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$\newcommand{\Q}{\mathbb Q}$ Let $F=\mathbb Q^{ab} \subset \mathbb C$, i.e. the algebraic numbers. Let $G$ be a finite group of order $n$ and let $\phi: G \rightarrow GL_m(F)$ be a representation. Denote the extension of $\Q$ generated by the entries of $\phi(g)$ for each $g \in G$ by $\Q(\phi)$. Note this is a finite extension. My question is whether or not $\Q(\phi)$ is contained in a cyclotomic extension of $\Q$ or if we can find a change of basis such that $\Q(\phi)$ is wrt to this basis.

Clearly this is the case if $\phi$ is a sum of $1$-dimensional representations, because then the $\phi(g)$ are simultaneously diagonalizable. Since each $\phi(g)$ has only roots of unity for eigenvalues it follows. Other than this rather simple case, I'm at somewhat of a loss. Mainly I'm unsure of how to show something is not contained in a cyclotomic extension. I would guess we would want to show $\Q(\phi)$ has some $\alpha$ in it such that the splitting field of $\alpha$'s minimum polynomial has non-abelian galois group.

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At least Curtis & Reiner (probably most books on representation theory of finite groups) contains the fact that $K=\mathbb{Q}(e^{2\pi i/m})$, where $m$ is the l.c.m. of orders of elements of $G$, is a splitting field of the group algebra of $G$. In other words, the Wedderburn decomposition of $KG$ has summand of the form $M_{n_i}(K)$ only, or all the absolutely irreducible representations can be defined over $K$. This seems to answer your question in the affirmative. –  Jyrki Lahtonen May 18 '12 at 5:57

2 Answers 2

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Firstly, $\mathbb{Q}^{ab}$ customarily denotes the maximal abelian extension of $\mathbb{Q}$. You probably wanted to say $F=\bar{\mathbb{Q}}$, which is the common notation for the algebraic closure.

To answer the actual question, yes any complex representation of a finite group can be defined over a cyclotomic field (and in particular over $\mathbb{Q}^{ab}$, but you probably didn't want to assume that at the beginning). You can first diagonalise one $\phi(g)$ over a cyclotomic extension. Then diagonalise some other $\phi(h)$ over some possibly bigger cyclotomic extension. In doing so, you will destroy the diagonalisation of $\phi(g)$, but that doesn't matter, since the entries of $\phi(g)$ will still lie in a cyclotomic extension (they were in a cyclotomic extension before you diagonalised $\phi(h)$, and you changed the basis by a matrix with entries in a cyclotomic extension). Keep going, until the entire representation is defined over a cyclotomic field by taking care of the elements one by one.

In analysing this process, you will also immediately see that we have defined any given representation over $\mathbb{Q}(e^{2\pi i/n})$, where $n$ is the exponent of the group, i.e. the least common multiple of the orders of all group elements.

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oops sorry, I missed the change basis part of your question, and I can't delete my answer; please ignore it

This should really be a comment (as you might want to edit the question), but I can't comment yet.

The answer is no. Suppose $n=2$, $G=\{\pm1\}$, and $\phi$ is $$\phi(\pm 1)=\begin{pmatrix} 1&0\\0&\pm1\end{pmatrix}.$$ Let now $$A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$$ and say $det A=1$, and let $\psi(g)=A\phi(g)A^{-1}$. Then the $1,1$-entry of $\psi(-1)$ is $ad+bc=1+2bc$. However, $bc$ can be any element of $F$.

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