Lemma 87.8.2. 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 definition of $B$, and $J'$ is open in $B$. $\square$
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