TITLE: What does a projective resolution mean geometrically?
QUESTION [50 upvotes]: For R a commutative ring and M an R-module, we can always find a projective resolution of M which replaces M by a sequence of projective R-modules. But as R is commutative, we can consider the affine variety X=Spec R and the sheaf of modules associated to M. What is the projective resolution doing geometrically to this sheaf?
Projectives are locally free sheaves, so if M itself is not projective then it must have some sort of "sharp twisting" or "pinching". In some way a projective resolution is "un-pinching" M. Geometrically, is this the same "un-pinching" that happens in a resolution of a singularity of a variety? Is there an example in low dimensions where one can actually draw a picture of this happening for modules?

REPLY [4 votes]: I think this important question deserves one more answer, even after all the excellent ones already given. 
One can think of a free resolution as "approximating the module by the ring". This tautology has some corollaries: 
If you want the resolutions to reveal geometric properties, you need the ring to be nice to start with. Examples of this have been explained in all the other answers. In fact, you get the most information if your ring is "smooth" (regular local or polynomial rings). 
Flipping the argument, this means that "good module + bad ring = bad resolution". For example, let $M$ be a field $k$, viewed as a module over the ring $R =k[x,y]/(x^2,xy)$. Then a resolution of $M$ is always bad, the ranks of the free modules will increase exponentially, and no nice information about $M$ can be learned from those numbers. In fact, the information now flows the other way, it tells you how bad $R$ is. For example, it follows that $R$ can not be a complete intersection, because if it was, a result by Eisenbud implies that the Betti numbers would have had polynomial growth.