Proof.
Choose a covering $\{ Y_ j \to Y\} $ as in Definition 87.11.1. For each $j$ choose a covering $\{ X_{ji} \to Y_ j \times _ Y X\} $ as in Definition 87.11.1. For each $j$ choose a covering $\{ Z_{jk} \to Y_ j \times _ Y Z\} $ as in Definition 87.11.1. Observe that $X_{ji} \times _{Y_ j} Z_{jk}$ is an affine formal algebraic space by Lemma 87.16.4. Hence
\[ \{ X_{ji} \times _{Y_ j} Z_{jk} \to X \times _ Y Z\} \]
is a covering as in Definition 87.11.1. Thus it suffices to prove (1), (2), (3), and (4) in case $X$, $Y$, and $Z$ are affine formal algebraic spaces.
Assume $X$ and $Z$ are countably indexed. Say $X = \mathop{\mathrm{colim}}\nolimits X_ n$ and $Z = \mathop{\mathrm{colim}}\nolimits Z_ m$ as in Lemma 87.10.1. Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda $ as in Definition 87.9.1. For each $n$ and $m$ we can find $\lambda _{n, m} \in \Lambda $ such that $X_ n \to Y$ and $Z_ m \to Y$ factor through $Y_{\lambda _{n, m}}$ (for example see Lemma 87.9.4). Pick $\lambda _0 \in \Lambda $. By induction for $t \geq 1$ pick an element $\lambda _ t \in \Lambda $ such that $\lambda _ t \geq \lambda _{n, m}$ for all $1 \leq n, m \leq t$ and $\lambda _ t \geq \lambda _{t - 1}$. Set $Y' = \mathop{\mathrm{colim}}\nolimits Y_{\lambda _ t}$. Then $Y' \to Y$ is a monomorphism such that $X \to Y$ and $Z \to Y$ factor through $Y'$. Hence we may replace $Y$ by $Y'$, i.e., we may assume that $Y$ is countably indexed.
Assume $X$, $Y$, and $Z$ are countably indexed. By Lemma 87.10.4 we can write $X = \text{Spf}(A)$, $Y = \text{Spf}(B)$, $Z = \text{Spf}(C)$ for some weakly admissible topological rings $A$, $B$, and $C$. The morphsms $X \to Y$ and $Z \to Y$ are given by continuous ring maps $B \to A$ and $B \to C$, see Lemma 87.9.10. By Lemma 87.16.4 we see that $X \times _ Y Z = \text{Spf}(A \widehat{\otimes }_ B C)$ and that $A \widehat{\otimes }_ B C$ is a weakly admissible topological ring. In particular, we see that $X \times _ Y Z$ is countably indexed by Lemma 87.4.12 part (3). This proves (1).
Proof of (2). In this case $X$ and $Z$ are countably indexed and hence the arguments above show that $X \times _ Y Z$ is the formal spectrum of $A \widehat{\otimes }_ B C$ where $A$ and $C$ are admissible. Then $A \widehat{\otimes }_ B C$ is admissible by Lemma 87.4.12 part (2).
Proof of (3). As before we conclude that $X \times _ Y Z$ is the formal spectrum of $A \widehat{\otimes }_ B C$ where $A$ and $C$ are weakly adic. Then $A \widehat{\otimes }_ B C$ is weakly adic by Lemma 87.7.6.
Proof of (4). Arguing as above, this follows from Lemma 87.4.12 part (4).
Proof of (5). To deduce case (5) from Lemma 87.4.12 part (5) we need to show the hypotheses match. Namely, with notation as in the first parapgrah of the proof, if $X_{red} \to Y_{red}$ is locally of finite type, then $(X_{ji})_{red} \to (Y_ j)_{red}$ is locally of finite type. This follows from Morphisms of Spaces, Lemma 67.23.4 and the fact that in the commutative diagram
\[ \xymatrix{ (X_{ji})_{red} \ar[d] \ar[r] & (Y_ j)_{red} \ar[d] \\ X_{red} \ar[r] & Y_{red} } \]
the vertical morphisms are étale. Namely, we have $(X_{ji})_{red} = X_{ij} \times _ X X_{red}$ and $(Y_ j)_{red} = Y_ j \times _ Y Y_{red}$ by Lemma 87.12.3. Thus as above we reduce to the case where $X$, $Y$, $Z$ are affine formal algebraic spaces, $X$, $Z$ are Noetherian, and $X_{red} \to Y_{red}$ is of finite type. Next, in the second paragraph of the proof we replaced $Y$ by $Y'$ but by construction $Y_{red} = Y'_{red}$, hence the finite type assumption is preserved by this replacement. Then we see that $X, Y, Z$ correspond to $A, B, C$ and $X \times _ Y Z$ to $A \widehat{\otimes }_ B C$ with $A$, $C$ Noetherian adic. Finally, taking the reduction corresponds to dividing by the ideal of topologically nilpotent elements (Example 87.12.2) hence the fact that $X_{red} \to Y_{red}$ is of finite type does indeed mean that $B/\mathfrak b \to A/\mathfrak a$ is of finite type and the proof is complete.
$\square$
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