**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.
$\square$

## Comments (2)

Comment #1560 by Matthew Emerton on

Comment #1579 by Johan on