Lemma 87.8.3. 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 definition of B, and J' is open in B. \square
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