TITLE: A non-noetherian ring with noetherian spectrum
QUESTION [33 upvotes]: Question 1: Does such a ring exist?
Note: The definition of a noetherian topological space is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite number of steps
(Question 2: is this equivalent to saying that every descending chain of opens stops?)

REPLY [2 votes]: Yet another example from the $p$-adic world: Let $p$ be prime $F = ℚ_p^{\mathrm{tr}}$ the maximal totally ramified extension of $ℚ_p$. Then $F$ is a non-archimedically valued field with a value $\lvert\, ·\,\rvert \colon F → [0..∞)$ that extends every value of every finite totally ramified extension of $ℚ_p$.
Hence, $\mathfrak o = \{x ∈ F;~\lvert x \rvert ≤ 1 \}$ is a local ring with its maximal ideal given by $\mathfrak m = \{x ∈ \mathfrak o;~\lvert x \rvert < 1 \}$. Let $\mathfrak a = p\mathfrak o = \{x ∈ \mathfrak o;~\lvert x \rvert ≤ 1/p \}$. Then $\operatorname{rad} \mathfrak a = \mathfrak m$, and so $\operatorname{Spec} \mathfrak o / \mathfrak a = \operatorname{Spec} \mathfrak o / \mathfrak m$ is a point. However, $\mathfrak o / \mathfrak a$ is not Noetherian, since there are infinite strictly ascending chains of ideals above $\mathfrak a$, for instance
$$\mathfrak a = (p) \subsetneq (\sqrt p) \subsetneq (\sqrt[4] p) \subsetneq (\sqrt[8] p) \subsetneq ….$$