# Finitely generated projective resolution of a module over a regular local ring

Let $$R$$ be a regular local ring and let $$M$$ be an $$R$$-module. Then there exists a finite projective resolution $$P_\bullet\to M\to 0$$. However, need there exist a finite projective resolution consisting of finitely generated projective modules? What if we require that $$M$$ be finitely generated?

• Of course if $M$ is not finitely generated, then there is no surjection $P_0\to M$ from a finitely generated projective $P_0$. – Angina Seng May 16 '19 at 4:10

A regular local ring $$R$$ is Noetherian by hypothesis. If $$M$$ is a finitely generated $$R$$-module, then there is a finite rank free module $$F$$ and a surjection $$F\to M$$ fitting into a short exact sequence $$0\to N\to F\to M\to0.$$ As $$N$$ is a submodule of the finitely generated free module $$F$$ over the Noetherian ring $$R$$ then $$N$$ is finitely generated. Iterating this, gives a resolution of $$N$$ by finitely generated free modules. As $$R$$ is regular, it has finite global dimension, and the iterated kernels are eventually projective (so free) and the resolution can be brought to an end.
• Thanks. I don't quite see while $R$ having finite global dimension implies that this process should eventually end. The global dimension is the supremum over $R$-modules of the infimum of the lengths of resolutions. Why does that imply that this particular resolution should be finite? – Anonymous May 16 '19 at 4:26
• In the exact sequence above, if $M$ is not projective, the projective dimension of $N$ is strictly less than that of $M$, so iterating one eventually gets a kernel of projective dimension zero, that is a projective module (indeed free as projectives over local rings are free). @Anonymous – Angina Seng May 16 '19 at 4:30