I know that $\operatorname{Spec}(A)$ noetherian implies a.c.c. on prime ideals of $A$. However, is the converse true?
I have to prove the yes, but could not. In fact, read some posts that leaded me to think that the answer is not... But in this case I need to present some counterexample.
Many thanks in advance!
Hint: The correct statement is that $\operatorname{Spec}(R)$ is Noetherian iff $R$ has the ascending chain condition on radical ideals.
This follows immediately from the one-to-one correspondence between closed subsets of the Zariski topology and radical ideals of $R$, which extends to a one-to-one correspondence between ascending chains of radical ideals and descending chains of closed subsets.
The statement you are asking about is in fact false, because there exist rings that have ACC on prime ideals but do not have ACC on radical ideals. The simples examples of these come from Von Neumann Regular rings (VNRs). Any VNR is $0$-dimensional and hence trivially satisfies the ACC on prime ideals. Also every ideal of a VNR is radical. Thus for $R$ a VNR, we see that $R$ is Noetherian iff $\operatorname{Spec}(R)$ is Noetherian iff $R$ has finitely many prime ideals. In particular, any infinite product of fields will have ACC on prime ideals but will not have Noetherian spectrum.