# Roots of the cyclotomic polynomial

Let $\Phi_p$ be a cyclotomic polynomial and $q,p$ different primes. I know how many roots $\Phi_p$ has in $\mathbb{Z}/q\mathbb{Z}[X]$ and that these roots are simple. Now I can lift these roots to roots in $\mathbb{Z}_q[X]$ according to Hensel.

Is it possible to find out if there are more roots of $\Phi_p$ in $\mathbb{Z}_q[X]$?

-
Surely the roots $\alpha$ are in $\mathbb{Z}_q$ whereas the factors $X-\alpha$ are in $\mathbb{Z}_q[X]$?? –  Jyrki Lahtonen Jan 9 '13 at 20:11
Reducing a putative $q$-adic root modulo $q$ gives a root in $\mathbb{Z}/q\mathbb{Z}$. The latter are known to be simple, so... All the roots of $\Phi_p(X)\in \mathbb{Z}/q\mathbb{Z}[X]$ belong to the finite field $GF(q^f)$, where $f$ is the order of $q$ in the group $\mathbb{Z}/p\mathbb{Z}^*$. Similarly all the $q$-adic roots belong to the unique unramified extension of $\mathbb{Q}_q(\mu_{q^f-1})$ of degree $f$. Hensel lifting works at the level of $\mathbb{Q}_q(\mu_{q^f-1})\leftrightarrow GF(q^f)$ as well. –  Jyrki Lahtonen Jan 9 '13 at 20:24

There are two cases. In case $p|(q-1)$, then the multiplicative group of $\mathbb F_q$, cyclic of order $q-1$, has a subgroup of order $p$, whose elements are the roots of $\Phi_p$, since these are the primitive $p$-th roots of unity. In this case, all the roots of $\Phi_p$ are in $\mathbb F_q$, and they lift to $\mathbb Z_q$ as $p-1$ distinct elements of the $q$-adic integers. They are all the roots of $\Phi_p$. There can’t be any others, in $\mathbb Q_q$ or any other integral domain containing $\mathbb Z_q$.
The other case is that $p$ doesn’t divide $q-1$, in which case neither $\mathbb F_q$ nor $\mathbb Z_q$ has primitive $p$-th roots of unity. This may be the case that was confusing James. Here, $\Phi_p$ has roots in a proper extension of $\mathbb F_q$, as @Jyrki remarks, and these lift to the corresponding unramified proper extension of $\mathbb Z_q$ but not to $\mathbb Z_q$ itself.
So, in both cases, there are just as many primitive $p$-th roots of unity in $\mathbb Z_q$ as in $\mathbb F_q$.