# A ring on which all finitely generated projectives modules are free but not all projectives are free?

Dear all,

Sorry if the question is naive: any nice example of such a ring or, better, of a class of such rings?

• Just curious - Is this motivated by the Quillen-Suslin theorem which says that every projective module over $k[x_1,\ldots,x_n]$, where $k$ is a PID, is free? – Somnath Basu Apr 1 '10 at 21:19
• (Answering Somnath Basu's question) Not really. My motivation is that I am trying to define a "categorically sound" definition of a "finitely related" algebra, i.e., one which is preserved under categorical equivalences. A first proposal would be the direct sum of a projective and a finitely presentable, but a better concept would be the retracts of those. I was curious to know when they are the same. Michel Hebert – Michel Hebert Apr 3 '10 at 7:13

Cher Michel, these rings are uncommon.

1) Over a local ring ALL projective modules are free : this is a celebrated theorem due to Kaplansky.

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

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.ijm/1255637479

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.

• Here's another reason why we expect such rings to be uncommon, but likely to exist. Another result of Kaplansky says that, over any ring, any projective right module is a direct sum of countably generated projective modules. Thus, a ring over which every finitely generated right module is free is, in a sense, "ever so close" to having all projective right modules free! – Manny Reyes Apr 2 '10 at 2:56