More specifically, let $G$ be a finite group and $\pi$ a complex representation of $G$ in $V=\Bbb{C}^n$. Then we can choose a basis of $V=\Bbb{C}^n$ s.t. if we write $\pi$'s operators in matrix form by this basis as $\pi(g)=(\pi_{ij}(g))_{i,j\leq n}$, then there exists a finite extension $\Bbb{Q}\subseteq K$ s.t. $\pi_{ij}(g) \in F$ for all $i,j,g$. Here is my take:
For $g\in G$ let $A=\pi(g)$ then if $|G|=m$ then $A^m=I$. Hence $A$'s minimal polynomial divides $x^m-1$ and since we are in $\Bbb{C}$ the polynomial $x^m-1$ has no double roots. From here $A$'s minimal polynomial splits into distinct linear factors hence $A$ is diagonalizable and all of its eigenvalues are roots of unity. So there exists a basis of $V$ s.t. the transformation $v\mapsto Av$ is represented by a matrix with entries in $K=\Bbb{Q}(e^{\frac{2i\pi}{m}})$.
Now, we have proved that for each $g\in G $ it is possible to find a basis s.t. $\pi(g)$ is represented as a matrix with entries in $K$. Here I am stuck, I know that every $g\in G$ has a basis s.t. $\pi(g)$ in that basis is represented by a matrix with entries in $K$ and that $K$ does not depend on the $g\in G$ we choose. The problem is finding a basis of $V$ s.t. all of $\pi$'s operators are represented as matrices with operators in $K$.
Any help is welcome