Let $\chi(\cdot)$ be an irreducible (over $\mathbb{C}$) character of an representation in a finite Group G:
Show that $\chi(g)$ is an algebraic integer in the cyclotomic field $\mathbb{Q}$($\zeta_n$), where $\zeta_n$ := $e^{2\pi i/n}$.
I already got an answer here: my old question
But I struggle with the part, where he follows the similarity of M(g) to some diagonal matrix.
So any help or even a new approach?
A fact from linear algebra: If the minimal polynomial of a matrix $M\in \mathrm{M}_n(\mathbb{C})$ has no repeated roots then $M$ is diagonalizable. You can prove this as follows. If $p,q$ are coprime polynomials then using Bézout's identity we have $\ker(pq(M))=\ker(p(M))\oplus \ker(q(M))$. So in particular if the minimal polynomial of $M$ is say $(z-\lambda_1)\times\cdots\times (z-\lambda_k)$ with the $\lambda_i's$ distinct, then $\mathbb{C}^n = \ker(M-\lambda_1 I_n)\oplus \cdots \oplus (M-\lambda_k I_n)$, so $M$ is diagonalizable.
Now if $\rho:G\rightarrow \mathrm{GL}_n(\mathbb{C})$ is a representation with character $\chi(g)=\mathrm{tr}\left(\rho(g)\right)$ then $I_n=\rho\left(g^{|G|} \right)= \rho\left(g \right)^{|G|}$. Hence the minimal polynomial of $\rho\left(g \right)$, which is a divisor of $X^{|G|}-1$, has distinct roots. Hence $\rho\left(g \right)$ is diagonalizable with eigenvalues (which are some of the roots of $X^{|G|}-1$) in $\mathbb{Q}(\zeta_{|G|})$ and so is their sum, which is $\chi(g)$.