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. 
 A: 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).)
A: 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 $\chi^{\sigma} = \mu.$
A: Well, wait a minute.  I guess we can agree that every complex representation $\rho$ of $G$ is defined over $\mathbb{C}$!  Let $\sigma$ be an automorphism of the character field $K$.  Then $\sigma$ extends to an automorphism of $\mathbb{C}$: for a proof, see for instance $\S 10$ of my field theory notes (the numbering of the results and the sections is not stable as yet, so search for automorphism extension theorem to find it quickly.) The map $g \mapsto \sigma( \rho(g))$ is clearly a representation of $G$ with character $\sigma(\chi)$, where $\chi$ is the character of $\rho$.  Aren't we done?
A: This is just a slightly more quantitative version of Matt E's answer.  I agree also that this is the obvious and implicit argument.
CG is a direct product of matrix rings of size χ(1).  The primitive central idempotents lie in KG, so KG is a direct product of the same number of matrix rings, but over division K-algebras and of size only dχ dividing χ(1).  Each division algebra has a splitting field of degree χ(1)/dχ, and so G has a splitting field of degree at most the product of those numbers (called indices of the division algebras and Schur indices of the characters).
Zhen Lin's answer is then the correct answer.
You may also keep in mind that if G is not p-solvable, then Galois conjugates of Brauer characters may or may not also be Brauer characters for a fixed maximal ideal of the algebraic integers.  The pth power obviously works, since the Frobenius automorphism works on the underlying representation.  The −1st power works by taking dual modules.  However, the powers not in this subgroup may fail to be automorphisms of the Brauer table.
A: 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 $\sigma: \zeta \mapsto \zeta^2$. 
