Minimal polynomial of root $\zeta_n$ over finite field $\mathbb{F}_p$

Question

Let $n>1$ a number and $\zeta_n$ the the $n$-th root. $p$ be a prime. I'm looking for strategys & theorems which allow to calculate the minimal poplynomial $m_{\zeta_n}$ of $\zeta_n$ over $\mathbb{F}_p$. (the case $n=p-1$ is boring).

first of all since the cyclotomic polynomial $\Phi_n(X)$ is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$ with coefficients in $\mathbb{Z}$ we apply the reduction modulo $p$ and see that $m_{\zeta_n}$ divides $\overline{\Phi_n(X)}$ with $\overline{\Phi_n(X)}$ the image of $\Phi_n(X)$ under reduction map $\mathbb{Z}[X] \to \mathbb{F}_p[X]$.

are there some irriducibility criterions? for example I found here: see Bruno Joyal's answer following statement:

the cyclotomic polynomial $\Phi_n(X)$ is irreducible over $\mathbf F_p$ precisely when $p$ has multiplicative order $\varphi(n)$ modulo $n$.

could anybody give the reference where it is proved or a sketch of it if it's not too deep?

do there exist another nice theorems treat this question?

Answer 1

For the first part of the question: recall that for any $\zeta$ in a (minimal) field extension $\mathbb{F}_{p^s}$ of $\mathbb{F}_p$, its minimal polynomial is $\prod_{t=0}^{s-1}{(X-\zeta^{p^t})}$.

Let $\zeta$ be a primitive $n$-th root of unity (with $p$ not dividing $n$). Let $\mathbb{F}_{p^s}$ be the field extension generated by $\zeta$. Then $s$ is the degree of the minimal polynomial $\mu(X) | \Phi_n(X)$ of $\zeta$ over $\mathbb{F}_p$.

We know thus that $\zeta^{p^s-1}=1$ so $n | p^s-1$. Since $\mu$ is the product of the $X-\zeta^{p^k}$, $0 \leq k < s$, and $\mu | X^n-1$ which is coprime with its derivative, $\mu$ has only simple roots in $\mathbb{F}_{p^s}$, hence, for all $1 \leq k < s$, $\zeta^{p^k} \neq \zeta$, so $p^k-1$ is not divisible by $s$. In other words, $s$ is the multiplicative order of $p$ mod $n$.

So $\Phi_n$ irreducible iff $\mu=\Phi_n$ iff $\mu$ has degree $\varphi(n)$ iff $p$ has multiplicative order $\varphi(n)$ mod $n$.

Note that if $p$ does not divide $n$ and has multiplicative order $\varphi(n)$ then $(\mathbb{Z}/(n))^{\times}$ is cyclic, so $n$ is one of $2,4,q^l,2q^l$, for $q$ an odd prime and $l \geq 1$ an integer.