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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 ?

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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.2012-03-01
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    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?2012-03-17
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    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.2012-03-17
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    @Mariano: historically, I don't think so. If I remember the introduction of Lam's marvellous book correctly, Serre's question was mainly motivated by its simple topological counterpart (a vector bundle on the affine space is trivial). Serre's result “projective => stably free” is (slightly) more recent.2012-03-19
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    @PseudoNeo, I'll look—but «projective implies stably free» for manifolds, for example, is just as natural :)2012-03-19
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    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 $k=rank(\mathfrak{P})-dim(\mathfrak{R}$).))2012-03-19
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    Thank you all for pay attention at my question. I also posted it on MO, however, may be MO do not feel it is interesting :|. @BrunoStonek and AndrewParker: Thank you very much for the information. I will find the book and check it out.2012-03-20
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    @AndrewParker A counterexample was given [as the first comment here](http://math.stackexchange.com/questions/214469/are-projective-modules-over-an-artinian-ring-free) that shows commutative artinian rings do not always have that property.2012-10-16
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    @rschwieb I have appended and expanded upon my previous comment as an answer to the question to which you've linked.2013-02-22
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    for completeness: the link to the [MO question](http://mathoverflow.net/questions/91349/condition-for-a-ring-on-projective-and-free-problem)2013-06-23
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    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 https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://projecteuclid.org/download/pdf_1/euclid.tmj/1178244175&ved=2ahUKEwje_5bv1f_aAhULp48KHYSNA6wQFjAAegQIARAB&usg=AOvVaw0Wg3pDYtBZHdtx_2aqGyrK . This generalizes Kaplansky's theorem for projective modules over local rings.2018-05-12

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