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Suppose $R$ is a commutative ring with $1$. It is well-known that if $R$ is Noetherian, then every irreducible ideal is primary (Lemma 7.12 in Atiyah & Macdonald). Is the converse true? That is:

If every irreducible ideal of $R$ is primary, then is it necessarily true that $R$ is Noetherian?

This is just to satisfy my own curiosity. :)

Note the following related problem. There, the question is about different converse (which is false).

Added. In order to make the post self-contained, I will add the relevant definitions. An ideal $\mathfrak{a}$ is called irreducible if $\mathfrak{a}=\mathfrak{b}\cap\mathfrak{c}$ implies $\mathfrak{a}=\mathfrak{b}$ or $\mathfrak{a}=\mathfrak{c}$. An ideal $\mathfrak{a}$ is called primary if $\mathfrak{a}\neq R$, and for all $x,y\in R$, $xy\in \mathfrak{a}$ implies $x\in\mathfrak{a}$ or $y^{n}\in\mathfrak{a}$ for some $n\in\mathbb{N}$.

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Prism
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    If $R$ satisfies the ascending chain condition on irreducible ideals then every irreducible ideal is primary. –  Aug 03 '13 at 23:47
  • @YACP: Thanks! I think you are indicating that there is a counter-example, that is: we just need to find a ring that satisfies ascending chain condition on irreducible ideals, but otherwise is non-Noetherian. – Prism Aug 04 '13 at 07:02

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If $R$ is a ring with the property that every ideal (excepting $R$ itself) has a (finite) primary decomposition, that is, $R$ is laskerian, then every irreducible ideal is primary. Now take a laskerian ring which is not noetherian.

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