[Math] A Module with $Ext^i(M,R) = 0$ for all $i > 0$

ac.commutative-algebracohen-macaulay-ringshomological-algebra

Let $M$ be a finitely generated module over a noetherian local ring $R$. We can take our ring to be Cohen-Macaulay. Suppose $M$ satisfies the condition $Ext^i(M,R) = 0$ for all $i > 0$. We want to know if $M$ is projective?

One can easily show from the given condition that for any module $N$ of finite projective dimension, we do have $Ext^1(M,N) = 0$. Thus if (for example) our ring $R$ were regular local ring (which means any f.g module will have finite projective dimension), then we get the desired result (that $M$ is projective).

Now, Regular local => Cohen-Macaulay. So my first question is can we say the same with only Cohen-Macaulay condition on $R$ (that $M$ is projective)?

If it helps, we may assume that $M$ itself has finite projective dimension.

My second question is that can we write any f.g. module on (say) a noetherian local ring, as direct limit of modules having finite projective dimension?

Best Answer

As regards the second question: consider for example $R = k[[x]]$ where $k$ is a field. By the structure theory for modules over a PID, indecomposable finitely generated $R$-modules are cyclic, of the form $R/(x^n)$ for $n\geq 0$ or $R$. No module of the form $R/(x^n)$ contains a free submodule, so they can't be a direct limit of free modules.

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