I'm having trouble verifying the following proposition after Lemma 4.6 in the paper PRIMES is in P:
Let $Q_r(X)$ be the $r^{th}$ cyclotomic polynomial over $\mathbb{F}_p$. The Polynomial $Q_r(X)$ divides $X^{r}−1$ and factors into irreducible factors of degree $o_r(p)$. Note that it has been shown previously in the said paper that $p$ and $r$ are coprime.
Thanking you in advance.
EDIT: $o_r(p)$ is the smallest $k$ such that $p^{k}\equiv 1 (\textrm{mod}\ r)$
For an irreducible polynomial $f$ of degree $n$ over $\mathbb{F}_q$, if $\alpha$ is a root in its splitting field, then $\alpha$, $\alpha^q$, $\dots$, $\alpha^{q^{n-1}}$ are $n$ distinct roots of $f$.
Note that $Q_r$ has no multiple roots, and its degree is $\phi(r)$. Let $\beta$ be one of its roots so that $\beta^r=1$, and $\beta^s=1$ if and only if $r$ divides $s$. If $k$ is the least positive integer such that $q^k\equiv 1\pmod r$, then $\beta^{q^k-1}=\left(\beta^r\right)^{(q^k-1)/r}=1$, but $\beta^{q^j-1}\ne1$ for all $j<k$. It follows that $k$ is the least positive integer such that $\beta\in\mathbb{F}_{q^k}$. The minimal polynomial of $\beta$ thus has degree $k$. Since $\beta$ is arbitrary, every irreducible factor of $f$ has degree $k$, and $f$ has $\phi(r)/k$ factors ($k$ is the order of $q$ in the multiplicative group modulo $r$, and $k$ divides the order of the group).