Proof.
Since the notion of a quasi-coherent module is intrinsic (Modules on Sites, Lemma 18.23.2) we see that the equivalence (35.11.0.1) 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}
and
\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
Comment #9352 by Stacks project on