The Stacks project

Lemma 75.13.1. Let $X$ be a scheme. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The following are equivalent

  1. $\mathcal{F}$ is of finite type as an $\mathcal{O}_ X$-module, and

  2. $\epsilon ^*\mathcal{F}$ is of finite type as an $\mathcal{O}_{\acute{e}tale}$-module on the small étale site of $X$.

Here $\epsilon $ is as in (75.4.0.1).

Proof. The implication (1) $\Rightarrow $ (2) is a general fact, see Modules on Sites, Lemma 18.23.4. Assume (2). By assumption there exists an étale covering $\{ f_ i : X_ i \to X\} $ such that $\epsilon ^*\mathcal{F}|_{(X_ i)_{\acute{e}tale}}$ is generated by finitely many sections. Let $x \in X$. We will show that $\mathcal{F}$ is generated by finitely many sections in a neighbourhood of $x$. Say $x$ is in the image of $X_ i \to X$ and denote $X' = X_ i$. Let $s_1, \ldots , s_ n \in \Gamma (X', \epsilon ^*\mathcal{F}|_{X'_{\acute{e}tale}})$ be generating sections. As $\epsilon ^*\mathcal{F} = \epsilon ^{-1}\mathcal{F} \otimes _{\epsilon ^{-1}\mathcal{O}_ X} \mathcal{O}_{\acute{e}tale}$ we can find an étale morphism $X'' \to X'$ such that $x$ is in the image of $X' \to X$ and such that $s_ i|_{X''} = \sum s_{ij} \otimes a_{ij}$ for some sections $s_{ij} \in \epsilon ^{-1}\mathcal{F}(X'')$ and $a_{ij} \in \mathcal{O}_{\acute{e}tale}(X'')$. Denote $U \subset X$ the image of $X'' \to X$. This is an open subscheme as $f'' : X'' \to X$ is étale (Morphisms, Lemma 29.36.13). After possibly shrinking $X''$ more we may assume $s_{ij}$ come from elements $t_{ij} \in \mathcal{F}(U)$ as follows from the construction of the inverse image functor $\epsilon ^{-1}$. Now we claim that $t_{ij}$ generate $\mathcal{F}|_ U$ which finishes the proof of the lemma. Namely, the corresponding map $\mathcal{O}_ U^{\oplus N} \to \mathcal{F}|_ U$ has the property that its pullback by $f''$ to $X''$ is surjective. Since $f'' : X'' \to U$ is a surjective flat morphism of schemes, this implies that $\mathcal{O}_ U^{\oplus N} \to \mathcal{F}|_ U$ is surjective by looking at stalks and using that $\mathcal{O}_{U, f''(z)} \to \mathcal{O}_{X'', z}$ is faithfully flat for all $z \in X''$. $\square$


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