**Proof.**
It is enough to show that (e) implies (a). Let $J' \subset B$ be a weak ideal of definition and let $I' \subset A$ be the closure of $J'A$. We have to show that $B/J' \to A/I'$ is of finite type. If the corresponding statement holds for the smaller weak ideal of definition $J'' = J' \cap J$, then it holds for $J'$. Thus we may assume $J' \subset J$. As $J$ is an ideal of definition (and not just a weak ideal of definition), we get $J^ n \subset J'$ for some $n \geq 1$. Thus we can consider the diagram

\[ \xymatrix{ 0 \ar[r] & I/I' \ar[r] & A/I' \ar[r] & A/I \ar[r] & 0 \\ 0 \ar[r] & J/J' \ar[r] \ar[u] & B/J' \ar[r] \ar[u] & B/J \ar[r] \ar[u] & 0 } \]

with exact rows. Since $I' \subset A$ is open and since $I$ is the closure of $J A$ we see that $I/I' = (J/J') \cdot A/I'$. Because $J/J'$ is a nilpotent ideal and as $B/J \to A/I$ is of finite type, we conclude from Algebra, Lemma 10.126.8 that $A/I'$ is of finite type over $B/J'$ as desired.
$\square$

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