Irreducible implies primary (Noetherian)

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 $R$ be a Noetherian ring and $P$ an irreducible ideal in $R$ . Suppose $a b \in P$ but $a \notin P$ . We need to show that there exists a $n$ such that $b^{n} \in P$ .

Define $P_{i}$ as the ideal of all elements $x$ such that $x b^{i} \in P$ . Clearly we get an ascending chain of ideals in $R$ :

$P = P_{0} \leq P_{1} \leq P_{2} \leq \dots$

Since $R$ is Noetherian, there exists a natural number $n$ such that $P_{n} = P_{m}$ for all $m \geq n$ . In other words, there exists a $n$ such that if $b^{n + 1} x \in P$ , then $b^{n} x \in P$ for any $x \in R$ .

Now let $Q = P + (a)$ and $R = P + (b^{n})$ . Clearly, $Q \cap R$ contains $P$ . We want to show that $Q \cap R = P$ .

For this, suppose $y \in Q \cap R$ . Then we can write:

$y = p_{1} + u a = p_{2} + v b^{n}$

where $p_{i} \in P$ . This tells us that:

$u a - v b^{n} \in P$ .

multiplying both sides by $b$ we get:

$u a b - v b^{n + 1} \in P$

By our assumption, $a b \in P$ and hence $u a b \in P$ .

This gives us:

$v b^{n + 1} \in P$

but from the property of $n$ , $v b^{n} \in P$ and hence $y = p_{2} + v b^{n} \in P$ .

Thus any element of $Q \cap R$ is in $P$ , so $Q \cap R = P$ .

Comparison with the proof for principal ideal rings

In the case of principal ideal rings, we employ a very similar proof, except that the $n$ is replaced by 1.