Characters are cyclotomic integers

Statement
Let $$G$$ be a finite group, and $$k$$ be any field. Then, the character of any finite-dimensional linear representation of $$G$$ over $$k$$ takes, at every element of the group, a value that is a cyclotomic integer over the prime subfield of $$k$$: in other words, it is in the subring generated by all the roots of unity.

(The statement, does not require the characteristic of $$k$$ to not divide the order of the group).

Related facts

 * Element of finite order is semisimple and eigenvalues are roots of unity (the semisimplicity requires that the order be relatively prime to the characteristic of the field, something we don't need here).

Applications

 * Characters are algebraic integers: Any cyclotomic integer is an algebraic integer, so every character takes values in algebraic integers.
 * If, over a field of characteristic zero, a character takes a rational value, that value must be an integer.

Other related facts

 * Size-degree-weighted characters are algebraic integers
 * Zero-or-scalar lemma

Proof
Given: A finite group $$G$$, field $$k$$, a finite-dimensional linear representation $$(V,\rho)$$ of $$G$$ over $$k$$. $$\chi$$ is the character of $$\rho$$.

To prove: For any $$g$$, $$\chi(g)$$ is a cyclotomic integer over the prime subfield of $$k$$.

Proof: $$\chi(g)$$ is the sum of eigenvalues of $$\rho(g)$$, counted with multiplicity, in the algebraic closure of $$k$$. Since $$G$$ is finite, $$g$$ has finite order, so $$\rho(g)$$ has finite order, and hence $$\rho(g)$$ satisfies a polynomial of the form $$x^n - 1$$. Hence, every eigenvalue of $$\rho(g)$$ is a $$n^{th}$$ root of unity, so $$\chi(g)$$ is a sum of $$n^{th}$$ roots of unity. Hence, $$\chi(g)$$ is a cyclotomic integer.