Question

Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$? I am going to be on the road soon, so pleas don't be offended if I don't respond quickly to a comment.

Answer 1

The genus class field of an extension $K/F$ is defined to be the largest extension $L/K$ with the following properties:

1. $L/K$ is unramified
2. $L$ is the compositum of $K/F$ and an abelian extension $A/F$.

Thus the quick answer to your question is: the Hilbert class field of $K$ is abelian over ${\mathbb Q}$ if and only if the Hilbert class field of $K$ coincides with its genus class field.

The not-so-quick answer would tell you more about the construction of the genus class field. For abelian extensions of the rationals, the construction is easy: everything you'd like to know should be contained in Frölich's book

• Central extensions, Galois groups, and ideal class groups of number fields AMS 1983

Basically you will have to look for the largest abelian extension of ${\mathbb Q}$ with the same conductor as $K/{\mathbb Q}$.

Answer 2

I think $K$ has to be an abelian extension of $\mathbb{Q}$ of class number one. Because if the Hilbert class field is abelian over $\mathbb{Q}$, then it is contained in a cyclotomic field, so $K$ itself is inside that cyclotomic field so $K$ is abelian, but an extension of fields inside a cyclotomic field is always ramified (not 100% sure here but I think it's OK) so the Hilbert class field cannot be a proper extension of $K$.

Edit: Definitely not OK. See comments below.

Answer 3

Obivously, if the Hilbert class field $H$ of $K$ is abelian over $\mathbb Q$, then $K$ (which is a subfield of $H$) must be abelian over $\mathbb Q$. So suppose that this is the case. In general, there is then a maximal subfield of $H$ that is abelian over $\mathbb Q$; call it $F$. It certainly contains $K$, and is called the genus field of $K$. By class field theory $Gal(F/K)$ is a quotient of the class group $Cl(K)$ of $K$. Which quotient? At least when $Gal(K/\mathbb Q)$ is cyclic, it is the maximal quotient on which $Gal(K/\mathbb Q)$ acts trivially.

If $K$ is quadratic then $Gal(K/\mathbb Q)$ acts on $Cl(K)$ by inversion, so $F/K$ corresponds to the maximal 2-elementary abelian quotient of $Cl(K)$. In particular, in this case $F = H$ if and only if $Cl(K)$ is an elementary abelian 2-group. (E.g. $\mathbb Q(\sqrt{-5})$, whose class group is of order 2.)

Answer 4

Is there a good/intuitive way to generate a stock of examples of non-abelian unramified extensions? The only examples I know of are a bit unintuitive (eg in Janusz's book). Intuition might suggest to start looking at Galois extensions with group equal to semi-direct products. I was originally interested in this question by the necessity of "ab" on the rhs of $H^{1} (X_{Zar}\,, \mathcal{O}_{X} ^ {*}) = \pi_{1} ^{ab} (X), \; \; X = Spec\; \mathcal{O}_K$ (re-interpretation of unramified global cft).

Answer 5

I happened to come across this question again today. In some cases at least, the Hilbert class field $H$ of an abelian extension $K$ of $\mathbf{Q}$ will have to be abelian over $\mathbf{Q}$ for purely algebraic reasons.

Let $F$ be any field, $K|F$ an abelian extension of group $G=\mathrm{Gal}(K|F)$ and containing a primitive $n$-th root of unity for some $n>1$, $\omega:G\to(\mathbf{Z}/n\mathbf{Z})^\times$ the cyclotomic character giving the action of $G$ on $\mu_n$, and $H|K$ an abelian extension of exponent dividing $n$. Then $H=K(\root n\of D)$ for some subgroup $D\subset K^\times/K^{\times n}$, by Kummer theory. It can be checked that $H|F$ is galoisian if and only if $D$ is $G$-stable. When such is the case, the conjugation action of $G$ on $\mathrm{Gal}(H|K)$ coming from the short exact sequence $$ 1\to\mathrm{Gal}(H|K)\to\mathrm{Gal}(H|F)\to G\to1 $$ is trivial if and only if $G$ acts on $D$ via $\omega$. In this situation ($H=K(\root n\of D)$ for some subgroup $D\subset(K^\times/K^{\times n})(\omega)$), a sufficient condition for $H$ to be abelian over $F$, is that the order of $G$ be prime to $n$, because then $\mathrm{Gal}(H|F)=\mathrm{Gal}(H|K)\times\mathrm{Gal}(K|F)$.

I'm sure this situation can be realised when $F=\mathbf{Q}$, for example when the finite abelian extension $K$ has odd degree $[K:\mathbf{Q}]$, $n=2$, the class group of $K$ has order ($1$ or) $2$, and $H$ is the Hilbert class field of $K$. In this case the extension $H|\mathbf{Q}$ will be necessarily abelian.