Lectures on The Theory of Algebraic Functions of One Variable


 Malcolm Gibson
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1 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, Bombay  1 Tata Institute of Fundamental Research Bombay 1959
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3 Contents 1 Lecture Introduction Ordered Groups Valuations, Places and Valuation Rings Lecture (Contd.) Lecture The Valuations of Rational Function Field Extensions of Places Lecture Valuations of Algebraic Function Fields The Degree of a Place Independence of Valuations Lecture Divisors Lecture The Space L(U ) The Principal Divisors Lecture The Riemann Theorem iii
4 iv Contents 13 Repartitions Lecture Differentials The RiemannRoch theorem Lecture Rational Function Fields Function Fields of Degree Two Fields of Genus Zero Fields of Genus One Lecture The Greatest Common Divisor of a Class The Zeta Function of Algebraic Lecture The Infinite Product forζ(s, K) The Functional Equation Lseries Lecture The Functional Equation for the Lfunctions Lecture The Components of a Repartition Lecture A Consequence of the RiemannRoch Theorem Lecture Classes ModuloF Lecture Characters ModuloF
5 Contents v 17 Lecture Lfunctions ModuloF The Functional Equations of the Lfunctions Lecture Extensions of Algebraic Function Fields Lecture Application of Galois Theory Lecture Divisors in an Extension Ramification Lecture Constant Field Extensions Lecture Constant Field Extensions Lecture Genus of a Constant Field Extension The Zeta Function of an Extension
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7 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
8 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) forα,β W, one and only one of the relationsα<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) 3 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 W 0 be the set{α :α Wα<1}. W 0 is seen to be a semi group by 0(2) and 0(3). Moreover, W= W 0 {1} W0 1 is a disjoint partitioning of W (where W0 1 means the set of inverses of elements of W 0 ). Conversely, if an Abelian group W can be partitioned as W 0 {1} W0 1, where W 0 is a semigroup, we can introduce an order in W by definingα<β to meanαβ 1 W 0 ; it is immediately verified that 0(1), 0(2) and 0(3) are fulfilled and that W 0 is precisely the set of elements<1 in this order. For an Abelian group W, the mapα α n (n any positive integer) is in general only an endomorphism. But if W is ordered, the map is a monomorphism; for ifαis greater than or less than 1,α n also satisfies the same inequality.
9 3. Valuations, Places and Valuation Rings 3 3 Valuations, Places and Valuation Rings We shall denote the nonzero 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 nontrivial; 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 everyα W( + =α+ = +α= ), (2) α>0 for everyα 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)=1by 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 a i K, (i=1,...n). Then an obvious induction on V(2) gives v( n a i ) max n v(a i), and equality holds if v(a i ) v(a j ) for i j. i=1 1 (d) If a i K, (i=1,...n) such that n a i = 0, then v(a i )=v(a j ) for at least one pair of unequal indices i and j. For let a i be such that 1
10 4 1. Lecture 1 v(a i ) v(a k ) for k i. Then v(a i )=v( n a k ) max n (v(a k ))=v(a j ) k=1 k i for some j i, which proves that v(a i )=v(a j ). Let be a field. By ( ), we shall mean the set of elements of together with an abstract element with the following properties. α+ = +α= for everyα. α. =.α= for everyα,α 0. + and 0. are not defined. k=1 k i 5 Definition. A place of a field K is a mappingϕ of K into U( ) (where may be any field ) such that 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 thatϕ(a) = andϕ(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), thatϕ(1)ϕ(b)=ϕ(b), so that ϕ(1) = 1, and similarly ϕ(0) = 0. Consider the set O ϕ of elements a Ksuch thatϕ(a). Then by P(1), P(2) and P(3), O ϕ is a ring which is neither zero nor the whole of K, andϕis a homomorphism of this ring into. Since is a field, the kernel of this homomorphism is a prime ideal Y of O ϕ Let b be an element in K which is not in O ϕ. We contend thatϕ( 1 b )= 0. For if this mere not true, we would get 1=ϕ(1)=ϕ(b).ϕ( 1 b )=, by P(2). Thus 1 Y, and thus Y is precisely the set of nonunits of b O ϕ. Since any ideal strictly containing Y should contain a unit, we see that Y is a maximal ideal and hence the image of O ϕ in is again a field. We shall therefore always assume that is precisely the image of O ϕ byϕ, or thatϕis a mapping onto U( ). The above considerations motivate the
11 3. Valuations, Places and Valuation Rings 5 Definition. Let K be a field. A valuation ring of K is a proper subring O of K such that if a K, at least one of the elements a 1 is in O. 6 a In particular, we deduce that O contains the unity element. Let Y be the set of nonunits in O. Then Y is a maximal ideal. For, let a O, b Y. If ab Y, ab would be a unit of O, and hence 1 O. This implies ab that a 1 ab = 1 O, contradicting our assumption that b is a nonunit of 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, at least one of b c or c b, say b c, is in O. Hence, b b c 1= O. If c c 1 b c were not in Y, b c O, and hence b c b c 1 = 1 c c O, contradicting our assumption that c Y. Finally, since every element outside Y is a unit of O, Y is a maximal ideal in O.
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13 Lecture 2 3 (Contd.) In this lecture, we shall establish the equivalence of the concepts of val 7 uation, place and valuation ring. Two placesϕ 1 : K 1 ( ) andϕ 2 : K 2 ( ) are said to be equivalent if there exists an isomorphismλ of 1, onto 2 such that ϕ 2 (a)=λ ϕ 1 (a) for every a, with the understanding thatλ( )=. This is clearly an equivalence relation, and thus, we can put the set of places of K into equivalence classes. Moreover, equivalent placesϕ 1 andϕ 2 obviously define the same valuation rings O ϕ1 and O 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 andηthe natural homomorphism of O onto. It is an easy matter to verify that the mapϕ : K U{ } defined by η(a) if a O ϕ(a)= if a O is a place, whose equivalence class corresponds to the given valuation ring O. Let v 1 and v 2 be two valuations of a field K in the ordered group W 1 and W 2. 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 v 1 and v 2 are equivalent if v 1 (a)>1 if and only if v 2 (a)>1. Let v 1 and v 2 be two equivalent valuations. From the definition, it follows, by taking reciprocals, that v 1 (a)<1 if and only if v 2 (a)<1, 7
14 8 2. Lecture 2 9 and hence (the only case left) v 1 (a)=1 if and only if v 2 (a)=1. Let α be any element of W 1. Choose a K such that v 1 (a)=α (this is possible since v 1 is onto W 1 ). Defineσ(α) = v 2 (a). The definition is independent of the choice of a since if b were another element with v 1 (b)=α, then v 1 (ab 1 )=1 so that v 2 (ab 1 )=1, i.e. v 2 (a)=v 2 (b). Thus,σis a mapping from W 1 onto W 2 (since v 2 is onto W 2 ). It is easy to see thatσis an order preserving isomorphism of W 1 onto W 2 and we have v 2 (a)=(σ.v 1 )(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 ring. Also, if a K, v(a)>1, then v < 1 and hence 1 O. Thus, O a a is a valuation ring. Also, if v 1 and v 2 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. Letη : K K /O Y be the natural group homomorphism. Thenη(O ) is obviously a semigroup 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 thatηis 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.
15 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. Letϕbe a place of K= k(x) onto ( ). We consider two cases Case 1. Letϕ(X)=ξ. Then, the polynomial ring k[x] is contained in O ϕ, 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 r(x) K, it can be written in the form r(x)=(p(x)) ρ g(x), 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,we clearly have 0 ifρ>o g(ξ) ϕ(r(x))= h(ξ) ifρ=0 ifρ<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
16 10 3. Lecture 3 If Z is the additive group of integers with the natural order, the valuation v associated with the placeϕabove is given by v(r(x))=ρ. Case 2. Suppose now thatϕ(x)=. Thenϕ( 1 )=0. Then since K= X k(x)=k( 1 ), we see thatϕis determined by an irreducible polynomial X p( 1 X ), and sinceϕ 1 )=0, p(y) should divide Y. Thus, p(y) must be Y X (except for a constant in k), and if r(x)= a 0+ a 1 x+ +a n x n b 0 + b 1 x+ +b m x m,a n, b m 0, a 0 ϕ(r(x))=ϕ ( X 1 )m n x + a n 1 o if m>n + +a x n 1 n b 0 x + b m 1 + +b = a n b m if m=n x m 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 numeratorthe 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 placeϕ L of L into, we wish to prove in this section that there exists a placeϕ K of K into 1, where 1 is a field containing and the restriction ofϕ K to L isϕ L. Such aϕ K is called an extension of the placeϕ L to K. For the proof of this theorem, we require the following Lemma (Chevalley). Let K be a field, O a subring andϕahomomorphism 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
17 5. Extensions of Places 11 K. Thenϕcan be extended into a homomorphismφof at least one of the rings O[q], O[ 1 ], such thatφrestricted to O coincides withϕ. q Proof. We may assume that ϕ is not identically zero. Since the image of O is contained in a field, the kernel{ ofϕis a prime ideal Y which is a } not the whole ring O. Let O 1 = O b a, b O, b Y. Clearly, O 1 is a ring with unit, ( andϕhas a unique extension ϕ to O 1 as a homomorphism, give by ϕ = a b) ϕ(a). The image by ϕ is then the quotient field ϕ(b) ofϕ(o). We shall denote ϕ(a) by ā for a O 1. Let X and X be indeterminates over O 1 and respectively. ϕ can be extended uniquely to a homomorphism ϕ of O 1 [X] onto [ X] which takes X to X by defining ϕ(a 0 + a 1 X+ +a n X n )=ā 0 + ā 1 X+ +ā n X n. Let U be the ideal O 1 [X] consisting of all polynomials which van 13 ish for X= q, and let U be the ideal ϕ(u ) in [ X]. We consider three cases. Case 1. Let U = (0). In this case, we defineφ(q) to be any fixed element of.φ is uniquely determined on all other elements of O 1 [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 O 1 vanishes for q, its image by ϕ vanishes forφ(q). But this is implied by our assumption. Case 2. Let U (0), [ X]. Then U = ( f ( X)), where f is a nonconstant polynomial over. Letαbe any root f ( X) in (there is a root in since is algebraically closed). Define Φ(q) = α. This can be extended uniquely to a homomorphism of O 1 [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 be extended to O 1 [q]. Suppose now that it cannot be extended to O 1 [ 1 q ]
18 12 3. Lecture 3 14 either. Then ifδdenotes the ideal of all polynomials in O 1 [X] which vanish for 1 q, and if δ is the ideal ϕ(δ)in [ X], we should have δ = [ X]. Hence, there exist polynomials f (X)=a 0 + a 1 X+ +a( n X) n and 1 b 0 +b 1 X+ +b m X m 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 the required conditions. Let us assume that m n. Then, we have ā 0 = b 0 = 1, ā i = b j = 1 for i, j>0. Let g 0 (X)=b 0 X m + +b m. Applying the division algorithm to the polynomials b n 0 f (X) and g 0(X), we obtain b n 0 f (X)=g 0(X)Q(X)+R(X), Q(X), R(X) O 1 [X], deg R<m. Substituting X=q, we obtain R(q)=0. Also, acting with ϕ, we have 1= b n 0 f ( X)=ḡ 0 ( X) Q( X)+ R( X)= Q( X) X m + 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. Letϕbe 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{ϕ α, O α }, where O α is a subring if K containing O andϕ α a homomorphism of O α in extendingϕon O. The family is nonempty, since it contains (ϕ, O). We introduce a partial order in this family by defining (ϕ α, O α )>(ϕ β, O β ) if O α O β andϕ α is an extension ofϕ β. 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
19 5. Extensions of Places 13 a valuation ring of K. For if not, there exists a q Ksuch that neither q nor 1 belongs to O. we may then extendφto 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 O; for if q were a nonunit andφ(q) 0, we may defineφ = 1 [ ] q Φ(q) 1 and extend this to a homomorphism of O, which again contradicts q the maximality of O. This proves thatφcan be extended to a place of K by defining it to be outside O. In particular, a placeϕof a subfield L of K, when considered as a homomorphism of its valuation ring O ϕ and extended to K, gives a place on K; for ifϕwhere 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). ButΦbeing an extension ofϕ, the kernel contains at least one nonzero 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 a place Y 1 in k(x) such that v Y1 (X)>0. If we extend this place Y 1 to a place Y of K, we clearly have v Y (X)>0.
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21 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 elementsα 1,...α n+1 of V, at least two lie in the same coset modulo V. Choose a i K such that v(a i )=α i (i=1,...n+1). Since there can be at most n linearly independent elements of K over L, we should have n+1 l i a i = 0, l i L, not all l i being zero. i=1 This implies that v(l i a i )=y(l j a j ) for some i and j, i j (see Lecture 1, 3). Hence, we deduce that v(l i )+v(a i )=v(l i a i )=v(l j a j )=v(l j )+v(a j ), α i α j = v(a i ) v(a j )=v(l j ) v(l i ) V 15
22 16 4. Lecture 4 andα i andα 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 elementξ>0 in V such that it is the least positive element; i.e.,α>0 α ξ. Proof. The necessity is evident. Now, letαbe 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. Letα,β be any two elements of V, withα>0. Then nα and nβ are in V, nα>0. 18 Since V is archimedean, there exists an integer m such that mnα> nβ, from which it follows that mα>β. Again, consider the set of positive elementsαin 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,ξis 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.
23 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 k Y. Since Y is an isomorphism when restricted to k, we may assume that k Y is an extension of k. We shall moreover assume that k Y is the quotient O Y /M Y where O Y is the ring of the place Y and M Y the maximal ideal. We now prove the Theorem. Let Y be a place of an algebraic function field. Then k Y/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<. Letα 1,...α n+1 be any (n+1) elements of 19 k Y. Then, we should haveα i = Y (a i ) for some a i K(i=1,...n+1). There therefore exist polynomials f i (X) in k[x] such that n+1 f i (X)a i = 0, not all f i (X) having constant term zero. i=1 Writing f i (X)=l i + Xg i (X), we have the Y image, n+1 i=1 n+1 n+1 l i a i = X a i g i (X), and taking i=1 i=1 l i α i = Y (X) n+1 i=1 a i g i (Y X)=0, l i k, not all l i being zero. Thus, we deduce that the degree of k Y / k is at most n. The degree f Y of k Y over k is called the degree of the place Y. Note that f Y is always 1. If the constant field is algebraically closed (e.g. in the case of the complex number field), f Y = 1, since k Y, being an algebraic extension of k, should coincide with k.
24 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 v Y (v Y is said to be a normed valuation at the place Y ). The ring of the place shall be denoted by O Y 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 v i (i=1,...n) a set of valuations on K with valuation rings O i such that O i O j if i j. There is then an element X Ksuch that v 1 (X) 0, v i (X)<0 (i=1,...n). Proof. We shall use induction. If n=2, since O 1 O 2, there is an X O 1, X O 2, 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 v 1 (Y) 0, v i (Y)<0 (i=2,...n 1). Since O 1 nsubseto n, we can find a Z K such that v 1 (Z) 0, v n (Z)<0 Let m be a positive integer. Put X= Y+ Z m. Then v 1 (Y+ Z m ) min(v 1 (Y), mv 1 (Z)) 0 Now suppose r is one of the integers 2, 3,...n. If v r (Z) 0, r cannot be n, and since v r (Y)<0, we have v r (Y+ Z m )=v r (Y)<0.
25 8. Independence of Valuations If v r (Z)<0 and v r (Y+ Z m r ) 0for some m r, for m>m r we have v r (Y+ Z m )=v r (Y+ Z mr + Z m Z m r )=min(v r (Y+ Z m r ), v r (Z m Z m r )), and v r (Z m Z m r )=v r (Z m r )+v r (1 Z m m r )=m r v r (Z)<0, Since v r (1 Z m mr )=v r (1)=0 Thus, v r (Y+ Z m )<0 for large enough m in any case. Hence X satisfies the required conditions. If we assume that the valuations v i of the theorem are archimedean, then the hypothesis that O i O j for i j can be replaced by the weaker one that the valuations are inequivalent (which simply states that O i O j for i j). To prove this, we have only to show that if v and v 1 are two archimedean valuations such that the corresponding valuation rings O and O 1 satisfy O O 1, then v and v 1 are equivalent. ( ) For, consider an element 1 a K such that v(a)>0. Then, v < 0, and consequently 1 is not a ( ) a 1 in O, and hence not in O 1. Thus, v 1 < 0, v 1 (a)>0. Conversely, a suppose a K and v 1 (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 integer, we have v(a n b)=nv(a)+v(b)<0, a n b O. But since v 1 is archimedean and v 1 (a)>0, for large enough n we have v 1 (a n b)=nv 1 (a)+v 1 (b)>0, a n b O 1. This contradicts our assumption that O 1 O, and thus, v(a) > 0. Hence v and v 1 are equivalent. Under the assumption that the v i archimedean, we can replace in the theorem above the first inequality V 1 (X) 0 even by the strict inequality v 1 (X)>0. To prove this let X 1 K satisfy v 1 (X 1 ) 0, v i (X 1 )<0, i>1. Let Y be an element in K with v 1 (Y)>0. Then, if X=X 1 m Y, where m is a sufficiently large positive integer, we have 22 v 1 (X)=v 1 (X m 1 Y)=mv 1(X 1 )+v 1 (Y)>0, v i (X)=v i (X m 1 Y)=mv i(x 1 )+v i (Y)<0, i=2,...,n.
26 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 v i (i=1,...n) are inequivalent archimedean valuations, andρ i are elements of the corresponding valuations group, we can find X i (i=1,...n) in K such that v i (X i 1)>ρ i, v j (X i )>ρ j, i j Proof. Choose Y i K such that v i (Y i )>0, v j (Y i )<0 for j i. 1 Put X i = 1+Yi m. Then, if m is chosen large enough, we have (since the valuation are archimedean) v j (X i )= v j (1+Y m i )= mv j (Y i )>ρ j, i j 23 and v i (X i 1)=v i ( Ym i 1+Yi m since v i(1+yi m )=0. )=mv i (Y i ) v i (1+Y m i )=mv i (Y i )>ρ i. A set of valuations v i (i=1,...n) are said to be independent if given any set of elements a i K and any set of elementsρ i in the respective valuation groups of v i, we can find an X Ksuch that We then have the following v i (X a i )>ρ i. Lemma 2. Any finite set of inequivalent archimedean valuations are independent. Proof. Suppose v i (i=1,...n) is a given set of inequivalent archimedean valuations. If a i K andρ i are elements of the valuation group of the v i, putσ i =ρ i min n v i(a j ). Choose X i as in Lemma 1 for the v i andσ i. j=1 Put X= n a i X i. Then 1 v i (X a i )=v i a j X j + a i (X i 1) >σ i+ min n v i(a j )=ρ i, j=1 j i
27 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 v i (i=1,...n) are inequivalent archimedean valuations, ρ i is an element of the value group of v i for every i, and a i are given 24 elements of the field, there exists an element X of the field such that v i (X a i )=ρ i Proof. Choose Y by lemma 2 such that v i (Y a i )>ρ i. Find b i K such that v i (b i )=ρ i and an Z Ksuch that v i (Z b i )>ρ i. Then it follows that v i (Z)=min(v i (Z b i ), v i (b i ))=ρ i. Put X= Y+ Z. Then, v i (X a i )=v i (Z+ Y a i )=v i (Z)=ρ i, 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 Y 1,...,Y n. Choose an X such that v Yi (X) > 0 (i = 1,...n). Then, v Yi (X+ 1)=v Yi (1)=0 for all the places, Y i, which is impossible since X an consequently X+ 1 is a transcendental element over k.
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