The Stacks project

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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