Let $A$ be a noetherian ring and $M$ be a finitely generated $A$-module, I want to prove the following

M is projective $\iff$ for all $P\subset A$ prime ideal the localisation $M_P$ is a free $A_P$-module

As far as I know since $A$ is noetherian and $M$ is finitely generated then it is also finitely presented. Moreover the localisation of a noetherian ring is still noetherian and so $A_P$ are all noetherian rings. It seems to me that I have to use Nakayama's lemma but I don't see how.

Is there a direct proof of this statement by using directly properties of projective modules? (Such as they are direct summand of a free module) Thanks.

  • $\begingroup$ This is Theorem 13.15 in Altman--Kleiman $\endgroup$
    – Kenny Lau
    Mar 6, 2021 at 11:23
  • $\begingroup$ Do you know that "(finitely generated and) projective over a noetherian local ring” implies free? $\endgroup$
    – egreg
    Mar 6, 2021 at 11:23
  • $\begingroup$ @egreg Do you have to use that a projective module is a direct summand of a free module and then using the noetherian hypothesis on $A$ prove that the other summand is zero? $\endgroup$
    – John117
    Mar 6, 2021 at 11:39
  • $\begingroup$ @John117 See math.stackexchange.com/q/3362463/62967 $\endgroup$
    – egreg
    Mar 6, 2021 at 11:41
  • $\begingroup$ @egreg Thank you. In my case if $M$ is projective, do I have that the localisation $M_P$ are still projective over the noetherian local ring $A_P$, right? $\endgroup$
    – John117
    Mar 6, 2021 at 11:43

1 Answer 1


If $M$ is projective, then there is module $K$ s.t. $M \oplus K \cong R^{(\Lambda)}$. For any multiplicative set $S \subset R$, one gets $S^{-1}M \oplus S^{-1} K \cong (S^{-1}R)^{(\Lambda)}$. Hence, any localization of a projective module is projective.

If $(R,m)$ is local and $M$ is finitely generated projective then $M$ is free. Here is a proof. The short exact sequence $$ 0 \to K \to R^n \to M \to 0, $$ splits, hence $K$ is finitely generated and sequence $$ 0 \to K \otimes R/m \to R^n \otimes R/m \to M \otimes R/m \to 0, $$ also splits. By Nakayama's lemma we can choose generators of $M$ in such way that the map $R^n \otimes R/m \to M \otimes R/m$ is an isomorphism, then $K \otimes R/m \cong K/mK=0$. By Nakayma's lemma $K=0$, since it is finitely generated.

Fact: if $M$ is finitely presented, then $S^{-1}Hom_R(M,N) \cong Hom_{S^{-1}R} (S^{-1}M, S^{-1}N)$.

To check that $M$ is projective is enough to check that for any surjection $N \to N'$ map $Hom(M,N) \to Hom(M, N')$ is surjective, but this can be checked stalkwise if $M$ is finitely presented (by the fact).

There are plenty of other proofs, but often they involve flat modules, in this proof we only use two ways to characterize projective modules and the fact. Notice that this proof doesn't use that $R$ is Noetherian.


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