Condition for a ring on projective and free modules problem

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 properties of projectives 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 ?

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There is a general theorem on projective modules: an $R$-module is projective iff it is a direct summand of a free $R$-module. From what little I've read on Quillen-Suslind, it seems that the chief difficulty was/is telling when something is a direct summand of a free module when our ring is more complicated. –  KReiser Mar 1 '12 at 7:08
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? –  lentic catachresis Mar 17 '12 at 13:01
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. –  Mariano Suárez-Alvarez Mar 17 '12 at 13:18
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 $k=rank(\mathfrak{P})-dim(\mathfrak{R}$).)) –  Andrew Parker Mar 19 '12 at 23:55
@AndrewParker A counterexample was given as the first comment here that shows commutative artinian rings do not always have that property. –  rschwieb Oct 16 '12 at 1:25