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Let $A$ be a commutative noetherian ring.

Suppose $M$ is a finitely generated $A$-module.

Let $n>0$ be an integer. It is well known that if $Ext^n(M,N) = 0$ for all $A$-modules $N$, then $M$ has a finite projective dimension.

What happens if we only know this for finitely generated $N$?

That is, suppose that for all fintiely generated $N$ we have that $Ext^n(M,N)=0$. Does it still follow that $M$ has a finite projective dimension?

Because $M$ is finitely generated, the $Hom(M,-)$ functor and taking cohomology are both operations which commute with infinite direct sum, the result is true for any free $A$-module. Is this enough to show that $Ext^n(M,N)$ vanishes for any $N$?

Edit: maybe going a little further is enough? as far as I understand, since $M$ is finitely generated, it follows (does it?) that $Hom(M,\varinjlim N_i) \cong \varinjlim Hom(M,N_i)$, and since direct limits commute with taking cohomology, expressing $N$ has the direct limit of finitely generated modules, the result follows. Is this correct?

Second Edit: so it turns out by the comments that if $M$ is not projective, the argument above is false. In my case, $M$ is actually not projective. So, anyone has another idea/counter example?

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By THM 5.51 in Rotman's "Introduction to Homological Algebra", an additive functor $F: \text{Mod}_A \rightarrow \text{Mod}_A$ preserves direct limits iff it is right exact and preserves direct sums. So, $\text{Hom}_A(M, -)$ preserves direct limits iff $M$ is projective. –  mbrown May 22 '13 at 14:37
    
I see. Thanks. Do you have an answer to the originial question? –  the L May 22 '13 at 18:04
    
Not at the moment. My guess is that the answer to your question is no, but I don't have a counterexample. –  mbrown May 22 '13 at 19:59

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