**Proof.**
By definition of degree and genus we have

\[ \dim _ k H^0(X, \mathcal{L}) - \dim _ k H^1(X, \mathcal{L}) = d + 1 \]

Let $s$ be a nonzero section of $\mathcal{L}$. Then the zero scheme of $s$ is an effective Cartier divisor $D \subset X$, we have $\mathcal{L} = \mathcal{O}_ X(D)$ and we have a short exact sequence

\[ 0 \to \mathcal{O}_ X \to \mathcal{L} \to \mathcal{L}|_ D \to 0 \]

see Divisors, Lemma 31.14.10 and Remark 31.14.11. Since $H^1(X, \mathcal{O}_ X) = 0$ by assumption, we see that $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_ D)$ is surjective. As $\mathcal{L}|_ D$ is generated by global sections (because $\dim (D) = 0$, see Varieties, Lemma 33.33.3) we conclude that the invertible module $\mathcal{L}$ is generated by global sections. In fact, since $D$ is an Artinian scheme we have $\mathcal{L}|_ D \cong \mathcal{O}_ D$^{1} and hence we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_ D$. The short exact sequence also shows that $H^1(X, \mathcal{L}) = 0$.

For $n \geq 1$ consider the multiplication map

\[ \mu _ n : H^0(X, \mathcal{L}) \otimes _ k H^0(X, \mathcal{L}^{\otimes n}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1}) \]

We claim this is surjective. To see this we consider the short exact sequence

\[ 0 \to \mathcal{L}^{\otimes n} \xrightarrow {s} \mathcal{L}^{\otimes n + 1} \to \mathcal{L}^{\otimes n + 1}|_ D \to 0 \]

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 $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_ D)$ is surjective and since $t^ n$ maps to a trivialization of $\mathcal{L}^{\otimes n}|_ D$ we see that $\mu _ n(H^0(X, \mathcal{L}) \otimes t^ n)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes n + 1})$ surjecting onto the global sections of $\mathcal{L}^{\otimes n + 1}|_ D$. This proves the claim.

Observe that $\mathcal{L}$ is ample by Varieties, Lemma 33.44.14. Hence Morphisms, Lemma 29.43.17 gives an isomorphism

\[ X \longrightarrow \text{Proj}\left( \bigoplus \nolimits _{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\right) \]

Since the maps $\mu _ n$ are surjective for all $n \geq 1$ we see that the graded algebra on the right hand side is a quotient of the symmetric algebra on $H^0(X, \mathcal{L})$. Choosing a $k$-basis $s_0, \ldots , s_ d$ of $H^0(X, \mathcal{L})$ we see that it is a quotient of a polynomial algebra in $d + 1$ variables. Since quotients of graded rings correspond to closed immersions of $\text{Proj}$ (Constructions, Lemma 27.11.5) we find a closed immersion $X \to \mathbf{P}^ d_ k$. We omit the verification that this morphism is the morphism of Constructions, Lemma 27.13.1 associated to the sections $s_0, \ldots , s_ d$ of $\mathcal{L}$.
$\square$

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