Algebraic function fields


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1 Algebraic function fields
2 1 Places Definition An algebraic function field F/K of one variable over K is an extension field F K such that F is a finite algebraic extension of K(x) for some element x F which is transcendental over K. K = {z F : z is algebraic over K} K is called the field of constants of F/K. K is algebraically closed in F if K = K. We have 1. K K F 2. F/K is an algebraic function field over K (why?)
3 2 Definition F/K is called rational, if F = K(x) for some x F transcendental over K. Definition A valuation ring of the function field F/K is a ring O F with the following properties. 1. K O F 2. For any z F, z O or z 1 O Example Let p(x) K[x] be an irreducible polynomial. Then { } f(x) O p(x) = : f(x), g(x) K[x] and p(x) g(x) g(x) is a valuation ring of K(x)/K.
4 3 Proposition Let O be a valuation ring of the function field F/K. Then (a) O is a local ring, i.e., O has a unique maximal ideal P = O \ O, where O is the group of units of O. (b) For 0 x F, x P if and only if x 1 O. (c) K O and K P = {0}. Proof (a) It suffices to show that P = O \ O is an ideal of O. Let x P and let z O. Then xz O (why?), implying xz P. Let x, y P. We may assume that x/y O. Then 1 + x/y O, implying x + y = y(1 + x/y) P. (b)...
5 4 (c) Let z K and assume to the contrary that z O. Then z 1 O. Since z 1 is algebraic over K, there are elements a 1,..., a r K such that a r (z 1 ) r + + a 1 z = 0, implying z 1 (a r (z 1 ) r a 1 ) = 1. Thus, z = (a r (z 1 ) r a 1 ) K[z 1 ] O which contradicts our assumption.
6 5 Theorem Let O be a valuation ring of the function field F/K and let P be its unique maximal ideal. Then (a) P is a principal ideal. (b) If P = to, then any 0 z F has a unique representation of the form z = t n u for some n Z and u O. (c) O is a principal ideal domain, and if P = to and {0} I O is an is an ideal, then I = t n O for some n N. Definition A ring possessing the above properties is called a discrete valuation ring. Lemma Let O be a valuation ring of the algebraic function field F/K, P be its maximal ideal, and let 0 x P. Let x 1,..., x n P be such that x 1 = x and x i x i+1 P, i = 1,..., n 1. Then n [F : K(x)] <.
7 6 Proof The inequality [F : K(x)] < is immediate (why?) and for the proof of n [F : K(x)] it suffices to show that x 1,..., x n are linearly independent over K(x). Assume to the contrary that n i=1 ϕ i x i = 0, where ϕ i K(x), not all equal zero. We may assume that all ϕ i K[x] and that x does not divide all of them (why). Let j = max{i : x ϕ i }. Then for all i > j there exists a polynomial g i K[x] such that ϕ i = xg i, and it follows from x j ϕ j = i j x i ϕ i. Therefore, implying ϕ j P (why?). ϕ j = i<j x i x j ϕ i + i>j x x j x i g i, n i=1 ϕ i x i = 0 that However, since ϕ j = ϕ j (0)+xg j with g j K[x] O, ϕ j (0) = ϕ j xg j P K (why?), which is impossible, because ϕ j (0) 0 (why?).
8 7 Corollary If t P, then n=1 t n O = {0}. Proof Assume to the contrary that for some z 0, z n=1 t n O. Then for all n N, the sequence x 1 = z, x 2 = t n 1, x 3 = t n 2,..., x n = t satisfies the lemma prerequisite x i x i+1 P, i = 1..., n 1. However, this is impossible for n > [F : K(x)].
9 8 Proof of the theorem (a) Assume to the contrary that P is not principal and let 0 x 1 P. Since P x 1 O, there is x 2 P \ x 1 O, implying x 2 x 1 1 O. Therefore, x 1 x 1 2 P and x 1 x 2 P. We proceed by induction and obtain an infinite sequence x 1, x 2,... P such that x i x i+1 P for all i = 1, 2,.... However, this contradicts the lemma. (b) The uniqueness of the representation is immediate (why?), and for the proof of the existence we may assume that z O (why?). If z is a unit, then z = t 0 z, i.e., u = z. Otherwise, z P (why?) and, by the corollary to the lemma, there is a maximal nonnegative integer n such that z t n O. Therefore, z = t n u, where u is a unit (why?).
10 9 (c) Let {0} I O be an ideal an let A = {r N : t r I}. Then A (why?) and let n = min(a). Since t n O I, the proof of part (c) of the theorem will be complete if we show the converse inclusion. Let 0 x I. Since x = t s u for some u O and s n, t s I as well. Therefore, x = t n t s n u t n O. Definition (a) A place P of the function field F/K is the maximal ideal of some valuation ring O of F/K. Any element t P such that P = to is called a prime element for P (or local parameter or uniformizing variable). (b) P F = {P : P is a place of F/K}.
11 10 Remark If O is a valuation ring of F/K and P is its maximal ideal, then O is uniquely determined by P : O = {z F : z 1 P }. We call O p = O the valuation ring of the place P. Definition (The arithmetic of Z { }) For all n Z, + = n + = + n > n
12 11 Definition A discrete valuation of F/K is a function v : F Z { } with the following properties. (1) v(x) = if an only if x = 0. (2) v(xy) = v(x) + v(y) for all x, y F. (3) v(x + y) min{v(x), v(y)} for all x, y F. (4) There exists an element z F with v(z) = 1. (5) v(a) = 0 for all 0 a K. It follows from (2) and (4) that v : F Z { } is surjective (how?). Also it follows from the definition that v(x) = v( x) for all x F (how?).
13 12 Lemma (Strict Triangle Inequality) Let v be a discrete valuation of F/K and let x, y F be such that v(x) v(y). Then v(x + y) = min{v(x), v(y)}. Proof Since v(x) v(y), we may assume that v(x) < v(y). Now, assume to the contrary that v(x + y) min{v(x), v(y)}. Then v(x + y) > v(x) (why?), implying a contradiction v(x) = v((x + y) y) min{v(x + y), v(y)} > v(x). Definition To a place P P F we associate a function v P : F Z { } such that v P (0) = and for 0 z F v P (z) is defined as follows. Let t be a prime element for P. Then for some u O and some n Z, z = t n u and this representation of z is unique (for a given t). We put v P (z) = n. This function v P is welldefined (why?).
14 13 Theorem Let F/K be a function field. (a) For any place P P F, the function v P F/K. In addition, is a discrete valuation of O P = {z F : v p (z) 0}, O P = {z F : v p (z) = 0}, P = {z F : v p (z) > 0}. An element z of F is a prime element for P if and only if v P (z) = 1. (b) Conversely, if v is a discrete valuation of F/K, then P v = {z F : v(z) > 0} is a place of F/K, and O Pv = {z F : v(z) 0} is the corresponding valuation ring. (c) Any valuation ring O of F/K is a maximal proper subring of F.
15 14 Proof (a) We shall prove the triangle inequality only. Let x, y F, v P (x) = n, and v P (y) = m, where n m <. Then x = t n u 1 and y = t m u 2, where u 1, u 2 O P and where z O P. If z = 0, then x + y = t n (u 1 + t m n u 2 ) = t n z, v P (x + y) = > min{n, m}. Otherwise, z = t k u, where k 0 and u O P, implying v P (x + y) = v P (t n+k u) = n + k n = min{v P (x), v P (y)}. The rest of (a)... (b)...
16 15 (c) Let O be a valuation ring of F/K and let z F \ O. We shall prove that O[z] = F. Let y F and let P be the maximal ideal of O. Then, for sufficiently large k, v P (yz k ) 0 (why?). Consequently, w = yz k O, implying y = wz k O[z]. Since P is a maximal ideal of O P, O P /P is a field, and for z O P we define z(p ) to be the residue class of z modulo P. If z F \ O P, we put z(p ) =. 1 Sometimes we shall also write z + P instead of z(p ) for z O P. Since K O P (K O P ) (why?) and K P = {0} (K P = {0}) (why?), the residue class map O P O P /P induces a canonical embedding of K (K) into O P /P. We shall always consider K (K) as a subfield of O P /P via this embedding. 1 This is not the same as in discrete valuations.
17 16 Definition Let P P F. (a) F P = O P /P is called the residue class field of P and the map x x(p ) from F to F P { } is called the residue class map with respect to P. (b) deg P = [F P : K] is called the degree of P. What happens when deg P = 1? Proposition Let P be a place of F/K and let 0 x P. Then deg P [F : K(x)] <. Proof We have already seen that [F : K(x)] < (where?), and for the proof of deg P [F : K(x)] it suffices to show that any elements z 1,..., z n of O P, whose residue classes z 1 (P ),..., z n (P ) are linearly independent over K, are linearly independent over K(x).
18 17 So, let z 1 (P ),..., z n (P ) be linearly independent over K, and assume to the contrary that there is a nontrivial linear combination with ϕ i K(x). n i=1 ϕ i z i = 0 We may assume that the ϕ i are polynomials in x and not all of them are divisible by x (why?). That is, ϕ = a i + xg i, where a i K and g i K[x], and not all a i = 0. Applying the residue class map to n i=1 ϕ i z i = 0 we obtain 0 = 0(P ) = n ϕ i (P )z i (P ) = i=1 n a i z i (P ), i=1 which contradicts the linear independence of z 1 (P ),..., z n (P ) over K.
19 18 Definition Let z F and let P P f. We say that P is a zero (respectively, a pole) of z if and only if v P (z) > 0 (respectively, v P (z) < 0). If v P (z) = m > 0 (respectively, v P (m) = m < 0), then P is a zero (respectively, pole) of P of order m. Theorem Let F/K be a function field, K R be a subring of F, and let I {0} be a proper ideal of R. Then there is a place P P F such that I P and R O P. Proof Let F = {S : S is a subring of F with R S and IS S}. 2 F, because R F, and we shall prove that F is inductively ordered by inclusion (what does it mean?). 2 IS is the ideal of S generated by I.
20 19 Indeed, if H F is a totally ordered subset of F, then T = {S : S H} is a subring of S with R T. We have to show that IT T. For this, assume to the contrary that n i=1 a is i = 1, with a i I and s i T. Since H is totally ordered, there is an S H such that s 1,..., s n S. Therefore, 1 = n i=1 a is i S as well, in contradiction with the definition of F (which contradiction?). By Zorn s lemma, F contains a maximal element O that, as we prove below, is a valuation ring of F/K. We have K O F (why?). Assume to the contrary that there exists an element z of F with z O and z 1 O. Then IO[z] = O[z] and IO[z 1 ] = O[z 1 ] (why?). Consequently, there are a 0,..., a n, b 0,..., b m IO such that 1 = a 0 + a 1 z + + a n z n 1 = b 0 + b 1 z a n z m
21 20 It follows that m, n 1 (how?) and we may assume that m and n are chosen minimally and m n (why?). Multiplying the first equality by 1 b 0 and the second by a n z n we obtain 1 b 0 = (1 b 0 )a 0 + (1 b 0 )a 1 z + + (1 b 0 )a n z n 0 = (b 0 1)a n z n + b 1 a n z n b m a n z n m Adding these equalities yields 1 = c 0 + c 1 z + + c n z n 1 with coefficients c i in IO (why?). This contradicts the minimality of n. That is, we have shown that for any z F, z O or z 1 O.
22 21 Corollary Let F/K be a function field and let z F be transcendental over K. Then z has at least one zero and one pole. 3 Proof Let R = K[z]. Then I = zk[z] is a proper ideal of R. Thus, there is a place P P F with z P (hence P is a zero of z). To prove that z has a pole we start with... Corollary The field K of constants of F/K is a finite extension of K. Proof Let P P F. Since K embeds into F P via the residue class map O P F P, it follows that 3 In particular, P F. [K : K] [F P : K] (< ).
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