Lemma 34.8.7. Let $S$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ S$-module. Let $\tau \in \{ Zariski, \linebreak fppf, \linebreak {\acute{e}tale}, \linebreak smooth, \linebreak syntomic\}$.

1. The sheaf $\mathcal{F}^ a$ is a quasi-coherent $\mathcal{O}$-module on $(\mathit{Sch}/S)_\tau$, as defined in Modules on Sites, Definition 18.23.1.

2. If $\tau = {\acute{e}tale}$ (resp. $\tau = Zariski$), then the sheaf $\mathcal{F}^ a$ is a quasi-coherent $\mathcal{O}$-module on $S_{\acute{e}tale}$ (resp. $S_{Zar}$) as defined in Modules on Sites, Definition 18.23.1.

Proof. Let $\{ S_ i \to S\}$ be a Zariski covering such that we have exact sequences

$\bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{S_ i} \longrightarrow \bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{S_ i} \longrightarrow \mathcal{F} \longrightarrow 0$

for some index sets $K_ i$ and $J_ i$. This is possible by the definition of a quasi-coherent sheaf on a ringed space (See Modules, Definition 17.10.1).

Proof of (1). Let $\tau \in \{ Zariski, \linebreak fppf, \linebreak {\acute{e}tale}, \linebreak smooth, \linebreak syntomic\}$. It is clear that $\mathcal{F}^ a|_{(\mathit{Sch}/S_ i)_\tau }$ also sits in an exact sequence

$\bigoplus \nolimits _{k \in K_ i} \mathcal{O}|_{(\mathit{Sch}/S_ i)_\tau } \longrightarrow \bigoplus \nolimits _{j \in J_ i} \mathcal{O}|_{(\mathit{Sch}/S_ i)_\tau } \longrightarrow \mathcal{F}^ a|_{(\mathit{Sch}/S_ i)_\tau } \longrightarrow 0$

Hence $\mathcal{F}^ a$ is quasi-coherent by Modules on Sites, Lemma 18.23.3.

Proof of (2). Let $\tau = {\acute{e}tale}$. It is clear that $\mathcal{F}^ a|_{(S_ i)_{\acute{e}tale}}$ also sits in an exact sequence

$\bigoplus \nolimits _{k \in K_ i} \mathcal{O}|_{(S_ i)_{\acute{e}tale}} \longrightarrow \bigoplus \nolimits _{j \in J_ i} \mathcal{O}|_{(S_ i)_{\acute{e}tale}} \longrightarrow \mathcal{F}^ a|_{(S_ i)_{\acute{e}tale}} \longrightarrow 0$

Hence $\mathcal{F}^ a$ is quasi-coherent by Modules on Sites, Lemma 18.23.3. The case $\tau = Zariski$ is similar (actually, it is really tautological since the corresponding ringed topoi agree). $\square$

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