Lemma 10.105.9. A Noetherian Cohen-Macaulay ring is universally catenary. More generally, if $R$ is a Noetherian ring and $M$ is a Cohen-Macaulay $R$-module with $\text{Supp}(M) = \mathop{\mathrm{Spec}}(R)$, then $R$ is universally catenary.

**Proof.**
Since a polynomial algebra over $R$ is Cohen-Macaulay, by Lemma 10.104.7, it suffices to show that a Cohen-Macaulay ring is catenary. Let $R$ be Cohen-Macaulay and $\mathfrak p \subset \mathfrak q$ primes of $R$. By definition $R_{\mathfrak q}$ and $R_{\mathfrak p}$ are Cohen-Macaulay. Take a maximal chain of primes $\mathfrak p = \mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n = \mathfrak q$. Next choose a maximal chain of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ m = \mathfrak p$. By Lemma 10.104.3 we have $n + m = \dim (R_{\mathfrak q})$. And we have $m = \dim (R_{\mathfrak p})$ by the same lemma. Hence $n = \dim (R_{\mathfrak q}) - \dim (R_{\mathfrak p})$ is independent of choices.

To prove the more general statement, argue exactly as above but using Lemmas 10.103.13 and 10.103.9. $\square$

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