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Jul 10, 2021 at 10:56 history edited Jeremy Rickard CC BY-SA 4.0
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Jun 15, 2020 at 7:27 history edited CommunityBot
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May 10, 2020 at 13:02 comment added Tim Campion Thanks so much -- I hadn't dared hope for such a complete answer! It's interesting that the pure semisimplicity conjecture would imply that if $Mod_R$ has a basis, then it has a finite basis of indecomposable modules. This would be surprising, but definitely in the spirit of Warfield's theorem, which says that if $R$ is commutative and $Mod_R$ has a basis, then it has a basis of cyclic modules.
May 10, 2020 at 12:43 vote accept Tim Campion
May 10, 2020 at 9:19 history answered Jeremy Rickard CC BY-SA 4.0