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

Proof. Suppose that we have in $\mathcal{O}[x]$ the factorisation $$f(x) = (b_0+b_1x+\ldots+b_sx^s)(c_0+c_1x+\ldots+c_tx^t)$$ 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 $b_sc_t = a_n$ . So, there is a certain smallest index $k$ such that $p \nmid c_k$ . Accordingly, in the sum $b_0c_k+b_1c_{k-1}+\ldots+b_kc_0$ , 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.