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Here is an unsolved problem for me in Kaplansky's "Commutative rings" p. 103, no. 18.

Let $R$ be a Cohen-Macaulay ring. Let $T$ be a ring containing $R$ and suppose that as an $R$-module it is free and finitely generated. Show that $T$ is also a Cohen-Macaulay ring.

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    $\begingroup$ You can reduce to the local case. Then, take a maximal regular sequence of $R$ and show that it is also regular on $T$. $\endgroup$
    – Hailong Dao
    Commented Jun 13, 2011 at 15:11
  • $\begingroup$ How do you deal with the equation $grade(I)=ht(I)$ for any ideal $I$ of $T$? $\endgroup$
    – Ehsan M. Kermani
    Commented Jun 14, 2011 at 9:21
  • $\begingroup$ @ehsanmo Rather than consider all ideals, I'd recommend simply using the equivalent definition of CM for local rings: A local ring is CM iff Krull dim(R) = depth(R) $\endgroup$
    – David White
    Commented Jun 14, 2011 at 15:04

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Let $\mathfrak p\in\mathrm{Spec}T$ and $\mathfrak q=\mathfrak p\cap R\in\mathrm{Spec}R$. Since $T$ is finite and hence integral over $R$, the dimension of $T_{\mathfrak p}$ is the same whether considered as a ring, a module over itself or a module over $R_{\mathfrak q}$. It will be denoted by $\dim T_{\mathfrak p}$. Consider the following inequalities: $$ \mathrm{depth}_{\mathfrak q}T_{\mathfrak p}\overset{\text{since $\mathfrak q\subseteq \mathfrak p$}}{\leq} \mathrm{depth}_{\mathfrak p}T_{\mathfrak p} \leq \mathrm{dim}T_{\mathfrak p} $$ Since $T$ is free over $R$, $T_{\mathfrak p}$ is free over $R_{\mathfrak q}$ and hence it is CM as an $R_{\mathfrak q}$-module, so the two ends of the above inequality are equal. Therefore the middle is also equal to that value and hence $T$ is a CM ring.

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