proof that the cyclotomic polynomial is irreducible

We first prove that $\Phi_n(x) \in \Z[x]$ . The field extension $\Q(\zeta_n)$ of $\Q$ is the splitting field of the polynomial $x^n - 1 \in \Q[x]$ , since it splits this polynomial and is generated as an algebra by a single root of the polynomial. Since splitting fields are normal, the extension $\Q(\zeta_n)/\Q$ is a Galois extension. Any element of the Galois group, being a field automorphism, must map $\zeta_n$ to another root of unity of exact order $n$ . Therefore, since the Galois group of $\Q(\zeta_n)/\Q$ permutes the roots of $\Phi_n(x)$ , it must fix the coefficients of $\Phi_n(x)$ , so by Galois theory these coefficients are in $\Q$ . Moreover, since the coefficients are algebraic integers, they must be in $\Z$ as well.

Let $f(x)$ be the minimal polynomial of $\zeta_n$ in $\Q[x]$ . Then $f(x)$ has integer coefficients as well, since $\zeta_n$ is an algebraic integer. We will prove $f(x) = \Phi_n(x)$ by showing that every root of $\Phi_n(x)$ is a root of $f(x)$ . We do so via the following claim:

Claim: For any prime $p$ not dividing $n$ , and any primitive $n^{\rm th}$ root of unity $\zeta \in \mathbb{C}$ , if $f(\zeta) = 0$ then $f(\zeta^p) = 0$ .

This claim does the job, since we know $f(\zeta_n) = 0$ , and any other primitive $n^{\rm th}$ root of unity can be obtained from $\zeta_n$ by successively raising $\zeta_n$ by prime powers $p$ not dividing $n$ a finite number of times ¹.

To prove this claim, consider the factorization $x^n - 1 = f(x) g(x)$ for some polynomial $g(x) \in \Z[x]$ . Writing $\O$ for the ring of integers of $\Q(\zeta_n)$ , we treat the factorization as taking place in $\O[x]$ and proceed to mod out both sides of the factorization by any prime ideal $\p$ of $\O$ lying over $(p)$ . Note that the polynomial $x^n - 1$ has no repeated roots mod $\p$ , since its derivative $n x^{n-1}$ is relatively prime to $x^n - 1$ mod $\p$ . Therefore, if $f(\zeta) = 0 \bmod \p$ , then $g(\zeta) \neq 0 \bmod \p$ , and applying the $p^{\rm th}$ power Frobenius map to both sides yields $g(\zeta^p) \neq 0 \bmod \p$ . This means that $g(\zeta^p)$ cannot be 0 in $\mathbb{C}$ , because it doesn't even equal $0 \bmod \p$ . However, $\zeta^p$ is a root of $x^n - 1$ , so if it is not a root of $g$ , it must be a root of $f$ , and so we have $f(\zeta^p) = 0$ , as desired.

Footnotes

... times¹

Actually, if one applies Dirichlet's theorem on primes in arithmetic progressions here, it turns out that one prime is enough, but we do not need such a sharp result here.