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Let $S$ be a Cohen-Macaulay (C-M) ring, and $R$ a ring containing $S$ such that as an $S$-module is finitely generated free. Could we deduce that $R$ is also C-M?

I guess probably we could use the fact that $S$ being C-M yields $S[x_1,\ldots,x_n]$ C-M.

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Let $S\subset R$ a finite extension of noetherian rings such that $S$ is CM and $R$ is a free $S$-module. Then $R$ is CM.

Let $P\in \operatorname{Spec}(R)$. We want to show that $R_P$ is CM. Let $p=P\cap S$. Then $R_p$ is a free $S_p$-module and thus $\operatorname{depth}_{S_p}R_p=\operatorname{depth}S_p$. Take ${\bf x}\subset S_p$ an $R_p$-sequence of length $\operatorname{depth}S_p=\dim S_p$. From the dimension formula we have $\dim R_P=\dim S_p+\dim R_P/pR_P$. But $\dim R_P/pR_P=0$, so $\dim R_P=\dim S_p$. Since $R_P$ is a localization of $R_p$ we get that $\bf x$ is also an $R_P$-sequence, hence $\operatorname{depth}R_P\ge\dim R_P$, that is, $R_P$ is CM.

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