Lemma 10.102.9. Let $R$ be a Noetherian local ring. Assume there exists a Cohen-Macaulay module $M$ with $\mathop{\mathrm{Spec}}(R) = \text{Supp}(M)$. Then any maximal chain of ideals $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n$ has length $n = \dim (R)$.

** In a local Cohen-Macaulay ring, any maximal chain of prime ideals has length equal to the dimension. **

**Proof.**
We will prove this by induction on $\dim (R)$. If $\dim (R) = 0$, then the statement is clear. Assume $\dim (R) > 0$. Then $n > 0$. Choose an element $x \in \mathfrak p_1$, with $x$ not in any of the minimal primes of $R$, and in particular $x \not\in \mathfrak p_0$. (See Lemma 10.14.2.) Then $\dim (R/xR) = \dim (R) - 1$ by Lemma 10.59.12. The module $M/xM$ is Cohen-Macaulay over $R/xR$ by Proposition 10.102.4 and Lemma 10.102.6. The support of $M/xM$ is $\mathop{\mathrm{Spec}}(R/xR)$ by Lemma 10.39.9. After replacing $x$ by $x^ n$ for some $n$, we may assume that $\mathfrak p_1$ is an associated prime of $M/xM$, see Lemma 10.71.8. By Lemma 10.102.7 we conclude that $\mathfrak p_1/(x)$ is a minimal prime of $R/xR$. It follows that the chain $\mathfrak p_1/(x) \subset \ldots \subset \mathfrak p_ n/(x)$ is a maximal chain of primes in $R/xR$. By induction we find that this chain has length $\dim (R/xR) = \dim (R) - 1$ as desired.
$\square$

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