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. $\square$