The Stacks project

Lemma 87.16.4. 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 87.9.1. 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 87.9.6 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 87.4.7). $\square$

Comments (2)

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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