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
\text{depth}(M'_{\mathfrak p''}) = \text{depth}((M_\mathfrak q \otimes _{A_\mathfrak q} A')_{\mathfrak p''}) = \text{depth}(M_\mathfrak p \otimes _{A_\mathfrak p} A'_{\mathfrak p''}) = \text{depth}(M_\mathfrak p)
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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