Lemma 68.17.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Assume $f$ is finite, surjective and $X$ locally Noetherian. Let $i : Z \to X$ be a closed immersion. Denote $i' : Z' \to Y$ the inverse image of $Z$ (Morphisms of Spaces, Section 66.13) and $f' : Z' \to Z$ the induced morphism. Then $\mathcal{G} = f'_*\mathcal{O}_{Z'}$ is a coherent $\mathcal{O}_ Z$-module whose support is $Z$.
Proof. Observe that $f'$ is the base change of $f$ and hence is finite and surjective by Morphisms of Spaces, Lemmas 66.5.5 and 66.45.5. Note that $Y$, $Z$, and $Z'$ are locally Noetherian by Morphisms of Spaces, Lemma 66.23.5 (and the fact that closed immersions and finite morphisms are of finite type). By Lemma 68.12.9 we see that $\mathcal{G}$ is a coherent $\mathcal{O}_ Z$-module. The support of $\mathcal{G}$ is closed in $|Z|$, see Morphisms of Spaces, Lemma 66.15.2. Hence if the support of $\mathcal{G}$ is not equal to $|Z|$, then after replacing $X$ by an open subspace we may assume $\mathcal{G} = 0$ but $Z \not= \emptyset $. This would mean that $f'_*\mathcal{O}_{Z'} = 0$. In particular the section $1 \in \Gamma (Z', \mathcal{O}_{Z'}) = \Gamma (Z, f'_*\mathcal{O}_{Z'})$ would be zero which would imply $Z' = \emptyset $ is the empty algebraic space. This is impossible as $Z' \to Z$ is surjective. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)