The Stacks project

Lemma 87.19.9. Let $S$ be a scheme. Let $Y$ be an affine formal algebraic space. Let $f : X \to Y$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ which is representable and affine. Then

  1. $X$ is an affine formal algebraic space,

  2. if $Y$ is countably indexed, then $X$ is countably indexed,

  3. if $Y$ is countably indexed and classical, then $X$ is countably indexed and classical,

  4. if $Y$ is weakly adic, then $X$ is weakly adic,

  5. if $Y$ is adic*, then $X$ is adic*, and

  6. if $Y$ is Noetherian and $f$ is (locally) of finite type, then $X$ is Noetherian.

Proof. Proof of (1). Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda $ as in Definition 87.9.1. Since $f$ is representable and affine, the fibre products $X_\lambda = Y_\lambda \times _ Y X$ are affine. And $X = \mathop{\mathrm{colim}}\nolimits Y_\lambda \times _ Y X$. Thus $X$ is an affine formal algebraic space.

Proof of (2). If $Y$ is countably indexed, then in the argument above we may assume $\Lambda $ is countable. Then we immediately see that $X$ is countably indexed too.

Proof of (3), (4), and (5). In each of these cases the assumptions imply that $Y$ is a countably indexed affine formal algebraic space (Lemma 87.10.3) and hence $X$ is too by (2). Thus we may write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for some weakly admissible topological $S$-algebras $A$ and $B$, see Lemma 87.10.4. By Lemma 87.9.10 the morphism $f$ corresponds to a continuous $S$-algebra homomorphism $\varphi : B \to A$. We see from Lemma 87.19.8 that $\varphi $ is taut. We conclude that (3) follows from Lemma 87.5.9, (4) follows from Lemma 87.7.5, and (5) follows from Lemma 87.6.5.

Proof of (6). Combining (3) with Lemma 87.10.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 87.10.5. First, it tells us the affine schemes $Y_\lambda $ are Noetherian. Then $X_\lambda \to Y_\lambda $ is of finite type, hence $X_\lambda $ is Noetherian too (Morphisms, Lemma 29.15.6). Then the criterion tells us $X$ is Noetherian and the proof is complete. $\square$

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Your email address will not be published.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