Eisenstein criterion in terms of divisor theory

The below theorem generalises Eisenstein criterion of irreducibility from UFD’s to domains with divisor theory.

Theorem.

Let $f(x):=a_{0}\\!+\\!a_{1}x\\!+\\ldots+\\!a_{n}x^{n}$ be a primitive polynomial over an integral domain $\\mathcal{O}$ with divisor theory (http://planetmath.org/DivisorTheory) $\\mathcal{O}^{*}\\to\\mathfrak{D}$ . If there is a prime divisor $\\mathfrak{p\\in D}$ such that

•
$\\mathfrak{p}\\mid a_{0},\\,a_{1},\\,\\ldots,\\,a_{n-1},$
•
$\\mathfrak{p}\mid a_{n},$
•
$\\mathfrak{p}^{2}\mid a_{0},$

then the polynomial is irreducible.

Proof. Suppose that we have in $\\mathcal{O}[x]$ the factorisation

f(x)=(b_{0}+b_{1}x+\\ldots+b_{s}x^{s})(c_{0}+c_{1}x+\\ldots+c_{t}x^{t})

with $s>0$ and $t>0$ . Because the principal divisor $(a_{0})$ , i.e. $(b_{0})(c_{0})$ is divisible by the prime divisor $\\mathfrak{p}$ and there is a unique factorisation in the monoid $\\mathfrak{D}$ , $\\mathfrak{p}$ must divide $(b_{0})$ or $(c_{0})$ but, by $\\mathfrak{p}^{2}\mid(a_{0})$ , not both of $(b_{0})$ and $(c_{0})$ ; suppose e.g. that $\\mathfrak{p}\\mid c_{0}$ . If $\\mathfrak{p}$ would divide all the coefficients $c_{j}$ , then it would divide also the product $b_{s}c_{t}=a_{n}$ . So, there is a certain smallest index $k$ such that $p\mid c_{k}$ . Accordingly, in the sum $b_{0}c_{k}+b_{1}c_{k-1}+\\ldots+b_{k}c_{0}$ , the prime divisor $\\mathfrak{p}$ divides (http://planetmath.org/DivisibilityInRings) every summand except the first (see the definition of divisor theory (http://planetmath.org/DivisorTheory)); therefore it cannot divide the sum. But the value of the sum is $a_{k}$ which by hypothesis is divisible by the prime divisor. This contradiction shows that the polynomial $f(x)$ is irreducible.