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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