Eisenstein criterion

For a primitive polynomialif a prime $p$ does not divide $a_n$, divides every $a_j$ for $j<n$, and has $p^2\nmid a_0$, then $F$ is irreducible. If $F=GH$, reduction modulo $p$ forces both $\overline{G}$ and $\overline{H}$ to be positive-degree monomials. Their constant terms are consequently divisible by $p$, making $p^2\mid F(0)$, a contradiction. Gauss lemma for polynomials transfers the result between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$.

If an algebraic integer $\alpha$ has a degree-$n$ minimal polynomial that is Eisenstein at $p$, then $p$ has a unique prime ideal $P$ above it in $\mathbb{Q}(\alpha )$, withThis conclusion concerns the full ring of integers of a number field and does not assume that it equals $\mathbb{Z}[\alpha ]$.

For $n>1$,is irreducible over the integers exactly when $n$ is prime. Composite $n$ gives a geometric factorization; for prime $p$, translation $X\mapsto X+1$ makes the polynomial Eisenstein at $p$.