Lemma 87.18.3. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Then $X$ is an affine formal algebraic space if and only if its reduction $X_{red}$ (Lemma 87.12.1) is affine.
Proof. By Lemmas 87.16.1 and 87.17.1 and Definitions 87.16.3 and 87.17.2 we see that $X$ is quasi-compact and quasi-separated. By Yasuda's lemma (Lemma 87.18.1) we can write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as a filtered colimit of thickenings of algebraic spaces. However, each $X_\lambda $ is affine by Limits of Spaces, Lemma 70.15.3 because $(X_\lambda )_{red} = X_{red}$. Hence $X$ is an affine formal algebraic space by definition. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)