Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recall that the cyclotomic polynomial of order $n$ is $$\Phi_n(X)=\prod_{gcd(k,n)=1}(X-e^{2ik\pi/n}).$$ Its degree is $\phi(n)$ (Euler's indicator). Inversion of the Moebius formula $$X^n-1=\prod_{d|n}\Phi_d(X)^{\mu(n/d)}$$ implies that $\Phi_n\in{\mathbb Z}[X]$. The height of $\Phi_N$ is the maximal modulus of its coefficients. Thus the height of $$\Phi_p(X)=X^{p-1}+\cdots+X+1$$ ($p$ odd prime) is $1$. A cyclotomic polynomial is flat if its height is $1$. We know that the height of $\Phi_{2n}$ equals that of of $\Phi_n$, and also that if a prime $p$ divides $n$, then the height of $\Phi_{pn}$ equals that of $\Phi_n$. Therefore it is enough to analyse the case where $n=p_1\cdots p_\ell$ is the product of distinct odd primes. When $\ell=2$ it is known that $\Phi_{p_1p_2}$ is flat. It is known that there are infinitely many flat $\Phi_{p_1p_2p_3}$, but $\Phi_{105}$ ($105=3\cdot 5\cdot 7$) is not flat (its height equals $2$).

Question. What is known about the growth of the height of $\Phi_n$ ? Is there a bound of the form $C_\ell$ (thus extending $C_1=C_2=1$) ?

share|cite|improve this question
The inversion formula should be $\Phi_n(X) = \prod_{d \mid n} (X^d-1)^{\mu(n/d)}$. –  user42355 Feb 26 '14 at 10:34

3 Answers 3 gives the smallest cyclotomic polynomial with a given coefficient, and the paper by Vaughan gives lower bounds for the growth of the max coefficient of the form exp(log 2 log n/log log n)

share|cite|improve this answer
Vaughan improves the previous results by factor $\log2$ (under exponential). These do not refer to bounds for particular $C_\ell$'s but it's reasonable to compare his results (using a completely different analytic method) with the ones coming in the studies of the ternary case. –  Wadim Zudilin Oct 15 '10 at 6:02

The problem of $C_3$ is discussed in the MPIM preprint "The family of ternary cyclotomic polynomials with one free prime" by Yves Gallot, Pieter Moree, and Robert Wilms. As far as I know Pieter works on the problem for several years; he once said that most of the conjectures in this area are false.

share|cite|improve this answer
From this paper, it seems that there is no finite bound $C_\ell$. Instead, even if $\ell=3$, one has a bound in terms of $p_1$ ($p$ in the literature). Correct ? –  Denis Serre Oct 13 '10 at 11:54
Yes, this is correct. –  Wadim Zudilin Oct 13 '10 at 12:04

In this book, a result due to Beiter is mentioned (see Thm. 2.7), according to which the height $C_3$ is bounded by $p-k$ or $p-k-1$ according as $p = 4k+1$ or $p = 4k+3$, where the three primes are $p$, $q$ and $r$ with $p < q < r$. Conjecturally, the best possible bound in this case is $C_3 = (p+1)/2$.

share|cite|improve this answer
Franz, Beiter's original conjecture was recently corrected in arXiv:0910.2770. –  Wadim Zudilin Oct 13 '10 at 10:18
The value $(p+1)/2$ is conjectured by Sister Marion Beiter (AMM 75 (1968)), who proved it when either $q$ or $r$ is $kp±1$. According to E. Leher's thesis, this conjecture is false. I expected that an accurate proof had been found since then. –  Denis Serre Oct 13 '10 at 11:52
The proof in arXiv:0910.2700 is false. There is a mistake on the page 11 $v_1 \in P \rightarrow 2v_0+r_p^*-v_1 \notin S$ is not correct. We would need $v_1 \in S$. –  Bruno Bolzen Sep 15 at 9:46

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.