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Suppose that $R$ is a ring (commutative, if it simplifies things), and that $M$ is a (left) $R$-module. Then $M$ has a projective resolution of length $n$ if and only if $\operatorname{Ext}_R^m(M,-)$ vanishes for all $m>n$. Additionally, if $M$ does have a projective resolution of length $n$, then any resolution of length $m>n$ can be truncated to a resolution of length $n$. For these and other reasons, being projective seems to be a much more natural condition than being free in the context of homological algebra.

Still, for the sake of computation, it is much easier to work with free resolutions, as free modules are much easier to describe. However, free resolutions do not appear to have the nice properties that projective resolutions do. Despite this, it is conceivable that they are at least reasonably behaved.

Along these lines, does anybody know if any of these statements are true? (some of which would imply others). For the ones which aren't, are there good counterexamples?

  • If $M$ has a finite projective resolution, then $M$ has a finite free resolution.
  • If $M$ has a projective resolution of length $n$, then $M$ has a free resolution of length $n$ (or $n+k$, for some $k$ depending only on $R$).
  • There exists a functorial invariant which will determine the length of a minimal length free resolution of $M$.
  • There is a computationally feasible way to determine the length of of a minimal length free resolution of $M$.

Added: In light of the links provided by Alon, I would like to amend the question to look specifically at the case when $R$ is noetherian, $M$ is finitely generated, and resolutions under consideration are finitely generated free resolutions. In this case, the answer to the first question is a resounding no (Alon's second link), but (Alon's first link) that there exist rings where the answer to question 2 is yes with $k=0$ (well, $k=1$ if we want to deal with projective modules).

For the third question, it is impossible with derived functors, as the LES from a SES allows a dimension shifting, and any functor which vanishes on free modules also vanishes on projective modules, but that does not preclude functors which are not as well behaved.

For local commutative noetherian rings, the last question is probably handled by the existence of minimal resolutions. However, it is not immediately clear that this generalizes (the obvious generalization clearly fails for rings with trivial jacobson radical).

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Related questions: mathoverflow.net/questions/27424/… and mathoverflow.net/questions/27406/free-resolution-dimension. The answer to your first question seems to be "no". –  Alon Amit Jun 15 '11 at 4:41
    
@Alon Thanks, the second link was exactly the kind of thing I was looking for. The answer to the first two questions is is yes if we aren't restricting ourselves to finitely generated modules (and if we are, I don't actually know what happens in the projective case in the absence of $R$ being noetherian). A slightly cleaner way to prove that finite projective resolution implies (same length, unless $M$ is projective) free resolution is to take a (possibly infinite) free resolution, truncate (taking the kernel at the end) to get a resolution ending in $0\to P \to F \to \ldots$ (continued) –  Aaron Jun 15 '11 at 5:48
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where $P$ is projective and $F$ is free. We can find a $Q$ such that $P\oplus Q$ is free, and by adding a direct sum $Q\oplus P \oplus Q \oplus P \oplus \cdots$ with the identity map to the last two terms, we get that all the terms in the resolution are free. This is a variant of the Eilenberg swindle. –  Aaron Jun 15 '11 at 5:50
    
Reading the chapter about homological dimension in any book can help you, I think. –  tetrapharmakon Jun 15 '11 at 19:24
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@tetrapharmakon: well, these specific questions are not answered in "any book (with a chapter on homological dimension)". If you have a specific suggestion which does deal with this answer, it would be best to actually make it explicit! –  Mariano Suárez-Alvarez Jun 15 '11 at 20:35
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