The Stacks project

Lemma 30.17.5. Let $i : Z \to X$ be a closed immersion of Noetherian schemes inducing a homeomorphism of underlying topological spaces. Let $\mathcal{L}$ be an invertible sheaf on $X$. Then $i^*\mathcal{L}$ is ample on $Z$, if and only if $\mathcal{L}$ is ample on $X$.

Proof. If $\mathcal{L}$ is ample, then $i^*\mathcal{L}$ is ample for example by Morphisms, Lemma 29.37.7. Assume $i^*\mathcal{L}$ is ample. We have to show that $\mathcal{L}$ is ample on $X$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the coherent sheaf of ideals cutting out the closed subscheme $Z$. Since $i(Z) = X$ set theoretically we see that $\mathcal{I}^ n = 0$ for some $n$ by Lemma 30.10.2. Consider the sequence

\[ X = Z_ n \supset Z_{n - 1} \supset Z_{n - 2} \supset \ldots \supset Z_1 = Z \]

of closed subschemes cut out by $0 = \mathcal{I}^ n \subset \mathcal{I}^{n - 1} \subset \ldots \subset \mathcal{I}$. Then each of the closed immersions $Z_ i \to Z_{i - 1}$ is defined by a coherent sheaf of ideals of square zero. In this way we reduce to the case that $\mathcal{I}^2 = 0$.

Consider the short exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}_ X \to i_*\mathcal{O}_ Z \to 0 \]

of quasi-coherent $\mathcal{O}_ X$-modules. Tensoring with $\mathcal{L}^{\otimes n}$ we obtain short exact sequences
\begin{equation} \label{coherent-equation-ses} 0 \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \to \mathcal{L}^{\otimes n} \to i_*i^*\mathcal{L}^{\otimes n} \to 0 \end{equation}

As $\mathcal{I}^2 = 0$, we can use Morphisms, Lemma 29.4.1 to think of $\mathcal{I}$ as a quasi-coherent $\mathcal{O}_ Z$-module and then $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} = \mathcal{I} \otimes _{\mathcal{O}_ Z} i^*\mathcal{L}^{\otimes n}$ with obvious abuse of notation. Moreover, the cohomology of this sheaf over $Z$ is canonically the same as the cohomology of this sheaf over $X$ (as $i$ is a homeomorphism).

Let $x \in X$ be a point and denote $z \in Z$ the corresponding point. Because $i^*\mathcal{L}$ is ample there exists an $n$ and a section $s \in \Gamma (Z, i^*\mathcal{L}^{\otimes n})$ with $z \in Z_ s$ and with $Z_ s$ affine. The obstruction to lifting $s$ to a section of $\mathcal{L}^{\otimes n}$ over $X$ is the boundary

\[ \xi = \partial s \in H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) = H^1(Z, \mathcal{I} \otimes _{\mathcal{O}_ Z} i^*\mathcal{L}^{\otimes n}) \]

coming from the short exact sequence of sheaves ( If we replace $s$ by $s^{e + 1}$ then $\xi $ is replaced by $\partial (s^{e + 1}) = (e + 1) s^ e \xi $ in $H^1(Z, \mathcal{I} \otimes _{\mathcal{O}_ Z} i^*\mathcal{L}^{\otimes (e + 1)n})$ because the boundary map for

\[ 0 \to \bigoplus \nolimits _{m \geq 0} \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes m} \to \bigoplus \nolimits _{m \geq 0} \mathcal{L}^{\otimes m} \to \bigoplus \nolimits _{m \geq 0} i_*i^*\mathcal{L}^{\otimes m} \to 0 \]

is a derivation by Cohomology, Lemma 20.25.5. By Lemma 30.17.4 we see that $s^ e \xi $ is zero for $e$ large enough. Hence, after replacing $s$ by a power, we can assume $s$ is the image of a section $s' \in \Gamma (X, \mathcal{L}^{\otimes n})$. Then $X_{s'}$ is an open subscheme and $Z_ s \to X_{s'}$ is a surjective closed immersion of Noetherian schemes with $Z_ s$ affine. Hence $X_ s$ is affine by Lemma 30.13.3 and we conclude that $\mathcal{L}$ is ample. $\square$

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