A ring on which all finitely generated projectives modules are free but not all projectives are free? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T04:11:30Z http://mathoverflow.net/feeds/question/20105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20105/a-ring-on-which-all-finitely-generated-projectives-modules-are-free-but-not-all-p A ring on which all finitely generated projectives modules are free but not all projectives are free? Michel Hebert 2010-04-01T20:27:05Z 2010-04-01T21:19:00Z <p>Dear all, </p> <p>Sorry if the question is naive: any nice example of such a ring or, better, of a class of such rings?</p> http://mathoverflow.net/questions/20105/a-ring-on-which-all-finitely-generated-projectives-modules-are-free-but-not-all-p/20109#20109 Answer by Georges Elencwajg for A ring on which all finitely generated projectives modules are free but not all projectives are free? Georges Elencwajg 2010-04-01T21:19:00Z 2010-04-01T21:19:00Z <p>Cher Michel, these rings are uncommon.</p> <p>1) Over a local ring ALL projective modules are free : this is a celebrated theorem due to Kaplansky.</p> <p>2) If $R$ is commutative noetherian and $Spec(R)$ is connected, every NON-finitely generated projective module is free. This is due to Bass in his article "Big projective modules are free" which you can download for free here </p> <p><a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.ijm/1255637479" rel="nofollow">http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.ijm/1255637479</a></p> <p>And now for the good news: the rings you are after are uncommon but they exist. Bass in the article just quoted shows that the ring $R=\mathcal C([0,1])$ of continuous functions on the unit interval has all its finitely generated projective modules free. Nevertheless the ideal consisting of functions vanishing in a neighbourhood of zero (depending on the function) is projective, not finitely generated and not free. Bass attributes the result to Kaplansky.</p>