In the answers to [Qiaochu's post](https://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers) on defining representations of finite groups over the algebraic integers, it came out that which fields a representation of a finite group is defined over might depend on whether you require your representation to be free or just projective. If your representation V is defined over a number field $K,$ it contains a projective submodule $V'$ for the integers $O$ of $K$ such that $V'\otimes\_O K=V,$ but it's not at all clear if this can be chosen to be free.


Luckily, there is a very nice theory of algebraic number theory that says that any projective module over the ring of integers of a number field becomes free when you extend scalars to the Hilbert class field. So, since all representations of finite groups are defined over cyclotomic fields (in fact, you just need the roots of unity for the orders of elements in G), every representation has as an integral basis in the Hilbert class field of a cyclotomic field. Which is.....?


Edit: while the class numbers are interesting that's not the question I asked. I want to actually what the Hilbert class field is, or something about it. For example, is it cyclotomic (seems unlikely, but cyclotomic fields are nice...)?



What is the Hilbert class field of a cyclotomic field?
Giving an "explicit" description of the Hilbert class field of a number field K (or, more generally, all abelian extensions of K) is Hilbert's 12th problem, and has only been solved for Q and for imaginary quadratic fields. The Hilbert class field H of Q(zeta) will only be contained in a cyclotomic field if H = Q(zeta) itself --- since one can explicitly compute that any abelian extension of Q properly containing Q(zeta) is ramified at some place.


