# Irreducible implies primary (Noetherian)

## Contents

## Statement

In a Noetherian ring, any irreducible ideal is primary.

## Relation with other results

- Irreducible implies prime (PID) uses a very similar approach
- Irreducible implies prime (Dedekind)

## Proof

### Hands-on proof

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.