most recent 30 from http://mathoverflow.net 2013-05-23T07:19:51Z http://mathoverflow.net/feeds/question/847 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers Qiaochu Yuan 2009-10-17T08:09:01Z 2011-04-26T19:45:57Z

<p>Apologies in advance if this is obvious.</p>

http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers/882#882 moonface 2009-10-17T15:03:06Z 2009-10-17T15:03:06Z

<p>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! </p>

http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers/975#975 Ben Webster 2009-10-18T01:46:25Z 2009-10-18T01:46:25Z

<p>By the way, <a href="http://www.ams.org/mathscinet-getitem?mr=1155753" rel="nofollow">this paper</a> 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 <a href="http://www.ams.org/mathscinet-getitem?mr=2381795" rel="nofollow">this one</a> has more examples.</p>

http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers/1071#1071 Seth Case 2009-10-18T20:51:23Z 2009-10-18T21:57:37Z

<p>-Misread stated question. Sorry about that.</p>

http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers/63068#63068 Geoff Robinson 2011-04-26T19:45:57Z 2011-04-26T19:45:57Z

<p>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.</p>