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Jan 14, 2017 at 4:19 comment added Mohan For $A=k[x]/x^n$ and a finitely generated $A$-module $M$, $\mathrm{Ext}^1(M,M)=0$ implies $M$ is free, as you say.
Jan 9, 2017 at 14:14 comment added Mare math.uni-bielefeld.de/~ringel/lectures/tachi/tachikawa/… and the overview article Frobenius algebras by Yamagata.
Jan 9, 2017 at 1:02 comment added wonderich Can you give a Ref? Thanks.
Jan 8, 2017 at 15:58 history asked Mare CC BY-SA 3.0