Question

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.

Answer 1

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.