# Cohen Macaulay, free and finitely generated module

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.

• You can reduce to the local case. Then, take a maximal regular sequence of $R$ and show that it is also regular on $T$. Jun 13 '11 at 15:11
• How do you deal with the equation $grade(I)=ht(I)$ for any ideal $I$ of $T$? Jun 14 '11 at 9:21
• @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) Jun 14 '11 at 15:04

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.