# Eisenstein criterion in terms of 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

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

• $\mathfrak{p}\nmid a_{n},$

• $\mathfrak{p}^{2}\nmid a_{0},$

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 $b_{0}c_{k}+b_{1}c_{k-1}+\ldots+b_{k}c_{0}$, 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.

Title Eisenstein criterion in terms of divisor theory EisensteinCriterionInTermsOfDivisorTheory 2013-03-22 18:00:45 2013-03-22 18:00:45 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 13A05 DivisorTheory