Cyclotomic polynomials coprime to a fixed polynomial - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:46:42Z http://mathoverflow.net/feeds/question/76193 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76193/cyclotomic-polynomials-coprime-to-a-fixed-polynomial Cyclotomic polynomials coprime to a fixed polynomial Pedro Martins Rodrigues 2011-09-23T09:34:28Z 2011-09-23T14:15:38Z <p>Let $f \in \mathbb{Z}[x]$ be monic, irreducible and hyperbolic (no roots of absolute value $1$), and such that $f(0)= \pm 1$. Denoting as $c_{p}(x)$ the cyclotomic polynomial $$c_{p}(x)=1+x+\cdots +x^{p-1},$$ my question is: how can be characterized the (certainly finite?) set of primes $p$ for which $f$ and $c_{p}$ are co-prime, ie there are polynomials $u$ and $v$ in $\mathbb{Z}[x]$ such that $$c_{p} u+f v=1.$$ As an obvious consequence, we have that if $\zeta$ is a primitive $p$-root of unit, $f(\zeta)$ is a unit in $\mathbb{Z}[\zeta]$ and so, for each $1\leq k\leq p-1$, $$f(\zeta^{k})=\zeta^{j} r$$ for some $j$, where $r$ is a real unit of that ring. </p> http://mathoverflow.net/questions/76193/cyclotomic-polynomials-coprime-to-a-fixed-polynomial/76200#76200 Answer by Felipe Voloch for Cyclotomic polynomials coprime to a fixed polynomial Felipe Voloch 2011-09-23T11:42:12Z 2011-09-23T14:15:38Z <p>Not an answer, but the set of $p$ that you define is quite likely infinite (Oops, see David's comment). Here is an argument. Let $K$ be the splitting field of the polynomial $f(x^2)$ and $\ell$ a prime that splits completely in $K$. There are lots of those primes and is a reasonable assumption (?) that infinitely many of them satisfy $\ell - 1 =2p$, $p$ prime. Given such an $\ell$ and a root $\alpha$ of $f \mod \ell$, by construction, $\alpha$ is a square, so $\alpha^p = \alpha^{(\ell-1)/2} = 1$ and $\alpha$ is a common root of $f$ and $c_p$ modulo $\ell$ and $p$ is not in your set.</p> <p>A better argument follows. The resultant of $f$ and $c_p$ grows like $a^{p-1}$ where $a$ is the Mahler measure of $f$, which is $>1$ under the hypotheses. So indeed the set of primes you define is finite.</p>