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.



surjectivity of irreducible representation
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.


