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This is a natural generalization of this question.

Let $f$ be a monic irreducible polynomial over $\mathbb Z$. Let $S_f$ be the set of natural numbers $n$ such that one of the three equivalent conditions hold:

a. There exists some prime $p$ such that a root of $f$ in $\bar{\mathbb F}_p$ is also a primitive $n$th root of unity.

b. The ideal in $\mathbb Z[x]$ generated by $f$ and the $n$th elementary cyclotomic polynomial is not the unit ideal.

c. The resultant of $f$ and the $n$th elementary cyclotomic polynomial is not $\pm 1$.

Is $S_f$ cofinite for all $f$ with Mahler measure larger than $1$?

What I know:

It is easy to check that $S_f$ is infinite, since for each prime a root of $f$ is the $n$th root of unity for some $n$, but each prime can only show up finitely many times for a fixed $n$, since the resultant is nonzero and thus has finitely many prime factors.

If $f$ has no complex roots of norm $1$, then $S_f$ is cofinite. This is an extension of the argument in the linked mathoverflow question. The resultant of $f$ and the $n$th elementary cyclotomic polynomial can be computed by inclusion-exclusion from the resultants of $f$ and $x^{n/d}-1$, for $d$ squarefree dividing $n$. Since the resultant of $f$ and $x^n-1$ is $a^{n+o(1)}$ for $a>1$ equal to the Mahler measure of the $f$, the result of $f$ and the elementary cyclotomic polynomial is $a^{\Phi(n)+O(2^k)}$, where $k$ is the number of primes dividing $n$. Since $2^k/\Phi(n)$ is small for large $n$, the exponential is larger than $1$ for large $n$, therefore the resultant equals $1$ only finitely many times.

But this argument fails if a root is norm 1. In fact, purely analytic methods of estimating the resultant should fail in this case. There are bad points on the unit circle that get too close to roots of unity too often. To prove $S_f$ cofinite, you would need to use some algebraic method to prove that those points are not algebraic integers.

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I think that it's not cofinite for $f(x)=x$ or $f(x)=x-1$ or $f(x)=x+1$ or $f(x)=x^2+1$ or... . What did you mean to ask? Or did I misunderstand? –  Kevin Buzzard May 27 '12 at 22:12
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If $f$ is not cyclotomic then its Mahler measure is $> 1$ and your analytic argument goes through. –  Felipe Voloch May 27 '12 at 22:23
    
@Kevin: Clarified. Ignoring $f(x)=x$ and every cyclotomic polynomial. @Felipe: Let $\alpha$ be a root of norm $1$. Couldn't $\alpha^n$ be extremely close to $1$? This would make the error in the estimate of the resultant very large. I don't see why the error should be small enough to ensure that the resultant is bigger than $1$. –  Will Sawin May 27 '12 at 22:34
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@Will: I haven't tried to write out the estimates, so I can't guarantee anything, it's just my feeling. If you want to have an estimate for $|\alpha^n -1|$ you can use linear forms in logarithms à la Baker (Gelfond-Schneider really since it's only one log). –  Felipe Voloch May 27 '12 at 23:54
    
If you take the resultant of $f$ with $x^n-1$ rather than with the cyclotomic polynomial, I think this follows from Skolem-Mahler-Lech. Let $\alpha_1$, ..., $\alpha_r$ be the roots of $f$, so the resultant above is $s_n:=\prod (\alpha_i^n-1)$. Then $s_n$ obeys a linear recursion with integer coefficients so the only way for it to be $1$ infinitely often is if it is $1$ on an entire arithmetic progression. I feel like there should be an easy contradiction here, although I haven't actually found one. The actual cyclotomic question strikes me as harder, although Felipe's suggestion sounds good. –  David Speyer May 28 '12 at 0:20

1 Answer 1

up vote 6 down vote accepted

For an algebraic number $\gamma$ which is not a root of unity, Baker's theorem gives a bound (uniform in $n$) of the form $|\gamma^n - 1| > n^{-C}$ for some constant $C$ (only depending on $\gamma$).

In particular, if $s_n:=\prod |\alpha^n_i - 1|$, then Baker's method gives the following estimate uniform in $n$ (for $\alpha_i$ not a root of unity): $$n \log \mathcal{M}(\alpha) + A \ge \log|s_n| \ge n \log \mathcal{M}(\alpha) - A - B \log|n|,$$ where $\mathcal{M}(\alpha)$ is the Mahler measure of $\alpha$. Proof: for $|\alpha_i| > 1$, one has $\log|\alpha^n_i - 1| \sim n \log|\alpha_i|$ up to $O(1)$, which gives rise to the term $n \log \mathcal{M}(\alpha)$; the term $\log|\alpha^n_i - 1|$ for $|\alpha_i| \le 1$ is trivial to bound from above and can be bound from below by Baker's Theorem. The logarithm of the Mahler measure is the sum of $\log|\alpha_i|$ for the roots $|\alpha_i|> 1$. By a theorem of Kronecker this sum is positive if $\alpha$ is not a root of unity.

OTOH, if $\Phi_n(x)$ is the $n$th cyclotomic polynomial and $t_n:=\prod |\Phi_n(\alpha_i)|$, then we deduce that $\sum_{d|n} \log(t_n) = \log(s_n)$, and hence $$\log(t_n) = \sum_{d|n} \log(s_{n/d}) \mu(d) \ge \varphi(n) \log \mathcal{M}(\alpha) - d(n)(A + B \log(n)),$$ where $d(n)$ is the number of divisors of $n$. The bounds $\phi(n) \gg n^{1-\epsilon}$ and $d(n) = n^{\epsilon}$ easily give the asymptotic relation $$\log(t_n) \sim \varphi(n) \log \mathcal{M}(\alpha) \gg 1$$ as $n$ goes to infinity.

FWIW, the bounds of Baker (and the other bounds used above) are effective, so for any particular $\alpha$ one could in principle find all $n$ with $t_n = 1$. (n.b. Gelfond's estimate would give $\epsilon \cdot n$ instead of $B \log|n|$ as an error term, which is enough to show that $s_n \rightarrow \infty$ but not $t_n$.)

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