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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 $\mathcal{O}^{*}\to\mathfrak{D}$. If there is a prime divisor $\mathfrak{p\in D}$ such that

$\mathfrak{p}\mid a_{0},\,a_{1},\,\ldots,\,a_{{n1}},$

$\mathfrak{p}\nmid a_{n},$

$\mathfrak{p}^{2}\nmid a_{0},$
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 $\mathfrak{p}^{2}\nmid(a_{0})$, 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 $p\nmid c_{k}$. Accordingly, in the sum $b_{0}c_{k}+b_{1}c_{{k1}}+\ldots+b_{k}c_{0}$, the prime divisor $\mathfrak{p}$ divides every summand except the first (see the definition of divisor theory); 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.
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