The Stacks project

Proof. Let $s$ be a regular global section of $\mathcal{L}$. Let $i : Z = Z(s) \to X$ be the zero scheme of $s$, see Divisors, Section 31.14. By condition (3) we see that $Z \not= \emptyset $ (small detail omitted). Consider the short exact sequence

Tensoring with $\mathcal{L}$ we obtain

Observe that $Z$ has dimension $0$ (Divisors, Lemma 31.13.5) and hence is the spectrum of an Artinian ring (Varieties, Lemma 33.20.2) hence $\mathcal{L}|_ Z \cong \mathcal{O}_ Z$ (Algebra, Lemma 10.78.7). The short exact sequence also shows that $H^1(X, \mathcal{L}^{\otimes 2}) = 0$ (for example using Varieties, Lemma 33.33.3 to see vanishing in the spot on the right). Using induction on $n \geq 1$ and the sequence

we see that $H^1(X, \mathcal{L}^{\otimes n}) = 0$ for $n > 0$ and that there exists a global section $t_{n + 1}$ of $\mathcal{L}^{\otimes n + 1}$ which gives a trivialization of $\mathcal{L}^{\otimes n + 1}|_ Z \cong \mathcal{O}_ Z$.

Consider the multiplication map

We claim this is surjective for $n \geq 3$. To see this we consider the short exact sequence

The sections of $\mathcal{L}^{\otimes n + 1}$ coming from the left in this sequence are in the image of $\mu _ n$. On the other hand, since $H^0(\mathcal{L}^{\otimes 2}) \to H^0(\mathcal{L}^{\otimes 2}|_ Z)$ is surjective (see above) and since $t_{n - 1}$ maps to a trivialization of $\mathcal{L}^{\otimes n - 1}|_ Z$ we see that $\mu _ n(H^0(X, \mathcal{L}^{\otimes 2}) \otimes t_{n - 1})$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes n + 1})$ surjecting onto the global sections of $\mathcal{L}^{\otimes n + 1}|_ Z$. This proves the claim.

From the claim in the previous paragraph we conclude that the graded $k$-algebra

is generated in degrees $0, 1, 2, 3$ over $k$. Recall that $X = \text{Proj}(S)$, see Morphisms, Lemma 29.43.17. Thus $S^{(6)} = \bigoplus _{n} S_{6n}$ is generated in degree $1$. This means that $\mathcal{L}^{\otimes 6}$ is very ample as desired. $\square$

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. $\square$