# prove cyclotomic polynomial is a minimal polynomial

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?

• – lhf
Commented Nov 3, 2019 at 11:03

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.