Is any representation of a finite group defined over the algebraic integers? Apologies in advance if this is obvious.
 A: By the way, this paper may be of interest. It shows that for solvable groups, one doesn't have to do the Hilbert class extension moonface suggests, but for some non-solvable ones you do.  Also this one has more examples.
A: Not a satisfying argument: We can, first of all, find a basis in which the entries lie in some algebraic number field $K$. Let $\mathcal{O}$ be the ring of integers of $K$.
Then there is a locally free $\mathcal{O}$-module $M$ of rank $n$ preserved by $G$: add up all the translates of $\mathcal{O}^n$ under $G$. Now, $M$ need not itself be free,  but it is isomorphic
as an $\mathcal{O}$-module to the sum of various ideals of $\mathcal{O}$. Now pass to an extension $L/K$ so that every ideal class of $K$ trivializes in $L$, e.g. the Hilbert class field; then $G$ preserves a free rank $n$ module
for the ring of integers of $L$.   Sorry!
A: This is not really an answer, but is too long for a comment. The proof given by Moonface above is given in more or less that form in the 1962 book of Curtis and Reiner. As far as I know, it is still open whether all irreducible representations of a finite group $G$ can be realized over 
$\mathbb{Z}[\omega]$, where $\omega$ is a complex primitive $|G|$-th roots of unity, though I think the paper of Cliff,Ritter and Weiss settles the questions for finite solvable groups. The paper
of Serre ( the three letters to Feit) give counterexamples to a slightly different question:
they show (among other things) that a representation of a finite group can be realised over 
some number fields, but might not be able to be realised over the ring of algebraic integers
of that field. Brauer's characterization of characters/Brauer's induction theorem show that 
all representations of the finite group $G$ may be realised over $\mathbb{Q}[\omega]$ for 
$\omega$ as above ( $|G|$ can be replaced by the exponent of $G$ if desired). As I said, 
realizability over $\mathbb{Z}[\omega]$ is a different matter.
