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.
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8$\begingroup$ There's not going to be a "definitive answer" to this question because it includes the question "when does a real quadratic field have class number 1", about which not much (in some sense) is known. For example it's still an open problem whether there are infinitely many real quad fields with class number 1. $\endgroup$– Kevin BuzzardCommented Feb 19, 2010 at 8:19
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$\begingroup$ FC--Is there some source you recommend to read up on this? I would be happy to learn more about the totally real case, as well as these finitely many examples. In particular, I would like to be able to have a class of such fields. It would be wonderful if [K:Q] > or = 3. Is it true that there are only finitely many fields that are not totally real, or only finitely many fields that are totally real with [K:Q] >2? $\endgroup$– ScarletCommented Feb 21, 2010 at 19:14
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5 Answers
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The genus class field of an extension $K/F$ is defined to be the largest extension $L/K$ with the following properties:
- $L/K$ is unramified
- $L$ 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 $K/{\mathbb Q}$.
answered Feb 19, 2010 at 7:28
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$\begingroup$ With respect to the last comment, are you trying to say that if K has conductor $f(K)$, then the genus class field is contained in $\Q(\zeta_f(K))?$ $\endgroup$– ScarletCommented Feb 19, 2010 at 14:25
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$\begingroup$ Sorry about the "unknown control sequence". I mean in the cyclotomic extension of $\mathbb{Q}$ one gets by adding the f(K)th roots of unity. $\endgroup$– ScarletCommented Feb 19, 2010 at 14:50
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1$\begingroup$ For abelian extensions of the rationals: yes. In David's example, the field K generated by a square root of -5 has discriminant 20, and its genus class field is contained in the field of 20th roots of unity. More exactly, it's the maximal subfield in which 2 and 5 have the same ramification indices as in K. $\endgroup$ Commented Feb 19, 2010 at 15:55
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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.)
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$\begingroup$ Two typos: "Hilbet' should be "Hilbert" and "which is a subfield of K" should be "which is a subfield of H" in the first sentence. $\endgroup$– user1073Commented Feb 19, 2010 at 4:18
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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).
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$\begingroup$ Yes. Of particular interest is Maire's paper "On Infinite Unramified Extensions", in which he explicitly constructs infinite unramified extensions over fields with trivial (or near-trivial) Hilbert class field. Such extensions are necessarily nonb-abelian. $\endgroup$ Commented Feb 25, 2010 at 2:18
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2$\begingroup$ There are the famous Golod--Shafarevich towers. $\endgroup$– EmertonCommented Feb 25, 2010 at 3:11
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$\begingroup$ In his paper "Unramified Quaternion Extensions of Quadratic Number Fields", Lemmermeyer constructs (using only elementary methods) unramified extensions with the group H_8 of quaternions as the Galois group. He is very explicit about his methods, so you should be able to use his ideas to generate a few easy to remember examples. $\endgroup$– user1073Commented Feb 25, 2010 at 13:13
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$\begingroup$ Thanks! explicit finite extensions was what I was looking for. In skimming the paper, it looks very well-written and with only elementary methods as you mentioned. jtnb.cedram.org/item?id=JTNB_1997__9_1_51_0 $\endgroup$ Commented Feb 26, 2010 at 4:43
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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.
answered Dec 28, 2012 at 7:11
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2$\begingroup$ If K is an abelian of odd degree over Q, then the 2-class group is never cyclic. This follows easily by looking at the action of the Galois group on the class group. The same thing holds for new-class groups over arbitrary base fields F. $\endgroup$ Commented Dec 28, 2012 at 11:26
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1$\begingroup$ All I can say is that while Algebra is of some use, Arithmetic is far too subtle. $\endgroup$ Commented Dec 28, 2012 at 13:28
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1$\begingroup$ Actually your argument may be completed as follows: if K has odd degree n and H is abelian of degree 2n, let F be the quadratic subfield. This extension is ramified at some prime p, and the ramification must survive when lifted to FK/K. Thus there is no such example. $\endgroup$ Commented Dec 28, 2012 at 17:50
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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.
answered Feb 19, 2010 at 3:59
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6$\begingroup$ That's not true. The class filed of Q(sqrt(-5)) is Q(sqrt(-5), sqrt(-1)). $\endgroup$ Commented Feb 19, 2010 at 4:01
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$\begingroup$ @Felipe: I think that if N has more than 1 prime divisor then Q(zeta_N) is unramified over its real subfield. $\endgroup$ Commented Feb 19, 2010 at 8:17
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1$\begingroup$ @Kevin: this is correct if you neglect infinite primes. What this means is that the corresponding class field corresponds to a subgroup of the class group in the strict sense. $\endgroup$ Commented Feb 19, 2010 at 9:25
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$\begingroup$ @Franz: I've never known whether Hilbert class Fields are allowed to be ramified at infinity! From what you say, they aren't, so my comment is wrong, but David's still stands as an explicit counterexample. David's example is the reason that you can use binary quadratic forms and the usual tricks to figure out precisely which primes are of the form 2x^2+2xy+3y^2 (it's a congruence condition: 3 or 7 mod 20), but you can't figure out which primes are of the form 2x^2+xy+3y^2 (because now the Hilbert class field isn't abelian). Of course you know this already :-) $\endgroup$ Commented Feb 19, 2010 at 14:36
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1$\begingroup$ @Kevin: The real quadratic field K=Q(sqrt(3)) has class number one, but its pretty obvious that the quadratic extension K(i) of K is unramified at all finite primes. So a Hilbert class field has to be everywhere unramified (infinite primes included). $\endgroup$– user1073Commented Feb 19, 2010 at 15:17