The Stacks project

Lemma 87.29.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces. Assume $Y$ countably indexed. The following are equivalent

  1. $f$ is locally of finite type,

  2. $f$ is of finite type,

  3. $f$ corresponds to a morphism $B \to A$ of $\textit{WAdm}^{count}$ (Section 87.21) satisfying the equivalent conditions of Lemma 87.29.6.

Proof. Since $X$ and $Y$ are affine it is clear that conditions (1) and (2) are equivalent. In cases (1) and (2) the morphism $f$ is representable by algebraic spaces by definition, hence affine by Lemma 87.19.7. Thus if (1) or (2) holds we see that $X$ is countably indexed by Lemma 87.19.9. Write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for topological $S$-algebras $A$ and $B$ in $\textit{WAdm}^{count}$, see Lemma 87.10.4. By Lemma 87.9.10 we see that $f$ corresponds to a continuous map $B \to A$. Hence now the result follows from Lemma 87.29.2. $\square$


Comments (2)

Comment #1974 by Brian Conrad on

The applicability of the Lemma at the end of the 2nd sentence of the proof requires knowing that is representable and affine. So it would help the reader if it is said that under (1) and (2) the map is representable in algebraic spaces by the definitions and thus it is also affine by Tag 0AKN.


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Your email address will not be published.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 0ANV. Beware of the difference between the letter 'O' and the digit '0'.