Irreducible implies primary (Noetherian)
Relation with other results
Let be a Noetherian ring and an irreducible ideal in . Suppose but . We need to show that there exists a such that .
Define as the ideal of all elements such that . Clearly we get an ascending chain of ideals in :
Since is Noetherian, there exists a natural number such that for all . In other words, there exists a such that if , then for any .
Now let and . Clearly, contains . We want to show that .
For this, suppose . Then we can write:
where . This tells us that:
multiplying both sides by we get:
By our assumption, and hence .
This gives us:
but from the property of , and hence .
Thus any element of is in , so .
Comparison with the proof for principal ideal rings
In the case of principal ideal rings, we employ a very similar proof, except that the is replaced by 1.