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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