# a.c.c. and d.c.c. on radical ideals in commutative ring of dimension zero

Let $$R$$ be a commutative ring with unity of dimension zero (i.e. every prime ideal is maximal). Does any of the following two conditions imply the other :

1) $$R$$ satisfies a.c.c. on radical ideals

2) $$R$$ satisfies d.c.c. on radical ideals

??

Motivation: For a zero dimensional ring, Artinian is equivalent to Noetherian ... now a.c.c. (resp. d.c.c. ) on radical ideals is same as saying the prime spectrum with Zariski topology is Noetherian (resp. Artinian) , where we mean a topological space to be Noetherian (resp. Artinian) iff a.c.c. (resp. d.c.c.) holds on open sets ... hence the question

• What's the context for this question?
– jgon
Dec 31, 2018 at 5:20
• @jgon: For a zero dimensional ring, Artinian is equivalent to Noetherian ... now a.c.c. (resp. d.c.c. ) on radical ideals is same as saying the prime spectrum with Zariski topology is Noetherian (resp. Artinian) , where we mean a topological space to be Noetherian iff a.c.c. holds on open sets ... etc. ... so a natural question to ask is ... the question I'm asking here... Dec 31, 2018 at 5:23
• fair enough, but that should probably be in the body of the question
– jgon
Dec 31, 2018 at 5:27
• @jgon: okay ... added now Dec 31, 2018 at 5:34

Yes. This follows from the fact that the spectrum of a zero-dimensional ring is Hausdorff. Since radical ideals correspond to closed sets in the spectrum (with the inclusion order reversed), it suffices to show that a Hausdorff space $$X$$ satisfies the descending chain condition on closed sets iff it satisfies the ascending chain condition on closed sets. In fact, I claim that both of these conditions are equivalent to $$X$$ being finite.
First, if $$X$$ is finite, it obviously satisfies both chain conditions. Conversely, suppose $$X$$ is infinite. Then $$X$$ has an infinite ascending chain of closed sets: just pick a sequence of distinct points $$(x_n)$$ and consider the sets $$\{x_0\}, \{x_0,x_1\},\{x_0,x_1,x_2\},\dots.$$ To find a descending chain of closed sets, it suffices to prove that any infinite Hausdorff space has an infinite closed proper subset, since then we can iterate this (start with $$C_0=X$$, and let $$C_{n+1}$$ be an infinite closed proper subset of $$C_n$$). To prove this, let $$U$$ and $$V$$ be two disjoint nonempty open sets in $$X$$ (which exist since $$X$$ is Hausdorff and has more than one point). Then the sets $$C=X\setminus U$$ and $$D=X\setminus V$$ cover $$X$$, so one of them must be infinite. Since $$C$$ and $$D$$ are both proper closed subsets of $$X$$, this completes the proof.
• I don't know. ${}$ Dec 31, 2018 at 6:12