
Lemma 10.71.9. Let $(R, \mathfrak m)$ be a local Noetherian ring and $M$ a finite $R$-module. For $\mathfrak p \in \text{Ass}(M)$ we have $\dim (R/\mathfrak p) \geq \text{depth}(M)$.

Proof. If $\mathfrak m \in \text{Ass}(M)$ then there is a nonzero element $x \in M$ which is annihilated by all elements of $\mathfrak m$. Thus $\text{depth}(M) = 0$. In particular the lemma holds in this case.

If $\text{depth}(M) = 1$, then by the first paragraph we find that $\mathfrak m \not\in \text{Ass}(M)$. Hence $\dim (R/\mathfrak p) \geq 1$ for all $\mathfrak p \in \text{Ass}(M)$ and the lemma is true in this case as well.

We will prove the lemma in general by induction on $\text{depth}(M)$ which we may and do assume to be $> 1$. Pick $x \in \mathfrak m$ which is a nonzerodivisor on $M$. Note $x \not\in \mathfrak p$ (Lemma 10.62.9). By Lemma 10.59.12 we have $\dim (R/\mathfrak p + (x)) = \dim (R/\mathfrak p) - 1$. Thus there exists a prime $\mathfrak q$ minimal over $\mathfrak p + (x)$ with $\dim (R/\mathfrak q) = \dim (R/\mathfrak p) - 1$ (small argument omitted; hint: the dimension of a Noetherian local ring $A$ is the maximum of the dimensions of $A/\mathfrak r$ taken over the minimal primes $\mathfrak r$ of $A$). Pick $n$ as in Lemma 10.71.8 so that $\mathfrak q$ is an associated prime of $M/x^ nM$. We may apply induction hypothesis to $M/x^ nM$ and $\mathfrak q$ because $\text{depth}(M/x^ nM) = \text{depth}(M) - 1$ by Lemma 10.71.7. We find $\dim (R/\mathfrak q) \geq \text{depth}(M/x^ nM)$ and we win. $\square$

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