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Let $F\subset\mathbb{C}$ by the subfield of algebraic numbers, and let $\varphi\colon G\to GL_n(F)$ be a representation of a finite group $G$, with character $\psi$.

Let $\mathbb{Q}(\varphi)$ denote the subfield of $F$ generated by adjoining the entries $\varphi(g)$ for all $g\in G$. Since this amounts to adjoining finitely many algebraic elements of finite degree, $\mathbb{Q}(\varphi)/\mathbb{Q}$ is finite.

Now let $K$ be a Galois extension of $\mathbb{Q}$ containing $\mathbb{Q}(\varphi)$. For $\sigma\in\operatorname{Gal}(K/\mathbb{Q})$, $\sigma$ extends to an automorphism of $F$, also denoted $\sigma$, and then we can define $\varphi^\sigma\colon G\to GL_n(F)$ by setting $\varphi^\sigma(g)$ to be the matrix $\varphi(g)$ with $\sigma$ applied to each entry. I was able to show that $\varphi^\sigma$ is a representation, and that the character of $\varphi^\sigma$ is $\sigma\circ\psi$.

Why is $\varphi$ irreducible if and only if $\varphi^\sigma$ is irreducible?

I know that a character is irreducible iff it has norm $1$, so I think it would be enough to show $\psi$ has norm $1$ iff $\sigma\circ\psi$ has norm $1$. Since the norm has formala $$ \|\psi\|^2=\frac{1}{|G|}\sum_i |C(g_i)||\psi(g_i)|^2 $$

where the sum is over a set of representatives $g_i$ of each conjugacy class $C(g_i)$, it would even be enough to show $\psi$ and $\sigma\circ\psi$ have the same norm on an arbitrary conjugacy class, but I'm not sure how to do that.

Is there a way to possibly show that $K$ contains the cyclotomic field $\mathbb{Q}(\zeta_n)$? I know that each $\varphi(g)$ is diagonalizable to a matrix whose diagonal consists of $n$th roots of unity. Then $K$ would be a Galois extension containing $\mathbb{Q}(\zeta_n)$, so its Galois group would be contained in the Galois group of $\mathbb{Q}(\zeta_n)$. Since every cyclotomic automorphism commutes with complex conjugation, I think this would imply $\sigma$ commutes with complex conjugation, and with that I could directly compute $\|\sigma\circ\psi\|$. Thanks.

This discussion is essentially 18.3.14-17 of Dummit and Foote.

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First of all, note that we only have to show that if $\varphi$ is not irreducible, then $\varphi^\sigma$ is not irreducible either (the other direction follows then from $(\varphi^{\sigma})^{\sigma^{-1}} = \varphi$).

If $\varphi$ is not irreducible, it admits a nontrivial subrepresentation $\varphi'$ and we find a matrix $A\in GL_n(F)$ such that $$A \varphi(g)A^{-1} = \pmatrix{\varphi'(g) & * \\ 0 & \varphi''(g)}$$ for all $g$. Because $\sigma$ is a ring automorphism we find that $$A^{\sigma} \varphi^{\sigma}(g){A^{\sigma}}^{-1} = (A \varphi(g)A^{-1})^{\sigma} = \pmatrix{\varphi'^\sigma(g) & * \\ 0 & \varphi''^\sigma(g)}$$ for all $g$, that is, $\varphi^{\sigma}$ has the nontrivial subrepresentation $\varphi'^\sigma$ and is thus not irreducible.

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