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Jul 6, 2014 at 19:42 comment added Jeremy Rickard If $R$ is an artinian ring of finite representation type (and I think it's still an open question whether this is equivalent to being pure semisimple?) then I think $K(R)$ is equivalent to $D(S)$ for $S$ the Auslander algebra of $R$ (i.e., the endomorphism ring of the direct sum of all indecomposable modules, or its opposite ring, depending on your choice of conventions). I'm confident of this if $R$ is an artin algebra (i.e., finitely generated as a module over an artinian centre), but I might be missing a problem for general artinian rings.
Jul 6, 2014 at 18:19 history answered Greg Stevenson CC BY-SA 3.0