Avoiding Minkowski's theorem in algebraic number theory. - MathOverflow

Avoiding Minkowski's theorem in algebraic number theory.

2010-03-22T15:00:54Z

For any course in algebraic number theory, one must prove the finiteness of class number and also Dirichlet's unit theorem. The standard proof uses Minkowski's theorem. Is there a way to avoid it?

The reasons I am asking this question are the following.

$1$. Minkowski lived long after Dirichlet and Dedekind(esp Dirichlet). So the original proof cannot likely have used Minkowski's theorem as such. If the original proof did use Minkowski's theorem, then it was of course found by someone else, most probably Dirichlet, and it is unfair to use the name Minkowski's theorem.

$2$. Even more importantly, the finiteness of classnumber and some version of unit theorem is true(at least I hope so) for all global fields. And there of course one cannot talk of Minkowski's theorem.

The objection I have for Minkowski's theorem is that it seems to be ad hoc, coming out of nowhere. And it seems that not much work is going on nowadays in the subject of geometry of numbers.

So it will be really nice to have a method which would feel more natural and is perhaps more general.

Answer by Felipe Voloch for Avoiding Minkowski's theorem in algebraic number theory.

2010-03-22T15:43:09Z

In the end all you need is the pigeonhole principle. Minkowski's theorem is just a sharpening of it. If you want an uniform proof for number fields and function fields for both the unit theorem and finiteness of class number which uses just the pigeonhole principle and is likely close to the original ones, see:

Axiomatic characterization of fields by the product formula for valuations E. Artin and G. Whaples, Bull. AMS 51 (1945) 469-492.

Answer by Keenan Kidwell for Avoiding Minkowski's theorem in algebraic number theory.

2010-03-22T15:55:56Z

Yes, there is a way to get the finiteness of class number as well as the S-unit theorem, avoiding using Minkowski's theorem explicitly. However, it doesn't give you the Minkowski bound. The idea is to carry out the work of Minkowski's theorem on convex bodies in the adele ring of your number field instead of in $\mathbb{R}^{r_1+2r_2}$. Basically you prove what Brian Conrad calls an ``adelic Minkowski lemma" involving the Haar measure of subsets of the adele ring. Using this, you can prove the compactness of the group $\mathbb{J}_K^1/K^\times$, where $\mathbb{J}_K^1$ is the kernel of the continuous idelic norm and $K^\times$ is the diagonally embedded discrete image of $K^\times$ in the idele group. The S-unit theorem and finiteness of class number are straightforward consequences of this compactness result. You can find a proof along these lines in Cassels and Frohlich. The finiteness of class number is easier, and comes from a natural surjective, continuous homomorphism from the idele group to the ideal class group (the ideal class group is given the discrete topology); you show that this gives you a continuous surjective map from a quotient of $\mathbb{J}_K^1/K^\times$ and you get that the ideal class group is compact and discrete, hence finite. Tom Weston used to have a writeup of this stuff on his website that I really liked, but I'm not sure if it's still there.

Answer by David Speyer for Avoiding Minkowski's theorem in algebraic number theory.

2010-03-22T15:57:56Z

You might want to read the early parts of Basic Number Theory, by A. Weil. Weil shows how to do all of these proofs in a very clean way which is uniform between the number field and the function field case; his proofs are all based on local compactness.

However, I don't think that Weil's proofs are morally different from the standard ones. In my, perhaps limited, experience, all proofs of these results are based on the pigeonhole principle (including the extension that, in a measure space of measure $1$, any two open sets whose measure add up to more than $1$ must meet.)

As a warning, remember that these results are false for function fields over $\mathbb{C}$. If $X$ is an affine algebraic curve over $\mathbb{C}$ with positive genus, then the class group of $\mathcal{O}_X$ is infinite and the unit group may have rank less than the number of punctures minus $1$.

Answer by KConrad for Avoiding Minkowski's theorem in algebraic number theory.

2010-03-22T17:09:13Z

For a course on algebraic number theory, you certainly can prove the finiteness of the class group without Minkowski's theorem. For example, if you look in Ireland-Rosen's book you will find a proof there which they attribute to Hurwitz. It gives a worse constant (which depends on a choice of $\mathbf Z$-basis for the ring of integers of the number field; changing the basis can shrink the constant, but it's still generally worse than Minkowski's) but it is computable and you can use it to show, say, that $\mathbf Z[\sqrt{-5}]$ has class number 2.

As for the history of the proof of the unit theorem, it was proved by Dirichlet using the pigeonhole principle. If you think about it, Minkowski's convex body theorem is a kind of pigeonhole principle (covering the convex body by translates of a fundamental domain for the lattice and look for an overlap). You can find a proof along these lines in Koch's book on algebraic number theory, published by the AMS. Incidentally, Dirichlet himself proved the unit theorem for rings of the form $\mathbf Z[\alpha]$; the unit theorem is true for orders as much as for the full ring of integers (think about Pell's equation $x^2 - dy^2 = 1$ and the ring $\mathbf Z[\sqrt{d}]$, which need not be the integers of $\mathbf Q(\sqrt{d})$), even though some books only focus on the case of a full ring of integers. Dirichlet didn't have the general conception of a full ring of integers.

One result which Minkowski was able to prove with his convex body theorem that had not previously been resolved by other techniques was Kronecker's conjecture (based on the analogy between number fields and Riemann surfaces, with $\mathbf Q$ being like the projective line over $\mathbf C$) that every number field other than $\mathbf Q$ is ramified at some prime.