The Stacks project

Lemma 35.11.1. Let $S$ be a scheme. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}_{\textit{Aff}}$-modules on $(\textit{Aff}/S)_{fppf}$. The following are equivalent

  1. for every morphism $U \to U'$ of $(\textit{Aff}/S)_{fppf}$ the map $\mathcal{F}(U') \otimes _{\mathcal{O}(U')} \mathcal{O}(U) \to \mathcal{F}(U)$ is an isomorphism,

  2. $\mathcal{F}$ is a sheaf on $(\textit{Aff}/S)_{Zar}$ and a quasi-coherent module on the ringed site $((\textit{Aff}/S)_{Zar}, \mathcal{O}_{\textit{Aff}})$ in the sense of Modules on Sites, Definition 18.23.1,

  3. same as in (2) for the étale topology,

  4. same as in (2) for the smooth topology,

  5. same as in (2) for the syntomic topology,

  6. same as in (2) for the fppf topology,

  7. $\mathcal{F}$ corresponds to a quasi-coherent module on $(\mathit{Sch}/S)_{Zar}$, $(\mathit{Sch}/S)_{\acute{e}tale}$, $(\mathit{Sch}/S)_{smooth}$, $(\mathit{Sch}/S)_{syntomic}$, or $(\mathit{Sch}/S)_{fppf}$ via the equivalence (,

  8. $\mathcal{F}$ comes from a unique quasi-coherent $\mathcal{O}_ S$-module $\mathcal{G}$ by the procedure described in Section 35.8.

Proof. Since the notion of a quasi-coherent module is intrinsic (Modules on Sites, Lemma 18.23.2) we see that the equivalence ( induces an equivalence between categories of quasi-coherent modules. Proposition 35.8.9 says the topology we use to study quasi-coherent modules on $\mathit{Sch}/S$ does not matter and it also tells us that (8) is the same as (7). Hence we see that (2) – (8) are all equivalent.

Assume the equivalent conditions (2) – (8) hold and let $\mathcal{G}$ be as in (8). Let $h : U \to U' \to S$ be a morphism of $\textit{Aff}/S$. Denote $f : U \to S$ and $f' : U' \to S$ the structure morphisms, so that $f = f' \circ h$. We have $\mathcal{F}(U') = \Gamma (U', (f')^*\mathcal{G})$ and $\mathcal{F}(U) = \Gamma (U, f^*\mathcal{G}) = \Gamma (U, h^*(f')^*\mathcal{G})$. Hence (1) holds by Schemes, Lemma 26.7.3.

Assume (1) holds. To finish the proof it suffices to prove (2). Let $U$ be an object of $(\textit{Aff}/S)_{Zar}$. Say $U = \mathop{\mathrm{Spec}}(R)$. A standard open covering $U = U_1 \cup \ldots \cup U_ n$ is given by $U_ i = D(f_ i)$ for some elements $f_1, \ldots , f_ n \in R$ generating the unit ideal of $R$. By property (1) we see that

\[ \mathcal{F}(U_ i) = \mathcal{F}(U) \otimes _ R R_{f_ i} = \mathcal{F}(U)_{f_ i} \]


\[ \mathcal{F}(U_ i \cap U_ j) = \mathcal{F}(U) \otimes _ R R_{f_ if_ j} = \mathcal{F}(U)_{f_ if_ j} \]

Thus we conclude from Algebra, Lemma 10.24.1 that $\mathcal{F}$ is a sheaf on $(\textit{Aff}/S)_{Zar}$. Choose a presentation

\[ \bigoplus \nolimits _{k \in K} R \longrightarrow \bigoplus \nolimits _{l \in L} R \longrightarrow \mathcal{F}(U) \longrightarrow 0 \]

by free $R$-modules. By property (1) and the right exactness of tensor product we see that for every morphism $U' \to U$ in $(\textit{Aff}/S)_{Zar}$ we obtain a presentation

\[ \bigoplus \nolimits _{k \in K} \mathcal{O}_{Aff}(U') \longrightarrow \bigoplus \nolimits _{l \in L} \mathcal{O}_{Aff}(U') \longrightarrow \mathcal{F}(U') \longrightarrow 0 \]

In other words, we see that the restriction of $\mathcal{F}$ to the localized category $(\textit{Aff}/S)_{Zar}/U$ has a presentation

\[ \bigoplus \nolimits _{k \in K} \mathcal{O}_{Aff}|_{(\textit{Aff}/S)_{Zar}/U} \longrightarrow \bigoplus \nolimits _{l \in L} \mathcal{O}_{Aff}|_{(\textit{Aff}/S)_{Zar}/U} \longrightarrow \mathcal{F}|_{(\textit{Aff}/S)_{Zar}/U} \longrightarrow 0 \]

With apologies for the horrible notation, this finishes the proof. $\square$

Comments (2)

Comment #8723 by Cameron Ruether on

Small typo, it seems points (3)-(6) should read "same as in (2)..."

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