Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There's a wonderful analogy I've been trying to understand which asserts that field extensions are analogous to covering spaces, Galois groups are analogous to deck transformation groups, and algebraic closures are analogous to universal covering spaces, hence the absolute Galois group is analogous to the fundamental group. (My vague understanding is that the machinery around etale cohomology makes this analogy precise.)

Does the Hilbert class field (of a number field) fit anywhere into this analogy, and how?

Phrased another way, what does the Hilbert class field of the function field of a nonsinguar curve defined over $\mathbb{C}$ (say) look like geometrically?

share|improve this question
add comment

3 Answers

up vote 11 down vote accepted

This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:

1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$\mathbb{Z}_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)

In the case of a curve over $\mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G.

This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.

2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $\mathbb{Z}_K$, which classifies line bundles on $\mathbb{Z}_K$. This generalizes nicely: the Picard group of $C$ is an exension of $\mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the Jacobian of $C$.

3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$.

So there are reasonable claims for calling either $G \cong \mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]

As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field.

Finally, let me note that things work out much more closely if you replace $\mathbb{C}$ with a finite field $\mathbb{F}_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(\mathbb{F}_q)$, the (finite!) group of $\mathbb{F}_q$-rational points on the Jacobian.

share|improve this answer
Minor point: Over a finite field, you still have arbitrarily large everywhere unramified extensions because can always just enlarge the base field. This corresponds to the degree in the Picard group. But if you pick a way of ruling these out, then you'll get Pic_0, which is finite. –  JBorger Jan 7 '10 at 7:02
I completely agree. I decided not to mention this for the sake of simplicity, since extending the base is a phenomenon which does not occur over C. –  Pete L. Clark Jan 7 '10 at 7:27
Thanks for the answer! Do you have a reference for that last statement you made? –  Qiaochu Yuan Jan 8 '10 at 5:08
add comment

The absolute Galois group is $\pi_1( Spec K)$ and the Galois group of the maximal unramified extension of $K$ is $\pi_1( Spec R)$, where $R$ is the ring of integers of $K$ and the class group of $K$ is the abelianization of $\pi_1( Spec R)$ and the Hilbert class field is the field corresponding to this quotient.

A curve has a function field and therefore an absolute Galois group, which is much bigger than the fundamental group and describes all branched covers of the curve. The fundamental group corresponds of course to unramified covers. Working over C gives you infinite groups even for the abelianization. If you work over a finite field, then the analogy is more precise.

share|improve this answer
add comment

I guess you're looking for the maximal unramified abelian extension. The fundamental group of a genus g curve has abelianization isomorphic to $\mathbb{Z}^{2g}$, so if g is positive, you get an infinite extension. The corresponding cover is not an algebraic curve in the usual sense of the word, although I suppose you can write down a scheme of infinite type.

Perhaps a better answer is: questions of this type fall into the domain of "geometric class field theory". If I'm not mistaken, Serre's Algebraic groups and class fields covers a lot the ideas.

share|improve this answer
Don't you feel, though, that the Jacobian has got to be better than some crazy inverse limit scheme? –  Pete L. Clark Jan 6 '10 at 20:04
I totally agree, although the two objects serve different purposes. As far as I can tell, the crazy inverse limit scheme is the fiber product of the curve with the universal cover of the Jacobian (which is also pretty crazy). I think the difference in tractability is that the finite connected covers of the Jacobian are easier to see. –  S. Carnahan Jan 6 '10 at 21:47
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.