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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.

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2 Answers 2

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

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    $\begingroup$ +1 Very direct way to clobber the ACC requirement and Noetherian requirement with a single stone. $\endgroup$ Commented Dec 8, 2014 at 20:25
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    $\begingroup$ Thanks for your hint. Now I see that it actually has krull dimension 0, so I have ACC on primes. About Noetherian requirement, if I take $I_n$ as the ideal generated by first n copies of $\mathbb{F}_2$ then $V(I_n)$ gives a descending chain of closed sets(every ideal in boolean ring is radical, as it has no nonzero nilpotent element). Am I right? $\endgroup$ Commented Dec 8, 2014 at 23:27
  • $\begingroup$ Hi, would be much to ask by a little bit more details...? I am beginner and could not understand Scream's comment. Thank you so much. $\endgroup$ Commented Nov 21, 2019 at 0:35
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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 $A$.

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    $\begingroup$ Spec A is noetherian consisting of one point. $\endgroup$ Commented Dec 3, 2017 at 8:44

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