Lemma 87.7.2. Let $A$ be a linearly topologized ring. The following are equivalent

1. $A$ is weakly pre-adic,

2. there exists a taut continuous ring map $A' \to A$ where $A'$ is a pre-adic topological ring, and

3. $A$ is pre-admissible and there exists an ideal of definition $I$ such that the closure of $I^ n$ is open for all $n \geq 1$, and

4. $A$ is pre-admissible and for every ideal of definition $I$ the closure of $I^ n$ is open for all $n \geq 1$.

The completion of a weakly pre-adic ring is weakly adic. If $A$ is weakly adic, then $A$ is admissible and has a countable fundamental system of open ideals.

Proof. Assume (1). Choose an ideal $I$ such that the closure of $I^ n$ is open for all $n$ and such that these closures form a fundamental system of open ideals. Denote $A' = A$ endowed with the $I$-adic topology. Then $A' \to A$ is taut by definition and we see that (2) holds.

Assume (2). Let $I' \subset A'$ be an ideal of definition. Denote $I$ the closure of $I'A$. Tautness of $A' \to A$ means that the closures $I_ n$ of $(I')^ nA$ are open and form a fundamental system of open ideals. Thus $I = I_1$ is open and the closures of $I^ n$ are equal to $I_ n$ and hence open and form a fundamental system of open ideals. Thus certainly $I$ is an ideal of definition such that the closure of $I^ n$ is open for all $n$. Hence (3) holds.

If $I \subset A$ is as in (3), then $I$ is an ideal as in Definition 87.7.1 and we see that (1) holds. Also, if $I' \subset A$ is any other ideal of definition, then $I'$ is open (see More on Algebra, Definition 15.36.1) and hence contains $I^ n$ for some $n \geq 1$. Thus $(I')^ m$ contains $I^{nm}$ for all $m \geq 1$ and we conclude that the closures of $(I')^ m$ are open for all $m$. In this way we see that (3) implies (4). The implication (4) $\Rightarrow$ (3) is trivial.

Let $A$ be weakly pre-adic. Choose $A' \to A$ as in (2). By Lemmas 87.5.3 and 87.5.4 the composition $A' \to A^\wedge$ is taut. Hence $A^\wedge$ is weakly pre-adic by the equivalence of (2) and (1). Since the completion of a linearly topologized ring $A$ is complete (More on Algebra, Section 15.36) we see that $A^\wedge$ is weakly adic.

Let $A$ be weakly adic. Then $A$ is complete and and pre-admissible by (1) $\Rightarrow$ (3) and hence $A$ is admissible. Of course by definition $A$ has a countable fundamental system of open ideals. $\square$

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