A classical result is
Theorem: Let $G$ be a abelian group and $(V, \rho)$ be a irreducible representation of $G$ over a algebraically closed field $k$. If $V$ is finite dimensional (more generally, if $\mathrm{dim}_k V < |k|)$ then $\mathrm{dim}_k V = 1$.
This is essencially Schur/Dixmier's lemma, the proof is based at $\rho(g)$ having a eingenvalue, $\forall g \in G$. But what if i don't suppose anything about the dimension of $V$?
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Question:
Are every irreducible $G$-module finite dimensional? Maybe the unitary ones? Do I really need to go topological world and suppose $G$ compact abelian and consider only strongly continuous representations or something like this?
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An idea:
$G$-modules are equivalent to $k[G]$-modules and if $G$ is abelian then this is a commutative algebra over $k$. If $G$ is finite, then $k[G]$ is finite dimensional over $k$ and $V = k[G]v$ for $v \in V \setminus 0$, we are done. More generally, this proves that all irredutible representations of a finite group are finite dimensional.
In abelian case, $k[G]^*$ is also a commutative cocommutative Hopf algebra. A $k[G]$-module $V$ gives a $k[G]^*$–comodule $V^*$ and simple $k[G]^*-$comodules are finite dimensional over $k$ (see here). But dualizing transforms injections into surjections, I don't know if still $V^*$ is a simple comodule.
Best Answer
It is not true in the infinite-dimensional case. An irreducible representation of an abelian group $A$ over a field $k$ is the same thing as a simple module over the commutative $k$-algebra $k[A]$. Since $k[A]$ is commutative, simple modules can be identified with quotients by maximal ideals. If $m$ is such a maximal ideal, then $k[A]/m$ is a field extension of $k$.
But it is possible for $k[A]/m$ to be a nontrivial field extension of $k$, even if $k$ is algebraically closed, if $A$ is large enough. For example we can arrange to have $k[A]/m \cong k(t)$ by choosing $A = k(t)^{\times}$, together with the natural map $k[k(t)^{\times}] \to k(t)$.
On the other hand, the following can be salvaged. I've seen this called Dixmier's lemma but I'm not sure it's the same statement other people call Dixmier's lemma.
Proof. Schur's lemma still holds in the sense that $\text{End}_G(V)$ is always a division algebra. Since $k$ is algebraically closed, in order for this division algebra to contain elements other than the elements of $k$ it must contain a transcendental element. So we observe that $k(t)$ itself has dimension at least the cardinality of $k$ since the rational functions $\frac{1}{t - a}, a \in k$ are linearly independent. Hence if $\dim V < |k|$ then $\text{End}_G(V)$, which has dimension at most $|V|^2$, also has dimension less than $|k|$ (which must be infinite), so it cannot contain a transcendental element. $\Box$
Proof. If $V$ is an irreducible representation of an abelian group $A$ then every $a \in A$ acts by endomorphisms on $V$, so the lemma $\text{End}_A(V) \cong k$ gives that every $a \in A$ acts by a scalar in $k$. Hence $V$ must be $1$-dimensional. $\Box$
So e.g. irreducible representations of countable abelian groups over $\mathbb{C}$ are $1$-dimensional (but not over $\overline{\mathbb{Q}}$!). This is true even though the eigenvalue argument is a priori unavailable in this setting: for example, in the regular representation of $\mathbb{Z}$ (the action of $k[x, x^{-1}]$ on itself) no nonzero element of the group has any eigenvalues.