in my algebra class, they give us as an exercise to prove that a finite group $G$ admits at most finitely many non-equivalent irreducible representations over an arbitrary field $F$.
Now, I showed that the number of non-equivalent irreducible representations of $G$ over $F$ is equal to the number of non-isomorphic simple $F[G]$-modules, where $F[G]$ denotes the group algebra of $G$ over $F$.
Also, I know that a consequence of Maschke's Theorem is that the statement above holds when $F$ has characteristic $0$ or not dividing $|G|$. However, here we look for an argument that could work for any $F$.
I believe the sense of the exercise was essentially proving the equivalence between simple $F[G]$-modules and irreducible representations (indeed it is a class in Group Theory). So the point is actually why there are finitely many such modules?
I know there are only finitely many non-isomorphic simple modules over semisimple ring. However, at the best of my knowledge $F[G]$ is a priori just Artinian. I cannot find any evidence in the literature about this specific case.
I would appreciate any suggestion, or reference that you can provide, thank you.
Best Answer
No specific property of the group algebra is needed beyond that it is finite-dimensional:
Proof. Recall that the Jacobson radical $J(A)$ is the intersection of the annihilators of all simple $A$-modules, so simple $A$-modules are in natural bijection with simple $A/J(A)$-modules. Now the key observation is that since $A$ is Artinian, $A/J(A)$ is semisimple (see e.g. here). So by the Artin-Wedderburn theorem it is isomorphic to a finite product
$$A/J(A) \cong M_{n_i}(D_i)$$
of matrix rings over division rings, and has one isomorphism class of simple module for each matrix ring. So there are finitely many. If $A$ is finite-dimensional over a field $K$ then the dimension of each algebra on the RHS over $K$ is at least $1$ so there can only be at most $\dim_K A$ of them (and this bound is sharp iff $A \cong K^n$). $\Box$