prove cyclotomic polynomial is a minimal polynomial

Question

I want to prove that the cyclotomic polynomial $Φ_n(x)$ is the minimal polynomial of the primitive $n$th root ζ of unity.

Is it enough to show that Φ is irreducible and in Z[x]?

Or is there a simpler proof for this?

Answer 1

Yes, as if this were not true, then we could write $\Phi_n(x) = f(x)g(x)$ where $f,g$ are irreducible monic polynomials. Then, take $\zeta$ to be a primitive $n^{th}$ root of unity and a root of $f(x)$ (thus $f(x)$ is the minimal polynomial for $\zeta$). Then, for any prime $p \not\mid n$, $\zeta^p$ is also a primitive $n^{th}$ root of unity. It must also be a root of $\Phi_n$, and so must be a root of either $f(x)$ or of $g(x)$. Suppose $g(\zeta^p) = 0$. Then $\zeta$ is a root of $g(x^p)$ and so $f(x)$ must divide $g(x^p)$ in $\mathbb{Z}[x]$. So, $g(x^p) = f(x)h(x)$ for some $h(x) \in \mathbb{Z}[x]$. Reducing mod $p$ gives $\tilde{g}(x^p) = \tilde{f}(x)\tilde{h}(x)$, in a finite field, we have that $\tilde{g}(x^p) = (\tilde{g}(x))^p$. So, $(\tilde{g}(x))^p = \tilde{f}(x)\tilde{h}(x)$. But then $\tilde{g}(x)$ and $\tilde{f}(x)$ share a factor in $\mathbb{F}_p[x]$. But then $\tilde{\Phi}_n(x)$ has a multiple root in $\mathbb{F}_p[x]$. But then $x^n - 1$, having $\Phi_n$ as a factor, must also have a multiple root in $\mathbb{F}_p[x]$, a contradiction. Hence, $f(\zeta^p) = 0$. But then for every root $\zeta$ of $f(x)$, $\zeta^q$ is also a root of $f(x)$ when $(p,q)=1$. Thus, every primitive $n^{th}$ root of unity is also a root of $f(x)$. But then $\Phi_n(x) = f(x)$, and thus $\Phi_n(x)$ is irreducible since $f(x)$ was irreducible by assumption.