proof that the cyclotomic polynomial is irreducible

We first prove that ${\mathrm{\Phi }}_n(x)\in \mathbb{Z}[x]$. 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 ${\mathrm{\Phi }}_n(x)$, it must fix the coefficients of ${\mathrm{\Phi }}_n(x)$, 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 $f(x)={\mathrm{\Phi }}_n(x)$ by showing that every root of ${\mathrm{\Phi }}_n(x)$ is a root of $f(x)$. We do so via the following claim:

Claim: For any prime $p$ not dividing $n$, and any primitive $n^{\mathrm{th}}$ root of unity $\zeta \in ℂ$, 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 $n^{\mathrm{th}}$ 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.

To prove this claim, consider the factorization $x^n-1=f(x)g(x)$ for some polynomial $g(x)\in \mathbb{Z}[x]$. Writing $𝒪$ for the ring of integers of $\mathbb{Q}({\zeta }_n)$, 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 $f(\zeta )=0mod𝔭$, then $g(\zeta )\ne 0mod𝔭$, and applying the $p^{\mathrm{th}}$ power Frobenius map to both sides yields $g({\zeta }^p)\ne 0mod𝔭$. This means that $g({\zeta }^p)$ cannot be 0 in $ℂ$, because it doesn’t even equal $0mod𝔭$. 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.