Lemma 53.7.1. Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume

1. $\mathcal{L}$ has a regular global section,

2. $H^1(X, \mathcal{L}) = 0$, and

3. $\mathcal{L}$ is ample.

Then $\mathcal{L}^{\otimes 6}$ is very ample on $X$ over $k$.

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

$0 \to \mathcal{O}_ X \xrightarrow {s} \mathcal{L} \to i_*(\mathcal{L}|_ Z) \to 0$

Tensoring with $\mathcal{L}$ we obtain

$0 \to \mathcal{L} \to \mathcal{L}^{\otimes 2} \to i_*(\mathcal{L}^{\otimes 2}|_ Z) \to 0$

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

$0 \to \mathcal{L}^{\otimes n} \xrightarrow {s} \mathcal{L}^{\otimes n + 1} \to i_*(\mathcal{L}^{\otimes n + 1}|_ Z) \to 0$

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

$\mu _ n : H^0(X, \mathcal{L}) \otimes _ k H^0(X, \mathcal{L}^{\otimes n}) \oplus H^0(X, \mathcal{L}^{\otimes 2}) \otimes _ k H^0(X, \mathcal{L}^{\otimes n - 1}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1})$

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

$0 \to \mathcal{L}^{\otimes n} \xrightarrow {s} \mathcal{L}^{\otimes n + 1} \to i_*(\mathcal{L}^{\otimes n + 1}|_ Z) \to 0$

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

$S = \bigoplus \nolimits _{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})$

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. $\square$

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