Are all (possibly infinite dimensional) irreducible representations of a commutative algebra one-dimensional? If $A$ is a commutative algebra over an algebraically closed field $k$, and $\rho:A \rightarrow End(V)$ is an irreducible representation of $A$ (where, a priori, $V$ may be infinite dimensional), can we conclude that $V$ must be one-dimensional? This is easy to show if we assume $V$ is finite dimensional. It is also true if $A$ is a $C^*$-algebra, $V$ is a (complex) Hilbert space, $End(V)$ denotes bounded operators, $\rho$ preserves conjugation (but is not required to be continuous), and irreducible means no closed subrepresentations, using some spectral theory. But is it true in the algebraic sense with no further restrictions on $A$ or $V$?
Edit: Given Dag Oskar Madsen's comment, I will need to place some restrictions on $A$... what if $A$ is finitely-generated?
Double Edit: Faisal's comment takes care of $A$ finitely-generated (and countably generated) over $\mathbb{C}$.
 A: If $A$ is finitely generated, as in your first edit, then this is much more elementary than the result of Dixmier that Faisal mentioned. A simple $A$-module would have the structure of a field extension of $k$ that is finitely generated as a $k$-algebra (since it's a quotient of $A$), but if $k$ is algebraically closed then there are no such extensions.
A: Here are the details on Faisal's suggestion in the comments.
Lemma (Dixmier): Let $k$ be an algebraically closed field, let $A$ be a $k$-algebra with $\dim A < |k|$, and let $V$ be a simple left $A$-module. Then $D = \text{End}_A(V) \cong k$.
Proof. By Schur's lemma, $D$ is a division algebra over $k$. Then $V$, as a $D$-module, is a sum of copies of $D$, and in particular $\dim D \le \dim V \le \dim A < |k|$. Since $k$ is algebraically closed, if $D$ is strictly larger than $k$ then it must contain a transcendental $t$, hence $k(t) \subseteq D$. But the elements $\frac{1}{t - a}, a \in k$ in $k(t)$ are linearly independent, hence $\dim k(t) \ge |k|$; contradiction. $\Box$
If $A$ is commutative, the image of $A$ in $\text{End}(V)$ is contained in $D$, so $A$ acts by scalars and the conclusion follows. (Incidentally, this gives a short proof of the weak Nullstellensatz over uncountable (algebraically closed) fields.) 
A: As John Wiltshire-Gordon suggested in a comment, this is a generalization of the invariant subspace problem. It follows that for Banach spaces the answer is negative and for Hilbert spaces the answer is open.
