Question: A commutative ring $R$ is called Noetherian if every ascending chain of ideals in $R$ stabilizes, that is,
$$
I\_1\subseteq I\_2\subseteq I\_3\subseteq\cdots
$$
implies the existence of $n\in\mathbb{N}$ such that $I\_n=I\_{n+1}=I\_{n+2}=\cdots$.

My question is the following: 

> 
> Does there exist a non-Noetherian ring $R$ such that every ascending chain of *primary* ideals stabilizes?
> 
> 
> 

**Remark.** Note that there exists non-Noetherian ring $R$ such that every ascending chain of *prime* ideals stabilizes. This happens exactly when $R$ is non-Noetherian and $\operatorname{Spec}(R)$ is Noetherian topological space. See [here](https://math.stackexchange.com/questions/7392/a-non-noetherian-ring-with-textspecr-noetherian) and Exercise 12 of Chapter 6 in *Introduction to Commutative Algebra* by Atiyah & Macdonald.

Answer: 
Yes, there do exist rings which aren't Noetherian but which do have ACC on primary ideals. An example is $\prod\_{i\in\Bbb N} F\_i$ where the $F\_i$ are fields.

This is clearly not Noetherian, and because it is commutative and von Neumann regular, [all of its primary ideals are maximal](http://www.mscand.dk/article.php?id=2024).

This is even more dramatic than the ACC really, since you cannot even have a chain of two primary ideals :)
In Commutative Algebra a suitable construction for counterexamples is given by the [*idealization*](https://math.stackexchange.com/questions/207406/irreducible-components-of-the-prime-spectrum-of-a-quotient-ring-and-primary-deco/207468#207468) of an $A$-module $M$, sometimes denoted by $I\_A(M)$. If one considers $R=I\_{\mathbb Z}(\mathbb Q)$, then $R$ is not noetherian (since $\mathbb Q$ is not finitely generated as $\mathbb Z$-module) and it satisfies ACC on primary ideals (since these are of the following form: $p^n\mathbb Z\times\mathbb Q$ with $p$ prime and $n\ge 1$, $\{0\}\times\mathbb Q$ or $\{0\}\times\{0\}$).