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  $f(x):=a_0+a_{1}x+\mathrm{\dots }+a_n{x}^n$  be a primitive polynomial over an integral domain $𝒪$ with divisor theory (http://planetmath.org/DivisorTheory)  ${𝒪}^{*}\to 𝔇$.  If there is a prime divisor  $𝔭\in 𝔇$  such that

• $𝔭\mid a_0,a_1,\mathrm{\dots },a_{n-1},$

• $𝔭\nmid a_n,$

• ${𝔭}^2\nmid a_0,$

then the polynomial is irreducible.

Proof.  Suppose that we have in $𝒪[x]$ the factorisation

$$f(x)=(b_0+b_{1}x+\mathrm{\dots }+b_s{x}^s)(c_0+c_{1}x+\mathrm{\dots }+c_t{x}^t)$$

with  $s>0$  and  $t>0$.  Because the principal divisor $(a_0)$, i.e. $(b_0)(c_0)$ is divisible by the prime divisor $𝔭$ and there is a unique factorisation in the monoid $𝔇$, $𝔭$ must divide $(b_0)$ or $(c_0)$ but, by ${𝔭}^2\nmid (a_0)$, not both of $(b_0)$ and $(c_0)$; suppose e.g. that $𝔭\mid c_0$.  If $𝔭$ 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 $b_0{c}_k+b_1{c}_{k-1}+\mathrm{\dots }+b_k{c}_0$, the prime divisor $𝔭$ 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
Canonical name	EisensteinCriterionInTermsOfDivisorTheory
Date of creation	2013-03-22 18:00:45
Last modified on	2013-03-22 18:00:45
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13A05
Related topic	DivisorTheory