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Aug 18, 2012 at 8:10 comment added W. Politarczyk I forgot to mention it, but for me $R$ is noetherian and I am mostly interested in the noncommutative case.
Aug 17, 2012 at 20:44 comment added Damian Rössler The ring $R$ has to be noetherian for this to be true. Also, I think W. Politarczyk is mainly interested in the non-commutative case (group rings mostly are).
Aug 17, 2012 at 19:20 history answered Steven Landsburg CC BY-SA 3.0