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I think this important question deserves one more answer, despite 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.

1

I think this important question deserves one more answer, despite 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.