[Lemma 3, Jongmin]

Lemma 53.21.3. In Situation 53.6.2 assume $X$ is integral and has genus $g$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. If $\deg (\mathcal{L}) \geq 2g$, then $\mathcal{L}$ is globally generated.

Proof. Let $Z \subset X$ be the closed subscheme cut out by the global sections of $\mathcal{L}$. By Lemma 53.21.2 we see that $Z \not= X$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal sheaf cutting out $Z$. Consider the short exact sequence

$0 \to \mathcal{I}\mathcal{L} \to \mathcal{L} \to \mathcal{O}_ Z \to 0$

If $Z \not= \emptyset$, then $H^1(X, \mathcal{I}\mathcal{L})$ is nonzero as follows from the long exact sequence of cohomology. By Lemma 53.4.2 this gives a nonzero and hence injective map

$\mathcal{I}\mathcal{L} \longrightarrow \omega _ X$

In particular, we find an injective map $H^0(X, \mathcal{L}) = H^0(X, \mathcal{I}\mathcal{L}) \to H^0(X, \omega _ X)$. This is impossible as

$\dim _ k H^0(X, \mathcal{L}) = \dim _ k H^1(X, \mathcal{L}) + \deg (\mathcal{L}) + 1 - g \geq g + 1$

and $\dim H^0(X, \omega _ X) = g$ by (53.8.1.1). $\square$

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).