Shortest way of proving that the Galois conjugate of a character is still a character

Let $G$ be a finite group and $\chi$ a character of $G$. The values of $\chi$ generate an abelian Galois extension $K$ of $\mathbb{Q}$, and so one can consider the conjugate $\sigma(\chi)$ of $\chi$ by any element $\sigma$ of the Galois group. What's the shortest way to prove that $\sigma(\chi)$ is also a character of $G$?

This follows from the fact that the corresponding representation $V$ is realizable over a finite extension of $K$, but this fact is somewhat annoying to prove, and on an exam I don't want to prove it from scratch if I can avoid it. Are there any shorter arguments?

Motivation: I'm looking at a representation theory exam question (for practice) which asks me to fill out a character table. If I could assume the above statement, I could conclude that the missing characters are integer-valued.

OK, I've thought about it a bit more and I'll give an alternate proof below the first horizontal line. However, I highly deny that this proof is better, and I am not sure it is actually shorter. To my mind, the morally correct proof is to show that, if $K \subseteq L$ with $L$ algebraically closed, and a system of polynomial equations has a root in $L$, then it has a root in a finite extension of $K$.

The point, which I am sure Qiaochu understands, is that he only knows a priori that the representation is defined over $\mathbb{C}$. Once he knows that the representation is definable over an algebraic extension $K'$ of $K$, he can replace $K'$ by its normal closure, note that $\mathrm{Gal}(K', \mathbb{Q}) \to \mathrm{Gal}(K, \mathbb{Q})$ is surjective, lift any element $\sigma$ of $\mathrm{Gal}(K, \mathbb{Q})$ to some $\tilde{\sigma}$ in $\mathrm{Gal}(K', \mathbb{Q})$, and apply $\tilde{\sigma}$ to the entries of his matrices.

Part of the problem is that the representation may honestly not be defined over $K$. For example, the two dimensional representation of the quaternion eight group has character with values in $\mathbb{Q}$, but can't be defined over $\mathbb{Q}$.

Fix $G$ and a representation $V$ of $G$. For $g \in G$, let $\lambda_1(g)$, $\lambda_2(g)$, ..., $\lambda_n(g)$ be the multiset of eigenvalues of $g$ acting on $V$. These are necessarily roots of unity, since $g^N=1$ for some $N$. For any symmetric polynomial $f$, with integer coefficients, define $\chi(f,g) = f(\lambda_1(g), \ldots, \lambda_n(g))$.

Lemma: With notation as above, $g \mapsto \chi(f,g)$ is a virtual character.

Proof: If $f$ is the elementary symmetric function $e_k$, then this is the character of $\bigwedge^k V$. Any symmetric function is a polynomial (with integer coefficients) in the $e_k$'s; take the corresponding tensor product and formal difference of virtual characters.

Any Galois symmetry $\sigma$ of $\mathbb{Q}(\zeta_N)$ is of the form $\zeta_N \mapsto \zeta_N^s$, for $s$ relatively prime to $N$. Consider the power sum symmetric function $p_s := \sum x_i^s$. So $\chi(p_s, \ )$ is the Galois conjugate $\chi^{\sigma}$, and we now know that it is a virtual character.

But $\langle \chi^{\sigma}, \chi^{\sigma} \rangle = \langle \chi, \chi \rangle =1$, because the inner product is built out of polynomial operations and complex conjugation, and complex conjugation is central in the Galois group. So this virtual character must correspond to $\pm W$, for some representation $W$. Since $\chi^{\sigma}(e) = \chi(e) = \dim V$, we conclude that the positive sign is correct.

It just occurred to me that actually writing this out for some specific small values of $s$ makes some nonobvious statements about representation theory. For example, if $G$ has odd order and $V$ is a $G$-irrep, then $\bigwedge^2 V$ has a $G$-equivariant injection into $\mathrm{Sym}^2 V$. Proof: The difference of their characters is the character of $V^{\sigma}$, where $\sigma: \zeta \mapsto \zeta^2$.

I think the natural proof (although I think that this is the one you want to avoid) is that $\mathbb Q[G]$ is a semi-simple $\mathbb Q$-algebra, hence when extended to $\overline{\mathbb Q}$ it splits as a product of matrix rings, hence it already so splits when extended over a finite extension $K$ of $\mathbb Q$, hence all representations of $G$ are defined over this $K$. (This $K$ is not the $K$ in the statement of your question, but rather the finite extension of it that you allude to.)

The reason I am writing this out despite your request not to is just to point out that it is not that annoying to prove; in fact it is quite natural. And all the arguments that I know of the type that certains systems of eigenvalues are closed under Galois conjugation (e.g. that if $f$ is a modular form which is a Hecke eigenform with system of eigenvalues $(a_p)$, then for any Galois element $\sigma$, the system of eigenvalues $(\sigma(a_p))$ is also attached to a Hecke eigenform) are proved in the same manner. (Namely, by showing that the natural $\mathbb C$-algebra that governs the situation, whether it be the group algebra or the Hecke algebra, actually has a model over $\mathbb Q$, with the same set of generators (group elements or Hecke operators, as the case may be).)

It's a bit late, but you could argue like this. By Schur's Lemma, when $\chi$ is an irreducible complex character of the finite group $G$, and $\sigma$ is a representation affording $\chi$, then for each $z \in Z(\mathbb{C}G$, we have $z\sigma = \lambda(z)I$ for some scalar $\lambda(z).$ Taking traces tells us that $\lambda(z) = \frac{\chi(z)}{\chi(1)}.$ Also, $\lambda$ defines an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}.$ Since the dimension of $Z(\mathbb{C}G)$ is $k= k(G)$, the number of conjugacy classes of $G$, and since there are $k$ different irreducible characters of $G$ (I am assuming the orthogonality relations as given), which give (using the orthogonality relations) $k$ distinct algebra homomorphisms from $Z(\mathbb{C}G)$ to $\mathbb{C}$, we see that all such algebra homomorphisms come from irreducible characters.

We use the natural "integral" basis of $Z(\mathbb{C}G)$ (that is, the basis of conjugacy class sums) to show that for $\lambda$ as above, $\lambda^{\sigma}$ is also an algebra homomorphism for each automorphism $\sigma \in {\rm Gal}(\mathbb{Q}[\omega]/\mathbb{Q}$, where $\omega$ is a primitive complex $|G|$-th root of unity. Notice that $\lambda(C) \in \mathbb{Q}[\omega]$ for each class sum $C$. Furthermore, for class sums $C_{r}$ and $C_{s}$, there are integers $a_{rst}$ such that $\lambda(C_r C_s) = \sum_{t=1}^{k}a_{rst}C_t$. That $\lambda$ is an algebra homomorphism is encapsulated precisely by the fact that we have $\lambda(C_r) \lambda(C_s) = \sum_{t=1}^{k} a_{rst}\lambda(C_t)$. Since the $a_{rst}$ are rational integers, we can apply $\sigma$ to this equation to conclude that $C_r \to \lambda^{\sigma}(C_r)$ for each class sum, (and extending by $\mathbb{C}$-linearity to $\mathbb{C}$-combinations of class sums) is an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}$. Associated to this is, as discussed above, is a complex irreducible character $\mu$ of $G$ such that $\frac{\mu(C)}{\mu(1)} = \frac{\chi^{\sigma}(C)}{\chi(1)}$ for each class sum $C$. Hence $\mu(g) = \mu(1)\frac{\chi^{\sigma}(g)}{\chi(1)}$ for each $g \in G.$ Since $\{ \chi(g): g \in G\}$ is $\sigma$-stable, we have $\sum_{g \in G}|\chi^{\sigma}(g)|^{2} = |G|$, so that $\chi(1) = \mu(1)$ and $\chi^{\sigma} = \mu.$