Lemma 59.31.2. Let $S$ be a scheme. Let $\mathcal{F}$ be an abelian sheaf on $S_{\acute{e}tale}$. Let $\sigma \in \mathcal{F}(U)$ be a local section. There exists an open subset $W \subset U$ such that

1. $W \subset U$ is the largest Zariski open subset of $U$ such that $\sigma |_ W = 0$,

2. for every $\varphi : V \to U$ in $S_{\acute{e}tale}$ we have

$\sigma |_ V = 0 \Leftrightarrow \varphi (V) \subset W,$
3. for every geometric point $\overline{u}$ of $U$ we have

$(U, \overline{u}, \sigma ) = 0\text{ in }\mathcal{F}_{\overline{s}} \Leftrightarrow \overline{u} \in W$

where $\overline{s} = (U \to S) \circ \overline{u}$.

Proof. Since $\mathcal{F}$ is a sheaf in the étale topology the restriction of $\mathcal{F}$ to $U_{Zar}$ is a sheaf on $U$ in the Zariski topology. Hence there exists a Zariski open $W$ having property (1), see Modules, Lemma 17.5.2. Let $\varphi : V \to U$ be an arrow of $S_{\acute{e}tale}$. Note that $\varphi (V) \subset U$ is an open subset and that $\{ V \to \varphi (V)\}$ is an étale covering. Hence if $\sigma |_ V = 0$, then by the sheaf condition for $\mathcal{F}$ we see that $\sigma |_{\varphi (V)} = 0$. This proves (2). To prove (3) we have to show that if $(U, \overline{u}, \sigma )$ defines the zero element of $\mathcal{F}_{\overline{s}}$, then $\overline{u} \in W$. This is true because the assumption means there exists a morphism of étale neighbourhoods $(V, \overline{v}) \to (U, \overline{u})$ such that $\sigma |_ V = 0$. Hence by (2) we see that $V \to U$ maps into $W$, and hence $\overline{u} \in W$. $\square$

There are also:

• 2 comment(s) on Section 59.31: Supports of abelian sheaves

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).