Lemma 87.37.1. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Let $T \subset |X_{red}|$ be a closed subset. Then the functor
is an affine formal algebraic space.
Lemma 87.37.1. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Let $T \subset |X_{red}|$ be a closed subset. Then the functor
is an affine formal algebraic space.
Proof. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. Then $X_{\lambda , red} = X_{red}$ and we may and do view $T$ as a closed subset of $|X_\lambda | = |X_{\lambda , red}|$. By Lemma 87.14.1 for each $\lambda $ the completion $(X_\lambda )_{/T}$ is an affine formal algebraic space. The transition morphisms $(X_\lambda )_{/T} \to (X_\mu )_{/T}$ are closed immersions as base changes of the transition morphisms $X_\lambda \to X_\mu $, see Lemma 87.14.4. Also the morphisms $((X_\lambda )_{/T})_{red} \to ((X_\mu )_/T)_{red}$ are isomorphisms by Lemma 87.14.5. Since $X_{/T} = \mathop{\mathrm{colim}}\nolimits (X_\lambda )_{/T}$ we conclude by Lemma 87.36.1. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)