Eisenstein criterion in terms of divisor theory
Proof. Suppose that we have in the factorisation
with and . Because the principal divisor , i.e. is divisible by the prime divisor and there is a unique factorisation in the monoid , must divide or but, by , not both of and ; suppose e.g. that . If would divide all the coefficients , then it would divide also the product . So, there is a certain smallest index such that . Accordingly, in the sum , the prime divisor 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 which by hypothesis is divisible by the prime divisor. This contradiction shows that the polynomial is irreducible.
|Title||Eisenstein criterion in terms of divisor theory|
|Date of creation||2013-03-22 18:00:45|
|Last modified on||2013-03-22 18:00:45|
|Last modified by||pahio (2872)|