Processing math: 100%

The Stacks project

Lemma 33.44.17. Let k be an algebraically closed field. Let X be a proper curve over k. Set g = \dim _ k H^1(X, \mathcal{O}_ X). For every invertible \mathcal{O}_ X-module \mathcal{L} with \deg (\mathcal{L}) \geq 2g - 1 we have H^1(X, \mathcal{L}) = 0.

Proof. Let \mathcal{N} be the invertible module we found in Lemma 33.44.16 part (2). The degree of \mathcal{N} is \chi (X, \mathcal{N}) - \chi (X, \mathcal{O}_ X) = 0 - (1 - g) = g - 1. Hence the degree of \mathcal{L} \otimes \mathcal{N}^{\otimes - 1} is \deg (\mathcal{L}) - (g - 1) \geq g. Hence \chi (X, \mathcal{L} \otimes \mathcal{N}^{\otimes -1}) \geq g + 1 - g = 1. Thus there is a nonzero global section s whose zero scheme is an effective Cartier divisor D of degree \deg (\mathcal{L}) - (g - 1). This gives a short exact sequence

0 \to \mathcal{N} \xrightarrow {s} \mathcal{L} \to i_*(\mathcal{L}|_ D) \to 0

where i : D \to X is the inclusion morphism. We conclude that H^0(X, \mathcal{L}) maps isomorphically to H^0(D, \mathcal{L}|_ D) which has dimension \deg (\mathcal{L}) - (g - 1). The result follows from the definition of degree. \square


Comments (0)

There are also:

  • 4 comment(s) on Section 33.44: Degrees on curves

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.