27
$\begingroup$

Apologies in advance if this is obvious.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Ben Webster
    Oct 17, 2009 at 14:51
  • $\begingroup$ 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. $\endgroup$
    – shenghao
    Oct 17, 2009 at 19:55
  • $\begingroup$ Yes, any particular subgroup. $\endgroup$ Oct 17, 2009 at 21:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – Ben Webster
    Oct 18, 2009 at 1:54
  • 1
    $\begingroup$ Somewhat related: groupprops.subwiki.org/wiki/… Unfortunately, not every ring of algebraic integers (or even cyclotomic integers) is a PID. $\endgroup$
    – Vipul Naik
    Apr 26, 2011 at 20:12

4 Answers 4

26
$\begingroup$

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!

$\endgroup$
17
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

I just stumbled across this ancient question, and I want to point out my notes here that prove the result in question. The proof is basically the same as moonface's accepted answer, but with two improvements: 1. I give a very soft proof that all representations can be defined over some number field; and 2. instead of invoking the Hilbert class field to trivialize all ideals, I give an elementary argument to show you can pass to an extension trivializing the finite collection of them needed for this proof.

Sorry for bumping such an old question, but the above did not quite fit into a comment.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.