The Stacks project

Proof. Choose basis $s_0, \ldots , s_ n$ of $H^0(X, \mathcal{L}^{\otimes 2})$ over $k$. By property (1) we see that $\mathcal{L}^{\otimes 2}$ is globally generated and 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. For each generic point $\eta _ i \in X$ let $V_ i \subset H^0(X, \mathcal{L})$ be the $k$-subvector space of sections vanishing at $\eta _ i$. Since $\mathcal{L}$ is globally generated, we see that $V_ i \not= H^0(X, \mathcal{L})$. Since $X$ has only a finite number of irreducible components and $k$ is infinite, we can find $s \in H^0(X, \mathcal{L})$ nonvanishing at $\eta _ i$ for all $i$. Then $s$ is a regular section of $\mathcal{L}$ (because $X$ is Cohen-Macaulay by Lemma 53.6.1 and hence $\mathcal{L}$ has no embedded associated points).

In particular, all of the statements given in the proof of Lemma 53.7.1 hold with this $s$. Moreover, as $\mathcal{L}$ is globally generated, we can find a global section $t \in H^0(X, \mathcal{L})$ such that $t|_ Z$ is nonvanishing (argue as above using the finite number of points of $Z$). Then in the proof of Lemma 53.7.1 we can use $t$ to see that additionally the multiplication map

is surjective. Thus

is generated in degrees $0, 1, 2$ over $k$. Arguing as in the proof of Lemma 53.7.1 we find that $S^{(2)} = \bigoplus _{n} S_{2n}$ is generated in degree $1$. This means that $\mathcal{L}^{\otimes 2}$ is very ample as desired. Some details omitted. $\square$