Apologies in advance if this is obvious.



Is any representation of a finite group defined over the algebraic integers?
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!


