Lemma 33.43.13. Let $k$ be a field. Let $X$ be a proper scheme over $k$ which is reduced, connected, and equidimensional of dimension $1$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. If $\deg (\mathcal{L}|_ C) \leq 0$ for all irreducible components $C$ of $X$, then either $H^0(X, \mathcal{L}) = 0$ or $\mathcal{L} \cong \mathcal{O}_ X$.

**Proof.**
Let $s \in H^0(X, \mathcal{L})$ be nonzero. Since $X$ is reduced there exists an irreducible component $C$ of $X$ with $s|_ C \not= 0$. But if $s|_ C$ is nonzero, then $s$ is nonwhere vanishing on $C$ by Lemma 33.43.12. This in turn implies $s$ is nowhere vanishing on every irreducible component of $X$ meeting $C$. Since $X$ is connected, we conclude that $s$ vanishes nowhere and the lemma follows.
$\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: