A ring with ACC on prime ideals whose spectrum is non-noetherian.

Question

I am currently working on the converse of the exercise #12 on chapter 6 of Atiyah-Macdonald's book on commutative algebra.

The problem is asking whether there is a ring $A$ which satisfies the ascending chain condition on prime ideals (that is, every chain of prime ideals of a ring $A$ stabilizes) such that $\mathrm{Spec}(A)$ is a non-noetherian space.

What I've tried so far was first, struggling with the proof that spec($A$) is noetherian when $A$ satisfies ACC on prime ideals, but I got stuck because I failed to extract ascending chains of prime ideals.

So I tried to find a counterexample. So, I looked up some examples of non-noetherian rings so that I can have non-noetherian spectrum. (It was tough for most cases since non-noetherian rings can have a noetherian spectrum. For example, the integral closure of $\mathbb{Z}$ in $\mathbb{C}$ is not noetherian. However, I couldn't understand whether it's spectrum is noetherian or not.) But, in every cases, I also couldn't find an appropriate counterexample.

Is there any hint or comment about what I should consider? Any help will be appreciated.

Answer 1

Hint. Take $A$ a countable direct product of copies of $\mathbb F_2$.

Answer 2

Another possibility is to take $A = k[x_1, x_2, x_3 , \dots]/(x_1, x_2^2, x_3^3, \dots)$, where $k$ is a field. The ideal $(x_1, x_2, x_3, \dots) / (x_1, x_2^2, x_3^3, \dots)$ is not finitely generated, but it is the unique prime ideal of $A$.