Lemma 33.44.13. 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 33.44.12. 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. $\square$

There are also:

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

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).