Csdrhrt

Question 1: Does such a ring can be found?

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?)

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 $R=koplus V$, 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

Two more examples I think work...

1. Let $k$ be a field, and $A = k[x_1,x_2,ldots]$
   a polynomial ring over $k$ in countably many indeterminates.
   Let $mathfrak{b}$ be the ideal generated by $x_1^2$ and
   $x_n - x_{n+1}^2$ for all $n geq 1$.
   Write $y_n = bar x_n$ in $B = A/mathfrak{b}$.
   Then $y_1^2 = 0$ and $y_n = y_{n+1}^2$ for all $n geq 1$,
   so $y_n^{2^n} = 0$.
   If $mathfrak{p} = (y_1,y_2,ldots)$,
   we have $B/mathfrak{p} cong k$, so $mathfrak{p}$ is maximal.
   On the other hand, the generators of $mathfrak{p}$ are nilpotent, so
   $mathfrak{p}$ is contained in the nilradical of $B$, and hence is the unique minimal prime as well.
   Since all primes then contain the maximal ideal $mathfrak{p}$
   $mathfrak{p}$ is the only prime of $B$.
   Thus $mathrm{Spec}(B) = {mathfrak{p}}$ is obviously Noetherian.
   But $(y_1) subsetneq (y_2) subsetneq (y_3) subsetneq cdots$
   is an infinite ascending chain of ideals, so $B$ is not Noetherian.




2. Let $k$ be a field, $mathbb{Q}$ the additive group of rational numbers, and $k[mathbb{Q}]$ the group algebra. If $K$ is its field of fractions, there is a naturally associated valuation $vcolon K^times twoheadrightarrow mathbb{Q}$; let $A = {0} cup {x in K : v(x) geq 0}$ be the associated valuation ring. It has an ideal $mathfrak{a}_q = {x in K : v(x) geq q}$ for each rational $q > 0$, with $mathfrak{a}_q subsetneq mathfrak{a}_r$ if $r < q$. Thus, taking an infinite decreasing sequence $(q_j)$, we see $A$ is not Noetherian. However, the only prime ideal of $A$ is $mathfrak{p} = {x in K : v(x) > 0}$.

$newcommand{Z}{mathbb{Z}}$
Just to supply another example. Let $R$ be the subring $Z/4+(2)subset(Z/4)[x_1,x_2,cdots]$. (In general, the sum of a subring and an ideal is a subring.) $R$ consists of all polynomials in $(Z/4)[x_1,x_2,cdots]$ with coefficients of non-constant terms nilpotent. The nilradical of $R$ is $(2)$, and $R/(2)=Z/2$ is a field, so $mathrm{Spec}(R)$ is Noetherian. However, $(2x_1)subset(2x_1,2x_2)subsetcdots$ is a strictly ascending chain.

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 $mathfrak a = pmathfrak o = {x ∈ mathfrak o;~lvert x rvert ≤ 1/p }$. 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
$$mathfrak a = (p) subsetneq (sqrt p) subsetneq (sqrt[4] p) subsetneq (sqrt[8] p) subsetneq ….$$