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 $\text{depth}(M)$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)