The Stacks project

Proof. Combining Varieties, Lemma 33.44.15 and Lemmas 53.22.2 and 53.23.2 we see that $X$ contains no rational tails or bridges. Then we see that $\omega _ X^{\otimes 3}$ is globally generated by Lemma 53.22.6. Choose a $k$-basis $s_0, \ldots , s_ n$ of $H^0(X, \omega _ X^{\otimes 3})$. We get a morphism

See Constructions, Section 27.13. The lemma asserts that this morphism is a closed immersion. To check this we may replace $k$ by its algebraic closure, see Descent, Lemma 35.23.19. Thus we may assume $k$ is algebraically closed.

Assume $k$ is algebraically closed. We will use Varieties, Lemma 33.23.2 to prove the lemma. Let $Z \subset X$ be a closed subscheme of degree $2$ over $Z$ with ideal sheaf $\mathcal{I} \subset \mathcal{O}_ X$. We have to show that

is surjective. Thus it suffices to show that $H^1(X, \mathcal{I}\mathcal{L}) = 0$. To do this we will use Lemma 53.21.6. Thus it suffices to show that

for every reduced connected closed subscheme $Y \subset X$. Since $k$ is algebraically closed and $Y$ connected and reduced we have $H^0(Y, \mathcal{O}_ Y) = k$ (Varieties, Lemma 33.9.3). Hence $\chi (Y, \mathcal{O}_ Y) = 1 - \dim H^1(Y, \mathcal{O}_ Y)$. Thus we have to show

which is true by Lemma 53.22.4 except possibly if $Y = X$ or if $\deg (\omega _ X|_ Y) = 0$. Since $\omega _ X$ is ample the second possibility does not occur (see first lemma cited in this proof). Finally, if $Y = X$ we can use Riemann-Roch (Lemma 53.5.2) and the fact that $g \geq 2$ to see that the inquality holds. The same argument with $Z = \emptyset $ shows that $H^1(X, \omega _ X^{\otimes 3}) = 0$. $\square$