The Stacks project

86.11 Fibre products

Obligatory section about fibre products of formal algebraic spaces.

Lemma 86.11.1. Let $S$ be a scheme. Let $\{ X_ i \to X\} _{i \in I}$ be a family of maps of sheaves on $(\mathit{Sch}/S)_{fppf}$. Assume (a) $X_ i$ is a formal algebraic space over $S$, (b) $X_ i \to X$ is representable by algebraic spaces and ├ętale, and (c) $\coprod X_ i \to X$ is a surjection of sheaves. Then $X$ is a formal algebraic space over $S$.

Proof. For each $i$ pick $\{ X_{ij} \to X_ i\} _{j \in J_ i}$ as in Definition 86.7.1. Then $\{ X_{ij} \to X\} _{i \in I, j \in J_ i}$ is a family as in Definition 86.7.1 for $X$. $\square$

Lemma 86.11.2. Let $S$ be a scheme. Let $X, Y$ be formal algebraic spaces over $S$ and let $Z$ be a sheaf whose diagonal is representable by algebraic spaces. Let $X \to Z$ and $Y \to Z$ be maps of sheaves. Then $X \times _ Z Y$ is a formal algebraic space.

Proof. Choose $\{ X_ i \to X\} $ and $\{ Y_ j \to Y\} $ as in Definition 86.7.1. Then $\{ X_ i \times _ Z Y_ j \to X \times _ Z Y\} $ is a family of maps which are representable by algebraic spaces and ├ętale. Thus Lemma 86.11.1 tells us it suffices to show that $X \times _ Z Y$ is a formal algebraic space when $X$ and $Y$ are affine formal algebraic spaces.

Assume $X$ and $Y$ are affine formal algebraic spaces. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ and $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ as in Definition 86.5.1. Then $X \times _ Z Y = \mathop{\mathrm{colim}}\nolimits X_\lambda \times _ Z Y_\mu $. Each $X_\lambda \times _ Z Y_\mu $ is an algebraic space. For $\lambda \leq \lambda '$ and $\mu \leq \mu '$ the morphism

\[ X_\lambda \times _ Z Y_\mu \to X_\lambda \times _ Z Y_{\mu '} \to X_{\lambda '} \times _ Z Y_{\mu '} \]

is a thickening as a composition of base changes of thickenings. Thus we conclude by applying Lemma 86.9.1. $\square$

Lemma 86.11.3. Let $S$ be a scheme. The category of formal algebraic spaces over $S$ has fibre products.

Proof. Special case of Lemma 86.11.2 because formal algebraic spaces have representable diagonals, see Lemma 86.7.2. $\square$

Lemma 86.11.4. Let $S$ be a scheme. Let $X \to Z$ and $Y \to Z$ be morphisms of formal algebraic spaces over $S$. Then $(X \times _ Z Y)_{red} = (X_{red} \times _{Z_{red}} Y_{red})_{red}$.

Proof. This follows from the universal property of the reduction in Lemma 86.8.1. $\square$

We have already proved the following lemma (without knowing that fibre products exist).

Lemma 86.11.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. The diagonal morphism $\Delta : X \to X \times _ Y X$ is representable (by schemes), a monomorphism, locally quasi-finite, locally of finite type, and separated.

Proof. Let $T$ be a scheme and let $T \to X \times _ Y X$ be a morphism. Then

\[ T \times _{(X \times _ Y X)} X = T \times _{(X \times _ S X)} X \]

Hence the result follows immediately from Lemma 86.7.2. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AJ0. Beware of the difference between the letter 'O' and the digit '0'.