The Stacks project

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. $\square$

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 87.7.

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 87.7.1. 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

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. $\square$

Proof. Assume (1). By Lemma 87.10.3 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 87.10.1 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 87.9.6 which moreover implies that $\{ I_ n\} $ is a fundamental system of open neighbourhoods of $0$. Thus we have a sequence

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. $\square$

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 87.9.6 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 87.10.3).

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 87.7.2. 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 87.7.4. Thus $X$ is adic* and (3) holds. $\square$