Let $K$ be an algebraically closed field with characteristic $0$ and $G$ be a finite group. Consider representations of $K[G].$
Theorem. Any character is uniquely represented as a linear combination of irreducible characters with integer non-negative coefficients. Representations are isomorphic iff their characters are equal.
Proving this we use the following statement:
Proposition. Let $R$ be a finite dimensional semisimple algebra over an algebraically closed field $F.$ Then $R\cong M_{n_1}(F)\times\dots\times M_{n_k}(F)$ for some $n_1,\dots,n_k.$
My question is: does the theorem hold for arbitrary fields with characteristic $0$ (not nessecary algebraically closed)? I believe that the answer is no but cannot find a counterexample.
Best Answer
Yes, the theorem still holds. Even if $K$ isn't algebraically closed Maschke's theorem still holds so we still have semisimplicity and the orthogonality relations still hold so characters of inequivalent irreducible representations are still linearly independent. The only modification is this: if $\chi_i$ are the irreducible characters of the irreducible representations $V_i$ then $\langle \chi_i, \chi_j \rangle = 0$ if $i \neq j$ (as usual) and if $i = j$ we have
$$\langle \chi_i, \chi_i \rangle = \dim_K \text{End}(V_i)$$
which can be greater than $1$. For example, if $K = \mathbb{R}$ then this dimension can be $1, 2$, or $4$, corresponding to $\text{End}(V_i)$ being $\mathbb{R}, \mathbb{C}$, or $\mathbb{H}$ (see Frobenius-Schur indicator).
However the theorem breaks if the characteristic of $K$ divides the order of $G$ due to lack of semisimplicity; you can find a counterexample with $\mathbb{F}_p[C_p]$.