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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....

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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
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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
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1 Answer 1

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.

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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
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