MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

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

Apologies in advance if this is obvious.

-
I'm pretty sure it's not obvious...my guess is there isn't, but I can't think of any counterexamples right off the bat. – Ben Webster Oct 17 2009 at 14:51
So you are asking if there exist a basis for every GIVEN finite subgroup, not one basis that works for every finite subgroup, right? Coz then the answer is no. – shenghao Oct 17 2009 at 19:55
Yes, any particular subgroup. – Qiaochu Yuan Oct 17 2009 at 21:39
The subset of GL_n(\bar{Q}) consisting of matrices such that the matrix AND its inverse have algebraic integer entries is a subgroup, and obviously a group representation with algebraic integer entries would have to land there. – Ben Webster Oct 18 2009 at 1:54
Somewhat related: groupprops.subwiki.org/wiki/… Unfortunately, not every ring of algebraic integers (or even cyclotomic integers) is a PID. – Vipul Naik Apr 26 2011 at 20:12
show 1 more comment

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!

-

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.

-

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.

-