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Dec 21, 2012 at 13:43 comment added Ralph Yes. The augmentation is given by $RG \to R,\;g \mapsto 1$ and $RG=\bigoplus_g Rg$ is $R$-free.
Dec 21, 2012 at 13:25 comment added Marc Palm I meant to say, eg RG is such an algebra? mathoverflow.net/questions/674/…
Dec 21, 2012 at 12:02 comment added Ralph Marc, I'm not sure what you mean by this equality.
Dec 21, 2012 at 11:46 comment added Marc Palm augmented R-projective R-algebra A = group ring RG?
Dec 21, 2012 at 9:10 vote accept Mark Opitz
Dec 21, 2012 at 0:16 history answered Ralph CC BY-SA 3.0