The nth cyclotomic polynomial is defined as follows \Phi_n(x)=\prod_{1\leq i\leq n, (n,i)=1}(x-\zeta_n^{i}) where zeta_n=e^{\frac{2\pi i }{n}}.

The polynomial x^n-1 has the following factorization x^n-1=\prod_{d|n}\Phi_d(x).

In this post, I will give a proof of irreducibility of nth cyclotomic polynomial. I will prove the following theorem.

Theorem: \Phi_n(x) is an irreducible polynomial in \mathbb{Q}[x] .

Proof: If \Phi_n(x) is not an irreducible polynomial then form Gauss’s lemma it has a non-trivial factorization \Phi_n(x)=f(x)g(x) where f(x),g(x)\in \mathbb{Z}[x].

As \Phi_n(x) has no repeated roots, f(x),g(x) are relatively prime to each other in \mathbb{Q}[x]. As \mathbb{Q}[x] is a Euclidean domain, there exist functions l(x), m(x) and a natural number D such that l(x)f(x)+m(x)g(x)=D.——–(1).

Without loss of generality, we can assume f(\zeta_n)=0. Let one root of g(x)=0 be \zeta_n^{r} where (r,n)=1. From Dirichlet’s theorem, there exists a prime number p>D satisfying p\equiv r\mod n.

Note that,  l(x)f(x)+m(x)g(x)\equiv D\neq 0\mod p.——-(2)

Hence, f(\zeta_n)=g(\zeta_n^{p})=0. Therefore, f(x),g(x^p) have a common root \zeta_n and therefore they both have a non trivial greatest common divisor in \mathbb{Q}[x].

Consider these polynomials f(x) and g(x^p) as polynomials in $Z_p(x)$.  As, g(x^p)=(g(x))^p in Z_p[x]. We have, f(x) and (g(x))^p have non-trivial common divisor in Z_p[x]–which implies f(x) and g(x) have a non trivial common divisor, d(x) , in Z_p[x]. This implies d(x)|D in \mathbb{Z}_p[x], which is absurd.


Welcome to! This is your very first post. Click the Edit link to modify or delete it, or start a new post. If you like, use this post to tell readers why you started this blog and what you plan to do with it.

Happy blogging!