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I have a problem like this

"Let $R$ be a Cohen-Macaulay local ring, $\dim R=d$. Given that every maximal Cohen-Macaulay $R$-module is free, prove that $R$ is a regular local ring."

My lecturer gave me a hint to use the Auslander-Buchsbaum theorem: $\operatorname{depth} R=\operatorname{depth}_R M +\operatorname{pd}_R M$ if $\operatorname{pd}_R M<\infty$. However, I am still stuck since I want to prove that $\operatorname{pd}_R R/\mathfrak{m}<\infty$, but I don't know how the hint works at all?

Thank you for your help

ziggurism
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    It is enough to prove that every finitely generated module has finite projective dimension. If $M$ is a finitely generated module and not free, then by assumption $\mathrm{depth}, M<\mathrm{depth} ,R$. Let $0\to N\to R^n\to M\to 0$ be exact. Can you say $\mathrm{depth} ,N>\mathrm{depth} , M$ and finish the rest of the proof? – Mohan Jan 10 '18 at 03:20
  • @long, please add a complete answer! – Mariano Suárez-Álvarez Jan 10 '18 at 03:58
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    @LongLêThiên One can prove that the $d$th syzygy of every finitely generated $R$-module ($R$ local CM of dimension $d$, of course) is or zero or a MCM. (See Bruns and Herzog, Exercise 2.1.26 or this thread.) This solves the problem immediately. – user26857 Jan 10 '18 at 10:14

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