proof that the cyclotomic polynomial is irreducible

We first prove that Φn(x)[x]. The field extension (ζn) of is the splitting fieldMathworldPlanetmath of the polynomialPlanetmathPlanetmath xn-1[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 extensionPlanetmathPlanetmath (ζn)/ is a Galois extensionMathworldPlanetmath. Any element of the Galois group, being a field automorphism, must map ζn to another root of unityMathworldPlanetmath of exact order n. Therefore, since the Galois group of (ζn)/ permutes the roots of Φn(x), it must fix the coefficients of Φn(x), so by Galois theoryMathworldPlanetmath these coefficients are in . Moreover, since the coefficients are algebraic integersMathworldPlanetmath, they must be in as well.

Let f(x) be the minimal polynomial of ζn in [x]. Then f(x) has integer coefficients as well, since ζn is an algebraic integer. We will prove f(x)=Φn(x) by showing that every root of Φn(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 nth root of unity ζ, if f(ζ)=0 then f(ζp)=0.

This claim does the job, since we know f(ζn)=0, and any other primitive nth root of unity can be obtained from ζn by successively raising ζn 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 xn-1=f(x)g(x) for some polynomial g(x)[x]. Writing 𝒪 for the ring of integersMathworldPlanetmath of (ζn), we treat the factorization as taking place in 𝒪[x] and proceed to mod out both sides of the factorization by any prime idealPlanetmathPlanetmath 𝔭 of 𝒪 lying over (p). Note that the polynomial xn-1 has no repeated roots mod 𝔭, since its derivativePlanetmathPlanetmath nxn-1 is relatively prime to xn-1 mod 𝔭. Therefore, if f(ζ)=0mod𝔭, then g(ζ)0mod𝔭, and applying the pth power Frobenius mapPlanetmathPlanetmath to both sides yields g(ζp)0mod𝔭. This means that g(ζp) cannot be 0 in , because it doesn’t even equal 0mod𝔭. However, ζp is a root of xn-1, so if it is not a root of g, it must be a root of f, and so we have f(ζp)=0, as desired.

Title proof that the cyclotomic polynomialMathworldPlanetmath 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