# Irreducible implies primary (Noetherian)

## Statement

In a Noetherian ring, any irreducible ideal is primary.

## Proof

### Hands-on proof

Let $R$ be a Noetherian ring and $P$ an irreducible ideal in $R$. Suppose $ab \in P$ 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 $xb^i \in P$. Clearly we get an ascending chain of ideals in $R$:

$P = P_0 \le P_1 \le P_2 \le \ldots$

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

$y = p_1 + ua = p_2 + vb^n$

where $p_i \in P$. This tells us that:

$ua - vb^n \in P$.

multiplying both sides by $b$ we get:

$uab - vb^{n+1} \in P$

By our assumption, $ab \in P$ and hence $uab \in P$.

This gives us:

$vb^{n+1} \in P$

but from the property of $n$, $vb^n \in P$ and hence $y = p_2 + vb^n \in P$.

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.