The Stacks project

Lemma 92.14.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$, $\mathcal{Y}$ be algebraic stacks over $S$. Then $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is an algebraic stack, and is a product in the $2$-category of algebraic stacks over $S$.

Proof. An object of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ over $T$ is just a pair $(x, y)$ where $x$ is an object of $\mathcal{X}_ T$ and $y$ is an object of $\mathcal{Y}_ T$. Hence it is immediate from the definitions that $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is a stack in groupoids. If $(x, y)$ and $(x', y')$ are two objects of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ over $T$, then

\[ \mathit{Isom}((x, y), (x', y')) = \mathit{Isom}(x, x') \times \mathit{Isom}(y, y'). \]

Hence it follows from the equivalences in Lemma 92.10.11 and the fact that the category of algebraic spaces has products that the diagonal of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is representable by algebraic spaces. Finally, suppose that $U, V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, and let $x, y$ be surjective smooth morphisms $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$, $y : (\mathit{Sch}/V)_{fppf} \to \mathcal{Y}$. Note that

\[ (\mathit{Sch}/U \times _ S V)_{fppf} = (\mathit{Sch}/U)_{fppf} \times _{(\mathit{Sch}/S)_{fppf}} (\mathit{Sch}/V)_{fppf}. \]

The object $(\text{pr}_ U^*x, \text{pr}_ V^*y)$ of $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ over $(\mathit{Sch}/U \times _ S V)_{fppf}$ thus defines a $1$-morphism

\[ (\mathit{Sch}/U \times _ S V)_{fppf} \longrightarrow \mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y} \]

which is the composition of base changes of $x$ and $y$, hence is surjective and smooth, see Lemmas 92.10.6 and 92.10.5. We conclude that $\mathcal{X} \times _{(\mathit{Sch}/S)_{fppf}} \mathcal{Y}$ is indeed an algebraic stack. We omit the verification that it really is a product. $\square$

Comments (2)

Comment #2537 by Anonymous on

Minor typo in line 8 of the proof: it should be .

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