precisely the field of constants of K. For later use, we formulate this in a more general form.

Lemma. Let K be a purely transcendental extension of a ﬁeld k. Then k is algebraically closed in K.

Proof. Let (xi)i∈I be any transcendence basis of K over k such that K = k(xi). Since any element of K is a rational combination of a ﬁnite number of xi, we may assume that I is a ﬁnite set of integers (1,... n).

We proceed by induction. Assume ﬁrst that n = 1. Let α be any f (x1) element of k(x1) algebraic over k. α may be written in the form , g(x1) where f and g are polynomials over k prime to each other. We then have f (x1) − αg(x1) = 0

This proves that if α were not in k, x1 is algebraic over k(α), (since 50 the above polynomial for x1 over K(α) cannot vanish identically ) and hence over k which is a contradiction. Hence α is in k.

45 46 9. Lecture 9

Suppose now that the lemma holds for n − 1 instead on n. If α were an element of k(x1,... xn) algebraic over k, it is algebraic over k(x1,... xn−1). By the ﬁrst part, it should be the k(x1,... xn−1), and hence by induction hypothesis in k. Our lemma is proved. Let us return to the rational function ﬁeld K. We have already seen 1 that the prime divisors are (i)NX, the prime divisor corresponding to , X and (ii)Yp(X), the prime divisors corresponding to irreducible polynomials p(X) in k[X]. If f (X) is a rational function of X over k having the e1 en unique decomposition p1 (X) ··· pr (X), the principal divisor ( f (X)) is clearly given by r = Y eγ − deg f ( f (X)) pγ (X)NX Yν=1 −t It follows that the space L(NX ) for t ≥ 0 consists precisely of all polynomials of degree ≤ t, and since there are t + 1 such polynomials independent over k, and generating all polynomials of degree ≤ t, (1, X,... Xt for example), we deduce that

t = + N(NX E) t 1.

t = But by the corollary to the Riemann - Roch theorem, if d(NX E) td(NX) = t > 2g − 2, we should have

t = t + = + N(NX E) d(NX E) 1 − g t 1 − g,

51 and hence, g = 0. Thus, there are no non-zero diﬀerentials of the ﬁrst kind. −2 = = −2 = = Since d(NX E) −2 2g−2 < 0, it follows that N(NX E) 0 g. = −2 and hence W nX E.

17 Function Fields of Degree Two Over a Rational Function Field

We start with a lemma which will be useful for later calculations. 17. Function Fields of Degree . . . 47

Lemma. Let K/k be any algebraic function ﬁeld and X ∈ K any tran- f (X) scendental element. If R(X) = 1 is any rational function of X, then f2(X) z f = 1 − deg R(X) z f1 and f2 being prime to each other (R(X)) NX and f1 z f2 and z f2 are prime to each other and both are prime to NX. Moreover, [k(X) : k(R(X))] = max(deg f1, deg f2). Proof. Let f (X) be any polynomial over k of the form

t f (X) = ao + a1X + ··· + atX .

If Y is a prime divisor not dividing NX, i.e., if vY (X) ≥ 0, we have vY ( f (X)) ≥ min(γvY (X)) ≥ o). If on the other hand, Y does occur in ν=o NX, we have vY (X) < o, and hence

vY ( f (X)) = min (νvY (X)) = tvY (X) ν=o−t = t Thus, we see that z f is prime to NX and N f (x) NX.

f1(X) Hence, if R(X) = , where ( f1, f2) = 1, we have f2(X) z = f1 − deg R(X) (R(X)) NX z f2

We assert that z f1 and z f2 are prime to each other. For if not, let z f1 52 Y and z f2 have a prime divisor in common. Then vY (X) ≥ o. Find polynomials g1 and g2 such that

f1g1 + f2g2 = 1.

Then, 0 = vY (1) = vY ( f1g1 + f2g2)

≥ min(vY ( f1) + vY (g1), vY ( f2) + vY (g2)) > 0,

a contradiction. Thus, z f1 and z f2 are prime to each other. To prove the last part of the lemma, we may assume without loss of generality that deg f1 ≥ deg f2 or that deg R(X) ≥ 0 (otherwise consider 1 ). It then follows that R(X) = = zR(X) z f1(X), [k(X) : k(R(X))] [K : k(R(X))]/[K : k(X)] 48 9. Lecture 9

d(zR(X)) d(z f1(X)) d(z f1(X)) = = = = deg f1. d(NX) d(NX) d(NX) Our lemma is proved. It follows in particular that k(x) = k(R(X)) if and only if max(deg f1, αX + β deg f ) = 1, or R(X) = ,α,β,γ,δ ∈ k and αδ − βγ , 0. 2 γX + δ Now, let k be a field of characteristic different from 2, X a transcendental element over k and K a field of degree two over k(X) which is not got from an algebraic extension of k (viz., K should not be got by the adjunction to k(X) of elements algebraic over k). Then k is the constant field of K, for if it were not, there exists an element α of K 53 which is algebraic over k but not in k. α cannot lie in k(X), since k is algebraically closed in k(X). Hence, k(X,α) should be an extension of degree at least two over k(X), and should therefore coincide with K. But this contradicts our assumption regarding K. Now, K can be get the adjunctions to k(X) of an element Y which satisfies a quadratic equation

Y2 + bY + c = 0, b, c ∈ k(X)

Completing the square ( note that characteristic k , 2), we get

b 2 b 2 Y + + c − = 0, 2! 4 ! b and hence k(X, Y) = k(X, Y1) where Y1 = Y + satisﬁes equation of 2 the form γ12 = R(X), R(X) being a rational function of X. Let R(X) = r eγ pγ (X), where pν(X) are irreducible polynomials in k[X] and eν are ν−1 integers.Q Putting eν = 2gν + ǫγ, where gν are integers and ǫν = 0 or 1 ′′ = Y = ′′ ′′ 1, and Y γ , we see that k(X, Y) k(X, Y ), and Y p ggamma(X) ν satisﬁes an equationQ of the form

r ′′2 = ǫν = Y pγ (X) D(X), Y1 17. Function Fields of Degree . . . 49 where D(X) is a polynomial which is a product of diﬀerent irreducible polynomials. We shall therefore assume without loss in generality that K = k(X, Y) with Y2 = D(X) of the above form. Let us assume that m is the degree of D.

Now, let σ be the automorphism of K over k(X) which is not the 54 σ identity. If Z = R1(X) + YR2(X) is any element of K, Z = R1(X) − YR2(X). To every prime divisor Y of K, let us associate a prime divisor Y σ by the deﬁnition

−1 vY σ (Z) = vY (σ Z)

This can be extended to an automorphism of the group ϑ of divisors (see Lecture 19). We shall denote this automorphism again by σ, and U U σ σ = σ = the image of a divisor by . Since X X, NX NX. = + −t Suppose now that Z R1(X) YR2(X) is any element of L(NX ) (t any integer). Applying σ, we deduce that R1(X) − YR2(X) should also −t −t −t be an element of L(NX ). Adding, 2R1(X) ∈ L(NX ), R1(X) ∈ L(NX ). f (X) If R (X) = 1 , where f and g are coprime polynomials, we have 1 g (X) 1 1 z 1 = f1 − deg R1 −t z = (R1(X)) NX divisible by NX , and hence we deduce that g1 zg1 n, and g1(X) is a constant. Hence R1(X) is a polynomial of degree ≤ t (if t < o, R1(X) = 0). σ −t Also, since both Z and Z are in L(NX ). This implies as before that 2 2 2 R1 − DR2 is a polynomial of degree ≤ 2t. Hence DR2 is a polynomial of degree ≤ 2t, and since D is square free, R2 is a polynomial of degree m ≤ t − . 2

Conversely, by working back, we see that if R1 is a polynomial of = m = + degree ≤ t and R2 a polynomial of degree ≤ t 2 , Z R1 YR2 ∈ 50 9. Lecture 9

−t L(NX ) . Hence, we obtain

0 if t < 0 t + 1 if 0 ≤ t ≤ m − 1 and m even  2 t = −t =  + m−1 N(ENX) l(NX ) t 1 if m and 0 ≤ t ≤  2 2t + 2 − m if m even and t ≥ m  2 2  + − m+1 ≥ m+1 2t 2 2 if m odd and t 2 .  55  = = t = Since d(NX) [K : k(X)] 2, for t > g − 1, we have d(ENX) 2t > 2g − 2 and hence t = t + = + N(ENX) d(NX) − g 1 2t − g 1 m Comparing with the above equations, we deduce that g is − 1 if 2 m − 1 m is even and if m is odd. 2 Thus, we obtain examples of fields of arbitrary genus over any constant field. g−1 The canonical class of K is ENX . For, g−1 = d(NX E) 2g − 2, g − 1 + 1 = g if g > 0 and N(Ng−1E) = X 0 = g if g = 0.   If gg > o, there exists a differential ω of the first kind with (ω) = g−1 ff g−1 NX . Then clearly the di erentials ω, Xω,..., X ω are all of the first kind and are linearly independent over k, and as they are g in number, they form a base over k for all differentials of the first kind.

18 Fields of Genus Zero

56 We shall find all fields of genus zero over a constant field k. First, notice that any divisor of degree zero of a field of genus zero is a principal divisor. For let C be a class of degree 0. Then since d(C) > −2 = 2g − 2, N(C) = d(C) − g + 1 = 1 and therefore C = E. 19. Fields of Genus One 51

Now,

d(W−1) = 2 > 2g − 2, N(W−1) = 2 − g + 1 = 3, and therefore there exists three linearly independent integral divisors −1 U1, U2, U3 in the class W (incidentally, this proves there exists integral divisors, and consequently prime divisors of degree at most two). U1 Let = (X). Then clearly NX divides U2, and hence we obtain U2

[K : k(X)] = d(NX) ≤ d(U2) = 2.

Thus, any field of genus zero should be either a rational function field or a quadratic extension of a rational function field. We have the following Theorem . The necessary and sufficient condition for a field of genus zero to be a rational function field is that it possess a prime divisor of degree 1.

Proof. If K = k(X), the prime divisor nX satisﬁes the requisite condition. Conversely, let Y be a prime divisor of degree 1 of K. Then, N(Y E) = d(Y ) + 1 = 2, and therefore there are elements X1, X2 in 57 K linearly independent over k such that X1Y = U1 and X2Y = U2 X1 X1 U1 are integral divisors. Thus if X = , (X) = ( ) = , and d(NX) ≤ X2 X2 U2 d(U2) = d(Y ) = 1. But X is not in k, and hence d(NX) = 1 = [K : k(X)], from which it follows that K = k(X). The theorem is proved.

19 Fields of Genus One

Let K/k be an algebraic function field of genus 1. Then, since N(W) = g = 1 and d(W) = 2g − 2 = 0, the canonical class W coincides with the principal class E. A function field of genus one which contains at least one prime divisor of degree one is called an elliptic function field. (The genus being one does not imply that there exists a prime divisor of degree one. In 52 9. Lecture 9

fact, it can be proved easily that the field R(X, Y), where R is the field of real numbers and X, Y transcendental over R and connected by the relation Y2 + X4 + 1 = 0 has every prime divisor of degree two). Let us investigate the structure of elliptic function fields. Let Y be a prime divisor of degree one. Then, since d(Y 2) = 2 > 2g − 2 = 0, we have l(Y −2) = 2. Let 1, X be a basis of L(Y −2) over k. 2 2 Since XY is integral, NX divides Y .NX cannot be Y , since if it were, we obtain [K : k(X)] = d(NX) = d(Y ) = 1, K = k(X) and hence g = 0. 2 Thus, NX should be equal to Y . Again, since l(Y −3) = 3, we may complete (l, X) to a basis (1, X, Y) −3 3 58 of L(Y ) over k. Nγ should divide Y , and since Y is not an element −2 2 3 of L(Y ), Nγ does not divide Y . Thus, Nγ = Y . The denominators of 1, X, Y, X2, XY, X3 and Y2 are respectively N, Y 2, Y 3, Y 4, Y 5, Y 6 and Y 6. Since the first six elements have different powers of Y in the denominator, they are linearly independent elements of L(Y −6). But l(Y −6) = 6, and the seventh element, being in L(Y −6) should therefore be a linear combination of the first six. We thus obtain

2 3 2 2 Y + γXY + δY = α3X + α2X + α2X + α1X + αo,γ,δ,αi ∈ k. f (X) Now, if Y where a rational function of X, writing Y = , f (X), g(X) g(X) ∈ k[X], ( f (X), g(X)) = 1, and substituting in the above equation, we easily deduce that g(X) should be a constant, and that Y should be a polynomial in X of degree ≤ 1. But this would mean that Y is divisible by Y 2, which we have already ruled out. Hence, [k(X, Y) : k(X)] = 2 = d(NX) = [K : k(X)], and consequently, K = k(X, Y). If the characteristic of k is diﬀerent from 2, we may as in §16 ﬁnd a Z such that K = k(X, Y), with

Z2 = f (X)

where f (X) is a cubic polynomial in X with non-repeating irreducible factors. A partial converse of the above result is valid. Suppose that K = 59 k(X, Z), where X is transcendental over k and Z satisfies an equation 19. Fields of Genus One 53 of the form Z2 = f (X), where f (X) is a cubic polynomial which we assume to be irreducible. Let ch(k) , 2. Then, the genus of K is one, = 2 −3 2 3 by §16. Also, since deg f (X) 3, Z ∈ L(NX ), and Nz divides NX. 2 = = But since X is of degree 3 over k(Z), we have d(Nz ) 2d(Nz) 2.[K : = = = = 3 k(Z)] 6 3.[K : k(X)] 3d(NX), and therefore Nz NX. From this, it is clear that there exists a prime divisor Y with d(Y ) = 1 such that 3 2 Nz = Y , NX = Y . Finally, suppose a field K is of genus greater than 1. Then, N(W) = g > o and d(W) = 2g − 2 > o, and hence we deduce the existence of an integral divisor , N of degree 2g − 2. Thus, we have proved that if g > 1, there always exists prime divisors of degree ≤ 2g − 2. The minimal degree of prime divisors for a field of genus one is not known.

Lecture 10

20 The Greatest Common Divisor of a Class

We wish to ﬁnd when the greatest common divisor of all integral divisors 60 of a class is diﬀerent from the unit divisor N. We assert that this is impossible when d(C) ≥ 2g. In fact, let U be an integral divisor of the class C. Then we obtain

d(C) − g + 1 = N(C) = N(CU −1) = d(C) − d(U ) + 1 − g + N(WC−1U ), d(U ) = N(WC−1U ) ≤ max(0, d(WC−1U ) + 1) and since 1 + d(WC−1U ) ≤ 2g − 2 − 2g + 1 + d(U ) = d(U ) − 1, we should have d(U ) = 0 and U = N. This is in a sense the best possible result. In fact, if there exists a prime divisor Y of degree 1 in the ﬁeld, we have d(WY ) = 2g − 2, and hence. N(WY ) = d(W) + d(Y ) − g + 1 = g = N(W), which proves that Y divides all integral divisor of the class WY . Nev- ertheless, if g > 0, we can prove that the greatest common divisor of the canonical class W is N. For, suppose U , N is an integral divisor of W. Then,

dim L(U −1) = N(EU ) = d(U ) + 1 − g + N(WU −1) = d(U ) + 1 − g + N(W) = d(U ) + 1 ≥ 2.

Thus, there exists a transcendental element X ∈ L(U −1). The divisor (X)U is then integral. Also, since N(W) = g > o, we can choose

55 56 10. Lecture 10

a differential ω of the first kind. Then, (Xω) = ((X)U ).((ω)U −1) is 61 also integral and therefore Xω is also first kind. By repetition of the argument, we obtain that Xnω is of the first kind for all positive integers n n. But since X is transcendental, NX , N , and hence X ω cannot be an integral divisor for large n. This is a contradiction. Our assertion is therefore proved. For fields of genus g > o, we shall improve the inequality N(C) ≤ max(o, d(C) + 1). Lemma. If g > o and d(C) ≥ o, C , E, we have N(C) ≤ d(C) Proof. We may obviously assume that N(C) > o. Then there exists an integral divisor U in C, and since C , E, U , N , and therefore d(C) = d(U ) > o. Since G > o, U cannot be a divisor of the class W,and therefore g = N(W) > N(WU −1) = N(WC−1) N(C) = d(C) − g + 1 + N(WC−1) ≤ d(C) − g + 1 + g − 1 = d(C). Our lemma is proved

21 The Zeta Function of Algebraic Function Fields Over Finite Constant Fields

In the rest of this lecture and the following two lectures, we shall always assume that k is a finite field of characteristic p > o and with q = pt 62 elements,and that K is an algebraic function field with constant field k. If Y is any prime divisor of K, we shall call the number of elements in the class field kY the norm of Y . Since [kY : k] = d(Y ), we see that norm of Y (which we shall denote by NY ) is given by d(Y ) NY = q We may extend this definition to all divisors, by putting NU = (NY )vY (U ) = qd(U ) YY 21. The Zeta Function of Algebraic Function . . . 57

Clearly we have for two divisors U and δ

NU δ = NU .Nδ.

Before introduction the zeta function, we shall prove an important Lemma. For any positive integer m, the number of prime divisors of degree ≤ m is finite. The number of classes of degree zero is finite. (The latter is called the class number of K and is denoted by h). Proof. To prove the first part of the lemma, choose any transcendental element X of K. Let Y be any prime divisor of K with d(Y ) ≤ m and which does not divide NX. (We may neglect those U which divide NX, since they are finite in number.) Let Y 1 be the restriction of Y to k(X). Y 1 must be a place on k(X). Since Y 1(X) , ∞, there corresponds a unique polynomial (p(X)) which gives rise to Y 1. Now, since kY ′ ⊂ kY , we obtain 63

deg(p(X)) = d(Y ′) ≤ d(Y ) ≤ m.

Since the number of polynomials of degree ≤ m over a finite field is finite, and since there the are only a finite number of prime divisors Y of K which divide zp(X) for a fixed p(X), the first part of our theorem is proved. To prove the second part, choose and fix an integral divisor Uo such that d(Uo) ≥ g. If C is any class of degree zero, we have

N(CUo) ≥ d(C) + d(Uo) − g + 1 ≥ 1, and hence there exists an integral divisor U in CUo such that d(U ) = d(CUo) = d(Uo). But since

d(U ) = d(Y )vY (U ), d(Y ) ≥ 1, vY (U ) ≥ 0, X and there are only a finite number of Y with d(Y ) ≤ d(Uo), there are only a finite of integral U with d(U ) = d(Uo) and consequently only a finite number of C with d(C) = 0. The lemma is completely proved. 58 10. Lecture 10

Remark 1. Let ρ denote the least positive integer which is the degree of a class. since C → d(C) is a homomorphism of the group R of divisor classes into the additive group Z of integers, we see that the degree of any class of the from νρ where ν is an integer, and that to any ν, there (γρ) (νρ) 64 correspond precisely h classes C1 ,..., Ch of degree νρ. Remark 2. The number of integral divisors in any C is precisely qN(C) − 1 . This is clear if N(C) = 0. If N(C) > 0, let U be any in- q − 1 tegral divisor of C. Then all integral divisors of C are of the from (X)U , X where X ∈ L(U −1), X , 0. Also (X)U = (Y)U if and only if = N Y or X = aγ, a ∈ k∗. Since the number of non zero elements of L(U −1) is qN(C) − 1 and the number of non zero elements of k is q − 1 our assertion follows.

Now, let s = σ + it be a complex variable. For σ> 1, we deﬁne the zeta function of the algebraic function ﬁeld K by the series

1 ζ(s, K) = , s = σ + it,σ> 1 (NU )s XU

the summation being extended over all integral divisors of the ﬁeld K. 1 1 Since = , and all the terms of the series are positive (NU )s (NU )σ

when s is real, the following calculations are valid ﬁrst for s > 1 and the for complex s with σ > 1. In particular, they prove the absolute convergence of the series τ(s, K)

= (number of integral divisors in C). q−sd(C), the last summation C beingP over all classes,

1 = (qN(C) − 1)q−sd(C); q − 1 XC

65 writing d(C) = νρ, and noticing that qN(C) −1 = 0 if d(C) < 0, the above 21. The Zeta Function of Algebraic Function . . . 59 expression becomes

∞ h ∞ 1 (νρ) h q−νρs qN(Cl ) − q−νρs q − 1 q − 1 Xν=o Xl=1 Xν=o Let us now put U = q−s. Then |U| = |q−s| = q−σ < 1 since σ > 1, and we can sum the second geometric series. Suppose now that g > o. We may then split the ﬁrst sum into two parts, the ranging over o ≤ 2g − 2 2g − 2 ν ≤ and the second over ν > . (Since d(w) = 2g − 2, ρ) ρ ρ 2g − 2 divides 2g − 2; or is an integer). In the second summation since ρ (νρ) = (νρ) = + d(Cl ) νρ > 2g − 2, we may substitute N(Cl ) νρ − g 1. We obtain the expression

2g−2 ρ h 1 −νρs N(C(νρ)) h −νρs νρ−g+1 h 1 q q l + q q − . q − 1 q − 1 q − 1 1 − qsρ Xν=o Xl=1 X2g−2 ν> ρ 2g−2 ρ h 1 νρ N(C(νρ)) = τ(s, K) = U q l q − 1 Xν=o Xl=1 hq1−g (Ug)2g−2+ρ h 1 + − (1) q − 1 1 − (Uq)ρ q − 1 1 − Uρ If g = o, N(C) = d(C) − g + 1 for all C with d(C) > o, and a similar computation gives hq 1 h 1 τ(S, K) = − (2) q − 1 1 − (Uq)ρ q − 1 1 − Uρ (1) and (2) may be combined as follows

(q − 1)τ(s, K) = F(U) + R(U), (3) where F(U) is the polynomial 66

F(U) = qN(C)Ud(C) (4) o≤d(CX)≤2g−2 60 10. Lecture 10 and R(U) the rational function

(Uq)max(o,2g−2+ρ) h R(U) = hq1−g − (5) 1 − (Uq)ρ 1 − Uρ These formulae provide the analytic continuation of τ(s, K) to the whole plane. The only possible poles are the values of s for which Uρ = 1 or (qU )ρ = 1. For a rational function ﬁeld K = k(X), since g = o, h = 1 and ρ = 1, we obtain q 1 q − 1 (q − 1)τ(s, K) = − = 1 − Uq 1 − U (1 − Uq)(1 − U) Lecture 11

22 The Inﬁnite Product for ζ(s, K)

Let K be an algebraic function field over a finite constant field k. Then, 67 the ζ function of K can be expressed as an infinite product taken over all prime divisors of K. Theorem. For s = σ + it, and σ> 1, we have 1 ζ(s, K) = , 1 − (NY )−s YY where the product is taken over all prime divisors Y of K. The product is absolutely convergent, and hence does not depend on the order of the factors. Proof. Since σ> 1, we have for any integer m > o, 1 1 1 = 1 + + + ··· , 1 − NY −s (NY )s (NY )2s ! NY≤m NY≤m and since there are only a finite number of factors in the product, each factor being an absolutely convergent series, we may multiply these to obtain 1 1 1 = + , 1 − (NY )−s (NU )s (NU )s NY≤m nUX≤m NXU >m where the first summation is over all integral divisors U with NU ≤ m, and the second over all integral divisors U which do not contain any

61 62 11. Lecture 11

prime divisor U with NU > m and which satisfy the inequality NU > m.

68 Hence

1 1 1 1 − s − s = 1 − (NY ) NU (NU )s Y≤ XU ≤ XU > N m N m N m 1 ≤ (NU )σ NXU >m 1 and letting m →∞, we obtain the asserted equality, since ( )σ, NU >m NU being the remainder of the convergent series for ζ(σ, K), tendsP to zero as m →∞. The absolute convergence of the product is deduced from the inequality 1 1 1 1 + + − ≤ + (NY )s (NY )2s (NY )σ (NY )2σ

As a corollary to this theorem, we see that ζ(s, K) has no zero for σ> 1.

23 The Functional Equation

In the last lecture, we obtained the following formula:

(q − 1)ξ(s, K) = F(U) + R(U), where U = q−s, F(U) = qN(C)Ud(C) o≤d(Xc)≤2g−2 (Uq)max(o,2g−2+ρ) h and R(U) = hq1−g − . 1 − (Uq)ρ 1 − Uρ Substituting for N(C) in F(U) from the theorem of Riemann-Roch, 69 we obtain

+ + −1 F(U) = qd(C)−g 1 N(WC )V d(C) o≤d(CX)≤2g−2 23. The Functional Equation 63

1 −1 + −1 = q1−g d(WC )−2g 2qN(WC ) qU ! o≤d(CX)≤2g−2 Writing WC−1 = C1, and noticing that C1 runs through the same set of classes as C in the summation, we have d(C1) 1 1 1 F(U) = qq−1U2g−2 qN(C ) = qg−1U2g−2F . qU ! qU ! o≤d(CX1)≤2g−2 We shall prove that a similar functional equation holds for R(U). First suppose that g > o. Then, (qU)2g−2+ρ h R(U) = hg1−g − 1 − (qU)ρ 1 − Uρ 2g+2 1 h.(q. 1 )ρ = 1−g qU + Uq −hg ρ − 1 1 1 qU ρ 1 − q. Uq 2g−2+ρ q 1 = 2g−2 g−1 1−g qU h U q hq ρ − ρ   1 − g. 1 q 1 − 1   U Uq     1  = qg−1U2g−2R . qU ! If g = o, the only divisor class of degree zero is E and hence h = 1. Also, since ρ divides 2g − 2 = −2, ρ = 1 or ρ = 2. If ρ = 1, we obtain q 1 q − 1 R(U) = − = 1 − qU 1 − U (1 − U)(1 − qU) q − 1 = q−1U−2 1 1 1 − qU 1 − q. qU 1 = q−1U−2R . qU ! If ρ = 2, we get 70 2 1 −2 q 1 −q. qU U R(U) = − = + 1 − (qU)2 1 − U2 1 2 1 2 1 − qU 1 − q. qU 64 11. Lecture 11

1 = q−1U−2R . qU ! Thus, in any case, we have the functional equation 1 R(U) = qg−1U2g−2R . qU ! Adding the equation for F(U) and R(U), we see that ζ(s, K) satisﬁes the functional equation (q − 1)ζ(s, K) = qg−1qs(2−2g)(q − 1)ζ(1 − s, K), or ζ(s, K) = qg−1qs(2−2g)ζ(1 − s.K)

We may also rewrite this in the from qs(q−1)ζ(s, K) = q(1−s)(g−1)ζ(1 − s, K). Thus, the function on the left is unaltered by the transformation s → 1 − s.

24 L-series

We wish to study the L-series associated to characters of the class group of an algebraic function with a ﬁnite constant ﬁeld.

71 Definition. A character X of finite order on the class group R is a homomorphism of R into the multiplicative group C∗ of non zero complex numbers such that there exists in integer N with XN(C) = 1 for all C in R. X(C) is therefore an Nth root of unity for all C. We may define X on the group v of divisors by comparing with the natural homomorphism v → R, i.e., by putting X(U ) = X(U E) for any divisor U . The L-function L(s, X, K) associated to a character X (which we shall always assume to be of finite order) is then defined for s = σ + it, σ> 1 by the series

L(s, X, K) = X(U )(NU )−s XU 24. L-series 65 where the summation is over all integral divisors U of the ﬁeld. The absolute value of the terms of this series is majorised by the correspond- 1 ing term of the series = ζ(σ, K) and is therefore absolutely (NU )σ convergent for σ> 1.P We may prove along exactly the same lines as in the case of the ζ -function the following product formula: 1 L(s, X, K) = 1 − X(Y )(NY )−s YY where Y runs through all prime divisors and σ> 1.

Lecture 12

25 The Functional Equation for the L-functions

Let X be a character of finite order on the class group of an algebraic 72 field K over a constant field k with q = p f elements. We consider two cases.

Case 1. X when restricted to the subgroup Ro of R of all classes of degree zero, is trivial; i.e., X(Co) = 1 for Co ∈ Ro.

Let ρ be as before the smallest positive integer which is the degree of a class, and let C(ρ) be a class of degree ρ. Then R is clearly the direct (ρ) (ρ) product of Ro and the cyclic group generated by C . Set X(C ) = e2πiξ. We then have

∞ L(s, X, K) = X(U )(NU )−s = e2πiξnq−nρs U n=−∞ X CXo∈Ro X n U ∈CoC(ρ)

− s− 2πiξ 2πiξ = (NU ) ρ log q = ζ s − . ρ log q XU !

Thus, the L-function reduces to a ζ- function in this case. We can therefore deduce a functional equation for L(s, X, K) from the functional equation for ζ(s, K). Deﬁne the character X¯ conjugate to X by putting X¯(C) = X(¯C). Clearly, X¯ is a character of ﬁnite order on the class group (ρ) −2πiξ and is trivial on Ro, also X¯C = e . Then, we obtain the following relation for L(s, X, K) by substituting from the functional equation for

67 68 12. Lecture 12

the ζ function.

2πiξ qs(g−1)L(s, X, K) = qs(g−1)ζ s − ρ log q! 2πiξ 2πiξ = q2(q−1) ζ 1 − s + q(1−s)(g−1) ρ log q ρ log q! (1−s)(g−1) = X(W)q L(1 − s, X¯, K)

2g − 2 2πiξ 73 since X(W) = (X(C(ρ))) = e (2g − 2) ρ ρ

Case 2. Suppose now that X when restricted to Ro is not identically 1. 1 1 Let Co be a ﬁxed class with X(Co) , 1. Then, 1 = 1 = X(Co) X(Co) X(CoCo) X(Co), CXo∈Ro CXo∈Ro CXo∈Ro

and therefore X(Co) = C ∈R Po Again, using the fact that N(C) = 0 if d(C) < 0, we obtain

∞ (ρ) n n (ρ) N(CoC ) −nρs (q − 1)L(s, X, K) = X(Co) X (C )(q − 1)q CXo∈Ro Xn=0 2g−2 ρ (ρ)n n (ρ) N(CoC ) = X(Co) X (C )(q − 1) = CXo∈Ro Xn o n (ρ) nρ−g+1 −nρs + X(Co) X (C )(q − 1)q CXo∈Ro X2g−2 n> ρ

The second sum vanishes, since X(Co) = 0. Co∈Ro Thus, we obtain P

(q − 1)L(s, X, K) = X(C)qN(C)Ud(C) o≤d(CX)≤2g−2

74 The coeﬃcient of U2g−2 is given by 25. The Functional Equation for the L-functions 69

N(C) N(CoW) X(C)q = X(W) X(Co)q d(CX)=2g−2 CXo∈Ro

= g−1 + g X(W)  X(Co)q q  ,   CXo,E      since N(CoW) = d(CoW) − g + 1 = g − 1 if Co , E and N(W) = g,

= g−1 + g g−1 X(W)  X(Co)q (q − q )   CXo∈Ro   g−1  = (q − 1)X(W)q , 0.  Thus, L(s, X, K) is a polynomial in U = qs of degree 2g − 2 and leading coeﬃcient qg−1X(W) Again, by substituting from the Riemann-Roch theorem, we have

(q − 1)L(s, X, K) = X(C)qN(C)Ud(C) o≤d(CX)≤2g−2 −1 = X(C)qd(C)−g+1+N(WC )Ud(C) o≤d(CX)≤2g−2 = qg−1U2g−2X(W) X(WC−1)qN(WC−1) o≤d(CX)≤2g−2 −1 1 d(WC ) qU !

Writing C1 for WC−1 and noting that C1 runs through exactly the same range of summation as does C, we obtain

(q − 1)L(s, X, K) = (q − 1)qg−1U2g−2X(W)L(1 − s, X¯, K), which is again the same functional equation that we got in Case 1. We have therefore proved the

Theorem. For any character X of ﬁnite order,

qs(g−1)L(s, X, K) = X(W)q(1−s)(g−1)L(1 − sX¯, K).

Lecture 13

26 The Components of a Repartition

Our next aim is to introduce the L-functions modulo an integral divisor 75 of an algebraic function field over a finite constant filed and to obtain their functional equation. We shall develop the necessary results for this in this and the next two lectures. Let K be an algebraic function field over an arbitrary (not necessarily finite) constant field k. For a repartition C ∈ X and a prime divisor Y , we define the component C Z of C at Z to be the repartition defined by

CZ if UY C Z (U ) = 0 if Y is a prime divisor other than Y .   The mapping is clearly a k-linear mapping of X into itself. This induces a linear mapping of the dual of X into itself, by which a diﬀer- ential ω is taken to a liner map ωZ : X → X given by

ωY (C ) = ω(C Z )

ωZ is called the component of the diﬀerential ω at Y .ωZ is not, in general, a diﬀerential. For X ∈ K, we have

(Xω)Y (C ) = (Xω)(C Z ) = ω(XC Z ) = ω((XC )Y ) = ωZ (XC )

We shall now prove a lemma which expresses a diﬀerential in terms of its components.

71 72 13. Lecture 13

Y Lemma. If ω is a diﬀerential and C a repartition, ω (C ) = 0 for all 76 but a ﬁnite number of prime divisors Y , and we have

ω(C ) = ωY (C ). XY Proof. Let U be a divisor which divides the diﬀerential ω. Let Y1,..., Yn be the ﬁnite number of prime divisors for which either U , C U , Y vYi ( ) 0 or vYi ( ) < 0. For any of the i, and any prime divisor Y , we have

= U vY (C U ) vY (C (Y ))

vC (YU ) if U = Y vU (C ) if Y = U = = ≥ vY (U ), vY (0) if , YU  ∞ if Y , U          and therefore C U ∈ Λ(U ), ωU(C )= ω(C U ) = 0.  n Y Also, if we put Y = C − C i , we have for any Y , i=1 P

vY (CY ) if Y is not any of theY1 Y Y = ≥ Y U , v ( )   v ( ) vY (0) = ∞ if Y is a certain Yi     and therefore Y ∈ Λ(U ). Hence, 

n n n Y Y Y ω(C ) = ω C − C i + ω C i = ω C i       Xi=1 Xi=1 Xi=1       n  n      Y  Y    = ω C i = ω i = ωY (C ). Xi=1 Xi=1 XY

Our lemma is proved. We shall now prove another useful

77 Lemma. Let ω be a diﬀerential and C a prime divisor. Then vY ((ω)) is the largest integer m such that whenever X ∈ K and vY (X) ≥ −m, we have ωY (X) = 0. 26. The Components of a Repartition 73

Proof. Suppose ﬁrst that X ∈ K with vY (X) ≥ −vY ((ω)). Then clearly the repartitions xY is in Λ((ω)−1) and therefore ωY (X) = ω(XY ) = 0.

Now, by the deﬁnition of (ω), ω does not vanish on the space Λ((ω)−1Y −1). Hence there exists a repartition C ∈ Λ((ω)−1Y −1) such that ω(C ) , 0. It is evident that for U , Y , C U ∈ Λ((ω)−1) and therefore ω(C ) , 0. Hence,

0 , ω(C ) = ω(C U ) = ω(C Y ) XU

Put X = C Y . Then,

ωY (X) = ω(XY ) = ω(C Y ) , 0 and vY (X) = vY (C ) ≥−vY ((ω)) − 1.

Thus, vY ((ω)) is the largest m for which vY (X) ≥ −m implies that ωY (X) = 0.

Lecture 14

27 A Consequence of the Riemann-Roch Theorem

In this lecture, we shall prove a theorem which will be crucial in the 78 proof of the functional equation of the L-functions modulo an integral divisor. Let F be an integral divisor and U1,..., Ur the distinct prime divisors occurring in it. We prove a series of lemmas leading upto the proof of the theorem we mentioned above. U U = Lemma 1. In any class C, there exists a divisor such that vUν ( ) 0(ν = 1, ··· r).

Proof. Let U◦ be any divisor of C. By the independence theorem for valuations, we may ﬁnd X ∈ K with = U vUν (X) −vUν ( ◦)

Then U = (X)U◦ ∈ C satisﬁes the required conditions.

Let us denote by R the vector space Γ(n/U1, Ur) and by i the subspace Γ(F /U1,... Ur). R is not only a vector space over k, but also an algebra. In fact, if X, Y ∈ R, = + = vUi (XY) vUi (X) vUi (Y) ≥ 0, XY ∈ R. (i 1,... r) i is an ideal of R, since X ∈ R, Y ∈ i implies that = + vUi (XY) vUi (X) vUi (Y) ≥ vUi (F ), XY ∈ i.

Thus, the quotient R¯ = R/i is an algebra of 79

75 76 14. Lecture 14

Γ(n/U r) U r) rank dimk = = d(F ) Γ(F /U1 Ur) over the field k. Now, choose a differential ω such that the divisor F (ω) is coprime to F . This is possible by Lemma 1. We shall assume this ω to be chosen and fixed throughout the discussion. We define a linear function S on the algebra R by r U S (X) = ω ν (X), X ∈ R. Xν=1

S vanishes on the ideal i, for if X ∈ i, vUν (X) ≥ vUν (F ), and U + = = since F (ω) is corprime to each ν, vUν (F ) vUν ((ω)) 0, vUν (F )

−vUν ((ω)). By the last lemma of the previous lecture, we deduce that

r U S (X) = ω ν (X) = 0. Xν=1

Thus, S induces a linear map from the quotient R¯ to the ﬁeld k. This in turn gives rise to a bilinear form on R¯ deﬁned by (X¯, Y¯) → S (XY¯ ). Our next lemma states that this is non - degenerate.

Lemma 2. If X¯ ∈ R,¯ X¯ , 0, there exist a Y¯ ∈ R¯ such that S (X¯.Y¯) = 1.

Proof. Let X be any element of the coset X¯. Since X¯ , 0, we have < = X i and vUν (X) < vUν (F ) −vUν ((ω)) for some ν. Thus, Xω is a ﬀ di erential with vUν (Xω) < 0. By the last lemma of the previous lecture,

we deduce that there exists an element Y1 ∈ K such that vUν (Y1) ≥ 0 and U (Xω) ν (Y1) , 0.

80 Find Y2 ∈ K such that

vUν (Y2 − Y1) ≥ max(0, −vUν ((Xω))), , vUν (Y2) ≥ max(0, −vUµ ((Xω))). for µ ν.

Since we also have

vUν (Y2) ≥ min(vUν (Y2 − Y1), vUν (Y1)) ≥ 0, 27. A Consequence of the Riemann-Roch Theorem 77

it follows that Y2 ∈ R¯. Also, for any µ , ν, we have by the second U condition, (Xω) µ (Y2) = 0. Again, it follows from the ﬁrst condition that Uν Uν Uν (Xω) (Y2 − Y1) = 0, (Xω) (Y2) = (Xω) (Y1) , 0 Thus,

r r Uλ Uλ Uν S (X¯Y¯2) = ω (XY2) = (Xω) (Y2) = (Xω) (Y2) = ̺ , 0, Xλ=1 Xλ=1 ¯ = ¯ and Y Y2̺¯ satisﬁes the conditions of our lemma. Now, let U be any divisor coprime to F . Then, we assert that L(U ) U U = is a subspace of R. In fact, we have for X ∈ L( ), vUν (X) ≥ vUν ( ) 0. Also, we assert that L(U ) ∩ i = L(U F ). This follows from the following argument X ∈ L(U F ) ⇐⇒ vU (X) ≥ vU (U ) + vU (F ) for U U U every ⇐⇒ vU (X) ≥ vU ( ) for every and vUv (X) ≥ vUν (F ) for (ν = 1, , r). Hence, if we denote by L(U ) the image of L(U ) under the natural homomorphism form R to R¯, we have

L(U ) + i L(U ) dimk(L(U )) = dimk = dim i ! L(U ) ∩ i L(U ) = dim = l(U ) − l(U F ). k L(U F ) 81 This proves that the dimension of L(U ) depends only on the class of U . Since there are divisors in any class prime to F , we may deﬁne −1 N0(C) for a class C to be dimk L(U ) for any U ∈ C coprime to F . For any class C, we shall call the class C∗ = WC−1F the complementary class of C modulo F . We then prove the following

Lemma 3. For any class C, we have

∗ N0(C) + N0(C ) = d(F ).

Proof. Let U ∈ C−1 and δ ∈ C∗−1 be prime to F . Then,

∗ N0(C) + N0(C ) = dim L(U ) + dim L(δ) 78 14. Lecture 14

= (l(U ) − l(U F )) + (l(δ) − l(δF ) = N(C) − N(CF −1) + N(WC−1F ) − N(WC−1) = (d(C) − g + 1) − (d(CF −1) − g + 1) = d(F )

by the Riemann-Roch theorem. Our lemma is proved.

Now, let V be a vector space on a ﬁeld k and B : V xV → K a non-degenerate bilinear form on V. ( A bilinear form is said to be non- degenerate if for every X , 0, there exist a X1 ∈ V such that B(X, X1) , 0 and for every Y , 0 there exists a Y1 ∈ V such that B(Y1, Y) , 0. Let V1 be a subspace of V. Then we deﬁne the complementary subspace of

V1 with respect to the bilinear form B to the space V1comp of all elements 82 Y ∈ V such that B(X, Y) = 0 for every X ∈ V1. We then have the Lemma 4. Let V be a finite dimensional vector space and B a; non- degenerate bilinear form on V. Then, if V1 is a subspace of V, we have + = dimk V1 dimk V1comp. dim V. Proof. Let V∗ be the dual of V. We can define a homomorphism ϕ : V → V∗ which takes an element X ∈ V ot the linear map ϕ(X) ∈ V∗ defined by ϕ(X)(Y) = B(Y, X) Since B is non-degenerate, ϕ(X) , 0 if X , 0, and ϕ is a monomorphism. Since V is finite dimensional, dim V∗ = dim V = dim ϕ(V). Thus, ϕ is also an epimorphism.

Now, let V2 be the complementary subspace of V1. Then, every element of ϕ(V2) is a linear map of V into k which vanishes on the ∗ subspace V1, and conversely every element of V which vanishes on V1 should be of the form ϕ(X), where X ∈ V2. Hence, we deduce that ϕ(V2) is isomorphic to the dual of the quotient space V/V1. Therefore,

∗ dimk V2 = dimk ϕ(V2) = dimk(V/V1) = dimk V − dimk V1,

dimk V1 + dimk V2 = dimk V.

The lemma is proved. 27. A Consequence of the Riemann-Roch Theorem 79

83 Now, let U be a divisor prime to F and ω the ( already chosen and fixed ) differential such that (ω)F is prime to F . We define the complementary divisor U ∗ of U as the divisor (ω)−1F −1U −1. It is clear that U ∗ is also prime to F and that if U −1 ∈ C, U ∗−1 ∈ C∗. We may then state our theorem as follows.

Theorem . The complementary space of L(U ) in R¯ with respect to the bilinear form B(X¯, Y¯) = S (XY) deﬁned on R¯ is L(U ∗).

Proof. Suppose X¯ ∈ L(U ) and Y¯ ∈ L(U ∗). Then, for any prime divisor Y , any of the Uν, we have

∗ −1 −1 vY (XY) ≥ vY (UU ) = vY ((ω) F ) = −vY ((ω)),

and therefore by the lemma of the previous lecture,

ωY (XY) = 0

Hence, we obtain,

n S (X¯Y¯ ) = S (XY) = ωU (XY) = ωY (XY) = ω(XY) = 0, Xi=1 XY since XY ∈ K. Thus, we deduce that

∗ L(U ) ⊂ ((L(U )compl.

Now, let C and C∗ be the classes of U −1 and U ∗−1. We then have

∗ ∗ dimk L(U ) = N0(C ) = d(F ) − N0(C)

= dimk R¯ − dimk L(U ) = dimk (L(U ))compl.

since B(X¯, Y¯ ) = S (X¯Y¯ ) is non-degenerate by lemma 2. Hence, it follows 84 = ∗ that (L(U )comp. L(U ).

Lecture 15

28 Classes Modulo F

Let K be an algebraic function field over an arbitrary constant field k and 85 F an integral divisor of K. We shall denote the distinct prime divisors of F by U1,..., Ur. We now define some groups associated to the integral divisor F . In all the cases, it is an easy matter to verify that the sets defined are closed under taking products or inverses. F F ϑ will be the group of divisors coprime to F , and ϑ0 the subgroup of all elements of ϑF whose degree is zero. K∗F is the multiplicative group of elements of K∗ which are coprime to F . EF is the group of principal divisors coprime to F . Clearly, k∗ ⊂ K∗F , and since two elements of K∗F define the same divisors if and only if their ratio is a constant, we have the isomorphism

F ∗F E ≃ K /k∗

Moreover, since every class contains a divisor coprime to F , the saturation of ϑF by E is the whole of ϑ, and consequently

F F F ϑ ϑ E ϑ ϑ / F = ≃ = = R, E E ∩ ϑF E E F and similarly ϑo /EF ≃ ϑ0. + = We shall say that X ≡ Y( mod F ) if vUi (X − Y) ≥ vUi (F ) (i 86 1, r). The condition is evidently equivalent to saying that X − Y ∈ Γ U U ∗ ∗ (F / 1, r). Let KF be the set of elements X of K which satisfy X ≡ + ∗ = 1( mod F ). If X ∈ KF , it follows that vUi (X) min(vUi (X − 1),

81 82 15. Lecture 15

= ∗ + vUi (1)) 0. Hence, if X, Y ∈ K , we have XY ≡ Y ≡ 1( mod F ) and ∗ 1 ∗ ∗ therefore XY ∈ K . Similarly, is also in K . Since X ∈ K ⇒U F X F F i = ∗ (X) 0, we deduce that KF . EF shall denote the group of principal ∗ ∗F ∗ divisors of KF ⊂ K . EF elements of KF . It follows from the above F inclusion that EF ⊂ E . EF is called the ray modulo F . , ∗ ∗ = If F N, it is easy to see that k ∩ KF 1 . Hence, we deduce, in K∗ k∗ ∗ = F K ∗ this case, that EF ≃ ∗ ∗ ≃ KF . k∗ kF ∩ k The class group modulo F is by definition the quotient group RF = F ϑ /EF and its elements are called classes modulo F . The subgroup R = F R F 0F ϑ0 /EF of F is called the group of classes modulo of degree zero and its order hF is called the class number modulo F . In the case of a finite field k, hF can be expressed in terms of the class number h. We have the following theorem.

Theorem. Let the constant field k be finite with q elements and let F be an integral divisor different from N. Then,

hNF r 1 hF = 1 − q − 1 NU ! Yν=1 ν Proof. From the isomorphisms

F ϑ /EF R0 ≃ ϑF /EF ≃ 0 , 0 F E /EF

87 it follows that = hF h F E : 1 EF Again,

F ∗F ∗ ∗F ∗F ∗ E K /k K K /KF ≃ ∗ ∗ ∗ ≃ ∗ ∗ ≃ ∗ ∗ ∗ , EF KF k /k KF k KF k /KF K∗ k∗ ∗ F k ∗ and ∗ ≃ ∗ ∗ ≃ k , kF k ∩ KF 28. Classes Modulo F 83

, ∗ ∗ = since F N and therefore k ∩ KF {1}. Hence, we obtain h h = K∗F : K∗ . F q − 1 F h i

∗F Now , K is precisely the set of elements of Γ(N/U1,... Ur) which ∗F do not lie in any of the spaces Γ(Ui/U1,... Ur). Also, if X, Y ∈ K , ∗ −1 they belong to the same coset modulo KF if and only if XY ≡ 1 (mod +F ) ⇐⇒ X ≡ Y (mod +F ) ( since Y is coprime to F ) ⇐⇒ X − Y ∈ Γ(F /U1 ··· Ur). We deduce from these facts that (with S = (U1,... Ur))

Γ Γ U Γ(U U /s) (n/s) ( i/s i j ··· ∗F ∗ = dim Γ(F /s) dim Γ(F /s) + dim Γ(F /s) K : KF q − q q h i Xi Xi, j U U U = qd(F ) − qd(F )−d( i) + qd(F )−d( i)−d( j) ··· Xi Xi, j r 1 = N(F ) 1 − NU ! Yν=1 ν

Substituting this expression in the value of hF , we get the required 88 result. Note that the theorem is not valid when F = N. In fact, we have = = R0N R0 and hN h. We shall end this lecture with a simple lemma asserting the existence of suﬃciently many divisors in any class modulo F .

Lemma. If U is any given divisor, any class CF modulo F contains a divisor δ prime to U .

Proof. Let U0 be any divisor of CF . Find a Y ∈ K such that = vUν (Y − 1) ≥ vUν (F ) (ν 1,..., r) and vU (Y) = −vU (Y0) if vY (U ) , 0 and Y , Uν

Then it is easy to verify that δ = U0(Y) satisﬁes the conditions of the lemma.

Lecture 16

29 Characters Modulo F

We shall now introduce the characters of an algebraic function ﬁeld K 89 modulo an integral divisor F

Deﬁnition. A chapter X modulo an integral divisor F is a homomorphism of RF into the multiplicative group of complex numbers with absolute value one.

If U is a divisor prime to F , we put X(U ) = X(U EF ). If U = δ/L , where δ and L are mutually coprime and integral and L coprime to F whereas δ is not, we define X(U ) = 0. X is this defined as a complex valued function on a subset of V , and is clearly multiplicative (i.e., if X(U )) and X(δ) are defined, X(U δ) = X(U X(δ)). Note that X is defined in particular on all integral divisors of K. Let X be a character mod F of K. An integral divisor F 1 is said to be a modulus of definition of X if for any X ∈ K coprime to F and X ≡ 1( mod +F ∞) we have X((X)) = 1. F itself is clearly a modulus of definition. The reason for our terminology is provided by the following

Theorem. If X is a character mod F and F 1 a modulus of deﬁnition of X, there exists a unique character X1 of K mod F 1 such that for any divisor U prime to FF 1,

X(U ) = X1(U )

Conversely, if X and X1 are two characters mod F and F 1 respec- 90

85 86 16. Lecture 16

tively such that whenever U is coprime to FF 1, we have X(U ) = X1(U ), then F ∞ is a modulus of deﬁnition of X and X1 is the character associated to X by the ﬁrst part of the theorem.

1 Proof. Suppose F is a modulus of definition of X. If CF 1 is any class 1 modulo F , we can find a divisor U in CF 1 coprime to F and we define

1 X (CF 1 ) = X(U ).

The deﬁnition is independent of the choice of U in CF 1 . For ,if −1 δ ∈ CF 1 and is coprime to F , there exists an X ∈ K such that U δ = (X), X ≡ 1( mod +F 1) and X coprime to F . Hence

X(U δ−1) = X((X)) = 1, X(U ) = X(δ).

So defined, X1 is evidently a character modulo F 1 which satisfies the condition X(U ) = X1(U ) if U is coprime to FF ′. The uniqueness follows from the fact that this definition of F 1 is forced upon us by the above condition. The first of our theorem is proved. 91 To prove the second part, suppose X and X1 are two characters modulo F and F 1 respectively satisfying the above condition. Then, if (X) is coprime to F and X ≡ 1(F 1), (X) is coprime to FF 1, and therefore

X((X)) = X1((X)) = 1

This proves that F 1 is a modules of deﬁnition of X and X1 the associated character mod F 1.

Corollary. F is a modulus of deﬁnition of X1.

Our next theorem states that to any given character, there exists a ‘smallest’ modulus of deﬁnition. To prove this, we require the following 29. Characters Modulo F 87

Lemma. If F 1 and F ′′ are two moduli of deﬁnition of a character modulo F , their greatest common divisor F ′′′ is also a modulus of deﬁni- tion.

Proof. Let X be coprime to F and X ≡ 1( mod F ′′′). Find a Y ∈ K such that

1 1 ′′′−1 vY (XY − 1) ≥ vY (F ) + vY (X) if vY (F F > 0, (1) ′′ + ′′ ′′′−1 vY (X(Y − 1)) ≥ vY (F ) vY (X ) if vF (F F > 0, (2)

′ ′ and vY (Y − 1) ≥ vY (FF F ”) + vY (X) if Y divides FF F ” but does not belong to the first two categories. (3) Since F ′′′ is the greatest common divisor of F ′ and F ′′ the first two 92 categories are mutually exclusive, and the third category is by definition exclusive of (1) or (2).

Now, one can easily verify that (Y) is coprime to F . Hence, (XY) is also coprime to F . We now assert that Y ≡ 1 ( mod +F ′′) and XY ≡ 1( mod +F ′). To verify the ﬁrst, suppose Y is a prime divisor dividing F ′′. Then Y must occur in one of the three categories. If Y belongs to (1),

vY (Y − 1) = vY (XY − 1 + 1 − X) − vY (X) ≥ min(vY (XY − 1), vY (X − 1)) − vY (X),

′′′ ′′ and since vY (X − 1) ≥ vY (F ) = vY (F ) > 0, vY (X) = 0, and the right hand side of the inequality becomes

′ ′′ ′′ ≥ min(vY (F ), vY (F )) = vY (F ).

If Y belongs to category (2), we get

′′ ′′ vY (Y − 1) ≥ vY (F ) + vY (X) − vY (X) ≥ vY (F ).

Finally, for Y in (3), we get again

′′ vY (Y − 1) ≥ vY (F ). 88 16. Lecture 16

The second congruence XY ≡ 1 (mod +F ′) can be proved similarly. Since F ′ and F ′′ are moduli of deﬁnition, we deduce that X(Y) = X(XY) = 1. Hence X(X) = 1. This completes the proof of the fact that F ′′′ is a modulus of deﬁnition. 93 The theorem we mentioned is a fairly easy consequence.

Theorem . Let X be a character of K mod F . Then there exists a unique integral divisor m such that it divides every modulus of deﬁnition of X and every integral divisor which is divisible by it is a modulus of deﬁnition. m is called the conductor of X.

If X1, is the associated character to m, the conductor of X1 is m itself.

Proof. By the previous lemma, the g.c.d. of all moduli of definition of X is an integral divisor which satisfies all the conditions of the first part of the theorem. Let X1, be the associated character mod m. If the conductor of X1, is m1, it is clear that m1 is also a modulus of definition of X. Hence m divides m1, and m1 divides m since m1 is the conductor of X1. Thus, m = m1. Our theorem is proved. A character F modulo F is said to be primitive or proper if F is the conductor of X. Lecture 17

30 L-functions Modulo F

Throughout this lecture, we shall assume that K is an algebraic function 94 field over a finite constant field k with q = p f elements. Let F be an integral divisor of K and X a character modulo F . For a complex variable s = σ+ it σ > 1, we define the L-function w.r.t. the character F by the infinite series X(U ) L(s, X) = , (NU )s XU the summation being over all integral divisors. The absolute convergence, etc. of the series do not offer any difficulty to prove, and we may also get the following product formula:

X(Y ) −1 L(s, X, K) = 1 − for σ> 1. (NY )s YY !

31 The Functional Equations of the L-functions.

Before proceeding to the proof of the functional equation of the L- functions, we shall prove some essential lemmas. Since it is possible to prove these in a general setting, we shall do so. Let A be a commutative algebra with unit element 1 of ﬁnite rank f over a ﬁnite k with q elements. We shall assume that a mapping X of A into the complex number is given with the following properties:

89 90 17. Lecture 17

95 1) χ(X) = 0 if and only if X is a zero divisor is A

2) χ(XY) = χ(X)χ(Y) for X, Y ∈ A

3) χ(αX) = χ(X) for X ∈ A and α ∈ k∗

4) If X is a zero divisor of A, there exists a Y in A such that XY = X and χ(Y) , 0, χ(Y) , 1.

We shall also assume that a k-linear mapping S of A in k is given which is such that if X ∈ A, X , 0, There exists a Y ∈ A such that S (XY) = 1. Lastly, we assume that a complex valued function ψ is given on k such that (a) ψ(0) = 1 and (b) ψ(α) = 0. (e.g, ψ(0)) = α∈k 1 P 1, ψ(α) = − for α , 0 satisﬁes the required conditions). We then q − 1 have the following lemmas.

Lemma 1. Let V be a vector subspace of A. Then,

qdim V if Y ∈ V . ψ(S (XY)) = comp 0 otherwise . X∈V   Proof. If Y ∈ Vcomp, S (X, Y) = 0 for every X ∈ V and the ﬁrst equality follows from condition (a) for ψ.

If Y < Vcomp, we can ﬁnd X1 ∈ V such that S (X1Y) = 1. Complete X1 to a basis X1,... Xd of V over k. Then the sum on the left is

d d = + = ψ S  αiXiY ψ α1 S  α1XiY 0, α ,...,αX∈K  Xi=1  α ,...,αX∈K αX∈K  Xi=2  1 d    1 d 1          + d since for ﬁxed α2,...αd, the sum α1 S ( i=2 αiXiY) runs through all elements of k when α1 does. Our lemma isP proved.

96 Lemma 2. Deﬁne the generalised Gaussian sum G(X, χ) for X ∈ A by

G(X, χ) = χ(Y)ψ(S (XY)). XY∈A 31. The Functional Equations of the L-functions. 91

Then we have G(X, χ) = χ(X)G(1, χ) Proof. Since A is a finite algebra every non-zero divisor is a unit. (In fact, if u is a non-zero divisor, ua , ub if a , b, and since A is a finite set, uA = A. In particular, there exists and element u1 ∈ A with uu1 = 1). Now, if X is a zero divisor, there exists an element u ∈ A by 4) with Xu = X and χ(u) , 0 or 1; thus, u, is a non-zero divisor. Hence we obtain G(X, χ) = χ(Y)ψ(S (XuY)) = χ(Yu−1)ψ(S (XY)) XY∈A XY∈A = χ(u−1)G(X, χ), and since χ(u−1) = χ−1(u) , 1, G(X, χ) = 0 = χ(X)G(1, χ). If X is not a zero divisor, G(X, χ) = χ(Y)ψ(S (XY)) = χ(X−1Y)ψ(S (Y)) = χ−1(X)G(1, χ). XY∈A XY∈A Now, the units of A form a finite group and therefore there exists an integer n , 0 with Xn = 1. Hence χn(X) = χ(Xn) = χ(1) = 1, |χ(X)| = 1 and χ−1(X) = χ(X). The lemma is proved. Lemma 3. Let V be a subspace of A and 97 M(V, χ) = χ(X). X∈V Then, − dim V M(Vcomp, χ¯) = q ∆(χ)M(V, χ), where ∆(χ) depends only on χ and |∆(χ)| = q f /2 Proof. By the first two lemmas, we have

− dim V dim V q M(Vcomp, χ¯) = χ¯(X)q = χ¯(X) ψ(S (XY))

X∈XVcomp X∈A X∈V = χ¯(X)ψ(S (XY)) G(Y, χ¯) = χ(Y)G(1, χ¯) = G(1, X¯)M(V, χ) XY∈V X∈A YX∈V XY∈V 92 17. Lecture 17

Put ∆(χ) = G(1, χ¯). It only remains to prove that |G(1, χ¯)| = q f /2. Now, k can be imbedded in A by the mapping α ∈ k → α1· ∈ A. Also, M(k, χ) χ(α) = χ(1) = q − 1 , 0. Xα∈k αX∈k∗ Choose ψ to be real. Taking V = k in the above formula, we obtain

dim kcomp+1 G(1, χ)G(1, χ¯)M(k, χ) = G(1, χ)qM(kcompχ¯) = q M(k, χ), G(1, χ)G(1, χ¯) = q f

But since ψ is real, G(1, χ¯) = G(1, χ), and therefore

|∆(χ)| = |G(1, χ¯)| = q f /2

98 Our lemma is proved. Since we had ∆(¯χ) = G(1, χ), we see that G(1, χ) does not depends on the function ψ. By lemma 2, it follows that G(X, χ) is independent of ψ. (This may also be proved directly.) We now proceed to the functional equation

Theorem . Let F be an integral divisor which is not the unit divisor and χ a proper character modulo F . Then L(s, χ, K) is a polynomial in U = q−s of degree 2g − 2 + d(F ) and satisﬁes the functional equation

s(g−1+ 1 d(F )) (1−s)(g−1+ 1 d(F )) q 2 L(s, χ, K) = ǫ(χ)q 2 L(1 − s, χ,¯ K)

where ǫ(χ) is a constant depending on χ, with |ǫ(χ)| = 1.

Proof. Let R be the algebra Γ(N /U1,... Ur), i the ideal Γ(F /U1, ... Ur) and R¯ the quotient algebra R/i.

Now, if X ∈ R, X , 0, NX is prime of F and χ((X)) is meaningful. We shall show that χ((X)) is constant on the cosets modulo the ideal i. Let X, Y ∈ R, X − Y ∈ i. If (X) is not coprime to F , (Y) cannot be coprime to F and therefore χ((X)) = χ((Y)) = 0. If X is coprime to F , we have Y vU i − 1 = vU i(Y − X) ≥ vU i(F ), X 31. The Functional Equations of the L-functions. 93

Y and hence ≡ 1( mod +F ), χ(X) = χ(Y). X Thus, we may deﬁne for X¯ ∈ R¯ a mapping which again we shall denote by χ by the equation

χ(X¯) = χ((X)).

99 Clearly we have χ(X¯)χ(Y¯) = χ(XY) and χ(αX¯) = χ(X¯) if α ∈ k∗. ¯ = Also, χ(X) 0 if and only if vUi (X) > 0 for some i. Find a Y ∈ K such that = vUi (Y) vUi (F ) − vUi (X) = , vUi (Y) vUi (F ) for j i. Then, Y¯ ∈ R¯, Y¯ , 0 and X¯Y¯ = 0. Thus χ(X¯) = 0 implies that X¯ is a zero divisor. Conversely, if X¯ is a zero divisor, we should have ¯ = vUi (X) > 0 for some i and therefore χ(X) 0. ¯ ¯ Now, suppose Z is a zero divisor in R. Then vUi (Z) > 0 for some i. Since χ is proper character modulo F , F −1 is not a modulus of defini- Ui tion of χ. Hence there exists an element X ∈ K∗, with (X) coprime to + U −1 , F , X ≡ 1( mod F i ), and χ((X)) 1. Then we have (X − 1)Z ≡ 0( mod +F ), X¯Z¯ = Z¯, and χ(X¯) , 0 or 1. Hence, the map χ on R¯ satisfies the conditions 1), 2), 3) and 4) stipulated at the beginning of this lecture. We have already seen (Lecture 14) that if ω is a differential such that F (ω) is coprime to F , S (X¯) = r U S (X) = ω i (X) has all the requisite properties. Now, i=1 P L(s, χ, K) = q−sd(c) χ(U ), XC UX∈C where the first summation is over all classes C and the second over all integral divisors in the class C. Choose a divisor UC in the class C 100 coprime to F . We have χ(U ) χ(U ) = C χ((X)), q − 1 UX∈C X∗ U −1 X∈L ( C ) 94 17. Lecture 17

since every integral divisor U ∈ C can be written in precisely (q − 1) U U −1 ways in the form (X) C, where X is a non-zero element of L( C ). U U −1 Again, since C is coprime to F , L( C ) ⊂ R and two elements of U −1 ﬀ L( C ) go to the same coset modulo i if and only if their di erence in U −1 L( C )F . We therefore have U ¯ χ( C) l(U −1F ) χ(X) χ(U ) = q C q − 1 ¯ ∈ U −1 UX∈C X PL( C )

= U −1 Applying lemma 3 with V L( C ), we have

−1 dim V χ(X¯) = M(V, χ) = ∆ (χ)q M(Vcomp, χ¯). ¯ XU −1 X∈L( C )

But by the theorem of lecture 14, we have

U −1 = U F L( C )compl. L( C (ω))

= U −1 = = Now, if d(C) < 0, N(C) l( C ) 0 and hence M(V, χ) 0. Also, −1 ∗ if d(C) > 2g − 2 + d(F ), d(C F W) < 0, and N(C ) = l(UCF (ω)) = 0; hence again M(V, χ) = 0. Substituting in the expression for L(s, χ, K), we see that it is a polynomial in U = q−s of degree at most 2g−2+d(F ). 101 We have

− U −1F = sd(C) U l( C ) U −1 (q − 1)L(s, x, K) q χ( C )q M(L( C ), χ) XC 1 = q−sd(C)+N(C)χ(U )M(L(U −1) χ¯) by lemma 3, ∆(χ) C c comp XC 1 + = q−sd(C) N(C)χ(U )M(L((U −1)∗)¯χ) ∆(χ) C c XC by the theorem of lecture 14,

1 −1 = (1−s)d(C)−g+1+N(WC ) U −1 U −1 ∗ ∆ q χ¯( C )M(L( C ) ), χ¯) (χ) X χ(ω)F + = q(2s−1)(1−g) (1−s)d(F ) ∆(χ) 31. The Functional Equations of the L-functions. 95

− ∗ − ∗− U −1 F ∗ − ∗ (1 s)d(C ) U 1 1 l( C ) ) U 1 × q χ¯(( C ) )q M(L(( C ) ), χ) XC χ((ω)F ) = (q − 1) q(2s−1)(1−g)+(1−s)d(F )L(1 − s, χ,¯ K), since C∗ ∆(χ) runs through all classes as C does. This gives

s(g−1+ 1 d(F )) (1−s)(g−1+ 1 d(F )) q 2 L(s, χ, K) = ǫ(χ)q 2 L(1 − s, χ,¯ K), 1 d(F ) χ((ω)F )q2 where ǫ(χ) = , |ǫ(χ)| = 1. ∆(χ)

That the degree in U of L(s, χ, K) is exactly 2g − 2 + d(F ) follows by comparing the coeﬃcients of highest powers on both sides of the equation. Our theorem is proved.

Lecture 18

32 Extensions of Algebraic Function Fields

In the next five Lectures, we shall investigate the relations between an al- 102 gebraic function field and an extension of it (see below for definition). In particular, we consider in the end the connection between the ζ-function of a function field and the ζ-function of a constant field extension. This gives in particular the interesting result that for algebraic function fields over a finite constant field, the smallest positive degree of a divisor is 1. We start with the definition of an extension. Definition . Let K be an algebraic function field with constant field k. An extension of K an algebraic function field L with constant field l such that L ⊃ K and l ∩ K = k. Now let L/l be an extension of K/k and K a prime divisor of L. If vK (X) = 0 for all X ∈ K is said to be variable over K or trivial on K. If this were not true, the restriction of vK to K defines a valuation of K trivial on k and should therefore correspond to a prime divisor Y of K. In this case, K is said to be fixed on K and is said to lie over the prime divisor Y of K. Also, since the restriction of vK to K and vY are equivalent valuations with values in Z, the latter having the whole of Z for its value group, there exist a positive integer eL/K (K ) such that

vK (X) = eL/KvY (X) for all X ∈ K eL/K(K ) is called ramiﬁcation index of K over K. 103 The relation between the residue ﬁelds of K and Y is given by the following

97 98 18. Lecture 18

Lemma . If L/l is an extension of K/k and a prime divisor K of L lies over a prime divisor Y of K, the residue ﬁeld KY of Y can be canonically imbedded in the residue ﬁeld LK of Y

Proof. Let O, K denote respectively the valuation ring and maximal ideal of K and O and Y those of Y . Since vK and vY are equivalent on K, we clearly have O ⊃ Y and Y = O ∩ K . Hence we have monomorphism Y = O/Y → O/K = LK and KY can be considered as imbedded as a subﬁeld of LK .

We now give condition for L/K to be a ﬁnite or an arbitrary algebraic extension.

Lemma . Let L/l be an extension of K/k. Then among the following statements, (1), (2) and (3) are equivalent and so are (1′), (2′) and (3′). (1) [l : k] < ∞ (1′) l is algebraic over k. (2) [L : K] < ∞ (2′) L is algebraic over K (3) If K is any prime divisor (3’) If K is any prime of L ly- of L lying over the prime ing over the prime divisor divisor Y of K, LK is algebraic Y of K[LK : KY ] < ∞ over KY Proof. We shall show that (1) ⇔ (2) ⇔ (3). The proof that (1)′ ⇔ (2)′ ⇔ (3)′ is similar, and even simpler

The equation (valid even when either side is inﬁnite)

[LK : k] = [LK : KY ][KY : k] = [LK : l][l : k]

104 show that (1) ⇔ (3) since [KY : k] and [LY : l] are both ﬁnite. We shall now prove that (1) ⇔ (2). Let X be any transcendental elements of K over k. Since X < k and L ∩ k = l, X < l and is therefore transcendental over 1. It follows that [K : k(X)] < ∞ and [L : l(X)] < ∞, and from the equalities

[L : k(X)] = [L : K][K : k(X)] = [L : l(X)][l(X) : k(X)],

it follows that [L : K] < ∞⇔ [l(X) : k(X)] < ∞. 32. Extensions of Algebraic Function Fields 99

We shall now prove that [l(X) : k(X)] = [l : k], which would finish the proof of the theorem. Suppose α1,...,αn are n linearly independent element of l over K. We assert that they are also linearly independent over k(X). For it there were a linear relation among these with coefficients in k(X) with at least one non-vanishing coefficient, we may clearly assume that it is of the form n αigi(X) = 0 Xi=1 at least one gi(X) with a non-zero constant term. Since X is transcendental over l, we may put X = 0 in the above equation to obtain a linear relation among the α1 over k with at least one non-zero co-efficient. But this is impossible since the αi are linearly independent by a assumption. Hence, [l(X); k(X)] ≥ [l : k]. To prove the reverse inequality, we may assume that [l : k] < ∞. 105 Hence, there exists a finite set β1,...βr of element of l such that l = k(β1,...βr). Then

[l(X) : k(X)] = [k(X, β1,...,βr) : k(X)]

= [k(X), β1,...βr] : k(X, β1,...,βr−1).[k(X, β1,...,βr−1)] :

k[X, β1,...,βr−2] ... [k(X, β1) : k(X)]

≤ [k, β1,...βr] : k(, β1,...,βr−1).[k(, β1,...,βr−2)] :

k[, β1,...,βr−2] ... [k(β1) : k(X)] since the degree of β1 over k(X, β1 ··· βi−1) is less than or equal to its degree over the smaller ﬁeld k(β1 ··· βi−1), [k(β1 ··· βr) : k] = [l : k]. The proof of the lemma is completed. We shall call an extension L/l over K/k satisfying any of the conditions (1), (2), (3) of lemma a ﬁnite extension. If K is a prime divisor of L lying over a prime divisor Y of K, the positive integer [LK : KY ] = dL/K(K ) is called the relative degree of K over K. It follows from the proof of the above lemma that

[LK : l][l : k] dL(K ) dL/K(K ) = = [l : k], [KY : k] dK(Y ) 100 18. Lecture 18

dL(K )[l : k] = dL/K (K )dK (Y )

106 (The suffix to d indicated the field in which the degree is taken) If L/l is an algebraic extension of K/k, there does not exist any prime divisor of L which is variable over K. For, suppose v is a valuation on L which is trivial on K. Any elements α ∈ L satisfies an irreducible equation

n n−1 α + a1α + ··· + an = 0, ai ∈ K, n−1 so that 0 = v(an) = v(α) + v(α + +an−1)

If v(α) > 0, we have

n−1 v(α + ··· + an−1) = min((n − 1)v(α), (n − 2)v(α), 0) = 0,

and we obtain a contradiction by substituting in the previous equation. Thus, v cannot be positive for any element of L, which is impossible. Now, let L/l be any extension of K/k. We shall prove that there are at most a ﬁnite number of prime divisors of L lying over a given prime divisor Y of K, and that there is at least one. Let g be the genus of K and let C be the class of the divisor Y g+1 . Since d(C) = d(Y g+1), ≥ g + 1, we have

N(C) ≥ d(C) − g + 1 ≥ 2,

and there exists at least one more integral divisor U in C. Then, Y g+1 −1 U = (X)K where X ∈ K and X transcendental over k. Hence X is also transcendental over l and the divisor (X)L has a decomposi- K a1 K ah 1 h 107 tion (X)L = , where h ≥ 1, and ai > 0. We assert that NX K1,... Kn are precisely the divisors of L lying over Y . For, if K lies over Y , vK (X) > 0 and hence should be one of the Ki; and since

vKi (X) > 0, the restriction of vKi to K should be a prime divisor occurring in the numerator of (X)K, and the only such prime divisor is Y . Our contention is proved. We now have the 32. Extensions of Algebraic Function Fields 101

Theorem . Suppose L/l is an algebraic extension of K/k. Let Y be a prime divisor of K and K1,... Kh all the prime divisors of L lying over Y . Then, h [L : K] = dL/K (Kν)eL/K(Kν) Xν=1 Proof. Choose an element X ∈ K as above. We have Y t (X)k = (NX)K vK (X) vK (X) K 1 K 2 K vK h(X) (zX)L 1 2 ... h and (X)L = = (NX)L (NX)L Therefore we have h h = = K = K K [L : l(X)] d((zX)L) vKν (X)dL( ν) vY (X) eL/K ( ν)dL( ν) Xν=1 Xν=1 and on the other hand

t [K : k(X)] = dK(Y ) = tdK(Y ) = vY (X)dK (Y ).

Hence, 108

[L : l(X)][l(X) : k(X)] [l : k] h [L : K] = = e (K )d (K ) [K : k(X)] d (Y ) L/K ν L ν K Xν=1 h = eL/K(Kν)dL/k(Kν), Xν=1 which is the formula we want. As corollaries, we deduce the inequalities h ≤ [L : K],

dL/K(Kν) ≤ [L : K]

and eL/K(Kν) ≤ [L : K].

Lecture 19

33 Application of Galois Theory

We shall now recall some basic facts of Galois theory which we shall 109 use in the sequel. Let A be a field and B a finite algebraic extension of A.Bs shall denote the subfields of all separable elements over A of B. The separable degree [B : A]s is then define to be the degree of the field Bs over A.B is a purely inseparable extension over Bs and its degree is called the degree of inseparability of B over A, and is denoted by [B : A]i. We obtained the relation

[B : A] = [B : Bs].[Bs : A] = [B : A]s.[B : A]i

If B/A is a normal extension, we define the Galois group G(B/A) to be the group of all automorphisms of B which leave all the elements of A fixed. The set A1 of all elements left fixed by every automorphism belonging to G(B/A) is clearly a field containing A. It is called the fixed field of G(B/A). A1 is a purely inseparable extension of A and B a separable extension of A1, and we have

[B : A1] = [B : B]s, [A1 : A] = [B : A]i.

We shall now apply these facts to the theory of algebraic functions. Let L/l ba a extension of K/k. We deﬁne the relative separable degree dL/K(K )s and the relative inseparable degree dL/K(K )i of a prime di- 110 visor K of L lying over the prime divisor Y of K to be respectively [LK : KY ]s and [LK : KY ]i. K is said to be separable, inseparable

103 104 19. Lecture 19

K or purely inseparable accordingly as dL/K ( )i is equal to one, greater K than one or equal to dL/K ( ). 1 Suppose now that L /l′ ⊃ L/l ⊃ K/k is a tower of algebraic function ﬁelds. Let K 1 be a prime divisor of L1 lying over a prime divisor K of L, which again lies over a prime divisor Y of K. It is easy to see that the following relation between the ramiﬁcation indices holds

1 1 eL′/K(K ) = eL′ (K ).eL/K (K ) If moreover we assume that [L1 : K]< ∞, we have

1 1 1 K = 1 K K dL /K ( ) dL /L ( )dL/K ( ), and similar relations for the separable and inseparable degrees. 1 Now, suppose L/l and L /l1 are two extensions of K/k and σ is an isomorphism of L onto L1 which maps l on l1 and ﬁxes every element of K. If K is any prime divisor of L, we deﬁne the prime divisor σK in L1 by the equation

−1 1 vσK (Y) = vK (σ Y) for Y ∈ L . Clearly, K → σK is a one-one and onto mapping of the set of prime divisors of L onto the prime divisors of L1. It is also immediate 111 that the isomorphism σ maps the valuation ring and the maximal ideal of K onto those of σK . Hence we have an induced isomorphismσ ¯ : 1 K LK → LσK of the residue class ﬁelds. If lies over a prime divisor Y of K, σK also lies over Y andσ ¯ ﬁxes every element of KY . From these facts, it follows that K = K eL1/K(σ ) eL/K ( ) and if [L : K] < ∞

1 K = K dL /K (σ ) dL/K ( ). We have the following theorem for ﬁnite normal extensions.

Theorem. Let Ll be a ﬁnite normal extension of K/k, and K a prime divisor of L lying over a prime divisor Y of K. Then every prime divisor of L lying over Y is of the form σK , where σ is an element of G(L/K) 33. Application of Galois Theory 105

Proof. We have already seen that σK lies over Y for every σ ∈ G(L/K). To prove the converse statement, let K = K1,..., Kh be all the prime divisors of L lying over Y . Find aY ∈ L such that

vK (Y) > 0 = = vK j (Y) 0 for j 2,... h.

Then ,

= = −1 vK (NL/K Y) [L : K]i vK(σY) [L : K]i vσ K (Y) > 0, σ∈GX(L/K) σ∈GX(L/K)

−1 since each σ K for σ ∈ G(L/K) is a certain Ki. Because K lies over 112 Y , we deduce that

vY (NL/K Y) > 0, and consequently for j = 2,..., h, we have

K = −1 v j (NL/K Y) [L : K]i vσ K j (Y) > 0. σ∈GX(L/K)

Since at least one term of the sum on the right must be positive, and since the only prime divisor lying over Y whose valuation on Y is K K = −1K positive is , there exists an automorphism σ j such that σ j j, K j = σ jK for every j. Our theorem is proved. For the rest of the lecture, we shall assume that L/l is a finite normal extension of K/k and G(L/K) the Galois group. If K is a prime divisor of L, we define the decomposition group (Zerlegungs gruppe) Z(K ) of K to be the subgroup of G(L/K) of all elements σ ∈ G(L/K) such that σK = K . It follows that if σ, σ1 ∈ G(L/K), σK = σ1K if and only if σ and σ1 belong to the same left coset of G(L/K) modulo Z(K ). Because of the above theorem, we are able to deduce that the number of prime divisors of L lying over a fixed prime divisor of K is equal to the index in G(L/K) of the decomposition group of any one of them. 106 19. Lecture 19

It is also easy to obtain the relation between the decomposition groups of two prime divisors K and σK lying over the same prime divisor of K. In fact, we have

τ ∈ Z(σK ) ⇔ τσK = σK ⇔ σγτσK = K ⇔ σ−1τσ ∈ Z(K )

113 and therefore Z(σK ) = σZ(K )σ−1

Theorem. Let K be a prime divisor of L lying over the prime divisor Y of K. Then LK /KY is a normal extension. Every element σ of Z(K ) induces an automorphism σ¯ of LK over KY , and every automorphism of LK over KY is got in this way.

Proof. Let K = K1, K2,..., Kh be all the prime divisors of L lying over Y . Ify ¯ ∈ LK , we can ﬁnd a representative ofy ¯ in the valuation ring OK of K such that

= vK j (y) > 0 for j 2,..., h.

In fact, if y1 is any representative ofy ¯, choose a y ∈ L such that

1 vK (y − y ) > 0

vK j(y) > 0 for j = 2,..., h.

y satisﬁes the required condition. The ﬁeld polynomial of y over K is given by

[L:K]i = − σ f (X)  (X y) σY∈G    −1   = Now, if σ < Z(K ),σ K , K and therefore vK (σy) vσ−1K (y) > 0. Passing to the quotient modulo K in the above equation, we get [L:K]i M f (X) =  (X − σy) X ,   σ∈YZ(K )      33. Application of Galois Theory 107

114 where M is a non-negative integer. But this is a polynomial over KY which is satisfied byy ¯, and which has all its roots lying in LK . LK is thus a normal extension of KY . Now if σ ∈ Z(K ),σK = K and by what we have already seen, σ induces a KY isomorphismσ ¯ of LK onto LK , i.e. an automorphism of LK . To prove the final part of the theorem, notice that LK is a separable extension of the fixed field (KY ) of the Galois group G(LK /KY ), and is therefore simple. Hence any automorphism of LK /KY is uniquely determined by its effect on a single elementy ¯. But the working above proves that every conjugate ofy ¯ io of the form σy = σ¯ (¯y) for some σ ∈ G(L/K). Hence every automorphism of LK over KY is of the form σ¯ with σ ∈ G(L/K). Our theorem is proved. We define the inertia group T(K ) of a prime divisor K of L to be the subgroup of all elements σ of Z(K ) for whichσ ¯ is the identity automorphism of LK . It is clearly a normal subgroup of Z(K ). The theorem proved above then establishes an isomorphism G(LK /KY ) ≃ Z(K ) . T(K ) We now given some consequences of the theorems of this lecture.

1. [G : (e)] = [L : K]s = h[Z(K ) : (e)]. = = 2. [Z(K ) : T(K )] [LK : KY ]s dL/K(K )s

h = K K = K K 3. [L; K] eL/K ( ν)dL/K ( ν) heL/K ( )dL/K ( ) ν=1 P

∴ K K = [L : K] K eL/K ( )dL/K ( ) [Z( ) : (e)] [L : K]s = [L : K]i[Z(K ) : (e)]

Hence, 115

e / (K )d / (K ) 4. [Z(K ) : (e)] = L K L K [L : K]i

e / (K )d / (K ) K 5. [T(K ) : (e)] = L K L K = e (K ) dL/K ( )i L/K [L:K]i [L : K]idL/K(K )s 108 19. Lecture 19

It follows from (5) that if L is separable over K, T(K ) , 1 if and only if at least one of eL/K(K ) or dL/K (K )i is greater than one. Lecture 20

34 Divisors in an Extension

Let L/l be an arbitrary extension of K/k. We wish to imbed the group 116 of divisors of K in the group of divisors of L in such a way that the principal divisor (X)K in K of an element X of K goes into the principal divisor of the element X in L. This condition may also be rewritten in the form m m hi vY (X) e (K )vY (X) Y i K L/K ij i i −→ i j Yi=1 Yi=1 Yj=1 where Yi(i = 1,..., m) are the prime divisors of K occurring in X and Ki j( j = 1,..., hi) are all the prime divisors of L lying over Yi . This motivates the following

m vY (U ) U = Y i Deﬁnition. If i is any divisor of K, we shall identity it i=1 m hQi K U K eL/K ( ij)vY i( ) with the divisor i j in L. i=1 j=1 Q Q It is easy to see that this accomplishes an isomorphic imbedding of the group vK of divisors of K in the group vL of divisors of L which takes the principal divisor (X)K to the principal divisor (X)L. Hence we also get a homomorphism of the class group KK of K the class group KL of L. From now on, we shall use the same for a divisor or class of K or its image as a divisor or class of L. We have the following theorem comparing the degree in K of a di- 117 visor of K and its degree in L

109 110 20. Lecture 20

Theorem. There exists a positive rational number λL/K depending only on L and K such that for any divisor U of K,

dL(U ) = dK(U )/λL/K .

Proof. Obviously it suﬃces to prove that for any two prime divisors Y and U of K, we have Y U dL( ) = dL( dK(Y ) dK(U )

Assume on the contrary that we have

d (Y ) d (U L < L dK(Y ) dK(U ) d (Y ) d (Y ) i.e., L < K dL(U ) dK(U ) Then there is a positive rational m/n such that

d (Y ) m d (Y ) L < < K dL(U ) n dK(U ) It follows that for suﬃciently large integral t, we have

nt −mt dK (Y U ) = t(ndk(Y ) − mdK(U )) > 2gK − 1 nt −mt and dL(Y U ) = t(ndL(Y ) − mdL(U )) < 0

118 It follows from the ﬁrst inequality and the Riemann-Roch theorem that there exists an element X , 0 in K divisible by the divisor U mtY −nt. Hence,

mt −nt (X)K = UU Y , U integral

and dL((X)) = dL(U ) − t(ndL(Y ) − mdL(U )) > 0

by the second inequality. But this is impossible and our theorem is proved. 34. Divisors in an Extension 111

[l : k] If [L : K] < ∞, the value of λ is . For, if Y be any prime L/K L : K divisor of K and Ki(i = 1,..., h) be the prime divisors of L lying over Y , we have

h dL(Y ) = dL(Ki)eL/K(Ki) Xi=1 1 h [L : K] = d (K )d (Y )e (K ) = d (Y ). [l : k] L/K i K L/K i [l : k] L Xi=1

Now let L/l be an extension of K/k of finite degree and L1 the smallest normal extension of K containing L. Let l1 be the algebraic closure of l in L1. Clearly L1/l1 is an algebraic function field with constant field l1. Let G = G(L1/K) be the Galois group of L1 over K and H be the 119 subgroup of all automorphisms of L1 fixing every element of L. Let G/H be the set of left cosets of G modulo H. If Y is an element of L, it is well-known that the norm of Y over K is given by

[L:K]i N (Y) = σY , L/K   σ¯ Y∈G/H    the product being over any set of representatives of the cosets in G/H. This suggests the following deﬁnition for the norm of a divisor of L. (As already explained, we shall use the same symbol for a divisor of L and its canonical image as a divisor of L1). If U is a divisor of L, we put

[L:K]i U = U = U NormL/K NmL/K  σ    σY∈G¯ /H    The definition is independent of the choice of the representative σ U = U ofσ ¯ , since σ if σ ∈ H. (More generally, if L1/l1 and L2/l2 are two extension of an algebraic function field and σ an isomorphism of L1 onto L2 mapping l1 onto l2 and fixing every element of K,σ maps every divisor U of K considered as a divisor of L1 onto U considered as a divisor of L2). We list below the essential properties of the norm 112 20. Lecture 20

1. The norm of a divisor of L is a divisor of K. Hence the norm mapping is a homomorphism of ϑL into ϑK

2. If K is a prime divisor of L lying over the prime divisor Y of K, we have d (K ) NmL/KK = Y L/K

120

3. If y ∈ L, NmL/K(y)L = (NmL/Ky)K

4. If U ∈ ϑK, [L:K] NmL/KU = U

5. If L1 ⊃ L ⊃ K is a tower of extensions of algebraic function ﬁelds, U and ∈ vL1 , U = U NmL1/K NmL/K(NmL1/L )

Proof. (3) and (4) are immediate consequences of the deﬁnition of the norm. It is also clear that the norm deﬁnes a homomorphism of ϑL into itself. We have only to prove that the image is contained in ϑK to complete the demonstration of (1). But this will follow if we can prove (2).

Again, it is enough to prove (5) for prime divisors K1 of L1 because of (1). But using (2), (5) reduces to the already proved equality

K = K K dL1/K( 1) dL1/L( 1)dL/K ( )

for a prime divisor K1 of L1 which lies over K of L. It only remains to prove (2). Let again G be the Galois group of the smallest normal extension L1 of K containing L. 121 Then L1 is also normal over L and its Galois group over L is the subgroup H of G of all automorphisms which ﬁx every element of L. Let K1 be any prime divisor of L1 lying over the prime divisor K of L. 34. Divisors in an Extension 113

We shall denote by ZK and ZL the decomposition group of K1 over K and L respectively. The, since L1 is normal over L, we have K eL1/L( 1) = K  σK1 ,   σ¯ ∈YH/ZL    and therefore   e (K)[L:K] [ZL:(e)] L1/L i [ZL:(e)] = (NL/KK )  τ  σK1  τ∈ / σ∈ /   ¯ YG H  ¯ YH ZL    K   eL /( 1)[L:K]i   eL / (K1)[L:K]i  1 L   1 L = τ σK = τK   1  1 τ∈YG/H σY∈H  Yτ∈G         eL / (K1)[L:K]i[ZK :(e)]     1 L    =  τK1   τ¯∈YG/ZK  e (K )[L:K] [Z :(e)]  L1/L 1 i K  e (K) [Z :(e)]d (K ) = Y L1/K 1 = Y L L/K , since e (K )[L : K] [Z : (e)] e (K ) [L : K] [Z : (e)] L1/L 1 i K = L/L 1 . 1 i K K K eL1/K( 1) [L1, : L]i eL1/K( 1) [Z : (e)] = L K = K .dL1/K( 1) [ZL : (e)]dL/K ( ) dL1/L(K1) 122 Since the group of divisors is free, it is also torsion free and our formula follows. Finally, for any divisor U of L, we have U = U dK(NL/K ) [l : k]dL( ). It is enough to prove this for a prime divisor K . But by the above result, we have

K = Y dL/K (K ) = K Y = K dK(NL/K ) dK( ) dL/K( )dK ( ) dL( )[l : k] which is the result we want. 114 20. Lecture 20

35 Ramiﬁcation

We wish to prove two theorems on ramiﬁcation. The ﬁrst one is easy.

Theorem. If L/l is an algebraic extension of K/k such that L is purely inseparable over K, there is exactly one prime divisor K of L lying over a given prime divisor Y of K and Y = Y pt where p is the characteristic of K and t a non-negative integer.

123 Proof. Let Y be any element of L. Since L is purely inseparable over K, pn there exists an integer n such that Y0 = Y ∈ K. Then, if K be any prime divisor lying over Y , we have

n p vK (Y) = vK (Yo) = eL/K(K )vY (Yo)

and therefore the value of vK (Y) is uniquely determined by vY (Y0). Hence K is unique.

If we choose Y such that vK (Y) = 1, we deduce that eL/K(K ) divides pn, and our theorem is proved. We say that a prime divisor K of an extension L of an algebraic function field K is ramified if eL/K (K ) > 1. We have the following Theorem. If L is separably algebraic over K, there are at most a finite number of prime divisors of L which are either ramified or inseparable over K.

Proof. We give the proof in three steps. We first prove the theorem for finite normal extension, then for finite separable extension, and finally in the general case.

First assume that L is finite and normal over K. A prime divisor K of L is either ramified or inseparable only if [T(K ) : (e)] = eL/K(K )dL/K (K )i > 1. This implies that there is an automorphism σ of L in the group T(K ) which is not the identity automorphism. Since L is finite and separable over K it is a simple extension K(Z) of K. Hence σZ , Z. 1 124 Al least one of the elements Z, lies in OK , and since σ ∈ T(K ) Z 35. Ramification 115 we deduce that one of the two inequalities 1 1 vK (Z − σZ) > 0 or vK − > 0 Z σZ ! should hold. Since there are only finitely many automorphisms σ of L over K and only finitely many prime divisors K for which one of the two inequalities above can be valid, the theorem is proved in this case. If L/K were finite and separable but not normal, let L1 be the smallest normal extension of K containing L. If a prime divisor K of L is ramified or inseparable over K, the same property should also hold for any prime divisor of L1 lying over K . Hence the theorem in this case follows from the first part. Finally, suppose L/K is any separably algebraic extension It l is the constant field of L, L is evidently a finite separable extension of the composite extension Kl. Also, a prime divisor of L in ramified (inseparable) over K if and only if it is either ramified (inseparable ) over Kl or the prime divisor of Kl over which it lies is ramified (inseparable) over K. Our theorem follows from what we have proved above and the following lemma. Lemma. If L/l is an algebraic function field which is separably algebraic over K/k and the that L = Kl, there are no prime divisors of L ramified or separable over K. Proof. If K is a prime divisor of L which is ramified (inseparable) over K, find an element Y ∈ Kl such that VK (Y) = 1(Y¯ ∈ LK is inseparable 125 over KY ). Since Y is a rational combination of a finite number of elements of K and l, Y lies in a finite extension K(α1,...αn) of K, where αi are elements of l. We may also assume that L1 = K(α1,...αn) is a normal separable extension of K; for it is already separable, and we have only to adjoin to it the conjugates of the αi (which are finite in number) to make it normal. Also by our choice of Y, we see that the prime divisor K1 of L1 over which K lies is ramified (inseparable) over K.

Let T(K1) be the inertia group of K1 in L1 over K. Then, we should have K = K K [T( 1) : (e)] eL1/K ( 1)dL1/K ( 1)i > 1, 116 20. Lecture 20

and there exists an element σ ∈ T(K1) which is not the identity. But since σ ∈ T(K1), = vK1 (αν − σαν) > 0 (ν 1, 2,... n) and the αν being constants, we should have

αν = σαν

Hence σ is the identity automorphism when restricted to K and ﬁxes each one of elements α1,...,αn. Hence σ should be the identity automorphism of L1 = K(α1,...αn). This is a contradiction and our theorem is proved. Lecture 21

36 Constant Field Extensions

An extension L/l of an algebraic function field K/k is said to be a con- 126 stant field extension If L is the composite extension Kl of K and l. The following question arises. Given an algebraic function field K/k and an extension l of k, is it possible to find a constant field extension L′/l′ of K/k such that l is k-isomorphic to l′ ? This is not possible in general, since the constant field of L′ = Kl′ will in general be larger than the isomorphic image l′ of l.. More precisely, we have the following

′ Theorem. Let K/k be an algebraic function ﬁeld and l0 an extension of k. Then there exists an algebraic function ﬁeld L/l which is an extension of K/k with the following properties:

(1) there exists a subﬁeld l0 of l containing k and a k-isomorphism λ : ′ l0 → l0

(2) L = Kl0.

∗ ∗ ∗ ∗ If L /l is another extension of K/k with a subﬁeld l0 of l and a k- ∗ ∗ ′ isomorphism λ : l0 → l0 having the properties (1) and (2), there exists a ∗ ∗ K-isomorphism ρ : L → L such that the restriction of ρ to l0 coincides −1 ∗ ∗ with the map λ0 λ of l0 onto l0. l is a purely inseparable ﬁnite extension of l0. 127

Proof. Construction of a composite ﬁeld L = Kl0.

117 118 21. Lecture 21

′ ′ Let {ui } be a transcendence basis of l0 over k. Take a {ui} of independent transcendental elements ui over K in one-one correspondence ′ ′ Ω ui ↔ ui with the set {ui } and let be the algebraic closure of K({ui}). ′ Then we have an isomorphism λ of k(ui ) onto k(ui) trivial on k, defined ′ = ′ ′ by λui ui.λ can be extended to an isomorphism λ of l0 onto subfield l0 of Ω, and Ω contains the composite field Kl0 = L. Let l be the algebraic closure of l0 in L. If X is a transcendental element of K over k, X is also transcendental over l0. For if it were not, there exists a finite subset (u1,... un) of the ui such that X is algebraic over k(u1,... un). Hence there exists a relation of the form

r fo(u1,... un)X + ······ + fr(u1,... un) = 0, fi ∈ k[u1,... un],

with at least one non-constant polynomial fi. But this would imply that the set (u1,... un) is algebraically related over K, a contradiction. Also, since L = Kl0, and l0 ⊃ k,

[L : l0(X)] ≤ [K : k(X)] < ∞,

and therefore L/l is an algebraic function field with constant field l ⊃ l0. 128 The conditions (1) and (2) are evidently fulfilled. Moreover, since, X is transcendental over l, we have

[l : l0] = [l(X) : l0(X)] ≤ [L : l0(X)] < ∞

Only the second part of the theorem asserting uniqueness upto isomorphism and the last part asserting that l is purely in-separable over l0 remain to be proved. Suppose L∗/l∗ is another extension of K/k satisfying the conditions of the theorem. We have to set up an isomorphism ρ : L∗ → L such ∗ that ρ ﬁxes the elements of K and ρ restricted to l0 is the isomorphism = −1 ∗ ∗ ∗ = ∗ λ1 λ0 λ of l0 onto l0. Since any element of L Kl0 can be written in the form ∗ kili ′ ∗ ′∗ ∗ ′ ′∗ ki, k j ∈ K, li , l j ∈ l0, Pk jl j P 36. Constant Field Extensions 119 we should necessarily have

∗ k l∗ k λ (l ) i i = i 1 i ρ ′ ′∗ ′ ′∗ Pk jl j Pk jλ1(l j ) P P We make this the definition of ρ. In order to prove that it is well ∗ defined, we have to verify that if 0 has the representation kili , then ∗ = kiλ1(li ) 0. To prove that the map is an isomorphismP (and also toP prove that the denominator of the right side does not vanish when ′ ′∗ , ∗ = ∗ = k jl j 0) we have to prove that kiλ(li ) 0 ⇒ kili 0. P Thus, we have set up the requiredP isomorphismP provided we can 129 prove that ∗ = ∗ = kili 0 ⇔ kiλ1(li ) 0. X X But since these expressions involve only a finite number of elements ∗ ∗ of l0 and l0, we may assume that l0 (and consequently l0) is finitely generated over k. A simple argument shows that to prove the pure inseparability of l over l0 we may also assume that l0 is finitely generated over k. First assume that l0 is a purely transcendental extension k(u1,... un) ∗ = ∗ ∗ ∗ = −1 of k. Then l0 k(u1,... un), where ui λ1 (ui). Then u1,... un are ∗ ∗ algebraically independent over K, and so are u1,..., un. Hence there ∗ = ∗ ∗ = exists an isomorphism ρ : L K(u1,..., un) → K(u1,... un) L. In this case, the constant field l coincides with l0 = k(u1,..., un). This follows from the following more general

Lemma . If A is a ﬁeld which is algebraically closed in another ﬁeld B, and if X1,..., Xn is a set of algebraically independent elements over B, A(X1,... Xn) is algebraically closed in B(X1,..., Xn).

Proof. We may clearly assume that n = 1, X1 = X. f (X) Let α = α be any element of B(X), where α , 0 is an element 0 g(X) 0 of B and f (X) and g(X) are coprime polynomials over B with leading 130 coeﬃcients 1. If α is algebraic over A(X), we have

r ϕr(X)α + ········· + ϕo(X) = 0,ϕi(X) ∈ A[X], 120 21. Lecture 21

ϕo,...,ϕr coprime r r + + r = i.e., ϕr(X)αo f (X) ······ ϕo(X)g (X) 0.

Let ξ be any root of f (X). Substituting X = ξ in the above equation (which we may do since X is transcendental over B) we obtain

r ϕo(ξ)g (ξ) = 0

and since g(ξ) , 0, f and g being coprime,

ϕ0(ξ) = 0

and so ξ is algebraic over A. Since every root of f is algebraic over A and f (X) has leading coefficient 1, the coefficients of f are algebraic over A and hence lie in A. Similarly, g is also a polynomial over A. Substituting for X a root δ of f (X) − g(X), we get, since f (δ) = g(δ) , r + + = 0,ϕr(δ)α0 ··· ϕ0(δ) 0, hence α0 is algebraic over A and therefore in A, since not all ϕi(δ) = 0(ϕ0,...ϕr being coprime). Our lemma is proved. We are therefore left with the case when l0 is a finite algebraic ex- 131 tension of k. Then we have l0 = k(α1,...,αm). We use induction on m. The result is trivial when m = 0. Suppose the result holds for m − 1 in the place of m. Put k1 = = = ∗ = −1 k(α1,...,αm−1) and K1 Kk1 K(α1,...,αm−1). Let αi λ1 (αi), ∗ = ∗ ∗ ∗ ∗ = ∗ = ∗ ∗ ∗ k1 k (α1 ··· αm−1) and K1 Kk1 K(α1,...αm−1). Let l1 and l1 be ∗ ∗ the algebraic closures of k1 and k1 in K1 and K1 respectively. By our induction hypothesis, we have

∗ ∗ (1) an isomorphism ρ1 : K1 → K1 such that ρ1 when restricted to k1 ∗ coincides with the restriction of λ1 to k1, and ∗ ∗ (2) l1 and l1 are purely inseparable extensions of k1 and k1 respectively. = ∗ = −1 = ∗ = ∗ = Put αm α and αm λ1 (αm) α . Then, L K1(α) and L ∗ ∗ K1(α ). We would be through if we can extend the isomorphism ρ1 to ∗ ∗ ∗ an isomorphism ρ : L → L such that ρ(α ) = λ1(α ) and if we prove that the constant ﬁeld l of L is purely inseparable over l0. 36. Constant Field Extensions 121

To prove that we can extend the isomorphism ρ1 to ρ, it is necessary and sufficient to show that if F∗(X) is the irreducible polynomial of α∗ ∗ ∗ over K1, ρ1F (X) is the irreducible polynomial of α over K1. Assume that F∗(X) has leading coefficient 1. Since one of its roots is ∗ ∗ ∗ algebraic over k1, all its roots are algebraic over k1 and F (X) is therefore ffi ∗ ∗ ∗ a polynomial with coe cients in the algebraic closure, l1 of k1 in K1. ∗ ∗ Also, since l1 is purely inseparable over k1, the irreducible polynomial of 132 ∗ ∗ ∗ ∗ pt α over k1 is a certain power of F (X) of the form (F (X)) , t ≥ 0. Since ∗ = ∗ ∗ = ∗ = λ1 is an isomorphism of lo k1(α ) onto lo k1(α) with λ1(α ) α, ∗ pt ∗ pt we deduce that λ1(F (X)) = ρ1(F (X)) is the irreducible polynomial of α over k1. ∗ ∗ Again, since ρ1 maps k1 onto k1, it maps the algebraic closure l1 of ∗ ∗ k1 in K1 onto the algebraic closure l1 of k1 in K1. This proves that the irreducible equation of α over K1 (or what is the same, l1) with leading ∗ ∗ coefficient 1 is equal to ρ1F (X) since ρ1F (X) is obviously the only ∗ pt irreducible factor of ρ1(F (X)) over l1. Hence ρ1 can be extended to an isomorphism ρ having the requisite properties. To prove that l is purely inseparable over l0, notice that since l1 is purely inseparable over k1, l1(α) is purely in-separable over k1(α) = l0. It is therefore sufficient to prove that l is purely inseparable over l1(α). Now since l1 is algebraically closed in K1, the irreducible polynomial of α over K1 with leading coefficient l coincides with its irreducible polynomial over l1. Therefore we have

K1(α) : K1 = l1(α) : l1 and similarly l1(α, X) : l1(X) = l1(α) : l1.

From these two equalities and the following one 133

K1(α) : l1(X) = K1(α) : K1 K1 : l1(X)

= K1(α) :l1(α, X) l1(α, X) : l1(X) . we deduce that

K1 : l1(X) = K1(α) : l1(α, X) . 122 21. Lecture 21

Now, let β be a constant of K1(α). Then there exists an integer t ≥ 1 pt such that β is separably algebraic over l1(α). By a well-known theo- pt rem, the extension l1(α, β ) is a simple extension l1(γ) of l1. We have

pt K1(α) : l1(α, β , X) = K1(γ) : l1(γ, X) = K1 : l1(X) by an argument which is familiar to us, and using our previous equality, we get pt K1(α) : l1(α, β , X) = K1(α) : l1(α, X) pt pt pt and hence l1(α, β , X) = l1(α, X) and β ∈ l1(α, X). Since β is alge- pt braic over l1(α) which is algebraically closed in l1(α, X), β ∈ l1(α) and β is purely inseparable over l1(α). Our theorem is completely proved. We shall give an example where l , l0. Let k0 be a field of characteristic p > 0 and u and v two algebraically independent elements over k0. Let k = k0(u, v) and X a variable 134 over k. Put K = k(X, Y) where Y satisfies the equation Y p = uX p + v. We shall show that the constant field is k. If it were not, let k1 be the constant field. Since K = k(X, Y) is of degree l or p over k(X) and since k1(X) : k(X) = k1 : k > 1, we deduce that K = k1(X). Hence 1/p 1/p 1 Y = u X + v ∈ k (X). But since X is transcendental over k, (and hence also over k(u1/p, v1/p), we deduce that u1/p and v1/p are both in k1. Hence

k1 : k ≥ k(u1/p, v1/p) : k = k(u1/p, v1/p) : k(u1/p) k(u1/p) : k = p2, while on the other hand

k1 : k = k1(X) : k(X) ≤ K : k(X) ≤ which is a contradiction. 1/p Now, take l0 = k(v ). Then Kl0 clearly contains the element Y − v1/p = u1/p and hence l = k(u1/p, v1/p), l , l and l : l = p. X o o Lecture 22

37 Constant Field Extensions

We require some preliminary lemmas. 135

Lemma 1. Let A, B and C be subfields of a given field, B ⊃ A and C algebraic of finite degree n over A. Then the composite extension BC is algebraic over B of degree at most n. Moreover if y1,... yn is a basis of C over A, BC is spanned by the same set of elements y1,... yn over B. The degree of BC over B is equal to n if and only if B and C are linearly disjoint over A.

Proof. Since y1,... yn span C over A, we have in particular C = A(y1, ..., yn) and since B ⊃ A, BC = B(y1,... yn). Hence any element of BC can be written as a polynomial in y1,... yn with coeﬃcients in B (since y1,... yn are algebraic over A), and since any monomial in y1,..., yn can be written as a linear combination of y1,..., yn with coeﬃcients in A, we deduce that BC is the vector space spanned by y1,.....yn over B. Hence [BC : B] ≤ n.

If [BC : B] = n, y1,... yn should also be linearly independent over B, and since (y1,... yn) is an arbitrary set of n elements of C linearly independent over A, B and C are linearly disjoint over A. The converse is also evident.

Lemma 2. (a) Let B be any purely transcendental extension of A and C any ﬁeld containing A and algebraically disjoint with B over A. 136 Then C and B are linearly disjoint over A.

123 124 22. Lecture 22

(b) Let A be algebraically closed in B and C = A(α) a simple algebraic extension of A. Then B and C are linearly disjoint over A.

Proof. (a) Let B = A(uλ)λ∈A where uλ is a set of algebraically independent elements over A. If B and C are not linearly disjoint, there is a set of elements c1,... cn ∈ C which are linearly independent over A and polynomials fi(u1,... um), ui, ∈ {ui}λ∈A not all of which vanish identically such that

c1 f1(u1, ··· um) + ··· + cn fn(u1,... um) = 0, fi ∈ A[u1,..., um]

Since u1,... um are algebraically independent over A, they are algebraically independent over C also. We may therefore equate to zero separately the coeﬃcients of each of the monomial expressions in u1,... um occurring in the left hand side of the above equation. At least one of these provides a non-trivial linear combination of the ci with coeﬃcients in A which vanishes. This is a contradiction.

(b) Since A is algebraically closed in B, the irreducible monic polynomial of α over B is actually a polynomial over A, as we have proved earlier. Hence [B(α) : B] = [A(α) : A], and our result follows from Lemma 1.

Lemma 3. Let A, B, C, D be subﬁelds of a given ﬁeld such that B ⊃ 137 A, D ⊃ C ⊃ A. Then B and D are linearly disjoint over A if and only if (i)B and C are linearly disjoint and (ii) D and the composite extension BC are linearly disjoint over C.

Proof. Suppose ﬁrst that B and D are linearly disjoint. Then (i) is evidently fulﬁlled. To prove (ii), suppose di(i = 1,... n) is a set of elements of D linearly independent over C. If they are linearly dependent over BC, there exists a relation of the form

di ci jb j = 0 , ci j ∈ C, b j ∈ B, Xi Xj 37. Constant Field Extensions 125

with b j linearly independent over C and not all ci j being zero. On inter- changing the orders of summation, we get

b j dici j = 0, Xj Xj and we deduce from our hypothesis that

ci jdi = 0 for all j. Xi

But since di are linearly independent over C, we have ci j = 0 for all i and j. This is a contradiction. Suppose conversely that (i) and (ii) are fulﬁlled. Then any set of elements of B linearly independent over A are, by (i), linearly independent over C, and (since they are also elements of BC) by (ii), linearly independent over D. Our lemma is proved. We can now prove the following

Theorem. Let L = Kl0 be a constant ﬁeld extension of K with the ﬁeld 138 of constants l ⊃ l0. Then the following conditions are equivalent (A) K and l are linearly disjoint over k.

′ 1 = 1 (B) For every finitely generated subfield l0 of l0 over k, and L Kl0, 1 1 the constant field of L coincides with 10. If these are fulfilled, (B) holds for any (not necessarily finitely gen- 1 = erated) subfield l0 of l0, in particular for l0 itself, i,e., l l0. 1 Proof. We shall first show that (A) implies (B) for any subfield l0 of l0. 1 1 = 1 1 Let l be the constant field of L Kl0. It follows from (A) that l and K are linearly disjoint over k. Let X be any transcendental element of K over k. Then by Lemma 3, K and 11 are linearly disjoint over k if and only if (i) k(X) and l1 are linearly disjoint over k and (ii) K and l1k(X) = l1(X) are linearly disjoint over k(X). 126 22. Lecture 22

By lemma 2, since k(X) are 11 are algebraically disjoint (i) is always 1 1 1 = 1 satisﬁed. By lemma 1, since 10(X) ⊂ 1 (X) and L K10(X), we have the inequalities

1 1 1 1 L : l (X) ≤ L : lo(X) ≤ K : k(X) .

Again by lemma 1, since L1 = Kl1(X), the equality

L1 : l1(X) = K : k(X) follows from the linear disjointness of K and l1(X) over k(X). 1 = 1 139 From these we deduce that if (A) holds, l (X) l0(X), and since X 1 1 = 1 is transcendental over l , l l0. Conversely suppose (B) holds for every finitely generated subfield 1 l0 of l0. To prove that 1 and K are linearly disjoint, it is enough to prove that any finitely generated subfield of 1 over k is linearly disjoint with K over k. But clearly a finitely generated subfield of l is contained 1 = 1 1 = 1 1 in the constant field l l0 of L Kl0, where l0 is a suitable finitely generated extension of k. It is therefore enough to prove that any finitely 1 generated subfield l0 of l0 over k is linearly disjoint with K over k. 37. Constant Field Extensions 127

1 = = = Let l0 k(α1,...αm). Put ki k(α1,...αi), and li Kki. To show 1 that l0 and K are linearly disjoint over k, it is enough to show that k1 and K are linearly disjoint over k, k2 and L1 = KK1 are linearly disjoint over 1 = = k1, etc., and ﬁnally l0 km and Lm−1 KKkm−1 are linearly disjoint over km−1 (Lemma 3). But by (B) each ki is algebraically closed in Li and since ki+1 is a simple extension of ki, our result follows from Lemma 2.

Corollary. If either K or l0 is separably generated over k, then l = l0.

Proof. By the above theorem, we may assume that l0 is ﬁnitely generated over k. Morover, since we have already seen that l = l0 for a purely 140 transcendental extension l0 of k (see Lecture 21), we may assume that l0 is ﬁnitely algebraic over k.

Suppose now that l0 is separably algebraic over k. Then L = Kl0 is separably algebraic over K. But if α ∈ l, the irreducible monic polynomial of α over K lies in k and is therefore separable over k. Hence l is separable over l0, and is also purely inseparable l = l0. Suppose next that K is separably generated over k. Let X ∈ K be transcendental over k and such that K/k(X) is separable. Then L = Kl0(X) is separable over l0(X). Hence any element of l is separably algebraic over l0(X), and since l0 is algebraically closed in l0(X), l is separable over l0. The result follows as before. Our next theorem runs as follows. 128 22. Lecture 22

Theorem. Let L = l0K be a constant ﬁeld extension and λL/K the rational number satisfying

λL/KdL(U ) = dK(U )

for any divisor U of K. Then λL/K is a power of the characteristic with non-negative exponent (λL/K = 1 if the characteristic is zero). It is equal to one if and only if K and l are linearly disjoint over k.

Proof. Let X be a transcendental element of K. Choose U to be the numerator of (X). Then, we see that

dK (U ) = [K : k(X)] and dL(U ) = [L : l(X)]

141 Hence,

λL/K = 1 ⇔ dL(U ) = dK(U ) ⇔ [L : l(X)] = [K : k(X)].

But as we have already seen (see proof of the ﬁrst theorem of this lecture) [L : l(X)] = [K : k(X)] if and only if K and l are linearly disjoint over k.

In particular, if the characteristic is zero, K is separably generated over k, and by the corollary of the first theorem, the condition (B) of our first theorem is satisfied. Hence (A) holds, i.e., K and l are linearly disjoint over k. Hence λL/K = 1. If the characteristic p > 0, let K0 be the largest separable extension of k(X) contained in K, and L0 = K0l0. Then as above, K0 and l0 are linearly disjoint over k. Hence we obtain

K0 : k(X) = L0 : l0(X) s Also, K/K0 is a purely inseparable extension of degree p , s ≥ 0 and s therefore so is L = Kl0 inseparable of degree p , where s0 ≤ s (lemma 1). Therefore we have

K : k(X) K : k(X) l(X) : l0(X) λL/K = = L : l(X) L : l0(X) 37. Constant Field Extensions 129

K : K K : k(X) 0 0 s−s0 = l : l0 = p l : lo L : L0 L0 : l0(X) 142 Thus, we see that λL/K is a power of p divisible by l : l0 . Our theorem is proved. In particular, we see that if K or l0 is separably generated over k, K and l are linearly disjoint over k and therefore λL/K = 1. The last theorem of this section relates to the residue class field of a prime divisor in a constant filed extension. Theorem . Let L = Kl, where I is separably generated over k. If K is a prime divisor of L lying over the prime divisor Y of K, LR is the composite of the two subfields KY and l. Proof. It is clearly sufficient to prove the theorem when (1) l is purely transcendental over k and (2) when l is separably algebraic over k.

Case 1. Since KY is algebraic over K, l and KY are linearly disjoint over k. Let us agree to denote the image in LK of any element Y in the valuation ring OK by Y.¯ AnyY ∈ OK can be written in the form n kili i=1 1 1 mP , ki, k j ∈ K, li, l j ∈ 1, 1 1 k j l j j=1 P 1 and the two sets of elements (li) and (l j ) being linearly independent over n k. Find elements a, b of K such that vY (a) = min vY (ki) and vY (b) = i=1 m 1 1 1 − min vY (k ). Then clearly for at least one j vY (k b) = 0, k , 0. The j=1 j j j aY image L of the element is then 143 R b

(kia)li P 1 1 (k j b)l j P Since the li are linearly independent over k, they are also linearly independent KY and therefore the numerator does not vanish. Similarly the 130 22. Lecture 22

aY denominator does not vanish, and therefore the image of in LK is a b b non-zero element of KY l. Since Y ∈ OK , it follows that ∈ OY and a b¯ aY¯ hence Y¯ = ∈ KY l. a¯ ! b¯ !

Since KY and l are linearly disjoint, the structure of LK is uniquely determined; in fact LK is purely transcendental over K, and a transcendence basis of l over k is also a transcendence basis of LK over KY . Since the map Y → Y¯ is also uniquely fixed, we see that there is exactly one prime divisor K of L lying over the prime divisor Y of K. Case 2. In this case we may not only assume that l is separably algebraic over k, but also that it is finite. In fact, any element α of LK is clearly the image by the place of K of an element of Kl1, where l1 is a finite extension of k. If we have proved the theorem for finite separa- 1 ble extensions, it would follow that α ∈ l KY ⊂ lKY and we would be through. Suppose l is separably algebraic and of finite degree over k. Then it is simple and we have l = k(α), where α is separably algebraic over k of 144 degree n, say. Let K = K1, K2,..., Kn be all the prime divisors of L lying over Y . Let L1 be the smallest normal extension of K containing 1 1 L, and K a prime divisor of L lying over K . Let σi(i = 1,..., m) be 1 1 all the automorphisms of L over K. Then since σiK again lies over the prime divisor of K, its restriction to L is one of the K j. Let Z¯ ∈ LK . Find an element C ∈ L such that

vK (C − Z) > 0,

and vK j(C) ≥ 0, ( j = 2,... h)

By the ﬁrst condition , C¯ = Z¯ ∈ LK . Since C ∈ K(α), it can be written uniquely in the form

n−1 C = ao + a1α + ······ + an−1α , ai ∈ K (the degree of α over K being the same as over k, according to a previous statement). 37. Constant Field Extensions 131

Taking conjugates in the above equation over K, we obtain a set of n equations (i) (i) (i)n−1 C = a0 + a1α + ······ + an−1α Since α is separable of degree n over K, the determinant |α(i) j|(i = 1,..., n; j = o,..., n − 1) has a non-zero value, and we may solve the above equations for ak to obtain (1) (1)k−2 (1)n−1 1α ··· α C1 ··· α (n) (n)k−2 n (n)n−1 1α ··· α C ··· α ak = (1) (1)n− 1 1 α ····················· α (n) (n)n−1 1α ····················· α 145

The denominator is a constant of the ﬁled L1. The numerator is a linear combination of the C(i) with constant coeﬃcients. But we have (i) −1 1 vK 1 (C ) = v −1 ′ (C) = e 1 (σ K )vK (C) ≥ 0 σν K L /L ν j 1 (i) where σν is an automorphism of L /K taking C to C and K j is the −1 1 prime divisor of L lying below σnu K . We may therefore conclude that

vK 1 (ak) ≥ 0, vK (ak) ≥ 0

This means that the ak are in OK and therefore n−1 Z¯ = C¯ = a¯o + a¯1α + ······ + an−1α ∈ lKY . Our theorem is proved. From the proof of the theorem when l0 is purely transcendental, the following fact emerges. If K is a prime divisor of L = Kl0, l0 being purely transcendental, and K lies over a prime divisor Y of K, we have n n vK liai = min(vY (ai))   i=1 X1    if li ∈ l0 and ai ∈ K. This follows in fact from the equation 146 n n liai = lia¯i, if ai ∈ OY . Xi=1 Xi=1

Lecture 23

38 Genus of a Constant Field Extension

The notations will be the same as in the previous lecture; in addition, 147 we shall denote by gR the genus of a function ﬁeld R and by FR(U ) the vector space over the constant ﬁeld of R of elements divisible by the divisor U . The dimension of FR(U ) will be denoted by NR(U ).ΛR(U ) shall denote the vector space of repartitions of R divisible by the divisor U .

Theorem 1. If λL/K = 1, i.e., if K and l are linearly disjoint over k, gL ≤ gK. For any divisor U of K, a base of FK(U ) is a part of a base of FL(U ), and hence NK(U ) ≤ NL(U ).

Proof. The last part immediately follows from the fact that FK(U ) ⊂ FL(U ), if we observe that K and l are linearly disjoint over k and that l = l0. Taking a divisor U of K such that

−dK(U ) > 2gK − 2, −dL(U ) > 2gL − 2, we have

NK(U ) = dK (U ) − gK + 1,

NL(U ) = dL(U ) − gL + 1, and it follows from the previous inequality that gL ≤ gK.

Theorem 2. If l0 is separably generated over k, gK = gL and a basis of FK(U ) is also a base of FL(U ) for any divisor U of K; hence NL(U ) = NK(U ).

133 134 23. Lecture 23

Proof. We ﬁrst consider the case l0 = k(u), u transcendental over k. Let 148 Z ∈ FL(U ). Then Z can be written uniquely in the from

n ν aνu F(u) ν=o Z = = P , aν, bµ ∈ K G(u) m−1 m µ u + bµu µ=o P with F and G coprime. We shall show that bµ ∈ K. Let K be a prime divisor of L lying over a prime divisor Y of K. Then by the remark at the end of the previous lecture,

m m−1 vK (G(u)) = vK (u + bm−1u + ··· + bm) = min(0, vY (b1),..., vY (bm)) ≤ 0

Hence the only possible prime divisors which occur in the numerator zG(U) of G(u) are those which over K. But now, since Z ∈ FL(U ), the divisor z N (Z)U −1 = F G NFzGU

is integral, and since zG and NG are coprime, any prime divisor occurring in zG must divide zF. But since F and G are coprime, there exist polynomials F1(u) and G1(u) with

F(u)F1(u) + G(u)G1(u) = 1.

t If F1(u) = co + c1u + ··· + ctu and U a prime divisor of L variable over K, we would have

vU (F1(u)) ≥ min(vU (cν) + γvU (u)) = 0 ν

149 since vU (u) = vU (VU (cν)) = 0. Similarly vU (G1(u)) ≥ 0. Hence we have

0 = v(1) ≥ min(vU (F(u)) + vU (F1(u)), vU (G(u)) + vU (G(u)) ≥ min(vU ((F(u))), vU (g(u))),

and therefore zG(u) and zF(u) can not have a common prime divisor. Therefore zG(u) = N and G(u) is constant. 38. Genus of a Constant Field Extension 135

It follows that for any prime prime divisor Y of K,

vY (Z) = vY (F(u)) = min vY (aν) ≥ vY (U ) ν and therefore aν ∈ FK(U ). Thus, we see that FL(U ) is the vector space generated over l = l0 by FK(U ). from the liner disjointness of K and l, we deduce that NK(U ) = NL(U ). Next suppose l = l0 = k(α) is ﬁnite separable (and therefore simple) over k. Any element Z ∈ FL(U ) can be written uniquely in the form

n−1 Z = c0 + c1α + α + cn−1α , ci ∈ K where n is the degree of α over k or K. Let L1 be the smallest normal extension of L over K. Taking conjugates in the above equation over K, we obtain

(i) (i) (i)n−1 Z = c0 + c1α + ··· + cn−1α (i = 1,... n)

We may solve for ck of obtain 150

− − 1α(1) ··· α(1)k 2 Z(1) ··· α(1)k ··· α(1)n 1 − − 1α(n) ··· α(n)k 2 Zn ··· α(n)k ··· α(n)n 1 = ak n− 1 1 α(1) ································· α(1) − 1α(n) ································· α(n)n 1

The denominator is a constant , 0 and the numerator is a linear combination of Z(i) with constant coefficients. Since U is a divisor of K, it may be easily verified that every conjugate Z(i) of Z is divisible by U in L1. Hence the ck are divisible by U in L1 and hence in K. We have proved in this case also that FL(U ) is generated by FK(U ) over i. The equality NL(U ) = NK(U ) again follows from the liner disjointness of K and l = lo over k. The case of any separably generated extension l0 now follows along 1 1 familiar lines. Any Z ∈ FL(U ) is contained in a field L = Kl , where l1 is a finitely separably generated extension of k; hence it is an element of FL1 (U ). But in the case of a finitely separably generated extension, 136 23. Lecture 23

the theorem follows by induction if we use the ﬁrst two cases. Hence we again obtain the fact that FL(U ) is generated by FK(U ) over l, and the equality NL(U ) = NK(U ). Choosing a divisor U with −dk(U ) > 2gK − 2, −dL(U ) > 2gL − 2

we have − dK(U ) = NK(U ) − gK + 1, 151 and − dL(U ) = NL(U ) − gL + 1.

But since λL/K = 1, it follows that dK(U ) = dL(U ) and therefore gL = gK. The theorem is completely proved.

Theorem 3. For any constant ﬁeld extension L of K, we have gLλL/K ≤ gK. (In particular gL ≤ gK). Proof. Since the genus is preserved for for a purely transcendental extension of the constant ﬁeld, and since λL/K = 1for such an extension, it = 1 follows from the obvious formula λL/L1 λL1/K λL/K L ⊃ L ⊃ K that it is enough to prove the theorem for algebraic extension l0 of k.

First assume that l0 is a ﬁnite extension of k with a basis α1,...,αn over k. By lemma 1 of the previous lecture, L/K is a ﬁnite extension of

degree no ≤ n, and we may assume that α1,...,αn0 form a basis of L over K. Let us denote by XK and XL the vector spaces of repartitions of K and L over k and l respectively. We define a map σ of the direct product no XK of XK by itself no times into the space XL by defining the image ν=1 Q n0 C C X X C of ( 1,..., n0 ) ∈ K in L to be the repartition of L defined by ν=1 Q no C (K ) = ανCν(Y ) Xν=1

152 for any prime divisor K of L, where Y is the prime divisor of K lying below K . It is easy to verify that C so deﬁned is a repartition, and that n0 σ is a k-isomorphism of XK into XL. Let the image under σ be ν=1 X o X XQo X the subspace L of L L is a vector subspace of L, if the latter is considered as a vector space over k. 38. Genus of a Constant Field Extension 137

n = 0 If y αν xν, xν ∈ K, is any element of L, the element (x1,..., xn0 ) 1 n0 P of XK (keeping in mind that we have identiﬁed K(L) with a subspace ν=1 of XQK(XL) and we may use the same symbol for an element of K(L) and the corresponding repartition in XK(XL)) clearly goes to the repartition X X 0 U y of L. Hence, L ⊃ L. We assert that for any divisor of K, we have X = X 0 +Λ U L L L( )

To prove this, consider a repartition C of L. Let K1,..., Kr be the prime divisors of L lying over a prime divisor Y of K. We can ﬁnd an element y(Y ) of L satisfying

Y C K U = vK ν (y( ) − ( ν)) ≥ vK ν ( ), (ν 1,... r)

Deﬁne a repartition Y of L by

Y (K ) = y(Y ) if K lies over Y and vK 1 (U ) < 0 or if K lies 1 over Y and vK (C ) , 0 for some prime divisor K lying over Y , Y (K ) = 0 otherwise. C Y Λ U Y X 0 Y = Clearly, − L( ). We shall show that ε L . Let y( ) 153 no 1 1 1 = ανyν(Y ), yν(Y ) ∈ K. Deﬁne repartitions Yν of K(ν 1,... n0) by ν=1 puttingP 1 = 1 Yν (Y ) y (Y ) if vY (U ) , 0 or vK (C ) < 0 for some K lying 1 = over Y , Yν (Y ) 0 otherwise. σ Y 1,..., Y 1 = Y ∈ X 0 We then have (( 1 no )) L . Our assertion in proved. Now, if N denotes the unit divisor, we have

Xk dimk = gK ΛK(N ) + K XL and diml = gL ΛL(N ) + L 138 23. Lecture 23

From the ﬁrst equation,

n0 XK ν=1 n0gK = dimk Q n0 n0 ΛK(N ) + K ν=1 ν=1 Q Q and applying σ X 0 n g = dim L , 0 K K Λ0 N + L( ) L

n0 Λ0 U Λ U X U 154 where L( ) is the k-subspace σ( K( )) of L for any divisor ν=1 of K. Q On the other hand,

XL XL ngL = n diml = dimk ΛL(N ) + L ΛL(N ) + L o o X +ΛL(N ) + L X = dim L = dim L k Λ N + k X o Λ N + L( ) L L ∩ ( L( ) L) 0 0 X X ∩ (ΛL(N ) + L) = dim L − dim L k Λo N + k Λo N + L( ) L L( ) L Λ0 N X 0 Λ N + ( L( ) is obviously a subspace of L ∩ ( L( ) L)). Hence we deduce that

ngL ≤ n0gK n g ≤ 0 g L n K But if X is any element of K transcendental over k, we obtain

dK(NX) K : k(X) K : k(X) λL/K = = = . l(X) : k(X) dL(NX) L : l(X) L : k(X) = l : k = n L : k no 155 and our result is proved in the case of a algebraic extension l0 of k. 38. Genus of a Constant Field Extension 139

To prove the theorem in the case of an arbitrary algebraic extension, 1 we shall show that there exists a ﬁnite extension l0 of k such that for 1 = 1 = L Kl0, we have λL/L1 1. It would then follow from theorem 1 that gL ≤ gL1 and our result would follow. Let X ∈ K be transcendental over k and NX the denominator of X. We have

m = dK (NX) = K : k(X) ,

m◦ = dL(NX) = L : l(X) . A base x1,...... xm of K/k(X) spans L over l(X), and hence we have m − m◦ relations m xνCνµ = 0, µ = 1, 2,...... , m − m◦ Xν=1 with coeﬃcients Cνµ in l(X) such that the m − m◦ vectors

(C1µ,...... Cmµ) (µ = 1,...... , m − m◦) are linearly independent over l(X). The rational functions Cνµ of X over ffi 1 l have coe cients in a finitely generated subfield l0 ⊃ k of l. Since 1 = 1 1 L Kl0 is spanned by x1,... xm over l0(X) and since the Cνµ are in 1 l0(X), we deduce that

N 1 1 = N dL1 ( X) ≤ L : lX(X) ≤ m0 dL( X), = and since we already have λL/L1 ≥ 1, we deduce that λL/L1 1. 156 Our theorem is completely proved.

Remark . If λL/K > 2, we can actually assert that λL/K gL < gK. For = ﬀ suppose λL/K gL gK . Let ω be a non-zero di erential of K. Then, we have dK ((ω)) 2gk − 2 dL((ω)) = = , λL/K λL/K and hence (since dL((ω)) is an integer) λL/K divides 2gK −2. But from the equation λL/KgL = gK, we deduce that λL/K divides gK and hence 2gK. 140 23. Lecture 23

This implies that λL/K divides 2, λL/K ≤ 2, which is a contradiction. Hence we have the strict inequality.

If however λL/K = 2, we may have 2gL = gK as the following example follows. Let k be a ﬁeld of characteristic 2 and α◦,α1 two elements of k 1 1 2 2 = such that k α0 ,α1 : k 4. Then it can be seen easily that if X is 2 2 a transcendental element over k, the polynomial Y − (α1 + α1X ) is an irreducible polynomial of Y over k(X). Hence, if Y is a root of the 2 2 equation Y = α0 + α1X , [k(X, Y) : k(X)] = 2. Put K = k(X, Y). It can be proved (see the example for l , l0 given in Lecture 21) that the constant ﬁelds of k(X, Y) is k. Now, it can be deduced by taking valuations in the equation Y2 = 2 2 n α0 +α1X that NY = NX. Hence the elements 1, X, X ,... X , Y, YX,..., n−1 N −n 157 YX are all elements of K divisible by X . Since they are linearly N −n + independent, we have l( X )] ≥ 2n 1, and the Riemann-Roch theorem gives gK = 0. 1 1 2 2 Now, let l0 be any extension of k such that l0 α0 ,α1 : l0 < 4, = = = and L Kl0. Since gLλL/K ≤ gK 0, we necessarily have gL 0. We 1 1 = 2 2 ≤ shall show that λL/K 2. In fact, since l0 α0 ,α1 : l0 2, there is a relation of the form 1 1 2 + 2 = βα◦ γα1 δ, β, γ, δ ∈ l0, not all zero. 1 1 2 2 We may solve for α0 and α1 from this and the equation

1 1 2 + 2 = α0 Xα1 Y,

1 1 1 1 , 2 2 = 2 2 since βX − γ 0, thus proving that α0 ,α1 ∈ l0(X, Y), l l0 α◦ ,α1 . Hence, L = l(X) and dL(NX) = 1, Since dK(NX) = 2, λL/K = 2. One can in fact show that the above example covers the general case = when gL gK and λL/K > 1. To prove this, we ﬁrst observe that we must have gL = gK = 0, for otherwise, we would obtain

gL < λL/KgL ≤ gK. 38. Genus of a Constant Field Extension 141

If now, W were the canonical class of K, d(W−1) = 2 and N(W−1) = 3. Hence there exists an integral divisor U in the class W−1 of degree 2, −1 −1 and NK(U ) = 3. Let 1, X, Y be a basis of FK(U ). Then X is not a 158 constant, and since NX divides U and d(U ) = 2, we see that U = NX. Hence,

K : k(X) = d(NX) = 2. Now, we assert that Y < k(X). For if it were, we can write f1(X) Y = , f1 and f2 being coprime polynomials. Then, f2(X) z f = 1 N deg f2−deg f1 N (Y) X , and since (Y) X is integral, we deduce that z f2 f2 is constant and deg f1 = 1. This contradicts our assumption that 1, X, Y are linearly independent. Hence, k(X, Y) , k(X), and since

K : k(X) = 2, K = k(X, Y). Also, since λL/K > 1, Y should be purely inseparable over k(X), and therefore satisfy an equation of the form

Y2 = R(X),

2 N −2 R(X) being a rational function of X. Since Y is divisible by X , we deduce by an argument similar to the one used above that R(X) is a polynomial of degree at most two. Thus,

2 2 Y = α◦ + α1X + α2X

Since X should also be purely inseparable over k(Y), we deduce that α1 = 0. 1 1 2 2 Now, if k α◦ ,α2 : k were not equal to 4, it is less than or equal to 2. Hence we have a relation of the form

1 1 2 + 2 = βα◦ γα2 δ, β, γ, δ ∈ k.

This together with the relation 159

1 1 2 + 2 = α◦ α2 X Y 142 23. Lecture 23

1 1 2 2 proves that α◦ ,α2 , are both in K and hence in k. This would imply that Y ∈ K(X), which if false. Hence,

1 1 2 2 = k α◦ ,α2 : k 4. 1 1 2 2 = 1 Finally, suppose l0 α0 ,α2 : l0 4. Then for any subﬁeld l0 of l0 1 1 1 2 2 1 = containing k, we have l0 α0 ,α2 : l0 4. Hence the constant ﬁeld of 1 1 = Kl0 is l0. This implies that (see Lecture 22) λL/K 1, a contradiction. Our assertion is proved. If gL = gK > 0, we deduce from the equation = = gL λL/K gL gK = that λL/K 1. We now prove the following = = U Theorem . If gL gK and λL/K 1, then for any divisor of K a basis of FK(U ) over k is also a basis of FL(U ) over l; in particular NL(U ) = NK(U ).

Proof. Let us denote by lFK(U ) the vector space generated over l by U U U = FK( ) in L. Clearly we have lFK( ) ⊆ FL( ). Since λL/K 1, l and K are linearly disjoint over k and we obtain

NK(U ) = diml lFK(U ) ≤ diml FL(U ) = NL(U )

160 Now, let U be any divisor with dK (U ) = dL(U ) < 2 − 2gK. Then we have

NK(U ) + dK(U ) = 1 − gK,

NL(U ) + dL(U ) = 1 − gL,

and since dK(U ) = dL(U ) and gK = gL, we obtain

NK(U ) = NL(U ),

and lFK(U ) = FL(U ). 38. Genus of a Constant Field Extension 143

To draw the same conclusion for an arbitrary divisor U , choose two divisors δ and L of K such that (i) the least common multiple of δ and L is U and (ii)d(δ) < 2 − 2gK , d(L) < 2 − 2gL. This is clearly possible. We then have FK(δ) ∩ FK(L) = FK(U ), FL(δ) ∩ FL(L) = FL(U ). Let α1,...,αm be a basis of FK(U ). Complete this to a basis β1,..., β1,α1,...,αm of FK(δ) and to a basis γ1,..., γn,α1,...,αm of FK(L). We assert that α1,...,αm, β1,...,β1, γ1,...,γn are linearly independent elements of K over k. In fact, if we had a linear relation

aiαi + b jβ j = ckγk, ai, b j, ck ∈ k, X X X since the left side is an element of FK(δ) and the right side an element of 161 FK(L), ckγk is an element of FK(U ) and therefore ck = 0, ai = 0 and = b j 0.P Hence β1,...,βl,α1,...,αm, γ1,...,γn is also a set of linearly independent elements over l. Now suppose y is an element of FL(U ) = FL(δ) ∩ FL(L). Since FL(δ) has for basis (α1,...,αm, β1,...βl) over 1 and y ∈ FL(δ), we have

y = aiαi + b jβ j, ai, b j ∈ l, Xi Xj and similarly, since y ∈ FL(L) and FL(L) has for basis (α1,...,αm, γ1, ...,γn), we have

y = c jα j + dkγk, c j, dk ∈ l. Xj XK

Equating the above two expressions for y, we obtain (since α1,..., αm, β1, ..., β1, γ1,...,γn are linearly independent over l)

ai = ci , b j = dk = 0.

Thus y ∈ lFK(U ) and hence we have FL(U ) = lFK(U ). Again by linear disjointness of l and K over k, we obtain

NK(U ) = NL(U )

Our theorem is proved. 144 23. Lecture 23

The converse of the above theorem is very easy to prove. If NK(U ) = NL(U ) for all divisors U , or even only for a sequence of 162 divisors U with dk(U ) → −∞, we have the following equations for −dk(U ) suﬃciently large

NK(U ) + dK(U ) = 1 − gK,

NL(U ) + dL(U ) = 1 − gL.

Hence we obtain

dK(U ) −NK(U ) + 1 − gK λL/K = = , dL(U ) −NL(U ) + 1 − gL

and letting dK(U ) → −∞, and observing that the right hand side has = U = U U limit 1, we obtain λL/K 1. Hence dK( ) dL( ) for any divisor , and we obtain gK = gL.

Corollary. If gL = gK and λL/K = 1, the natural homomorphism of the class group KK of K into the class group KL of L is an isomorphism. Under this isomorphism, the canonical class of K goes to the canonical class of L.

Proof. Let U be any divisor of K which is a principal divisor in L. Then dL(U ) = 0 and NL(U ) = 1. By the above theorem, dK(U ) = 0 and NK(U ) = 1. This proves that U is a principal divisor of K, and thus the kernel of the homomorphism of KK consists of the unit class alone. Thus, the map is an isomorphism. We shall use the same symbol for a class of K and its image as a class of L.

163 Also, if WK is the canonical class of K, dL(WK) = dK (WK) = 2gL −2 and NL(WK) = NK(WK) = gL, which proves that WK is the canonical class of L.

39 The Zeta Function of an Extension

Let K/k be an algebraic function field with a finite field of constants k containing q elements. Let k f be the extension of k of degree f . Since 39. The Zeta Function of an Extension 145

k is perfect, k f /k is a separable extension, and hence the constant ﬁeld of L f = Kk f is k f . Let Y be a prime divisor of K and K1,..., Kh the prime divisors of L f lying over Y . Then, since k f /k is separable, the Ki are unramiﬁed over K, and we have

Y = K1 ...... Kh. = Now, since k f /k is separable, we know that LKi kY k f , and since the degrees of KY and k f over k are respectively dK (Y ) and f , the Y degree of LKi over k is l.c.m. [dK ( ), f ]. Thus, f d (Y K = K f dL f ( i) ( f, dK (Y )) But we know that

h = = = K = K f k f : k L f : K dL f /K ( i) hdL f /K ( 1), h i h i Xi=1 and using the relation

dL/K(K1)d(Y ) = dL(K )[l; k], we deduce the formula 164 h = ( f, dK (Y )) An immediate consequence of the above formula is the following

Theorem . For algebraic function fields with finite constant fields, the least positive value of the degree of its divisors is 1.

Proof. Let ρ be this least value. Take f = ρ in the above formula.

We obtain, since ρ divides each dK(Y ),

h = ρ d (Y ) and also d (K ) = K . L f i ρ 146 23. Lecture 23

Substituting in the Euler product for ζ(s, L f ), we obtain h K −1 −1 −s f dL ( ) −sdK (Y ) ζ(s, L f ) = 1 − q f = 1 − q YK YY Yi=1 = (ζ(s, K))h = (ζ(s, K))ρ

Since both ζ(s, L f ) and ζ(s, K) have a pole of order 1 at s = 1, we deduce that ρ = 1. Finally, we prove a theorem expressing the zeta function of a finite constant field extension in term of the L-series of the ground field. Theorem. With the same notation as above, we have f ζ(s, L f ) = L(s, χ f,ν, K) Yν=1

2πiν 165 where χ f,ν is the character on the class group taking the value e f on all classes of degree 1. Proof.

−1 −s f d (K ) −1 = K −s = L f ζ(s, L f ) 1 − NL f 1 − q YK YY KY/Y −( f,dK (Y )) −s f = (1 − N Y ( f,dK (Y ))  K  YY   f   2πiν  −1 dK (Y ) −s = 1 − e f NKY YY Yν=1 r 2πiν s (the last follows from the easily established formula 1 − e r z = ν=1 r (r,s) Q 1 − z (r,s) for positive integral r, s) f f 2πiν = ζ s − , K = L(s, χ f,ν, K) f log q ! Yν=1 Yν=1 The proof of the theorem is complete. Bibliography

[1] Artin, E. - Algebraic numbers and algebraic functions. Princeton 1950-51.

[2] Artin, E. - Quadratische korper im gebiet der hoheren kongruenzen. I and II. Math. Zeit. 19 (1924) p. 153-206, 207-246.

[3] Chevalley, C. - Introduction to the theory of algebraic functions of one variable. Mathematical Surveys. Number VI. Amer. Math. Soc. 1951.

[4] Dedekind, R. & Weber, H. - Theorie der algebraischen funktionen einer veranderlichen J. reine angew. math. 92(1882) p. 181-290.

[5] Hensel, K. & Landsberg, G. - Theorie der algebraischen funktionen einer variablen und ihre anwendung auf algebraische kurven und Abelsche Integrale. Leipzing. 1902.

[6] Kronecker, L. - Grundzuge einer arithmetischen theorie der algebraischen grossen. Werke Band 2.

[7] Riemann, B. - Theorie der Abelschen functionen. J. Reine angew. Math. 54 (1857). Werke Zweite Auﬂage Erste Abtheilung. VI p. 88-142.

[8] Roch, G. - Uber¨ die Anzahl der wilkurlicher Konstanten in algebraischen Funktionen. J. reine angew. Math. 64 (1865) p. 372-376.

147 148 BIBLIOGRAPHY

[9] Schmid, H. L. & Teichmuller, O. - Ein neuer bewies fur die Funk- tion algeichung der L-reihen. Abh. Math. sem. Hans. univer. 15 (1943) p. 85-96.

[10] Schmidt, F. K. - Analytische zahlentheorie in korpern der charak- teristik p. Math. Zeit. 33 (1931) p. 1-32.

[11] Schidt, F.K. - zur arithmetischen theorie algebrische funktionen I. Math. Zeit. 41 (1936) p. 415-438.

[12] Tate, J. - Genus change in inseparable extensions of function ﬁelds. Proc. Amer. Math. Soc. 3 (1952) p. 400-406.

[13] Weil, A. - Zur algebraischen theorie der algebraischen funktionen. J. reine angew. Math. 179 (1938) p. 129-133.

[14] Weissinger, J. - Theorie der divieoren Kongruenzen. Abh. Math. sem. Hans. Univer. 12 (1938) p. 115-126.