Let be a ring. An ideal in is said to be if, whenever is an intersection of two ideals: , then either or .
If is a gcd domain, and is an irreducible element, then is an irreducible ideal.
If is a unit, then and we are done. So we assume that is not a unit for the remainder of the proof.
Let and suppose and . Then for some . Let be a gcd of and . So
for some . Since is irreducible, either is a unit or is. The proof now breaks down into two cases:
is a unit. So is an associate of . Because divides , we get that as well, or , which is again impossible by assumption.
Therefore, the assumption that and is false, which is the same as saying or . But and , either or , or is irreducible. ∎
|Date of creation||2013-03-22 18:19:47|
|Last modified on||2013-03-22 18:19:47|
|Last modified by||CWoo (3771)|