A non-noetherian ring with noetherian spectrum

It is easy to come up with examples when you keep in mind that $X$ and $X_{\text{red}}$ are homeomorphic: The topology is not changed when you divide out (all) nilpotent sections.

For example, $A = k[x_1,x_2,...] / (x_1^2,x_2^2,...)$ satisfies $A/\mathrm{rad}(A)=k$, so that $\mathrm{Spec}(A)$ consists of exactly one point, namely the maximal ideal $(x_1,x_2,...)$, which is not finitely generated.

In general, $\mathrm{Spec}(A)$ is a noetherian topological space iff $A$ satisfies the ascending chain condition for radical ideals.

If the non-noetherianness of the ring is hidden inside the nilradical, then $\mathrm{Spec}$ won't see it.

Let $k$ be any ring, and let $V$ be a free $k$-module of infinite rank. Consider $R=k\oplus V$, and turn it into a ring by defining $$(a,v)\cdot(b,w)=(ab,aw+bv).$$ (Representation-people call this a trivial extension) Then $R$ is not-noetherian, because every $k$-submodule of $V$ is an ideal in $R$. Yet $V$ is contained in the nilradical of $R$: if you look at $\mathrm{Spec}\;R$ and at $\mathrm{Spec}\;k$, you'll see that they are very similar.

As for your second question: no. If $k$ is an infinite field, then $\mathrm{Spec}\;k[X]$ is noetherian, yet you'll easily find a decreasing chain of open sets init which does not stop. (What examples did you consider before asking the question? :) )

Answer to question 1: Yes.

Take a non-noetherian valuation domain $R$ of finite Krull dimension. The spectrum of $R$ is totally ordered under inclusion hence finite.

H