Irreducible representations of finite abelian groups over an arbitrary field

Question

Let $A$ be a finite abelian group and let $F$ be a field. If $F$ is algebraically closed, or if it is a finite field, then every irreducible $F$-representation of $A$ has cyclic image. Is this true without any assumption on $F$?

Answer 1

Every irreducible representation of $A$ over a perfect (this hypothesis may be dropped) non-algebraically closed field $F$ is constructed by the following:

1. Choose a character $\chi\colon A\to\overline{F}^\times$, where $\overline F$ is an algebraic closure
2. For $\sigma\in\mathrm{Gal}(\overline F/F)$, the twist $\chi^\sigma:=\sigma\circ\chi$ is also a character of $A$. Let $H:=\{\sigma\in\mathrm{Gal}(\overline F/F):\chi^\sigma=\chi\}$.
3. Now, $\bigoplus_{\sigma\in\mathrm{Gal}(\overline F/F)/H}\chi^\sigma$ is an irreducible representation of $A$.

Now, the image of the above representation is isomorphic to $$A/\bigcap_{\sigma\in\mathrm{Gal}(\overline F/F)/H}\ker(\chi^\sigma)=A/\ker(\chi)=\mathrm{im}(\chi).$$ Now $\mathrm{im}(\chi)\subset\overline F^\times$ is a finite abelian subgroup, which by a well-known theorem must be cyclic.