Prove that $Φ_{nm}(x) = Φ_n(x^m)$ if every prime divisor of m is also divisor of n

Question

Let $m$ and $n$ be natural numbers that every prime divisor of $m$ is also a divisor of $n$. We can define $Φ_{ab}(x)$ for every prime $a>0$ like this: $$Φ_{ab}(x) = \begin{cases} Φ_b(x^a), & \text{if a|b} \\ \frac{Φ_b(x^a)}{Φ_b(x)}, & \text{if (a,b)=1} \end{cases}$$ How can I prove $Φ_{nm}(x) = Φ_n(x^m)$? Thanks!

Answer 1

If you know that $\deg \Phi_n = \varphi(n)$, then notice that for every primitive $nm$-th root $\zeta$ the number $\zeta^n$ is a primitve $m$-th root, so $\Phi_{mn}(x) \vert \Phi_m(x^n)$ since every root of the first polynomial is a root of the second and the multiplicity is always one. Now, $\varphi(nm) = n\varphi(m)$ if all prime divisors of $n$ also divide $m$.* So in this case equality holds because the degrees are the same.

*just use the formula $\varphi(n) = n \prod_{p\vert n} (1-\frac 1 p)$.