9
$\begingroup$

Let $R$ be a ring. Then we know that a free module over $R$ is projective. Moreover, if $R$ is a principal ideal domain then a module over $R$ is free if and only if it is projective or if $R$ is local then a projective module is free.

We also have a very big question on free property of projective module over a polynomial ring, that was Serre's conjecture, and now is Quillen-Suslin's theorem.

I wonder, do we have a general condition for a ring $R$ so that every projective $R$-module is free which involves all of the cases mentioned above ?

$\endgroup$
  • 1
    $\begingroup$ I think this question is very interesting! Do you know of Lam's book "Serre's problem on projective modules"? It seems to be a standard reference on the subject, and a thorough account on the problem. The eigth chapter is called "New developments (since 1977)", and might include what you're looking for. It would be nice to have a characterization of the rings where projective implies free, eh? $\endgroup$ – Bruno Stonek Mar 17 '12 at 13:01
  • 1
    $\begingroup$ A weaker question is, for which rings are all projectives stably free? There are many classes of rings which have this property. In fact, Serre's problem (solved by Suslin and Quillen) was motivated by a theorem of Serre that polynomial rings have it. $\endgroup$ – Mariano Suárez-Álvarez Mar 17 '12 at 13:18
  • 3
    $\begingroup$ I'll throw another condition on the fire: By a theorem of Serre, if $\mathfrak{R}$ is a commutative artinian ring, every projective module is free. ((The theorem states that for any commutative noetherian ring $\mathfrak{R}$ and projective module $\mathfrak{P}$, if $rank(\mathfrak{P}) \gt dim(\mathfrak{R})$ then there exists a projective $\mathfrak{Q}$ with $rank(\mathfrak{Q})=dim(\mathfrak{R})$ such that $\mathfrak{P} \simeq \mathfrak{R}^k \oplus \mathfrak{Q}$ where $k=rank(\mathfrak{P})-dim(\mathfrak{R}$).)) $\endgroup$ – Andrew Parker Mar 19 '12 at 23:55
  • 2
    $\begingroup$ @AndrewParker A counterexample was given as the first comment here that shows commutative artinian rings do not always have that property. $\endgroup$ – rschwieb Oct 16 '12 at 1:25
  • 2
    $\begingroup$ Over a commutative (with unity) semilocal (i.e. finitely many maximal ideals) ring with no non-trivial idempotents , every projective module is free. This is proved in google.co.in/url?sa=t&source=web&rct=j&url=https://… . This generalizes Kaplansky's theorem for projective modules over local rings. $\endgroup$ – user495643 May 12 '18 at 7:46

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.