Baker’s Theorem and the Class Number Problem (Part 2)

Disclaimer: This post is the second of a set of three(ish) about Baker’s Theorem and the Class Number Problem. It is intended to be informative and not too rigorous. Find part one here!

In this post, I’ll introduce the Class Number Problem, before (eventually) pulling these posts together and seeing how Baker’s Theorem is relevant to the problem. First, I’ll give some definitions, before stating the problem.

Given an algebraic number field K , consider the ring of algebraic integers of K , denoted by \mathcal{O}_{K} . Recall that the algebraic integers of K are the elements of K that is the root of some monic polynomial with coefficients in \mathbb{Z} .

We next consider the fractional ideals of \mathcal{O}_{K} ; that is, the \mathcal{O}_{K}- submodules of K . It is sometimes easier to think of these in terms of the following lemma.

Lemma A subset \mathfrak{a} \subset K is a fractional ideal if and only if there is an ideal \mathfrak{b} of \mathcal{O}_{K} and an algebraic integer \alpha such that \mathfrak{a} = \frac{1}{\alpha} \mathfrak{b} .

This kind of explains why we call them fractional ideals; they are “fractions” of ideals. Proving the above isn’t too hard; we can show the submodules are finitely generated, clear denominators and the job is basically done; for the proof, see almost any book on algebraic number theory.

These fractional ideals from a group under multiplication, which I will denote by I_{K}. . The set of principal fractional ideals forms a normal subgroup of I_{K}, which we’ll call P_{K}. We define the class group of K to be the quotient group I_{K}/P_{K} . We define the class number of K to be the number of elements in the class group (we can show that the class number is finite using some clever tricks from Minkowski; think lattices!). The class number is 1 when all fractional ideals are principal, thus all ideals of \mathcal{O}_{K} are principal, thus \mathcal{O}_{K} is a unique factorisation domain. We can also show that if \mathcal{O}_{K} is a UFD, then it has class number 1.

The Class Number Problem was first formulated by Gauss in 1801 in terms of quadratic forms; I’ll write the problem in the modern phrasing, before discussing what it all means. Given an algebraic number field K, we denote the class number of K by h or h(K), or in the case of a quadratic field K=\mathbb{Q}(\sqrt{d}), by h(d).

The Class Number Problem: Conjectures
  1. h(d) \rightarrow \infty as d \rightarrow - \infty tends to - \infty ;
  2. There are 9 imaginary quadratic fields with Class Number 1, namely \mathbb{Q}(\sqrt{d}) where d =-3,\, -4,\,-7,\, -8,\, -11,\, -19,\, -43,\, -67,\, -163. ;
  3. There are infinitely many real quadratic fields with class number 1.

The first two of these have been solved; indeed, the proof of the second part is where Baker’s Theorem comes in. I’ll save discussion of this till the next part in this series, and for now will just heuristically discuss some things related to the problem which I find interesting.

Part 1 was proved by Heilbronn in 1934; the proof here is very interesting, as it first assumes the truth of the Riemann Hypothesis and showing the statement holds, before assuming the falsity of the Riemann Hypothesis and showing it still holds. Indeed, there is a close connection between the Riemann Hypothesis and the class number problem. Indeed, it was shown that finding a 10th imaginary quadratic field with class number 1 would show that the Riemann Hypothesis is not true; unfortunately we now know this isn’t a route to the Riemann Hypothesis.

Part 3 is still open, in no small part due to fundamental units. By Dirichlet’s Unit Theorem, the group of units of \mathcal{O}_{K} is finitely generated, and we use these to define the regulator, which is an important part of the proof of part 2 of the problem. However, the ring of units for the case of an imaginary quadratic field consists of finitely many roots of unity (we can explicitly find which these are!), whereas we have infinitely many units in the real quadratic case. Finding these fundamental units depends on solving Diophantine equations, which is famously hard. This will be a topic of future posts, yet to come!

In the next post, I’ll discuss the proof of part 2 and why Baker’s Theorem can be used. There’ll also probably be a postscript, of stuff I’ve left out the main posts but that I still find interesting.

Any questions or corrections, leave them in the comments or drop me a tweet!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: