Definition 87.7.1.reference Let $A$ be a linearly topologized ring.
87.7 Weakly adic rings
We suggest the reader skip this section. The following is a natural generalization of adic rings.
For complete linearly topologized rings we have the following implications
\[ \xymatrix{ \text{adic + Noetherian} \ar@{=>}[d] \\ \text{adic + finitely generated ideal of definition} \ar@{=>}[d] \\ \text{adic} \ar@{=>}[d] \\ \text{weakly adic} \ar@{=>}[d] \\ \text{admissible + first countable} \ar@{=>}[d] \ar@{=>}[r] & \text{admissible} \ar@{=>}[d] \\ \text{weakly admissible + first countable} \ar@{=>}[r] & \text{weakly admissible} } \]
where “first countable” means that our topological ring has a countable fundamental system of open ideals. There is a similar diagram of implications for noncomplete linearly topologized rings (i.e., using the notions of pre-adic, weakly pre-adic, pre-admissible, and weakly pre-admissible). Contrary to what happens with pre-adic rings the completion of a weakly pre-adic ring is weakly adic as the following lemma characterizing weakly pre-adic rings shows.
Lemma 87.7.2. Let $A$ be a linearly topologized ring. The following are equivalent
$A$ is weakly pre-adic,
there exists a taut continuous ring map $A' \to A$ where $A'$ is a pre-adic topological ring, and
$A$ 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
$A$ 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. $\square$
We give two criteria that guarantee that a weakly adic ring is adic and has a finitely generated ideal of definition.
Lemma 87.7.3. Let $A$ be a complete linearly topologized ring. Let $I \subset A$ be a finitely generated ideal such that the closure of $I^ n$ is open for all $n \geq 0$ and these closures form a fundamental system of open ideals. Then $A$ is adic and has a finitely generated ideal of definition.
Proof. Denote $A'$ the ring $A$ endowed with the $I$-adic topology. The assumptions tells us that $A' \to A$ is taut. We conclude by Lemma 87.6.5 (to be sure, this lemma also tells us that $I$ is an ideal of definition). $\square$
Lemma 87.7.4. Let $A$ be a weakly adic topological ring. Let $I$ be an ideal of definition such that $I/I_2$ is a finitely generated module where $I_2$ is the closure of $I^2$. Then $A$ is adic and has a finitely generated ideal of definition.
Proof. We use the characterization of Lemma 87.7.2 without further mention. Choose $f_1, \ldots , f_ r \in I$ which map to generators of $I/I_2$. Set $I' = (f_1, \ldots , f_ r)$. We have $I' + I_2 = I$. Then $I_2$ is the closure of $I^2 = (I' + I_2)^2 \subset I' + I_3$ where $I_3$ is the closure of $I^3$. Hence $I' + I_3 = I$. Continuing in this fashion we see that $I' + I_ n = I$ for all $n \geq 2$ where $I_ n$ is the closure of $I^ n$. In other words, the closure of $I'$ in $A$ is $I$. Hence the closure of $(I')^ n$ is $I_ n$. Thus the closures of $(I')^ n$ are a fundamental system of open ideals of $A$. We conclude by Lemma 87.7.3. $\square$
A key feature of the property “weakly pre-adic” is that it ascents along taut ring homomorphisms of linearly topologized rings.
Lemma 87.7.5. Let $\varphi : A \to B$ be a continuous homomorphism of linearly topologized rings. If $\varphi $ is taut and $A$ is weakly pre-adic, then $B$ is weakly pre-adic.
Proof. Let $I \subset A$ be an ideal such that the closure $I_ n$ of $I^ n$ is open and these closures define a fundamental system of open ideals. Then the closure of $I^ nB$ is equal to the closure of $I_ nB$. Since $\varphi $ is taut, these closures are open and form a fundamental system of open ideals of $B$. Hence $B$ is weakly pre-adic. $\square$
Lemma 87.7.6. Let $B \to A$ and $B \to C$ be continuous homomorphisms of linearly topologized rings. If $A$ and $C$ are weakly pre-adic, then $A \widehat{\otimes }_ B C$ is weakly adic.
Proof. We will use the characterization of Lemma 87.7.2 without further mention. By Lemma 87.4.12 we know that $A \widehat{\otimes }_ B C$ is admissible. Moreover, the proof of that lemma shows that the closure $K \subset A \widehat{\otimes }_ B C$ is an ideal of definition, when $I \subset A$ and $J \subset C$ of $I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C)$ are ideals of definition. Then it suffices to show that the closure of $K^ n$ is open for all $n \geq 1$. Since the ideal $K^ n$ contains $I^ n(A \widehat{\otimes }_ B C) + J^ n(A \widehat{\otimes }_ B C)$, since the closure of $I^ n$ in $A$ is open, and since the closure of $J^ n$ in $C$ is open, we see that the closure of $K^ n$ is open in $A \widehat{\otimes }_ B C$. $\square$
[2] By our conventions this includes separated.
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