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Let $A$ be an associative algebra over a field $K$ and let $\rho:A \to \operatorname{End}_K(V)$ be a representation of $A$. Is it true that if $\rho$ is faithful then it's also an isomorphism?

Note: I know it's ture if $A$ is a finite group algebra $K[G]$ which relies on the first Isomorphism theorem. But How about an arbitrary algebra? Even is it true for finite diemensional algebra?

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up vote 2 down vote accepted

The claim is never true.

Suppose $V$ is a faithful representation and $\rho:A\to\operatorname{End}_K(V)$ the corresponding morphism. If $\rho$ is an isomorphism, then $A$ is in fact semisimple. Since non-semisimple algebras exist, there are counterexamples.

Even if $A$ is semisimple, the claim is false. Indeed, an algebra has faithful modules of arbitrary large dimension (the direct sum of any faithful module with any module is itself faithful) and $A$ cannot be isomorphic to the endomorphism ring of each of them.

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Well, the statement would be true if $A=\operatorname{End}_K(V)$ and $\rho$ is the identity... But of course a false statement is never true. – Marc van Leeuwen Sep 5 '12 at 7:04

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