Let $A$ be a commutative ring such that $\operatorname{Spec} A$ is a Noetherian topological space. Does it follow that $A$ is a Noetherian ring?
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(Correct me if wrong) in other words, if prime ideals of $A$ satisfies the a.c.c, does it follow that $A$ is Noetherian? – aaa acb Mar 09 '23 at 17:12
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For what it’s worth, Theorem 3.4 in Matsumura’s Commutative Ring Theory says if all prime ideals of $A$ are finitely generated then $A$ is Noetherian. – aaa acb Mar 09 '23 at 17:13
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2As shown in the linked duplicate, no. There are non-noetherian rings with spectrum a single point. – KReiser Mar 09 '23 at 17:19
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@KReiser I think that finite type algebras over $\mathcal{O}_{\mathbb{C}_p}$ (which come up constantly in rigid geometry) addresses the OP's question and are more natural than the answers in the linked 'duplicates' (for this specific question). Is there no way to put this answer here? – Alex Youcis Mar 10 '23 at 05:30
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1@AlexYoucis I think the desired way to do this from a SE standpoint would be to record it on the root of the duplicate tree. I see you've done that, +1 over there and thank you for some nice reading for someone like me who's not as experienced with rigid geometry. – KReiser Mar 10 '23 at 21:56