Lemma 87.8.4. 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 87.7.2 without further mention. By Lemma 87.8.3 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

$J_\lambda \subset \bigcap \nolimits _\mu (J^ n + J_\mu )$

because the right hand side is the closure of $J^ n$ by Lemma 87.4.2. 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. $\square$

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