This is a natural generalization of [this question](https://mathoverflow.net/questions/76193/cyclotomic-polynomials-coprime-to-a-fixed-polynomial).


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.



The resultant of an arbitrary polynomial and a cyclotomic polynomial
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 $t\_n$.)


