The Stacks project

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

Hence it follows from the equivalences in Lemma 94.10.11 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

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

which is the composition of base changes of $x$ and $y$, hence is surjective and smooth, see Lemmas 94.10.6 and 94.10.5. 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. $\square$