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?)

Shubhrajit Bhattacharya
  • 5,493
  • 1
  • 13
  • 36
  • 10,608
  • 1
  • 33
  • 62
  • Question 2 arises because of **Z**, noetherian but non-artinian: definitions given dually deceive me every time... – fosco Oct 21 '10 at 09:00
  • 4
    If $m$ is a maximal ideal of a commutative ring $R$, then $R / m^r$ has only $m/m^r$ as prime ideal, since any prime has the form $p/m^r$, so that $m = \sqrt{m^r} \subset p \implies m=p$, since $m$ is maximal. So it is sufficient to find an example where $m/m^r$ is not finitely generated as $R/m^r$-module, which implies that $R/m^r$ is not Noetherian. $\tag*{}$ Moreover, the spectrum of a commutative ring is Noetherian iff the ring satisfies a.c.c. on radical ideals. – Watson Jan 25 '17 at 20:55

6 Answers6


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.

  • 1
  • 13
  • 62
  • 125
Martin Brandenburg
  • 146,755
  • 15
  • 248
  • 458

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? :) )

Mariano Suárez-Álvarez
  • 126,294
  • 9
  • 223
  • 348

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.


Hagen Knaf
  • 8,202
  • 17
  • 21

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 $v\colon K^\times \twoheadrightarrow \mathbb{Q}$; 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 $\mathfrak{p} = \{x \in K : v(x) > 0\}$.

  • 4,517
  • 17
  • 34

$\newcommand{\Z}{\mathbb{Z}}$ 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, $(2x_1)\subset(2x_1,2x_2)\subset\cdots$ is a strictly ascending chain.

Junyan Xu
  • 567
  • 4
  • 12

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 $$\mathfrak a = (p) \subsetneq (\sqrt p) \subsetneq (\sqrt[4] p) \subsetneq (\sqrt[8] p) \subsetneq ….$$

  • 17,653
  • 4
  • 31
  • 64