The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.104.9. 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 10.103.7, 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 10.103.3 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 10.102.13 and 10.102.9. $\square$


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