# proof of Riemann-Roch theorem

For a divisor^{} $D$, let $\U0001d50f(D)$ be the associated line bundle^{}. By Serre duality, ${H}^{0}(\U0001d50f(K-D))\cong {H}^{1}(\U0001d50f(D))$, so $\mathrm{\ell}(D)-\mathrm{\ell}(K-D)=\chi (D)$, the Euler characteristic^{} of $\U0001d50f(D)$. Now, let $p$ be a point of $C$, and consider the divisors $D$ and $D+p$. There is a natural injection $\U0001d50f(D)\to \U0001d50f(D+p)$. This is an isomorphism^{} anywhere away from $p$, so the quotient $\mathcal{E}$ is a skyscraper sheaf supported at $p$. Since skyscraper sheaves are flasque, they have trivial higher cohomology, and so $\chi (\mathcal{E})=1$. Since Euler characteristics add along exact sequences^{} (because of the long exact sequence in cohomology) $\chi (D+p)=\chi (D)+1$. Since $\mathrm{deg}(D+p)=\mathrm{deg}(D)+1$, 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. ${\mathcal{O}}_{X}$ is a line bundle of degree 0. $\mathrm{\ell}(0)=1$ and $\mathrm{\ell}(K)=g$. 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 |