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Mar 14, 2019 at 22:39 vote accept Sebastien Palcoux
Mar 14, 2019 at 20:35 comment added Jeremy Rickard @SebastienPalcoux For any finite dimensional algebra, there’s a bijection between (isomorphism classes of) indecomposable projective modules and simple modules, given by taking the quotient by the radical in one direction and taking the projective cover in the other.
Mar 14, 2019 at 17:07 comment added Sebastien Palcoux Why the projective cover of a simple $kH$-module $S$ must be indecomposable and direct summand of $kH$?
Mar 14, 2019 at 12:39 comment added Jeremy Rickard @SebastienPalcoux Not with group algebras, of course, as they'd be semisimple. But for abstract algebras I wouldn't expect the characteristic of $k$ to matter. It might just be harder to construct the examples because you don't have the groups to play with.
Mar 14, 2019 at 12:25 comment added Sebastien Palcoux Do you expect the existence of a counter-example if the field is algebraically closed of characteristic zero? If yes, what about $k=\mathbb{C}$?
Mar 14, 2019 at 9:15 history answered Jeremy Rickard CC BY-SA 4.0