The Stacks project

Proof. Choose $\mathfrak q' \in T$ with $\mathfrak p \subset \mathfrak q' \subset \mathfrak q$ such that there is no prime in $T$ strictly in between $\mathfrak p$ and $\mathfrak q'$. To prove the lemma we may and do replace $\mathfrak q$ by $\mathfrak q'$. Next, let $\mathfrak p' \subset A_\mathfrak q$ be the prime corresponding to $\mathfrak p$. After doing this we obtain that $V(\mathfrak p') \cap V(IA_\mathfrak q) = \{ \mathfrak q A_\mathfrak q\} $ because of the nonexistence of a prime $\mathfrak r$ as in the lemma. Let $A', I', \mathfrak m', M'$ be the $I$-adic completions of $A_\mathfrak q, I_\mathfrak q, \mathfrak qA_\mathfrak q, M_\mathfrak q$. Since $A_\mathfrak q \to A'$ is faithfully flat (Algebra, Lemma 10.97.3) we can choose $\mathfrak p'' \subset A'$ lying over $\mathfrak p'$ with $\dim (A'_{\mathfrak p''}/\mathfrak p' A'_{\mathfrak p''}) = 0$. Then we see that

by flatness of $A \to A'$ and our choice of $\mathfrak p''$, see Algebra, Lemma 10.163.1. Since $\mathfrak p''$ lies over $\mathfrak p'$ we have $V(\mathfrak p'') \cap V(I') = \{ \mathfrak m'\} $. Thus condition (6) in Situation 52.10.1 implies $\text{depth}(M'_{\mathfrak p''}) > s$ which finishes the proof. $\square$