The Stacks project

Lemma 95.14.4. Up to a replacement as in Stacks, Remark 8.4.9 the functor

\[ p : \mathcal{G}/\mathcal{B}\textit{-Torsors} \longrightarrow (\mathit{Sch}/S)_{fppf} \]

defines a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$.

Proof. This proof is a repeat of the proof of Lemma 95.14.2. The reader is encouraged to read that proof first since the notation is less cumbersome. The most difficult part of the proof is to show that we have descent for objects. Let $\{ U_ i \to U\} _{i \in I}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Suppose that for each $i$ we are given a pair $(b_ i, \mathcal{F}_ i)$ consisting of a morphism $b_ i : U_ i \to \mathcal{B}$ and a $U_ i \times _{b_ i, \mathcal{B}} \mathcal{G}$-torsor $\mathcal{F}_ i$, and for each $i, j \in I$ we have $b_ i|_{U_ i \times _ U U_ j} = b_ j|_{U_ i \times _ U U_ j}$ and we are given an isomorphism $\varphi _{ij} : \mathcal{F}_ i|_{U_ i \times _ U U_ j} \to \mathcal{F}_ j|_{U_ i \times _ U U_ j}$ of $(U_ i \times _ U U_ j) \times _\mathcal {B} \mathcal{G}$-torsors satisfying a suitable cocycle condition on $U_ i \times _ U U_ j \times _ U U_ k$. Then by Sites, Section 7.26 we obtain a sheaf $\mathcal{F}$ on $(\mathit{Sch}/U)_{fppf}$ whose restriction to each $U_ i$ recovers $\mathcal{F}_ i$ as well as recovering the descent data. By the sheaf axiom for $\mathcal{B}$ the morphisms $b_ i$ come from a unique morphism $b : U \to \mathcal{B}$. By the equivalence of categories in Sites, Lemma 7.26.5 the action maps $(U_ i \times _{b_ i, \mathcal{B}} \mathcal{G}) \times _{U_ i} \mathcal{F}_ i \to \mathcal{F}_ i$ glue to give a map $(U \times _{b, \mathcal{B}} \mathcal{G}) \times \mathcal{F} \to \mathcal{F}$. Now we have to show that this is an action and that $\mathcal{F}$ becomes a $U \times _{b, \mathcal{B}} \mathcal{G}$-torsor. Both properties may be checked locally, and hence follow from the corresponding properties of the actions on the $\mathcal{F}_ i$. This proves that descent for objects holds in $\mathcal{G}/\mathcal{B}\textit{-Torsors}$. Some details omitted. $\square$


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