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Aug 6, 2022 at 19:04 history edited Glorfindel CC BY-SA 4.0
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
Apr 2, 2010 at 5:41 vote accept Michel Hebert
Apr 2, 2010 at 2:56 comment added Manny Reyes 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!
Apr 1, 2010 at 21:19 history answered Georges Elencwajg CC BY-SA 2.5