Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is well known that we have many different definitions of noetherianity for rings. Namely, given a ring $R$, the following are equivalent:

1) every ideal of $R$ is finitely generated.

2) $R$ satisfies a.c.c. (ascending chain condition) on ideals.

3) every non-empty set of ideals of $R$ has a maximal element.

I would like to state something similar for topological spaces. First a definiton: we call a closed subset of a topological space irreducible if it cannot be decomposed in the union of two proper closed subsets. Now i want to prove:

Let $X$ be a topological space. Then the following are equivalent:

1') every closed subset $Y$ of $X$ has a decomposition $$Y=Y_1\cup\ldots\cup Y_r$$ where $Y_j$ is irreducible and $Y_j\nsubseteq Y_m$ for $j\neq m$.

2') $X$ satisfies d.c.c. (descending chain condition) on closed subsets.

3') every non-empty set of closed subsets of $X$ has a minimal element.

I have a proof for $2')\rightarrow 3')$ and for $3')\rightarrow 1')$, but not for $1')\rightarrow 2')$, maybe because it is false? Can you help me?

EDIT: condition $Y_j\nsubseteq Y_m$ for $j\neq m$ in 1') is needed only for unicity of decomposition, but i'm non intersted in, so we can skip it. I would like 1') to be "similar" to 1) in the sense that 1) tells every ideal is finitely generated, and 1') that every closed subset has finitely many irreducible components....

share|cite|improve this question
Why did you go from a.c.c. for the Nötherian ring case to d.c.c in the proposition? – awllower Feb 23 '13 at 13:29
because if the topological space is the affine space $A^n$ with zariski topology, then a descending chain of closed $X_1\supseteq X_2\supseteq\ldots$ corresponds to the ascending chain of ideals of polynomials vanishing at $X_i's$:$I(X_1)\subseteq I(X_2)\subseteq\ldots$ – Federica Maggioni Feb 23 '13 at 13:33
I see. Sorry for the ignorant comment. – awllower Feb 23 '13 at 13:36
I wonder why it isn't mentioned yet that there is a well-known notion of noetherian topological spaces (equivalent to 2,3). – Martin Brandenburg Feb 27 '13 at 10:02
up vote 4 down vote accepted

Let $R$ be a commutative ring. It is known that $X=\operatorname{Spec}(R)$ is noetherian iff $R$ satisfies ACC on radical ideals.

If $R$ is a valuation ring, then every proper radical ideal is prime; see Geraschenko, Commutative Rings, Theorem 30.5. Thus $X=\operatorname{Spec}(R)$ satisfies the condition $1')$. Furthermore, in this case $X=\operatorname{Spec}(R)$ is noetherian iff $R$ satisfies ACC on prime ideals. Now take $R$ a valuation ring which has an infinite chain of prime ideals.

share|cite|improve this answer
Maybe there are simpler examples when $X$ is an arbitrary topological space, but I was mostly interested in the particular case when $X$ is the prime spectrum of a commutative ring. – user26857 Feb 23 '13 at 22:03
Nice counterexample ! – user18119 Feb 23 '13 at 22:16
@QiL'8 Thanks! I took the opportunity that you decided to answer only after $8$ hours :) – user26857 Feb 23 '13 at 22:23
and if I know the answer :). – user18119 Feb 23 '13 at 22:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.