From now on we assume that K = K.

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Posy Marsh
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1 Divisors

2 From now on we assume that K = K. Definition The (additively written) free abelian group generated by P F is denoted by D F and is called the divisor group of F/K. The elements of D F are called divisors. That is, a divisor is a formal sum P P F n p P with n p Z, where almost all n P = 0. The support of a divisor D is defined by supp D = {P P f : n P 0}. 1

3 We shall also write n p P, P S where S is a finite subset of P F including supp D. A divisor of the form D = P, where P P F is called a prime divisor. For Q P F and D = n p P D F we define v Q (D) = n Q. Therefore, supp D = {P P f : v P (D) 0} and D = v P (D)P. P supp D We also define a partial ordering on D F by D 1 D 2 if and only if v P (D 1 ) v P (D 2 ) for all P P F. A divisor D 0 is called positive (or efficient). 2

4 The degree of a divisor is defined by deg D = P P F v P (D) deg P and induces a homomorphism deg : D F Z. Definition Let 0 x F. We denote by Z (respectively, by N) the set of all zeros (respectively, poles) of x and define the zero, pole, and principal divisors of x by (x) 0 = P Z v P (x)p, (x) = P N v P (x)p, and (x) = (x) 0 (x), respectively (why these divisors are well-defined?). 3

5 Then (x) 0, (x) 0, and (x) = v P (x)p. P P F Also, x K if and only if (x) = 0. Definition The group P F = {(x) : 0 x F } (why this is a group?) is called the group of principal divisors of F/K. The factor group C F = D F /P F is called the divisor class group. For a divisor D D F, the corresponding element of C F class of D and is denoted [D]. is called the divisor Divisors D and D are called equivalent, denoted D D, if [D] = [D ] (what does it mean?). Definition For a divisor A D F we define L(A) = {x F : (x) A}. 4

6 If A = r n i P i s i=1 j=1 m j Q j, where n i, m j > 0, then x L(A) is and only if x has zeros of order m j at Q j, j = 1,..., m, and x may have poles only at P 1,..., P n, with the pole order at P i being at most n i, i = 1,..., n. Remark Let A D F. Then (a) x L(A) if and only if v P (x) v P (A) for all P P F (why?), and (b) L(A) {0} if and only if there is a divisor A A such that A 0 (why?). 5

7 Lemma Let A D F. Then (a) L(A) is a vector space over K, and (b) if A A, then L(A ) L(A) (isomorphic as vector spaces over K). Proof (a) Let x, y L(A) and let a K. Then, for all P P F, v P (x + y) min{v P (x), v P (y)} v P (A), and v P (ax) = v P (a) + v P (x) = v P (x) v P (A). 6

8 (b) Let A = A + (z), 0 z F. Let x L(A). Since for all P P F, v P (zx) = v P (z) + v P (x) v P (z) v P (A) = v P (A (z)) = v P (A ), the function ϕ z : L(A) F defined by ϕ z (x) = zx is a linear mapping whose image is included in L(A ). Similarly, the function ϕ z 1 : L(A ) F defined by ϕ z 1(x) = z 1 x is a linear mapping whose image is included in L(A). Since ϕ z and ϕ z 1 are inverse to each other, both of them are isomorphisms between L(A ) and L(A). Lemma (a) L(0) = K. (b) If A < 0, then L(A) = {0}. 7

9 Proof (a) If 0 x K, then (x) = 0, implying K L(0). If 0 x L(0), then (x) 0. Thus, x has no pole, and, therefore, is in K (why?). (b) Assume to the contrary that there exists an 0 x L(0). Then (x) A > 0. Therefore, x has a zero, but has no pole, which is impossible (why?). 8

10 Lemma Let A, B D F be such that A B. Then L(A) L(B) and dim(l(b)/l(a)) deg B deg A. Proof The inclusion L(A) L(B) is trivial, and for the second assertion we may assume that B = A + P for some P P F (why?). Let t F be such that v P (t) = v P (B) = v P (A) + 1 (why there is such a t?). Then, for x L(B) we have v P (x) v P (B) = v P (t). Therefore, xt O P and we have a K-linear map ψ : L(B) F P defined by ψ(x) = xt(p ). This map is well defined (why?) and Ker ψ = L(A), because x Ker ψ if and only if v P (xt) > 0, which is equivalent to v P (x) v P (A) (why?). Thus, ψ induces an injection of L(B)/L(A) into F P, implying dim(l(b)/l(a)) dim F P = deg P = deg B deg A. 9

11 Proposition Let A = A + A with positive divisors A + and A. Then dim L(A) deg A In particular, L(A) is a finite-dimensional vector space over K. Proof Since L(A) L(A + ) (why?), it suffices to show that dim L(A + ) deg A We have dim(l(a + )/L(0)) deg A + (why?). Since L(0) = K (why?) dim L(A + ) = dim(l(a + )/L(0)) + 1, and the proof is complete. Definition For a divisor A D F, dim L(A) is called the dimension of A and is denoted dim A. 10

12 Proposition Let x F \ K. Then deg(x) 0 = deg(x) = [F : K(x)]. 1 In particular, any principal divisor has degree zero. Corollary (a) Let A A. Then dim A = dim A and deg A = deg A. (b) If deg A < 0, then dim A = 0. (c) If deg A = 0, then the following three assertions are equivalent. (1) A is principal. (2) dim A 1. (3) dim A = 1. 1 Here we use our assumption that K = K (how?). 11

13 Proof of the corollary (a) is immediate (how?). (b) Assume to the contrary that dim A > 0. Then there is a divisor A A such that A 0 (why?). Therefore, deg A = deg A 0, which contradicts our assumption. (c) (1) (2): If A = (x), then x 1 L(A) (why?), implying dim A 1. (2) (3): Again, there is a divisor A A such that A 0. This, together with deg A = 0, implies A = 0. Hence dim A = dim A = dim 0 = 1. (3) (1): There exists a 0 z L(A) (why?), and we have (z) + A 0. Since deg((z) + A) 0, (z) + A = 0. Thus, A = (z) = (z 1 ) is principal. 12

14 Proof of the proposition Let n = [F : K(x)] and let B = (x) = r v Pi (x)p i, i=1 where P 1,..., P r are all the poles of x. Then deg B = deg(x) = r v Pi (x 1 ) deg P i. i=1 We have already seen that deg B n (where?). Thus the proof will be complete when we show that deg B n as well. 13

15 Let u 1,..., u n be a basis of F/K(x) and let a divisor C 0 be such that (u i ) C, i = 1,..., n. Let l 0 and let 0 i l. Then for all j = 1,..., n, x i u j L(lB + C) (why?), and, since x i u j are linearly independent over K (why?), Let deg C = c. Then (why?), implying for all l N. Therefore, deg B n. dim L(lB + C) n(l + 1). l deg B + c + 1 dim L(lB + C) n(l + 1), l(deg B n) n c 1 Finally, since (x) 0 = (x 1 ), deg(x) 0 = deg(x 1 ) = [F : K(x 1 )] = [F : K(x)]. 14

16 Example Let 0 z K(x). Then z = a f(x), where a K and f(x), g(x) g(x) K[x] are monic and relatively prime. Let r s f(x) = p i (x) n i and g(x) = q j (x) m j i=1 j=1 with pairwise distinct irreducible polynomials p i (x), q j (x) K[x]. Then (z) = r n i P i i=1 s m j Q j + (deg g deg f)p, j=1 where P i and Q j are places corresponding to p i (x) and q j (x), respectively. 15

17 Proposition There is a constant γ Z such that for all divisors A D F, deg A dim A γ. Proof Let x F \ K and let B = (x). Then there exists a divisor C 0 such that for all l N, (why?). In addition, (why?). Therefore, dim L(lB + C) (l + 1) deg B dim L(lB) + deg C dim L(lB + C) dim lb (l + 1) deg B deg C = deg(lb) + ([F : K(x)] deg C) (why?). Thus, for some γ Z (which one?), for all l > 0 (why?). deg(lb) dim(lb) γ 16

18 That is, the proposition holds when A is of the form l(x), l 0 and x F \ K, and we continue to prove that it holds when we substitute lb with any A D F, with the above γ. For this we shall need the claim below. Claim For any A D F there exist divisors A 1 and D and a non-negative integer l such that A A 1, A 1 D, and D lb. Proof of the claim Let A 1 max{0, A}, e.g., A 1 = A +. Then dim(lb A 1 ) dim(lb) deg A 1 deg(lb) γ deg A 1 > 0 for sufficiently large l. Let 0 z L(lB A 1 ) and let D = A 1 (z). Then A 1 D and D A 1 (A 1 lb) = lb 17

19 Proof of the proposition (continued) Let A 1, D, and l be as in the claim. Then deg A dim A deg A 1 dim A 1 = deg D dim D deg(lb) dim(lb) γ Definition The genus g of F/K is defined by g = max{deg A dim A + 1 : A D F }. Remark The genus of F/K is a non-negative integer (why?). 18

20 Theorem (Riemann s theorem) Let F/K be a function field of genus g. Then (a) For all A D F, dim A deg A + 1 g. (b) There is a constant c (that depends on F/K) such that for all A D F with deg A c, dim A = deg A + 1 g. Proof (a) is immediate. (b) Let A 0 D F be such that g = deg A 0 dim A We put c = deg A 0 +g. Then dim(a A 0 ) deg(a A 0 ) + 1 g c deg A g 1 and there is a non-zero z L(A A 0 ). Let A = A + (z). Then A A 0, implying deg A dim A = deg A dim A deg A 0 dim A 0 = g 1. 19

21 Example The genus of the rational function field K(x)/K is zero. Indeed, for r 0, 1, x,..., x r L(rP ), implying r + 1 dim(rp ) = deg(rp ) + 1 g }{{} for sufficiently large r = r + 1 g. 20

Some consequences of the Riemann-Roch theorem

Some consequences of the Riemann-Roch theorem Proposition Let g 0 Z and W 0 D F be such that for all A D F, dim A = deg A + 1 g 0 + dim(w 0 A). Then g 0 = g and W 0 is a canonical divisor. Proof We have