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 $ab\in P$ 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 $R$:

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

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:

$y=p_{1}+ua=p_{2}+vb^{n}$

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.