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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.