Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Thm (Kronecker).- If all conjugates of an algebraic integer lie on the unit circle, then the integer is a root of unity.

Question: Can one provide a good effective version of this? That is: given that we have an algebraic integer alpha of degree <=d, can we show that alpha has a conjugate that is at least epsilon away from the unit circle, where epsilon depends only on d? It actually isn't hard to do this (from the standard proof of Kronecker, viz.: alpha, alpha^2, alpha^3... are all algebraic integers, and their minimal polynomials would eventually repeat (being bounded) if all conjugates of alpha lied on the unit circle) with epsilon exponential on d, i.e., epsilon of the form epsilon = 1/C^d; what we actually want is an epsilon of the form 1/d^C, say.

(Question really due to B. Bukh.)

share|improve this question
You are not original in your question, as it was asked by D. Lehmer many years ago (see the answer below). The best known estimate is due to E. Dobrowolski. His paper [On a question of Lehmer and the number of irreducible factors of a polynomial, Acta Arith. 34 (1979) 391-401] can be downloaded from matwbn.icm.edu.pl/ksiazki/aa/aa34/aa34411.pdf . –  Wadim Zudilin Jul 24 '10 at 9:21
add comment

1 Answer

up vote 13 down vote accepted

If $M(\alpha)$ is the Mahler measure of $\alpha$, then the largest conjugate of $\alpha$ has absolute value at least

$$1 + \frac{\log(M(\alpha))}{d}.$$

Lehmer's conjecture implies that this at least $O(d^{-1})$ away from one. Dobrowolski's lower bound for $M(\alpha)$ shows that there is a conjugate at least $O(d^{-1-\epsilon})$ away from $1$ for any $\epsilon > 0$. Better bounds are available if one has more information, for example, the signature of $\mathbf{Q}(\alpha)$, or whether $\alpha$ is conjugate to $\alpha^{-1}$ or not.

For explicit references, see


share|improve this answer
Thanks! I should have remembered this... –  H A Helfgott Sep 8 '10 at 2:38
NB. Dobrowolski's Theorem is totally explicit. One knows that when $\alpha$ is neither $0$ nor a root of unity, then $M(\alpha)=d h(\alpha) \geq (\log\log (3d)/\log(2d))^3 / 4$. (The constant $1/4$ is due to Voutier.) –  ACL Feb 8 '12 at 13:04
add comment

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.