Lemma 86.8.1. In the situation above, if $B$ has a countable fundamental system of open ideals, then $A$ has a countable fundamental system of open ideals.

## 86.8 Descending properties

In this section we consider the following situation

$\varphi : A \to B$ is a continuous map of linearly topologized topological rings,

$\varphi $ is taut, and

for every open ideal $I \subset A$ if $J \subset B$ denotes the closure of $IB$, then the map $A/I \to B/J$ is faithfully flat.

We are going to show that properties of $B$ are inherited by $A$ in this situation.

**Proof.**
Choose a fundamental system $B \supset J_1 \supset J_2 \supset \ldots $ of open ideals. By tautness of $\varphi $, for every $n$ we can find an open ideal $I_ n$ such that $J_ n \supset I_ nB$. We claim that $I_ n$ is a fundamental system of open ideals of $A$. Namely, suppose that $I \subset A$ is open. As $\varphi $ is taut, the closure of $IB$ is open and hence contains $J_ n$ for some $n$ large enough. Hence $I_ nB \subset IB$. Let $J$ be the closure of $IB$ in $B$. Since $A/I \to B/J$ is faithfully flat, it is injective. Hence, since $I_ n \to A/I \to B/J$ is zero as $I_ nB \subset IB \subset J$, we conclude that $I_ n \to A/I$ is zero. Hence $I_ n \subset I$ and we win.
$\square$

Lemma 86.8.2. In the situation above, if $B$ is weakly pre-admissible, then $A$ is weakly pre-admissible.

**Proof.**
Let $J \subset B$ be a weak ideal of definition. Let $I \subset A$ be an open ideal such that $IB \subset J$. To show that $I$ is a weak ideal of definition we have to show that any $f \in I$ is topologically nilpotent. Let $I' \subset A$ be an open ideal. Denote $J' \subset B$ the closure of $I'B$. Then $A/I' \to B/J'$ is faithfully flat, hence injective. Thus in order to show that $f^ n \in I'$ it suffices to show that $\varphi (f)^ n \in J'$. This holds for $n \gg 0$ since $\varphi (f) \in J$, the ideal $J$ is a weak ideal of defintion of $B$, and $J'$ is open in $B$.
$\square$

Lemma 86.8.3. In the situation above, if $B$ is pre-admissible, then $A$ is pre-admissible.

**Proof.**
Let $J \subset B$ be a weak ideal of definition. Let $I \subset A$ be an open ideal such that $IB \subset J$. Let $I' \subset A$ be an open ideal. To show that $I$ is an ideal of definition we have to show that $I^ n \subset I'$ for $n \gg 0$. Denote $J' \subset B$ the closure of $I'B$. Then $A/I' \to B/J'$ is faithfully flat, hence injective. Thus in order to show that $I^ n \subset I'$ it suffices to show that $\varphi (I)^ n \subset J'$. This holds for $n \gg 0$ since $\varphi (I) \subset J$, the ideal $J$ is an ideal of defintion of $B$, and $J'$ is open in $B$.
$\square$

Lemma 86.8.4. In the situation above, if $B$ is weakly pre-adic, then $A$ is weakly pre-adic.

**Proof.**
We will use the characterization of weakly pre-adic rings given in Lemma 86.7.2 without further mention. By Lemma 86.8.3 the topological ring $A$ is pre-admissible. Let $I \subset A$ be an ideal of definition. Fix $n \geq 1$. To prove the lemma we have to show that the closure of $I^ n$ is open. Let $I_\lambda \subset A$ be a fundamental system of open ideals. Denote $J \subset B$, resp. $J_\lambda \subset B$ the closure of $IB$, resp. $I_\lambda B$. Since $B$ is weakly pre-adic, the closure of $J^ n$ is open. Hence there exists a $\lambda $ such that

because the right hand side is the closure of $J^ n$ by Lemma 86.4.2. This means that the image of $J_\lambda $ in $B/J_\mu $ is contained in the image of $J^ n$ in $B/J_\mu $. Observe that the image of $J^ n$ in $B/J_\mu $ is equal to the image of $I^ nB$ in $B/J_\mu $ (since every element of $J$ is congruent to an element of $IB$ modulo $J_\mu $). Since $A/I_\mu \to B/J_\mu $ is faithfully flat and since $I_\lambda B \subset J_\lambda $, we conclude that the image of $I_\lambda $ in $A/I_\mu $ is contained in the image of $I^ n$. We conclude that $I_\lambda $ is contained in the closure of $I^ n$ and the proof is complete. $\square$

Lemma 86.8.5. In the situation above, if $B$ is adic and has a finitely generated ideal of definition and $A$ is complete, then $A$ is adic and has a finitely generated ideal of definition.

**Proof.**
We already know that $A$ is weakly adic and a fortiori admissible by Lemma 86.8.4 (and Lemma 86.7.2 to see that adic rings are weakly adic). Let $I \subset A$ be an ideal of definition. Let $J \subset B$ be a finitely generated ideal of definition. Since the closure of $IB$ is open, we can find an $n > 0$ such that $J^ n$ is contained in the closure of $IB$. Thus after replacing $J$ by $J^ n$ we may assume $J$ is a finitely generated ideal of definition contained in the closure of $IB$. By Lemma 86.4.2 this certainly implies that

Consider the finitely generated $A$-module $M = (J + IB)/IB$. The displayed equation shows that $JM = M$. By Lemma 86.4.9 (for example) we see that $J$ is contained in the Jacobson radical of $B$. Hence by Nakayama's lemma, more precisely part (2) of Algebra, Lemma 10.20.1, we conclude $M = 0$. Thus $J \subset IB$.

Since $J$ is finitely generated, we can find a finitely generated ideal $I' \subset I$ such that $J \subset I'B$. Since $A \to B$ is continuous, $J \subset B$ is open, and $I$ is an ideal of definition, we can find an $n > 0$ such that $I^ nB \subset J$. Let $J_{n + 1} \subset B$ be the closure of $I^{n + 1}B$. We have

Since $A/I^{n + 1} \to B/J_{n + 1}$ is faithfully flat, this implies $I^ n \cdot (A/I^{n + 1}) \subset I' \cdot (A/I^{n + 1})$ which in turn means

This implies $I^ n \subset I' + I^{n + k}$ for all $k \geq 1$ which in turn implies that $I^{nm} \subset (I')^ m + I^{nm + k}$ for all $k, m \geq 1$. This implies that the closure of $(I')^ m$ contains $I^{nm}$. Since the closure of $I^{nm}$ is open as $A$ is weakly adic, we conclude that the closure $(I')^ m$ is open for all $m$. Since these closures form a fundamental system of open ideals of $A$ (as the same thing is true for the closures of $I^ n$) we conclude by Lemma 86.7.3. $\square$

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