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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?

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

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  • $\begingroup$ 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? $\endgroup$
    – Anonymous
    May 16 '19 at 4:26
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    $\begingroup$ 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 $\endgroup$ May 16 '19 at 4:30
  • $\begingroup$ Interesting. Is it a well known fact that the projective dimension should decrease like that? $\endgroup$
    – Anonymous
    May 16 '19 at 4:33
  • $\begingroup$ Pretty much all texts on homological algebra cover this ... $\endgroup$ May 16 '19 at 4:35

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