Baker’s Theorem and the Class Number Problem (Part 1)

Disclaimer: This post is the first 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.

Over roughly 3 posts, I aim to introduce Baker’s Theorem, the Class Number Problem, and then discuss how Baker’s methods solved part of the problem. In this post, I’ll be discussing Baker’s Theorem (from Baker’s “Linear Forms in the Logarithms of Algebraic Numbers”, which earned Baker the Fields Medal) and giving some examples of it’s use. I will be heavily borrowing from Baker’s “Transcendental Number Theory” and Natarajan and Thangadurai’s “Pillars of Transcendental Number Theory“, both books I highly recommend to people interested in the area.

As usual, we take the complex logarithm \log z = \log\lvert z \rvert + i \arg z, where we take - \pi < \arg z \leq \pi. I denote effectively computable constants with c, and won’t make too much effort to differentiate between these constants. I will sometimes give in brackets what the constants depend on; for example c(n,m) would depend on numbers n and m. For an algebraic number \alpha, we define the height of \alpha to be the maximum of the absolute values of the coefficients of the minimal polynomial of \alpha (we note we can compare this to other heights such as absolute logarithmic height and so on).

There are a few ways to state Baker’s Theorem, we will use the following.

Baker’s Theorem Let \alpha_{1},\, \dots,\, \alpha_{n} \in \mathbb{A} \backslash \{0,\, 1\} be such that \log \alpha_{1},\, \dots,\, \log\alpha_{n} are linearly independent over \mathbb{Q}. Let \beta_{0},\,\beta_{1},\, \dots ,\,\beta_{n} \in \mathbb{A}, not all 0. Set \Lambda := \beta_{0} + \beta_{1} \log \alpha_{1} + \dots + \beta_{n} \log \alpha_{n} and let B= \max \{h(\beta_{0}),\, h(\beta_{1}),\dots,\, h(\beta_{n})\} . Assume that \Lambda \neq 0 . Then there exists an effectively computable constant c:=c(n,\,\alpha_{1},\, \dots ,\, \alpha_{n}) such that \lvert \Lambda \rvert \geq (eB)^{-c}  .

From this, we can immediately deduce that 1,\, \log \alpha_{1},\, \dots,\, \log \alpha_{n} are linearly independent over the algebraic numbers \mathbb{A}. This is essentially the order Baker writes these in his original paper; in many places including Baker’s book we prove the non-vanishing of \Lambda and thus linear independence, before giving a lower bound.

I do not aim to prove this theorem here; the proof is too long but essentially is a proof by contradiction (the proof has a nice use of Siegel’s Lemma too, another favourite result of mine!). For a discussion of proof using non-standard analysis, see Tao’s excellent post here.

First I show some applications of our deduction of linear independence, before using the qualitative result to solve a nice little problem.

Corollary 1 Let \alpha_{1}, \dots , \, \alpha_{n} \in \mathbb{A} \backslash \{ 0 \}  and \beta_{1},  \dots,\, \beta_{n} \in \mathbb{A} . Then \beta_{1} \log \alpha_{1} + \dots + \beta_{n} \log \alpha_{n}  is either zero or transcendental.

When n=1 this is the Hermite-Lindemann-Weierstrass Theorem. To prove the claim we induct on n. Assume the corollary is true for all m < n, but assume it is not true for n. Thus there exist algebraic numbers \beta_{0}, \, \beta_{1},\dots,\, \beta_{n} such that \beta_{1} \log \alpha_{1} + \cdots +\beta_{n} \log \alpha_{n} = \beta_{0} (*).

By the contrapositive of our deduction, \log \alpha_{1}, \, \dots,\, \log \alpha_{n} are linearly dependent over \mathbb{Q}; thus, there exist rational numbers a_{1},\, \dots,\, a_{n} such that a_{1} \log \alpha_{1} + \cdots a_{n} \log \alpha_{n} = 0 (**).

Without loss of generality, we mas assume that c_{n} is not 0. We combine (*) and (**) to eliminate \log \alpha_{n}, obtaining (c_{n} \beta_{1} - c_{1}) \log \alpha_{1} + (c_{n}\beta_{2}-c_{2}) \log \alpha_{2} + \cdots + (c_{n}\beta_{n-1}-c_{n-1}) \log \alpha_{n-1} = c_{n} \beta_{0}, where c_{n} \beta_{0} is non-zero and algebraic. By the inductive hypotheses, the left hand side is transcendental, giving us a contradiction and proving the claim.

An immediate consequence of this corollary allows us to show a large set of numbers are transcendental.

Corollary 2 Let \alpha_{1},\, \dots,\, \alpha_{n},\, \beta_{0},\, \beta_{1},\, \dots,\, \beta_{n} \in \mathbb{A} \backslash \{0\}. Then the number e^{\beta_{0}} \alpha^{\beta_{1}}\cdots \alpha_{n}^{\beta_{n}}  is transcendental.

To see this, suppose e^{\beta_{0}} \alpha^{\beta_{1}}\cdots \alpha_{n}^{\beta_{n}} equals an algebraic number \alpha_{n+1}. Then \alpha_{n+1} \neq 0, and we can take logarithms, obtaining \beta_{1} \log \alpha_{1} + \cdots +\beta_{n} \log \alpha_{n} - \log \alpha_{n+1} = - \beta_{0}. We recall \beta_{0} \neq 0, so this contradicts corollary 1; the left hand side must be transcendental!

Corollary 2 implies numbers such as e2^{\sqrt{2}} and \pi + \log \alpha for algebraic, non-zero \alpha are transcendental.

We now directly apply Baker’s Theorem with the lower bound to a specific case of Catalan’s conjecture; indeed, Baker’s Theorem can be used (in principle) to resolve Catalan’s conjecture in all but a finite number of cases.

Catalan’s Conjecture The only solution in the natural numbers of x^{a}-y^{b}=1  for a,\, b>1,\, x,\, y>0  is 3^{2}-2^{3}=1 .

We shall consider a much weaker example; we shall show there are finitely many solutions in natural numbers to the equation 3^{m}-2^{n}=1

With minimal work, we obtain the following corollary of Theorem 1:

Corollary 3 Let \alpha_{1},\, \dots,\, \alpha_{n} \in \mathbb{A} \backslash \{0,\, 1\}  , and b_{1},\, \dots,\, b_{n} \in \mathbb{Z}  such that \alpha_{1}^{b_{1}} \dots \alpha_{n}^{b_{n}} -1 \neq 0  . Then \lvert \alpha_{1}^{b_{1}} \dots \alpha_{n}^{b_{n}} -1 \rvert \geq (eB)^{c}  , where B=\max \{\lvert b_{1}\rvert,\, \dots,\, \lvert b_{n} \rvert\} and c:=c(n,\, \alpha_{1},\, \dots,\, \alpha_{n}).

We apply this to 3^{m}-2^{n}=1. I claim that \max \{m,\, n\} \leq c(2,3) for some effectively computable constant c(2,3). Let B = \max \{m,\, n \}, and assume without loss of generality that 3^{m} \geq 2^{n}. Throughout this proof, I just use c for multiple different effectively computable constants; I hope it doesn’t cause too much confusion!

By Corollary 3, \lvert 1 - 3^{-m}2^{n} \rvert \geq (eB)^{-c}. Thus, \lvert 3^{m}-2^{n} \rvert \geq \frac{3^{m}}{(eB)^{c}}. By the definition of B and assumption that 3^{m} \geq 2^{n}, 3^{m} \geq 2^{B}. Thus, 1= \lvert 3^{m}-2^{n} \rvert \geq \frac{2^{B}}{(eB)^{c}}.

It follows that 2^{B} \leq (eB)^{c}, so \frac{B}{\log B}\leq \frac{c}{\log 2}. Recalling that if \frac{a}{\log a}<x then a<2x \log x for real x >1, we obtain that B < c(2,3). Thus, we have an upper bound for the maximum exponent m or n, and thus have finitely may solutions to 3^{m}-2^{n}=1, as we wanted to show. As a side-note, showing this was one of the problems I was given in my PhD interview, and has stuck with me since! With some more work we can show that there are only finitely many solutions to 3^{m}-2^{n}=1 in integers, but I omit the details here.

Something I’ve not discussed at all in this post is how we compute this constant; there has been a lot of work done in optimising this constant. As should be expected, the constant comes from the proof of the theorem, and this has undergone many refinements. For the best bound as of writing, see Matveev’s result.

So far, I’ve not mentioned the Class Number Problem, and if you know the statement, it’s not immediately clear (at least to me!) how we can apply Baker’s result. In the next post I will introduce the Class Number Problem, and discuss some work that was done on the way to proof of part of the problem.

Any questions, mistakes noticed or anything else, please comment or drop me a tweet!

One thought on “Baker’s Theorem and the Class Number Problem (Part 1)

Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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: