Lemma 86.16.4. Let $S$ be a scheme. Let $X \to Z$ and $Y \to Z$ be morphisms of formal algebraic spaces over $S$. Assume $Z$ separated.

1. If $X$ and $Y$ are affine formal algebraic spaces, then so is $X \times _ Z Y$.

2. If $X$ and $Y$ are McQuillan affine formal algebraic spaces, then so is $X \times _ Z Y$.

3. If $X$, $Y$, and $Z$ are McQuillan affine formal algebraic spaces corresponding to the weakly admissible topological $S$-algebras $A$, $B$, and $C$, then $X \times _ Z Y$ corresponds to $A \widehat{\otimes }_ C B$.

Proof. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ and $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu$ as in Definition 86.9.1. Then $X \times _ Z Y = \mathop{\mathrm{colim}}\nolimits X_\lambda \times _ Z Y_\mu$. Since $Z$ is separated the fibre products are affine, hence we see that (1) holds. Assume $X$ and $Y$ corresponds to the weakly admissible topological $S$-algebras $A$ and $B$ and $X_\lambda = \mathop{\mathrm{Spec}}(A/I_\lambda )$ and $Y_\mu = \mathop{\mathrm{Spec}}(B/J_\mu )$. Then

$X_\lambda \times _ Z Y_\mu \to X_\lambda \times Y_\mu \to \mathop{\mathrm{Spec}}(A \otimes B)$

is a closed immersion. Thus one of the conditions of Lemma 86.9.6 holds and we conclude that $X \times _ Z Y$ is McQuillan. If also $Z$ is McQuillan corresponding to $C$, then

$X_\lambda \times _ Z Y_\mu = \mathop{\mathrm{Spec}}(A/I_\lambda \otimes _ C B/J_\mu )$

hence we see that the weakly admissible topological ring corresponding to $X \times _ Z Y$ is the completed tensor product (see Definition 86.4.7). $\square$

Comment #1559 by Matthew Emerton on

In statement (3), I think you want to say that they are McQuillan affine formal algebraic spaces (just as you do in (2)).

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