Dear all,

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

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    $\begingroup$ 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? $\endgroup$ – Somnath Basu Apr 1 '10 at 21:19
  • $\begingroup$ (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 $\endgroup$ – 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


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.

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    $\begingroup$ 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! $\endgroup$ – Manny Reyes Apr 2 '10 at 2:56

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