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This is a basic question-

Let $K$ be an algebraic number field, $\Gamma$ a finite group, and $R(\Gamma)$ the ring of virtual characters of $\Gamma$ with values in the algebraic closure $\mathbb{Q}^c$ of $\mathbb{Q}$ in $\mathbb{C}$.

Why is it true that the values of the characters of $\Gamma$, i.e. the numbers $\chi(\gamma)$ $(\chi\in R(\Gamma), \gamma\in \Gamma)$ lie in some number field $F \subset\mathbb{Q^c}$ containing $K$?

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Can't one just adjoin said numbers $\chi(\gamma)$ to $K$ to obtain $F$? –  anon Aug 21 '13 at 23:55

1 Answer 1

up vote 1 down vote accepted

If $\Gamma$ is a finite group of order $n$ then homomorphic images of its elements must have order dividing the number $n$, in which case the eigenvalues of the corresponding linear transformations are $n$th roots of unity, so any $\Bbb Z$-linear sum of these eigenvalues (virtual characters in particular) must be contained in ${\Bbb Q}(\zeta_n)$ and hence in $F=K(\zeta_n)\supseteq K$ for any number field $K$.

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