94.19 Change of base scheme
In this section we briefly discuss what happens when we change base schemes. The upshot is that given a morphism S \to S' of base schemes, any algebraic stack over S can be viewed as an algebraic stack over S'.
Lemma 94.19.1. Let \mathit{Sch}_{fppf} be a big fppf site. Let S \to S' be a morphism of this site. The constructions A and B of Stacks, Section 8.13 above give isomorphisms of 2-categories
\left\{ \begin{matrix} 2\text{-category of algebraic}
\\ \text{stacks }\mathcal{X}\text{ over }S
\end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} 2\text{-category of pairs }(\mathcal{X}', f)\text{ consisting of an}
\\ \text{algebraic stack }\mathcal{X}'\text{ over }S'\text{ and a morphism}
\\ f : \mathcal{X}' \to (\mathit{Sch}/S)_{fppf}\text{ of algebraic stacks over }S'
\end{matrix} \right\}
Proof.
The statement makes sense as the functor j : (\mathit{Sch}/S)_{fppf} \to (\mathit{Sch}/S')_{fppf} is the localization functor associated to the object S/S' of (\mathit{Sch}/S')_{fppf}. By Stacks, Lemma 8.13.2 the only thing to show is that the constructions A and B preserve the subcategories of algebraic stacks. For example, if \mathcal{X} = [U/R] then construction A applied to \mathcal{X} just produces \mathcal{X}' = \mathcal{X}. Conversely, if \mathcal{X}' = [U'/R'] the morphism p induces morphisms of algebraic spaces U' \to S and R' \to S, and then \mathcal{X} = [U'/R'] but now viewed as a stack over S. Hence the lemma is clear.
\square
Definition 94.19.2. Let \mathit{Sch}_{fppf} be a big fppf site. Let S \to S' be a morphism of this site. If p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf} is an algebraic stack over S, then \mathcal{X} viewed as an algebraic stack over S' is the algebraic stack
\mathcal{X} \longrightarrow (\mathit{Sch}/S')_{fppf}
gotten by applying construction A of Lemma 94.19.1 to \mathcal{X}.
Conversely, what if we start with an algebraic stack \mathcal{X}' over S' and we want to get an algebraic stack over S? Well, then we consider the 2-fibre product
\mathcal{X}'_ S = (\mathit{Sch}/S)_{fppf} \times _{(\mathit{Sch}/S')_{fppf}} \mathcal{X}'
which is an algebraic stack over S' according to Lemma 94.14.3. Moreover, it comes equipped with a natural 1-morphism p : \mathcal{X}'_ S \to (\mathit{Sch}/S)_{fppf} and hence by Lemma 94.19.1 it corresponds in a canonical way to an algebraic stack over S.
Definition 94.19.3. Let \mathit{Sch}_{fppf} be a big fppf site. Let S \to S' be a morphism of this site. Let \mathcal{X}' be an algebraic stack over S'. The change of base of \mathcal{X}' is the algebraic stack \mathcal{X}'_ S over S described above.
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