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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