Lemma 52.10.2. In Situation 52.10.1 let $\mathfrak p \subset \mathfrak q$ be primes of $A$ with $\mathfrak p \not\in V(I)$ and $\mathfrak q \in T$. If there does not exist an $\mathfrak r \in V(I) \setminus T$ with $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ then $\text{depth}(M_\mathfrak p) > s$.

**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. $\square$

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