# 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 $\zeta \in \u2102$, 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^{1}^{1}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..

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^{} $\U0001d52d$ of $\mathcal{O}$ lying over $(p)$. Note that the polynomial ${x}^{n}-1$ has no repeated roots mod $\U0001d52d$, since its derivative^{} $n{x}^{n-1}$ is relatively prime to ${x}^{n}-1$ mod $\U0001d52d$. Therefore, if $f(\zeta )=0mod\U0001d52d$, then $g(\zeta )\ne 0mod\U0001d52d$, and applying the ${p}^{\mathrm{th}}$ power Frobenius map^{} to both sides yields $g({\zeta}^{p})\ne 0mod\U0001d52d$. This means that $g({\zeta}^{p})$ cannot be 0 in $\u2102$, because it doesn’t even equal $0mod\U0001d52d$. 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 |
---|---|

Canonical name | ProofThatTheCyclotomicPolynomialIsIrreducible |

Date of creation | 2013-03-22 12:38:04 |

Last modified on | 2013-03-22 12:38:04 |

Owner | djao (24) |

Last modified by | djao (24) |

Numerical id | 9 |

Author | djao (24) |

Entry type | Proof |

Classification | msc 12E05 |

Classification | msc 11R60 |

Classification | msc 11R18 |

Classification | msc 11C08 |