# Character of Representations over C are Algebraic Integers

## Theorem

Let $G$ be a finite group.

Let $\chi$ be the character of any $\C \left[{G}\right]$-module $\left({V, \rho}\right)$.

Then for all $g \in G$, it follows that $\chi \left({g}\right)$ is an algebraic integer.

## Proof

By the definition of character:

$\chi \left({g}\right) = \operatorname{Tr} \left({\rho_g}\right)$

where:

$\rho \in \hom \left({\C \left[{G}\right], \operatorname{Aut} \left({V}\right)}\right): \vec {e_g} \mapsto \rho_g$

by definition.

Fix an arbitrary $g \in G$.

Let $\left\vert{g}\right\vert$ denote the order of $g$.

The trace $\operatorname{Tr} \left({\rho_g}\right)$ of $\rho_g$ is defined as the sum of the eigenvalues of $\rho_g$.

From Eigenvalues of G-Representation are Roots of Unity, we have that any eigenvalue $\lambda$ of $\rho_g$ is a root of unity whose order is $\left\vert{g}\right\vert$.

We have that $\lambda$ satisfies the monic polynomial $x^{\left\vert{g}\right\vert} - 1$

Hence we have that $\lambda$ is an algebraic integer.

From Ring of Algebraic Integers, we have that the sum of the eigenvalues is also an algebraic integer.

Thus $\chi \left({g}\right)$ is an algebraic integer.

$\blacksquare$