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Dec 16, 2011 at 7:05 comment added Vladimir Dotsenko True. Though in this case, you rather consider $A/I$ as an $A$-module, of course.
Dec 15, 2011 at 18:12 comment added darij grinberg I think this is one of a long list of cases when proving a result about algebras is harder than proving the same property of modules over these algebras. It seems to tell us that algebras are just modules that accidentally happen to be over themselves...
Dec 15, 2011 at 10:17 history answered Vladimir Dotsenko CC BY-SA 3.0