# Proving that $\Phi_{n}$ is irreducible (a problem with the proof)

I am trying to follow the proof in the book Abstract Algebra by Dummit and Foote (Theorem 41, pg. 554) that $\Phi_n$ is an irreducible monic polynomial in $\mathbb{Z}[x]$ of degree $\varphi(n)$.

What I understand is that if it is not irreducible, than we can factor $\Phi_{n}=fg$ where $f$ is the minimal polynomial of $\zeta_{n}$ (a primitive $n$-th root of unity).

I also agree that if $p$ is a prime s.t. $(p,n)=1$ then $f(\zeta_{n}^{p})=0$. The question is why if $(a,n)=1$ then $f(\zeta_{n}^{a})=0$ (I don't understand why this is true even for $a=p_{1}p_{2}$).

Can someone please explain why we can move from primes to their product?

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$\zeta_n$ is an arbitrary primitive $n$th root of unity which is a root for $f.$ Given that $f(\zeta_n^p ) =0,$ we know that $\zeta_n^p$ is also a primitive $n$th root of unity, so that for any second prime $q$ not dividing $n$ we can again apply the proposition to $\zeta^p$; $f(\zeta^{pq})=f((\zeta^p)^q)=0.$ If $(a,n)=1,$ we can write $a=\displaystyle\prod_{i=1}^n p_i$ (not necessarily distinct primes) and apply an easy inductive argument.