Lemma 51.9.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $\mathfrak p \in V(I)$ be a prime ideal. Assume $e = \text{depth}_{IA_\mathfrak p}(M_\mathfrak p) < \infty $. Then there exists a nonempty open $U \subset V(\mathfrak p)$ such that $\text{depth}_{IA_\mathfrak q}(M_\mathfrak q) \geq e$ for all $\mathfrak q \in U$.

**Proof.**
By definition of depth we have $IM_\mathfrak p \not= M_\mathfrak p$ and there exists an $M_\mathfrak p$-regular sequence $f_1, \ldots , f_ e \in IA_\mathfrak p$. After replacing $A$ by a principal localization we may assume $f_1, \ldots , f_ e \in I$ form an $M$-regular sequence, see Algebra, Lemma 10.67.6. Consider the module $M' = M/IM$. Since $\mathfrak p \in \text{Supp}(M')$ and since the support of a finite module is closed, we find $V(\mathfrak p) \subset \text{Supp}(M')$. Thus for $\mathfrak q \in V(\mathfrak p)$ we get $IM_\mathfrak q \not= M_\mathfrak q$. Hence, using that localization is exact, we see that $\text{depth}_{IA_\mathfrak q}(M_\mathfrak q) \geq e$ for any $\mathfrak q \in V(I)$ by definition of depth.
$\square$

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