# proof that the cyclotomic polynomial is irreducible

We first prove that $\Phi_{n}(x)\in\mathbb{Z}[x]$. The field extension $\mathbb{Q}(\zeta_{n})$ of $\mathbb{Q}$ is the splitting field  of the polynomial  $x^{n}-1\in\mathbb{Q}[x]$, 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  $\mathbb{Q}(\zeta_{n})/\mathbb{Q}$ 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 $\mathbb{Q}(\zeta_{n})/\mathbb{Q}$ permutes the roots of $\Phi_{n}(x)$, it must fix the coefficients of $\Phi_{n}(x)$, so by Galois theory  these coefficients are in $\mathbb{Q}$. Moreover, since the coefficients are algebraic integers  , they must be in $\mathbb{Z}$ as well.

Let $f(x)$ be the minimal polynomial of $\zeta_{n}$ in $\mathbb{Q}[x]$. 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 times11Actually, 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..

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

Title proof that the cyclotomic polynomial  is irreducible ProofThatTheCyclotomicPolynomialIsIrreducible 2013-03-22 12:38:04 2013-03-22 12:38:04 djao (24) djao (24) 9 djao (24) Proof msc 12E05 msc 11R60 msc 11R18 msc 11C08