Question

Apologies in advance if this is obvious.

Answer 1

Not a satisfying argument: We can, first of all, find a basis in which the entries lie in some algebraic number field K. Let O be the ring of integers of K. Then there is a locally free O-module M of rank n preserved by G: add up all the translates of O^n under G. Now, M need not itself be free, but it is isomorphic as an O-module to the sum of various ideals of 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!

Answer 2

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.

Answer 3

-Misread stated question. Sorry about that.

Answer 4

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.