Lemma 94.5.1. The functor $p_{fg} : \mathcal{QC}\! \mathit{oh}_{fg} \to (\mathit{Sch}/S)_{fppf}$ satisfies conditions (1), (2) and (3) of Stacks, Definition 8.4.1.

Proof. We will verify assumptions (1), (2), (3) of Stacks, Lemma 8.4.3 to prove this. By Lemma 94.4.1 a morphism $(Y, \mathcal{G}) \to (X, \mathcal{F})$ is strongly cartesian if and only if it induces an isomorphism $f^*\mathcal{F} \to \mathcal{G}$. By Modules, Lemma 17.9.2 the pullback of a finite type $\mathcal{O}_ X$-module is of finite type. Hence assumption (1) of Stacks, Lemma 8.4.3 holds. Assumption (2) holds trivially. Finally, to prove assumption (3) we have to show: If $\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ X$-module and $\{ f_ i : X_ i \to X\}$ is an fppf covering such that each $f_ i^*\mathcal{F}$ is of finite type, then $\mathcal{F}$ is of finite type. Considering the restriction of $\mathcal{F}$ to an affine open of $X$ this reduces to the following algebra statement: Suppose that $R \to S$ is a finitely presented, faithfully flat ring map and $M$ an $R$-module. If $M \otimes _ R S$ is a finitely generated $S$-module, then $M$ is a finitely generated $R$-module. A stronger form of the algebra fact can be found in Algebra, Lemma 10.83.2. $\square$

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