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. 



Cyclotomic polynomials coprime to a fixed polynomial 
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.


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.


