Eisenstein criterion

Theorem (Eisenstein criterion).

Let $f$ be a primitive polynomial over a commutative unique factorization domain $R$, say

$$f(x)=a_0+a_{1}x+a_2{x}^2+\mathrm{\dots }+a_n{x}^n.$$

If $R$ has an irreducible element $p$ such that

$$p\mid a_m\mathit{\hspace{1em}\hspace{1em}}0\le m\le n-1$$

$$p^2\nmid a_0$$

$$p\nmid a_n$$

then $f$ is irreducible.

Proof.

Suppose

$$f=(b_0+\mathrm{\dots }+b_s{x}^s)(c_0+\mathrm{\dots }+c_t{x}^t)$$

where $s>0$ and $t>0$. Since $a_0=b_0{c}_0$, we know that $p$ divides one but not both of $b_0$ and $c_0$; suppose $p\mid c_0$. By hypothesis, not all the $c_m$ are divisible by $p$; let $k$ be the smallest index such that $p\nmid c_k$. We have $a_k=b_0{c}_k+b_1{c}_{k-1}+\mathrm{\dots }+b_k{c}_0$.We also have $p\mid a_k$, and $p$ divides every summand except one on the right side, which yields a contradiction. QED∎