Lemma 33.44.14. Let $k$ be a field. Let $X$ be a proper curve over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then $\mathcal{L}$ is ample if and only if $\deg (\mathcal{L}) > 0$.

**Proof.**
If $\mathcal{L}$ is ample, then there exists an $n > 0$ and a section $s \in H^0(X, \mathcal{L}^{\otimes n})$ with $X_ s$ affine. Since $X$ isn't affine (otherwise by Morphisms, Lemma 29.44.11 $X$ would be finite), we see that $s$ vanishes at some point. Hence $\deg (\mathcal{L}^{\otimes n}) > 0$ by Lemma 33.44.12. By Lemma 33.44.7 we conclude that $\deg (\mathcal{L}) = 1/n\deg (\mathcal{L}^{\otimes n}) > 0$.

Assume $\deg (\mathcal{L}) > 0$. Then

grows linearly with $n$. Hence for any finite collection of closed points $x_1, \ldots , x_ t$ of $X$, we can find an $n$ such that $\dim _ k H^0(X, \mathcal{L}^{\otimes n}) > \sum \dim _ k \kappa (x_ i)$. (Recall that by Hilbert Nullstellensatz, the extension fields $\kappa (x_ i)/k$ are finite, see for example Morphisms, Lemma 29.20.3). Hence we can find a nonzero $s \in H^0(X, \mathcal{L}^{\otimes n})$ vanishing in $x_1, \ldots , x_ t$. In particular, if we choose $x_1, \ldots , x_ t$ such that $X \setminus \{ x_1, \ldots , x_ t\} $ is affine, then $X_ s$ is affine too (for example by Properties, Lemma 28.26.4 although if we choose our finite set such that $\mathcal{L}|_{X \setminus \{ x_1, \ldots , x_ t\} }$ is trivial, then it is immediate). The conclusion is that we can find an $n > 0$ and a nonzero section $s \in H^0(X, \mathcal{L}^{\otimes n})$ such that $X_ s$ is affine.

We will show that for every quasi-coherent sheaf of ideals $\mathcal{I}$ there exists an $m > 0$ such that $H^1(X, \mathcal{I} \otimes \mathcal{L}^{\otimes m})$ is zero. This will finish the proof by Cohomology of Schemes, Lemma 30.17.1. To see this we consider the maps

Since $\mathcal{I}$ is torsion free, these maps are injective and isomorphisms over $X_ s$, hence the cokernels have vanishing $H^1$ (by Cohomology of Schemes, Lemma 30.9.10 for example). We conclude that the maps of vector spaces

are surjective. On the other hand, the dimension of $H^1(X, \mathcal{I})$ is finite, and every element maps to zero eventually by Cohomology of Schemes, Lemma 30.17.4. Thus for some $e > 0$ we see that $H^1(X, \mathcal{I} \otimes \mathcal{L}^{\otimes en})$ is zero. This finishes the proof. $\square$

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