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

## Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B90. Beware of the difference between the letter 'O' and the digit '0'.