If every ascending chain of primary ideals in $R$ stabilizes, is $R$ a Noetherian ring?

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 and Exercise 12 of Chapter 6 in Introduction to Commutative Algebra by Atiyah & Macdonald.

Answer 1

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.

This is even more dramatic than the ACC really, since you cannot even have a chain of two primary ideals :)

Answer 2

In Commutative Algebra a suitable construction for counterexamples is given by the idealization 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 $\{0\}\times\{0\}$).