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….
Best Answer
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.