proof of Riemann-Roch theorem

For a divisorPlanetmathPlanetmath D, let 𝔏(D) be the associated line bundleMathworldPlanetmath. By Serre duality, H0(𝔏(K-D))H1(𝔏(D)), so (D)-(K-D)=χ(D), the Euler characteristicMathworldPlanetmath of 𝔏(D). Now, let p be a point of C, and consider the divisors D and D+p. There is a natural injection 𝔏(D)𝔏(D+p). This is an isomorphismMathworldPlanetmathPlanetmathPlanetmath anywhere away from p, so the quotient is a skyscraper sheaf supported at p. Since skyscraper sheaves are flasque, they have trivial higher cohomology, and so χ()=1. Since Euler characteristics add along exact sequencesPlanetmathPlanetmathPlanetmath (because of the long exact sequence in cohomology) χ(D+p)=χ(D)+1. Since deg(D+p)=deg(D)+1, we see that if Riemann-Roch holds for D, it holds for D+p, and vice-versa. Now, we need only confirm that the theorem holds for a single line bundle. 𝒪X is a line bundle of degree 0. (0)=1 and (K)=g. Thus, Riemann-Roch holds here, and thus for all line bundles.

Title proof of Riemann-Roch theorem
Canonical name ProofOfRiemannRochTheorem
Date of creation 2013-03-22 13:51:36
Last modified on 2013-03-22 13:51:36
Owner bwebste (988)
Last modified by bwebste (988)
Numerical id 6
Author bwebste (988)
Entry type Proof
Classification msc 14H99