## surjectivity of irreducible representation

I don't know how to show the following: Let $A$ be an associative algebra (not necessary finite-dimensional) and $p\colon A\to End(V)$ be it irreducible finite-dimensional representation. Then $p$ in general is not surjective.

The standard textbooks on representation theory don't contain answer on this and googling doesn't help.

The interesting case for me is an irreducible representation of universal enveloping of semisimple Lie algebra.

Why am I asking: I am reading articles (of G.I. Olshaskiy) on centralizers of Lie subalgebras and trying to understand if $U(gl_{n+m})^{gl_m}$ is a sufficient object to consider or not.

Upd: I've forgot: the ground field is algclosed of char 0.

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What's the ground field? Of course if it's $\mathbb{R}$ and $A=\mathbb{R}[t]/(t^2+1)$, then the regular module is irreducible, but the corresponding $p$ is not surjective. Over an algebraically closed field it's true even for infinite-dimensional $A$ though.

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Why this is true for algebraically closed field? – zroslav Dec 17 2011 at 15:15
Because it's called Jacobson's density theorem. Check Theorem 2.5 (a) in arxiv.org/abs/0901.0827 for a proof. – darij grinberg Dec 17 2011 at 15:48
Actually for algebraically closed fields it is Burnside's theorem of which Jacobson's theorem is a generalization. – Benjamin Steinberg Dec 17 2011 at 17:00

Let $D \ne k$ be a central division algebra of finite dimension over a (commutative) field $k$ say with $\dim_k D = n^2$. Now view $V = D$ as a left module over itself; then $V$ is an irreducible $D$-module.

Since $\dim \operatorname{End}_k(D) = n^4 > n^2$, the image of the mapping $D \to \operatorname{End}_k(V) = \operatorname{End}_k(D)$ is a proper subalgebra.

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