1
$\begingroup$

In some notes I'm following, there is a statement

(Clearly) over a regular local ring each finitely generated module has a finite free resolution

Is the finitely generated assumption neccessary? If it is infact clear (for finitely generated modules), why?

The furthest I can get with what I know is as follows.

A regular local ring has finite (right) global dimension, so every module has finite projective dimension. Let $M$ have $\mathrm{pd}(M) = n$. Therefore, if we write down an exact sequence:

$0 \rightarrow K \rightarrow F_n \rightarrow \dots \rightarrow F_0 \rightarrow M \rightarrow 0$

with the $F_i$ free. By Schanuel's Lemma, $K$ is projective.

I would aim to produce a free resolution from this, but cannot see an obvious path. With the finitely generated assumption, the $F_i$ and $K$ would be finitely generated.

$\endgroup$
  • $\begingroup$ Every projective module over a local ring is free. This statement is easier to prove for finitely generated projective modules, but in general (that is for arbitrary projective modules) it was proved by Kaplansky here: jstor.org/stable/1970252?seq=1#page_scan_tab_contents $\endgroup$ – Must Apr 26 '18 at 3:33
  • $\begingroup$ @Must I think that is what I was looking for for the finite case, thanks! $\endgroup$ – chilliBeanDream Apr 26 '18 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.