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Oct 20, 2010 at 15:11 comment added Todd Trimble I was. If OP meant to consider infinite-dimensional $V$ as well (and it seems there's no indication he doesn't), then I'd have to rethink the problem. The Morita business I had in mind does assume finite-dimensionality.
Oct 20, 2010 at 14:34 comment added t3suji I am confused: are you assuming that $V$ is finite-dimensional?
Oct 17, 2010 at 16:18 comment added Martin Brandenburg It surprises me that this shows that every algebra endomorphism of $End(V)$ is actually an automorphism.
Oct 16, 2010 at 3:33 comment added user5810 Thanks. (I would have accepted it even before you added that, it just took me a while to give up trying to comment first since the site wouldn't let me.)
Oct 16, 2010 at 3:27 vote accept CommunityBot
Oct 16, 2010 at 3:26 comment added Todd Trimble I have updated my answer to point you to a slightly different proof.
Oct 16, 2010 at 3:25 history edited Todd Trimble CC BY-SA 2.5
Added a reference
Oct 16, 2010 at 3:16 comment added Todd Trimble That's not an algebra homomorphism -- it doesn't preserve the identity.
Oct 16, 2010 at 3:07 comment added user5810 Okay, I'm lost in the category theory, but how could we possibly get the zero homomorphism by conjugating the identity with an isomorphism?
Oct 16, 2010 at 2:19 comment added Todd Trimble I wrote it that way in remembrance of the canonical bijection between linear maps $M \to \hom_F(V, V)$ and linear maps $M \otimes_F V \to V$, and (by specializing) between algebra maps and module structures. I was trying to segue from a question about algebras to a module-theoretic question.
Oct 16, 2010 at 2:12 comment added user5810 Am I missing a reason for you writing $M \to \hom(V, V)$ instead of $M \to M$?
Oct 16, 2010 at 1:58 history answered Todd Trimble CC BY-SA 2.5