Lemma 59.75.6. Let $S$ be a scheme such that every quasi-compact open of $S$ has finite number of irreducible components (for example if $S$ has a Noetherian underlying topological space, or if $S$ is locally Noetherian). Let $\Lambda$ be a Noetherian ring. Let $\mathcal{F}$ be a sheaf of $\Lambda$-modules on $S_{\acute{e}tale}$. The following are equivalent

1. $\mathcal{F}$ is a finite type, locally constant sheaf of $\Lambda$-modules, and

2. all stalks of $\mathcal{F}$ are finite $\Lambda$-modules and all specialization maps $sp : \mathcal{F}_{\overline{s}} \to \mathcal{F}_{\overline{t}}$ are bijective.

Proof. The proof of this lemma is the same as the proof of Lemma 59.75.5. Assume (2). Let $\overline{s}$ be a geometric point of $S$ lying over $s \in S$. In order to prove (1) we have to find an étale neighbourhood $(U, \overline{u})$ of $(S, \overline{s})$ such that $\mathcal{F}|_ U$ is constant. We may and do assume $S$ is affine.

Since $M = \mathcal{F}_{\overline{s}}$ is a finite $\Lambda$-module and $\Lambda$ is Noetherian, we can choose a presentation

$\Lambda ^{\oplus m} \xrightarrow {A} \Lambda ^{\oplus n} \to M \to 0$

for some matrix $A = (a_{ji})$ with coefficients in $\Lambda$. We can choose $(U, \overline{u})$ and elements $\sigma _1, \ldots , \sigma _ n \in \mathcal{F}(U)$ such that $\sum a_{ji}\sigma _ i = 0$ in $\mathcal{F}(U)$ and such that the images of $\sigma _ i$ in $\mathcal{F}_{\overline{s}} = M$ are the images of the standard basis element of $\Lambda ^ n$ in the presentation of $M$ given above. Consider the map

$\varphi : \underline{M} \longrightarrow \mathcal{F}|_ U$

on $U_{\acute{e}tale}$ defined by $\sigma _1, \ldots , \sigma _ n$. This map is a bijection on stalks at $\overline{u}$ by construction. Let us consider the subset

$E = \{ u' \in U \mid \varphi _{\overline{u}'}\text{ is bijective}\} \subset U$

Here $\overline{u}'$ is any geometric point of $U$ lying over $u'$ (the condition is independent of the choice by Remark 59.29.8). The image $u \in U$ of $\overline{u}$ is in $E$. By our assumption on the specialization maps for $\mathcal{F}$, by Remark 59.75.3, and by Lemma 59.75.4 we see that $E$ is closed under specializations and generalizations in the topological space $U$.

After shrinking $U$ we may assume $U$ is affine too. By Descent, Lemma 35.16.3 we see that $U$ has a finite number of irreducible components. After removing the irreducible components which do not pass through $u$, we may assume every irreducible component of $U$ passes through $u$. Since $U$ is a sober topological space it follows that $E = U$ and we conclude that $\varphi$ is an isomorphism by Theorem 59.29.10. Thus (1) follows.

We omit the proof that (1) implies (2). $\square$

Comment #7948 by Owen on

This lemma holds without any noetherian hypothesis on $\Lambda$ if the stalks of $\mathcal F$ are supposed of finite presentation. Also it holds with no assumptions on $S$ or $\Lambda$ if $\mathcal F$ is supposed constructible (SGA 4 IX Proposition 2.11 – the definition of constructible there requires the stalks to be of finite presentation).

Comment #8188 by on

Thanks! Those would indeed be statements we could add in other lemmas. One of the uses of this lemma is to show something is constructible. And in the treatment of constructible sheaves in the Stacks project we have, for better or worse, assumed the coefficient ring $\Lambda$ is Noetherian. Thus we cannot even make the statement (for constructible guys) you are talking about without a significant revision of all of this material. So I will leave this for now.

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