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

As far as I know there are two proofs of the finiteness of the ideal class group of a number field. One is due to Minkowski using the "geometry of numbers" and another one is due Chevalley using "ideles". My question is divided into two parts:

1: Is there any other proof?

2: Second question needs some preliminary background. Let $K$ be a number field and suppose $\mathcal{O}_K$ is its ring of integers. The group $SL_2(\mathcal{O}_K)$ acts on $\mathbb{P}^1(K)$, and one can show the ideal class number is equal to the number of orbits of this action. So proving the finiteness of orbits implies the finiteness of the ideal class number. Is there any proof for this?

share|improve this question
12  
For (1): Yes, there is a third way. Proof by historical record: Minkowski's proof was in the 1890s and Chevalley's was in the 1930s, so Dedekind had to have some other argument for his exposition in the 1870s! Proof by example: see the proof of finiteness of the class number in Ireland and Rosen's book. They say their argument is due to Hurwitz, but that's wrong. It goes back earlier to Kronecker. See the comments to my answer of the question mathoverflow.net/questions/19021/… –  KConrad Nov 6 '10 at 17:26
    
Your question #2 is confusing. Instead of asking "Is there any proof for this" (at first it's not clear what "this" means), I think you meant to ask, more precisely, "Is there any proof that the class number is finite by this method?" –  KConrad Nov 6 '10 at 17:28
1  
Regarding 2: A few years ago, a student in a course I was teaching gave a presentation of this fact based on very fine notes of Keith Conrad. At the time, I commented that it seemed that the proof worked with $\mathcal{O}_K$ replaced by any Dedekind domain $R$ and its fraction field. In this level of generality, the class group need not be finite. So a proof along these lines needs at least an additional idea: i.e., some special property of number rings must be exploited. –  Pete L. Clark Nov 6 '10 at 17:58
1  
Saul: The essence of the argument is that the norm-one idele class group is compact (Fujisaki's lemma, which uses a pigeon-hole principle) and naturally maps onto the ideal class group. Then note a) the kernel of this map is open, so the ideal class group is discrete b) the ideal class group is the continuous image of a compact set, so it is compact. Compact and discrete is finite. Peter Clark's website has links to several proofs of this (at the bottom of the page): math.uga.edu/~pete/MATH8410.html –  B R Nov 6 '10 at 22:41
1  
Dedekind's proof is in his book on algebraic numbers. This was translated into English (Richard Dedekind - Theory of Algebraic Integers - translated by John Stillwell, CUP 1996). The proof uses the pigeon hole principle, which is hardly surprising. –  engelbrekt Nov 10 '10 at 10:51
show 4 more comments

2 Answers

In principle, this follows from Borel and Serre's compactification of arithmetic orbifolds. Let $K$ be a field with $r$ real places and $s$ complex places, and $H_{r,s}=(\mathbb{H}^2)^r\times(\mathbb{H}^3)^s$. Then $SL_2(K)\leq PSL_2(\mathbb{R})^r\times PSL_2(\mathbb{C})^s$ by taking the product of the various Galois embeddings, and acts on $H_{r,s}$. Then via this embedding, $SL_2(\mathcal{O}K)$ acts discretely on $H_{r,s}$, with finite covolume. There are finitely many cusps of this orbifold $H_{r,s}/SL_2(\mathcal{O}_K)$, corresponding to the orbits of $PSL_2(\mathcal{O}_K)$ acting on $\mathbb{P}^1(K)$, which Borel and Serre provide a compactification for. When $K=\mathbb{Q}$, this compactifies $\mathbb{H}^2/PSL_2(\mathbb{Z})$ by a circle, and for $K=\mathbb{Q}(\sqrt{-D}), D\in \mathbb{N}$, $\mathbb{H}^3/PSL_2(\mathcal{O}_K)$ is compactified by Euclidean 2-orbifolds. In the real quadratic case, the compactification is by solv 3-orbifolds.

One may also deduce this from the fact that $H_{r,s}/SL_2(\mathcal{O}_K)$ is finite volume and from the Margulis lemma, which describes the structure of the cusps. I'm not sure who originally proved this, but Borel gave explicit formulae for the volume (although these formulae involve the class number).

This answer is not meant to indicate that this is how one should prove that the class group is finite, but to show how it fits into a certain mathematical context.

share|improve this answer
add comment

In the usual proof of the class number formula, i.e. the computation of the residue of the Dedekind zeta function at $s = 1$, it is used that the class number is finite and that the unit group has the right number of generators. But the proof essentially works also if you do not use this fact, and in the end you get both results for free - the price you have to pay is a presentation which is a bit messier than usual because you have to allow for the possibility that h is infinite and that you have too few units. It is a good exercise to go through the proof in the real quadratic case, though.

Edit. The situation is not as simple as I thought it is. The problem is the following: by counting lattice points you can easily prove that the number of ideals with norm $< X$ in any given ideal class $C$ is equal to $cX + O(\sqrt{x})$. I would have thought that the existence of infinitely many ideal classes quickly produces nonsense, but this is wrong. In fact, the constants in the O-term may depend on the ideal classes. The usual proof of the finiteness of the ideal class group shows that there is a finite constant $c'$ such that the error term is less than $c' \sqrt{X}$. The problem is what to do without this information.

It follows, if I am right, quite easily that if $h$ is not finite, then the number of ideals with norm $< X$ is not of the form $O(x)$, i.e. grows fast than $cX$ for any constant $c$. In the quadratic case, using the fact that the number of ideals with norm $m$ can be expressed in terms of Legendre symbols implies that the number of ideals with norm $< X$ is $L(1,\chi) X + O(\sqrt{X})$, and now we get a contradiction plus a proof that $L(1, \chi)$ does not vanish (which in turn implies the fact that the Dedekind zeta function of the quadratic field has a pole of order $1$ in $s = 1$, if you know that it is analytic).

I have not yet seen what to do for general number fields.

share|improve this answer
    
Potentially dumb question: If you don't assume that the class group is finite or that there are enough units, how do you know that the pole of the zeta function is simple? –  David Speyer Nov 9 '10 at 16:17
    
@David: You can use the Poisson summation formula to give a Mellin transform identity for the Dedekind zeta function, from which the nature of the singularity at $s = 1$ is clear. Harold M. Stark wrote an expository paper in which he sketched a proof of the Dirichlet Unit Theorem by this approach, among other things. The application of the Poisson summation formula is called the Hecke theta formula, and is central to Hecke's proof of the functional equation of the Dedekind zeta function. –  engelbrekt Nov 10 '10 at 10:45
    
Do I need that? The proof shows that the number of ideals in an ideal class with norm < x is O(x); for showing that h is finite I need the same result for all ideals. Doesn't this follow from unique prime ideal factorization? –  Franz Lemmermeyer Nov 10 '10 at 11:48
add comment

Your Answer

 
discard

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.