Ideal Class Number

Question

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?

Answer 1

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.

Answer 2

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.