Schönemann’s Proof of Irreducibility of Cyclotomic Polynomials.

Problem: Show that the cyclotomic polynomial $\displaystyle \varPhi_p(x)$ is irreducible.

$\displaystyle \varPhi_p(x)= \frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots x^2+x+1$

The standard presentation of irreducibility is by Eisenstein’s criterion:

Consider the shift

$\displaystyle \varPhi_p(x+1)= \frac{(x+1)^p-1}{(x+1)-1}=x^{p-1}+\left(\begin{array}{l} p \\ 1 \end{array}\right) x^{p-2}+\cdots+p$

Now observing that every term is ${0 \mod p}$ and the last term is ${\not \equiv 0 \mod p^2}$ , we are done by Eisenstein’s criterion.

We give an alternate proof by Schonemann.

Schonemann’s proof of irreducibility of ${\varPhi_p(x)=x^{p-1}+x^{p-2}+\cdots x+1}$

For a prime ${l}$ consider the factorisation of ${\varPhi_p(x) \mod l}$ into irreducible polynomials.

$\displaystyle \varPhi_p(x) = f_1(x)f_2(x)\dots f_k(x) \mod l$

Now the degree of ${f_i}$ equal the degree of the extension of ${\mathbb{F}_l}$ which contains ${p-th}$ root of unit. That is

$\displaystyle deg(f_i)=m \implies \mathbb{F}_{l^m} ~~ \text{contains a} ~~p\text{-th root of unity}$

This implies that ${l^m =1 \mod p.}$ That is ${m}$ is the order of ${l}$ in ${\mathbb{Z}/{p\mathbb{Z}}.}$

By Dirichlet’s theorem in arithmetic progression it’s possible to choose a prime ${l}$ such that ${l \mod p}$ can be any of the residue classes ${\left(\mathbb{Z}/{p\mathbb{Z}}\right)^{*}.}$ If we choose ${l}$ be a primitive element ${\mod p}$ , we get that the order of ${l}$ is ${p-1.}$ Hence the degree of ${f_i}$ is ${p-1}$ . This establishes that ${\varPhi_p(x)}$ is irreducible ${\mod l}$ and hence irreducible over ${\mathbb{Q}}$ . (Actually over ${\mathbb{Z}}$ , applying Gauss’s lemma gives us over ${\mathbb{Q}}$ )

Here is another proof- closely related to proof by Eisenstein’s criterion is by Schönemann’s Irreducibility Criterion:
Let $\displaystyle f(x) \in \mathbb{Z}[x]$
If there is a prime $p$ and an integer $a$ such that
$\displaystyle f(x)=(x-a)^{n}+p g(x), \quad g(x) \in \mathbb{Z}[x]$
If $g(a) \neq 0 \bmod p,$ then $f(x)$ is irreducible modulo $p^{2} .$