## 86.10 Countably indexed affine formal algebraic spaces

These are the affine formal algebraic spaces as in the following lemma.

Lemma 86.10.1. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. The following are equivalent

1. there exists a system $X_1 \to X_2 \to X_3 \to \ldots$ of thickenings of affine schemes over $S$ such that $X = \mathop{\mathrm{colim}}\nolimits X_ n$,

2. there exists a choice $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 86.9.1 such that $\Lambda$ is countable.

Proof. This follows from the observation that a countable directed set has a cofinal subset isomorphic to $(\mathbf{N}, \geq )$. See proof of Algebra, Lemma 10.86.3. $\square$

Definition 86.10.2. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. We say $X$ is countably indexed if the equivalent conditions of Lemma 86.10.1 are satisfied.

In the language of [BVGD] this is expressed by saying that $X$ is an $\aleph _0$-ind scheme.

Lemma 86.10.3. Let $X$ be an affine formal algebraic space over a scheme $S$.

1. If $X$ is Noetherian, then $X$ is adic*.

2. If $X$ is adic*, then $X$ is adic.

3. If $X$ is adic, then $X$ is weakly adic.

4. If $X$ is weakly adic, then $X$ is classical.

5. If $X$ is weakly adic, then $X$ is countably indexed.

6. If $X$ is countably indexed, then $X$ is McQuillan.

Proof. Statements (1), (2), (3), and (4) follow by writing $X = \text{Spf}(A)$ and where $A$ is a weakly admissible (hence complete) linearly topologized ring and using the implications between the various types of such rings discussed in Section 86.7.

Proof of (5). By definition there exists a weakly adic topological ring $A$ such that $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ where the colimit is over the ideals of definition of $A$. As $A$ is weakly adic, there exits in particular a countable fundamental system $I_\lambda$ of open ideals, see Definition 86.7.1. Then $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I_ n)$ by definition of $\text{Spf}(A)$. Thus $X$ is countably indexed.

Proof of (6). Write $X = \mathop{\mathrm{colim}}\nolimits X_ n$ for some system $X_1 \to X_2 \to X_3 \to \ldots$ of thickenings of affine schemes over $S$. Then

$A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_ n, \mathcal{O}_{X_ n})$

surjects onto each $\Gamma (X_ n, \mathcal{O}_{X_ n})$ because the transition maps are surjections as the morphisms $X_ n \to X_{n + 1}$ are closed immersions. Hence $X$ is McQuillan. $\square$

Lemma 86.10.4. Let $S$ be a scheme. Let $X$ be a presheaf on $(\mathit{Sch}/S)_{fppf}$. The following are equivalent

1. $X$ is a countably indexed affine formal algebraic space,

2. $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological $S$-algebra which has a countable fundamental system of neighbourhoods of $0$,

3. $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological $S$-algebra which has a fundamental system $A \supset I_1 \supset I_2 \supset I_3 \supset \ldots$ of weak ideals of definition,

4. $X = \text{Spf}(A)$ where $A$ is a complete topological $S$-algebra with a fundamental system of open neighbourhoods of $0$ given by a countable sequence $A \supset I_1 \supset I_2 \supset I_3 \supset \ldots$ of ideals such that $I_ n/I_{n + 1}$ is locally nilpotent, and

5. $X = \text{Spf}(A)$ where $A = \mathop{\mathrm{lim}}\nolimits B/J_ n$ with the limit topology where $B \supset J_1 \supset J_2 \supset J_3 \supset \ldots$ is a sequence of ideals in an $S$-algebra $B$ with $J_ n/J_{n + 1}$ locally nilpotent.

Proof. Assume (1). By Lemma 86.10.3 we can write $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological $S$-algebra. For any presentation $X = \mathop{\mathrm{colim}}\nolimits X_ n$ as in Lemma 86.10.1 part (1) we see that $A = \mathop{\mathrm{lim}}\nolimits A_ n$ with $X_ n = \mathop{\mathrm{Spec}}(A_ n)$ and $A_ n = A/I_ n$ for some weak ideal of definition $I_ n \subset A$. This follows from the final statement of Lemma 86.9.6 which moreover implies that $\{ I_ n\}$ is a fundamental system of open neighbourhoods of $0$. Thus we have a sequence

$A \supset I_1 \supset I_2 \supset I_3 \supset \ldots$

of weak ideals of definition with $A = \mathop{\mathrm{lim}}\nolimits A/I_ n$. In this way we see that condition (1) implies each of the conditions (2) – (5).

Assume (5). First note that the limit topology on $A = \mathop{\mathrm{lim}}\nolimits B/J_ n$ is a linearly topologized, complete topology, see More on Algebra, Section 15.36. If $f \in A$ maps to zero in $B/J_1$, then some power maps to zero in $B/J_2$ as its image in $J_1/J_2$ is nilpotent, then a further power maps to zero in $J_2/J_3$, etc, etc. In this way we see the open ideal $\mathop{\mathrm{Ker}}(A \to B/J_1)$ is a weak ideal of definition. Thus $A$ is weakly admissible. In this way we see that (5) implies (2).

It is clear that (4) is a special case of (5) by taking $B = A$. It is clear that (3) is a special case of (2).

Assume $A$ is as in (2). Let $E_ n$ be a countable fundamental system of neighbourhoods of $0$ in $A$. Since $A$ is a weakly admissible topological ring we can find open ideals $I_ n \subset E_ n$. We can also choose a weak ideal of definition $J \subset A$. Then $J \cap I_ n$ is a fundamental system of weak ideals of definition of $A$ and we get $X = \text{Spf}(A) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/(J \cap I_ n))$ which shows that $X$ is a countably indexed affine formal algebraic space. $\square$

Lemma 86.10.5. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space. The following are equivalent

1. $X$ is Noetherian,

2. $X$ is adic* and for every closed immersion $T \to X$ with $T$ a scheme, $T$ is Noetherian,

3. $X$ is adic* and for some choice of $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 86.9.1 the schemes $X_\lambda$ are Noetherian, and

4. $X$ is weakly adic and for some choice $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 86.9.1 the schemes $X_\lambda$ are Noetherian.

Proof. Assume $X$ is Noetherian. Then $X = \text{Spf}(A)$ where $A$ is a Noetherian adic ring. Let $T \to X$ be a closed immersion where $T$ is a scheme. By Lemma 86.9.6 we see that $T$ is affine and that $T \to \mathop{\mathrm{Spec}}(A)$ is a closed immersion. Since $A$ is Noetherian, we see that $T$ is Noetherian. In this way we see that (1) $\Rightarrow$ (2).

The implications (2) $\Rightarrow$ (3) and (2) $\Rightarrow$ (4) are immediate (see Lemma 86.10.3).

To prove (3) $\Rightarrow$ (1) write $X = \text{Spf}(A)$ for some adic ring $A$ with finitely generated ideal of definition $I$. We are also given that the rings $A/I_\lambda$ are Noetherian for some fundamental system of open ideals $I_\lambda$. Since $I$ is open, we can find a $\lambda$ such that $I_\lambda \subset I$. Then $A/I$ is Noetherian and we conclude that $A$ is Noetherian by Algebra, Lemma 10.97.5.

To prove (4) $\Rightarrow$ (3) write $X = \text{Spf}(A)$ for some weakly adic ring $A$. Then $A$ is admissible and has an ideal of definition $I$ and the closure $I_2$ of $I^2$ is open, see Lemma 86.7.2. We are also given that the rings $A/I_\lambda$ are Noetherian for some fundamental system of open ideals $I_\lambda$. Choose a $\lambda$ such that $I_\lambda \subset I_2$. Then $A/I_2$ is Noetherian as a quotient of $A/I_\lambda$. Hence $I/I_2$ is a finite $A$-module. Hence $A$ is an adic ring with a finitely generated ideal of definition by Lemma 86.7.4. Thus $X$ is adic* and (3) holds. $\square$

## 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).

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 0AIH. Beware of the difference between the letter 'O' and the digit '0'.