MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I saw that there are many "applications" questions in Mathoverflow; so hopefully this is an appropriate question. I was rather surprised that there were only five questions at Mathoverflow so far with the tag diophantine-approximation, while there are almost 900 questions on number theory overall. It is my intention to promote the important subject a little bit by asking one more question.


What are some striking applications of Baker's theorem on lower bounds for linear forms on logarithms of algebraic numbers?

If, for example, I were in a discussion with a person who has no experience with diophantic approximation, to impress upon the person the importance of Baker's theorem I would cite the following two examples:

  1. Giving effective bounds for solutions of (most of the time exponential) diophantine equations under favorable condition. For example, Tijdeman's work on the Catalan conjecture, or giving effective bounds for Siegel's theorem, Fermat's last theorem, Falting's theorem, etc., in certain cases.

  2. Transcendence results which are significant improvements over Gelfond-Schneider. In particular, the theorem that if $\alpha_1, \ldots, \alpha_n$ are $\mathbb{Q}$-linearly independent, then their exponentials are algebraically independent over $\mathbb Q$. I would cite the expose of Waldschmidt for more details.

These are, to me, quite compelling reasons to study Baker's theorem. But as I do not know much more on the subject, I would run out of arguments after these two. I would appreciate any more striking examples of the power of Bakers' theorem. This is 1. for my own enlightenment, 2., for future use if such an argument as I hypothesized above actually happens, 3. To promote the subject of diophantine approximation in this forum, especially in the form of Baker's theorem.

share|cite|improve this question
I don't think "I want to promote X" is an appropriate question (indeed, it isn't a question at all) but the embedded "Tell me what theorem Y is good for" question seems fine. – JSE Jul 27 '10 at 20:17
@JSE: I didn't say that the question was "I want to promote Baker's theorem". I specifically highlighted the question and the rest was background and motivation, which MO seems to require before answering a question. – Anweshi Jul 27 '10 at 21:06
Anweshi, all applications of Baker's (and not only!) estimates for linear forms in logs are quite striking. What strikes me more is that certain diophantine equations cannot be "finalised" (ie, reduced to a finite amount of solutions) using Baker. – Wadim Zudilin Jul 27 '10 at 22:13
@Wadim: Aha! You are more surprised when Baker's theorem fails to settle a diophantine equation! I was under the impression that Baker's theorem is mainly good for exponential diophantine equations and also that for fixed diophantine equations of low degree Baker may not always work. Anyway the original question could be phrased again as: Are there important applications of Baker's theorem besides bounding the solutions of diophantine equations, and results on transcendence? Are you aware of any besides these two? – Anweshi Jul 27 '10 at 22:32
Anweshi, I think that the only ones I have in mind are these two, no surprise. :-) I'll post an example of applications in diophantine approximations. An exponential diophantine equation where Baker's theorem is of no help is presented in – Wadim Zudilin Jul 27 '10 at 23:16

Surely the most striking application was the solution of the class number one problem.

share|cite|improve this answer
Oh, yes! I had forgotten it as my mind was full of the work of Heegner, Dorian Goldfeld, etc! Surely, it is the most striking. – Anweshi Jul 28 '10 at 11:11
Except that it was unneeded there, as pointed out in the aftermath by Stark. You only needed two logs, and then its just Gelfond (and effective) which is much simpler. Similar to other historical cases, what is the bottleneck is not always the first to seem to be the crux in retrospect.… – Junkie Jul 28 '10 at 12:29
@Junkie. Thanks. I didn't know that it could be obtained using merely Gelfond-Schneider. But it seems the case that Baker's theorem got its fame firstly through this achievement, if I am not mistaken. – Anweshi Jul 28 '10 at 12:31
@Junkie, my reading of the Stark paper is that one also needs a result on L-functions that was only published (by Stark) in 1968/69. But perhaps this result, too, could have been proved by Gelfond and Linnik in 1949. Anyway, thanks for the pointer to Stark's paper. – Gerry Myerson Jul 29 '10 at 6:14
I agree with Gerry. The last sentence of Stark's states: "Gelfond and Linnik's idea coupled with Gelfond's effective theorem on linear forms in two logarithms would have settled the class-number one problem in 1949 had only the expansion in (2) been available with characters to nonprime moduli." But this expansion was not available, it was proved by Stark in the 1960s. – Micah Milinovich Jul 30 '10 at 17:38

Brumer's proof of Leopoldt's conjecture for abelian extensions of Q.

share|cite|improve this answer

A quantitative example of applications of linear forms in logarithms is the following result from [S.D. Adhikari, N. Saradha, T.N. Shorey, and R.Tijdeman, Transcendental infinite sums, Indag. Math. (N.S.) 12:1 (2001) 1--14; Theorem 4 and Corollary 4.1] which is cited in many other articles.

Theorem. Let $P(x)$ and $Q(x)$ be two polynomials with algebraic coefficients such that $Q(x)$ has simple rational zeros and no others. Let $\alpha$ be an algebraic number. Then, assuming the convergence of the series $$ S=\sum_{n=1}^\infty\frac{P(n)}{Q(n)}\alpha^n, $$ the number $S$ defined by it is either rational or transcendental. Furthermore, if all zeros of $Q(x)$ lie in $-1\le x<0$, then either $S=0$ or $S$ is transcendental.

The theorem gives an elegant criterion for deciding whether a number of this particular form is transcendental or not (like continued fractions allow us to decide whether a given number is from a quadratic field or not). However it is not applicable for the sums like $$ \zeta(3)=\sum_{n=1}\frac1{n^3}, $$ as factorisation of the denominator polynomial, $Q(x)=x^3$, involves multiple rational zeros.

share|cite|improve this answer
It turns out the book (called Transcendental Number Theory) I was going to recommend as applicable was written by Alan Baker, so I imagine everybody knows the book if they know this topic. and – Will Jagy Jul 28 '10 at 1:00
This is nice. Thanks for the answer. – Anweshi Jul 28 '10 at 11:13

Your Answer


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.