Lemma 30.3.3. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume that

1. $X$ is quasi-compact,

2. for every quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ there exists an $n \geq 1$ such that $H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$.

Then $\mathcal{L}$ is ample.

Proof. This is proved in exactly the same way as Lemma 30.3.1. Let $x \in X$ be a closed point. Let $U \subset X$ be an affine open neighbourhood of $x$ such that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Write $U = \mathop{\mathrm{Spec}}(A)$ and let $\mathfrak m \subset A$ be the maximal ideal corresponding to $x$. Set $Z = X \setminus U$ and $Z' = Z \cup \{ x\}$. By Schemes, Lemma 26.12.4 there are quasi-coherent sheaves of ideals $\mathcal{I}$, resp. $\mathcal{I}'$ cutting out the reduced closed subschemes $Z$, resp. $Z'$. Consider the short exact sequence

$0 \to \mathcal{I}' \to \mathcal{I} \to \mathcal{I}/\mathcal{I}' \to 0.$

For every $n \geq 1$ we obtain a short exact sequence

$0 \to \mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to \mathcal{I}/\mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to 0.$

By our assumption we may pick $n$ such that $H^1(X, \mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = 0$. Since $x$ is a closed point of $X$ and $x \not\in Z$ we see that $\mathcal{I}/\mathcal{I}'$ is supported at $x$. In fact, the restriction of $\mathcal{I}/\mathcal{I'}$ to $U$ corresponds to the $A$-module $A/\mathfrak m$. Since $\mathcal{L}$ is trivial on $U$ we see that the restriction of $\mathcal{I}/\mathcal{I}' \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}$ to $U$ also corresponds to the $A$-module $A/\mathfrak m$. Hence we see that $\Gamma (X, \mathcal{I}/\mathcal{I'} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = A/\mathfrak m$. By our choice of $n$ we see there exists a global section $s \in \Gamma (X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$ which maps to the element $1 \in A/\mathfrak m$. Clearly we have $x \in X_ s \subset U$ because $s$ vanishes at points of $Z$. This implies that $X_ s = D(f)$ where $f \in A$ is the image of $s$ in $A \cong \Gamma (U, \mathcal{L}^{\otimes n})$. In particular $X_ s$ is affine.

Consider the union $W = \bigcup X_ s$ over all $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ for $n \geq 1$ such that $X_ s$ is affine. Obviously $W$ is open in $X$. By the arguments above every closed point of $X$ is contained in $W$. The closed subset $X \setminus W$ of $X$ is also quasi-compact (see Topology, Lemma 5.12.3). Hence it has a closed point if it is nonempty (see Topology, Lemma 5.12.8). This would contradict the fact that all closed points are in $W$. Hence we conclude $X = W$. This means that $\mathcal{L}$ is ample by Properties, Definition 28.26.1. $\square$

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