Lemma 59.73.14. Let $X$ be a quasi-compact and quasi-separated scheme. Let $\eta \in X$ be a generic point of an irreducible component of $X$.

1. Let $\mathcal{F}$ be a torsion abelian sheaf on $X_{\acute{e}tale}$ whose stalk $\mathcal{F}_{\overline{\eta }}$ is zero. Then $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ is a filtered colimit of constructible abelian sheaves $\mathcal{F}_ i$ such that for each $i$ the support of $\mathcal{F}_ i$ is contained in a closed subscheme not containing $\eta$.

2. Let $\Lambda$ be a Noetherian ring and $\mathcal{F}$ a sheaf of $\Lambda$-modules on $X_{\acute{e}tale}$ whose stalk $\mathcal{F}_{\overline{\eta }}$ is zero. Then $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ is a filtered colimit of constructible sheaves of $\Lambda$-modules $\mathcal{F}_ i$ such that for each $i$ the support of $\mathcal{F}_ i$ is contained in a closed subscheme not containing $\eta$.

Proof. Proof of (1). We can write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{F}_ i$ with $\mathcal{F}_ i$ constructible abelian by Lemma 59.73.2. Choose $i \in I$. Since $\mathcal{F}|_\eta$ is zero by assumption, we see that there exists an $i'(i) \geq i$ such that $\mathcal{F}_ i|_\eta \to \mathcal{F}_{i'(i)}|_\eta$ is zero, see Lemma 59.71.8. Then $\mathcal{G}_ i = \mathop{\mathrm{Im}}(\mathcal{F}_ i \to \mathcal{F}_{i'(i)})$ is a constructible abelian sheaf (Lemma 59.71.6) whose stalk at $\eta$ is zero. Hence the support $E_ i$ of $\mathcal{G}_ i$ is a constructible subset of $X$ not containing $\eta$. Since $\eta$ is a generic point of an irreducible component of $X$, we see that $\eta \not\in Z_ i = \overline{E_ i}$ by Topology, Lemma 5.15.15. Define a new directed set $I'$ by using the set $I$ with ordering defined by the rule $i_1$ is bigger or equal to $i_2$ if and only if $i_1 \geq i'(i_2)$. Then the sheaves $\mathcal{G}_ i$ form a system over $I'$ with colimit $\mathcal{F}$ and the proof is complete.

The proof in case (2) is exactly the same and we omit it. $\square$

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