The Riemann-Roch Theorem

The Riemann-Roch Theorem

In this lecture F/K is an algebraic function field of genus g. Definition For A D F, is called the index of specialty of A. i(a) = dim A deg A + g 1 Definition An adele of F/K is a mapping α : P F F such that α(p ) = α P O P for almost all P P F. We regard an adele as an element of the direct product use the notation α = (α P ) P PF or α = (α P ). The set A F = {α : α is an adele of F/K} P P F F and, therefore, is called the adele space of F/K. It is regarded as a vector space over K. 1

The principal adele of an element x F is the adele all of whose components are equal to x. This gives an embedding of F into A F. A valuation v P of F/K extends naturally to A F by setting v P (α) = v P (α P ). By definition, v P (α) 0 for almost all P P F. Definition For A D F we define A F (A) = {α A F : v P (α) v P (A) for all P P F }. Obviously, this is a K-subspace of A F. Theorem Let A D F. Then Corollary Proof of the corollary i(a) = dim(a F /(A F (A) + F )) g = dim(a F /(A F (0) + F )) i(0) = dim(0) deg(0) + g 1 = 1 0 + g 1 = g 2

Lemma Let A 1, A 2 D F be such that A 1 A 2. Then (1) A F (A 1 ) A F (A 2 ) and (2) dim(a F (A 2 )/A F (A 1 )) = deg A 2 deg A 1 dim((a F (A 2 )+F )/(A F (A 1 )+F )) = (deg A 2 dim A 2 ) (deg A 1 dim A 1 ) Proof (1) The inclusion A F (A 1 ) A F (A 2 ) is immediate, and it suffices to prove the equality in the case of A 2 = A 1 + P, where P P F (the general case follows by induction). Let t F be such that v P (t) = v P (A 1 ) + 1 = v P (A 2 ). Then the K-linear map ϕ : A F (A 2 ) F P defined by ϕ(α) = (tα P )(P ) is surjective (why?) and Ker ϕ = A F (A 1 ) (why?). Therefore, dim(a F (A 2 )/A F (A 1 )) = [F P : K] = deg P = deg A 2 deg A 1. 3

(2) We contend that the sequence of linear mappings below is exact. 0 L(A 2 )/L(A 1 ) σ 1 A F (A 2 )/A F (A 1 ) σ 2 (A F (A 2 ) + F )/(A F (A 1 ) + F ) 0 The only non-trivial part is to show that Ker σ 2 Im σ 1. Let α A F (A 2 ) be such that σ 2 (α + A F (A 1 )) = 0. Then α A F (A 1 ) + F, i.e., for some x F, α x A F (A 1 ) A F (A 2 ). Therefore, implying x A F (A 2 ) F = L(A 2 ), α + A F (A 1 ) = x + A F (A 1 ) = σ 1 (x + L(A 1 )) Im σ 1. Form the exactness of the above sequence and part (1) of the lemma we obtain dim(a F (A 2 ) + F )/(A F (A 1 ) + F ) = dim(a F (A 2 )/A F (A 1 )) dim(l(a 2 )/L(A 1 )) = (deg A 2 dim A 2 ) (deg A 1 dim A 1 ). 4

Corollary Let B, B 1 D F be such that dim B = deg B + 1 g and B 1 B. Then dim B 1 = deg B 1 + 1 g. Proof It follows from 0 dim(a F (B 1 ) + F )/(A F (B) + F ) = (deg B 1 dim B 1 ) (deg B dim B) that dim B 1 deg B 1 + dim B deg B = deg B 1 + 1 g. Since also dim B 1 deg B 1 + 1 g (why?), the corollary follows. Lemma Let B D F be such that dim B = deg B + 1 g. Then A F = A F (B) + F. Proof Let α A F such a B 1?). Since and let B 1 B be such that α A F (B 1 ) (why there is dim(a F (B 1 ) + F )/(A F (B) + F ) = (deg B 1 dim B 1 ) (deg B dim B) = 0, A F (B 1 ) + F = A F (B) + F, and, therefore, α A F (B) + F. 5

Proof of the theorem Let A 1 A be such that dim A 1 = deg A 1 + 1 g (why there is such an A 1?). Then A F = A F (A 1 ) + F, implying dim(a F /(A F (A) + F )) = dim((a F (A 1 ) + F )/(A F (A) + F )) = (deg A 1 dim A 1 ) (deg A dim A) = (g 1) + dim A deg A = i(a). 6

Definition A Weil differential of F/K is a K-linear map ω : A F K vanishing on A F (A) + F for some divisor A D F. The set of all differentials is denoted by Ω F and for a divisor A D F the set of differentials Ω F (A) is defined by Ω F (A) = {ω Ω F : ω vanishes on A F (A) + F }. Remark (1) If ω Ω F (A) and a K, then aω Ω F (A). (2) If A 1, A 2, A 3 D F are such that A 1, A 2 A 3 and ω i Ω F (A i ), i = 1, 2, then ω 1 + ω 2 Ω F (A 3 ). Therefore, Ω F is a vector space over K and Ω F (A) is its subspace. 7

Lemma Let A D F. Then dim Ω F (A) = i(a). Proof By definition, Ω F (A) is isomorphic to the space of linear forms on A F /(A F (A) + F ) whose dimension equals to that of A F /(A F (A) + F ), which is i(a). Corollary Ω F {0}. Proof Let A D F be such that deg A 2. Then dim Ω F dim Ω F (A) = dim A deg A + g 1 1. 8

Definition Let x F and let ω Ω F. The mapping xω : A F K is defined by (xω)(α) = ω(xα), where x(α P ) P PF = (xα P ) P PF. Remark If ω vanishes on A F (A) + F, then xω vanishes on A F (A + (x)) + F. Thus, xω is again a Weil differential, and the above definition makes Ω F a vector space over F. Proposition Ω F is a one-dimensional vector space over F. 9

Proof Let 0 ω 1 Ω F and let ω 2 Ω F. We have to show that for some z F, ω 2 = zω 1. We may assume that ω 2 0 as well (why?). Let A i D F be such that ω i Ω F (A i ), i = 1, 2 and let B D F. Consider maps ϕ i : L(A i +B) Ω F ( B) defined by ϕ i (x) = xω i, i = 1, 2. These maps are K-linear and injective (why?). We contend that for some B D F, ϕ 1 (L(A 1 + B)) ϕ 2 (L(A 2 + B)) {0}. If this is indeed so, then there are x 1 L(A 1 + B) and x 2 L(A 2 + B) such that x 1 ω 1 = x 2 ω 2 0, and ω 2 = (x 1 /x 2 )ω 1, as desired (z = x 1 /x 2 ). 10

It remains to prove the contention: Let B D F, B > 0, be such that dim(a i + B) = deg(a i + B) + 1 g, and let U i = ϕ i (L(A i + B)) Ω F ( B), i = 1, 2. Since dim Ω F ( B) = i( B) = dim( B) deg( B) + g 1 = deg B + g 1, we have im(u 1 U 2 ) dim U 1 + dim U 2 dim Ω F ( B) = deg(a 1 + B) + 1 g + deg(a 2 + B) + 1 g (deg B + g 1) = deg B + (deg A 1 + deg A 2 + 3(1 g)). Thus, if deg B is sufficiently large, U 1 U 2 {0}. 11

For a non-zero differential ω we define the set of divisors M(ω) by M(ω) = {A D F : ω Ω F (A)}. Lemma A W. There is a unique divisor W M(ω) such that for all A M(ω), Proof The uniqueness of W, if exists, is obvious, and for the proof of existence, let c Z be such that for all A D F, deg A c implies i(a) = 0 (why there is such a c?). Since for all A M(ω), (why?), for all A M(ω), deg A < c. i(a) = dim(a F /A F (A) + F )) 0 12

Let W be an element of M(ω) of the maximum degree and assume to the contrary that there are A 0 M(ω) and Q P F such that v Q (A 0 ) > v Q (W ). We shall prove that W + Q M(ω), in contradiction with maximality of deg W (what is the contradiction?). Let α A F (W + Q). Then α = α + α, where α P = { αp, if P Q 0, if P = Q and α P = { 0, if P Q α P, if P = Q. Then α A F (W ) and α A F (A 0 ), implying ω(α) = ω(α ) + ω(α ) = 0. That is, ω vanishes on A F (W + Q) + F and the proof is complete. 13

Definition (a) The divisor (ω) of a Weil differential ω 0 is the unique divisor such that (1) ω vanishes on A F ((ω)) + F, and (2) if ω vanishes on A F (A) + F, then A (ω). (b) For 0 ω Ω F and P P F, v P (ω) is defined by v P (ω) = v P ((ω)). (c) A place P is zero (respectively, pole) of ω, if v P (ω) > 0 (respectively, v P (ω) < 0). ω is regular at P, if v P (ω) 0, and is regular (or holomorphic), if it is regular at all P P F. (d) A divisor W is canonical, if W = (ω) for some ω Ω F. 14

Remark Ω F (A) = {ω Ω F : (ω) A} {0}. Ω F (0) = {ω Ω F : ω is regular}. dim Ω(0) = g Proposition (a) If 0 x F and 0 ω Ω F, then (xω) = (x) + (ω). (b) All canonical divisors are equivalent. Proof (a) If A M(ω), then A + (x) M(xω), implying (x) + (ω) (xω), and, similarly, (x 1 ) + (xω) (x 1 xω) = (ω). Therefore, (x) + (ω) (xω) (x 1 ) + (ω) = (x) + (ω) (b) follows from (a) and dim Ω F = 1. 15

Corollary The set of all canonical divisors is a class in C F. This class is called the canonical class of F/K. Theorem Let A D F and let W = (ω) be a canonical divisor. Then the mapping µ : L(W A) Ω F (A) defined by µ(x) = xω is an isomorphism of K-vector spaces. In particular, Proof Let x L(W A). Then i(a) = dim(w A) (µ(x)) = (xω) = (x) + (ω) (W A) + W = A, implying µ(x) Ω F (A) (how?). Clearly, µ is linear and injective (why?), and it remains to show that µ is surjective. Let ω 1 Ω F (A). There is an x F such that ω 1 = xω (why?). Since (x) + W = (x) + (ω) = (xω) = (ω 1 ) A, (x) (W A). That is, x L(W A), and ω 1 = µ(x). 16

Corollary (Riemann-Roch Theorem) Let W be a canonical divisor. Then for all A D F, dim A = deg A + 1 g + dim(w A) Corollary Let W be a canonical divisor. Then deg W = 2g 2 and dim W = g. Proof Substitution of 0 for A in the Riemann-Roch theorem results in dim W = g. Then substitution of W for A results in deg W = 2g 2. Theorem Let A D F be such that deg A 2g 1. Then dim A = deg A + 1 g. Proof Let W be a canonical divisor. Then i(a) = dim(w A) (why?). Since deg W = 2g 2 and deg A 2g 1, dim(w A) = 0 (why?). 17

Remark It follows from the las corollary that for a canonical divisor W, i(w ) = 1. Therefore, c = 2g 1 is the minimum integer for which the Riemann theorem holds (what is the Riemann theorem?). 18