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Jul 21, 2014 at 9:47 history edited tj_ CC BY-SA 3.0
added 118 characters in body
Jul 21, 2014 at 9:32 comment added tj_ Oh, I missed that it's generated as an algebra. Fixed it. Still, I think it should be pointed out that the basic property of $I$ (for any $B$) is $I=(b\otimes 1 - 1 \otimes b \mid b \in B)$. The proof is also implicitly in Julian's answer.
Jul 21, 2014 at 8:45 comment added AYK The elements $b_i$ generate $B$ as an algebra over $A$, not as a module. The solution given above (by Rosen) is correct. Anyway, thanks for the effort!
Jul 21, 2014 at 5:01 history answered tj_ CC BY-SA 3.0