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

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?

• There are many counterexamples, mostly (but not all) having to do with the ring being Gorenstein. If that's not a familiar word, you might look it up; it fits between regular and CM. Also, if M has finite projective dimension and all the Ext's vanish, then M is free -- this is a relatively easy exercise. – Graham Leuschke Dec 20 '11 at 15:32
• I think if you assume $M$ of finite projective dimension, then the answer is yes over any ring (you don't need CM), because if $M$ is not free and you take a minimal free resolution of $M$ and apply $\mathrm{Hom}_R(\:\cdot\:,R)$ to it, the last map on the left after you apply Hom cannot be surjective and will have a cokernel. It cannot be surjective because the resolution is minimal. – Mahdi Majidi-Zolbanin Dec 20 '11 at 15:32
• Sorry Graham, looks like I answered the exercise! – Mahdi Majidi-Zolbanin Dec 20 '11 at 15:34
• Compare with the Auslander-Reiten conjecture that $Ext^i(M,M\oplus R)=0$ for all $i>0$ implies $M$ is projective. They were originally interested in Artin algebras but people have considered things like Gorenstein rings. – Benjamin Steinberg Dec 20 '11 at 16:12
• @Mahdi, thanks. @Benjamin, thanks. I was not aware of this conjecture. – User3568 Dec 21 '11 at 16:33

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.
• Actually, this is a terrible example. All these modules have finite projective dimension! I should have used $k[x]/(x^2)$; the residue field has infinite projective dimension and is not a direct limiit of modules of finite projective dimension. – Graham Leuschke Dec 21 '11 at 11:33