Lemma 10.72.10. Let $R$ be a local Noetherian ring and $M$ a finite $R$-module. For a prime ideal $\mathfrak p \subset R$ we have $\text{depth}(M_\mathfrak p) + \dim (R/\mathfrak p) \geq \text{depth}(M)$.
Proof. If $M_\mathfrak p = 0$, then $\text{depth}(M_\mathfrak p) = \infty $ and the lemma holds. If $\text{depth}(M) \leq \dim (R/\mathfrak p)$, then the lemma is true. If $\text{depth}(M) > \dim (R/\mathfrak p)$, then $\mathfrak p$ is not contained in any associated prime $\mathfrak q$ of $M$ by Lemma 10.72.9. Hence we can find an $x \in \mathfrak p$ not contained in any associated prime of $M$ by Lemma 10.15.2 and Lemma 10.63.5. Then $x$ is a nonzerodivisor on $M$, see Lemma 10.63.9. Hence $\text{depth}(M/xM) = \text{depth}(M) - 1$ and $\text{depth}(M_\mathfrak p / x M_\mathfrak p) = \text{depth}(M_\mathfrak p) - 1$ provided $M_\mathfrak p$ is nonzero, see Lemma 10.72.7. Thus we conclude by induction on $\text{depth}(M)$. $\square$
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