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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

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  • $\begingroup$ What's the context for this question? $\endgroup$ – jgon Dec 31 '18 at 5:20
  • $\begingroup$ @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... $\endgroup$ – user521337 Dec 31 '18 at 5:23
  • $\begingroup$ fair enough, but that should probably be in the body of the question $\endgroup$ – jgon Dec 31 '18 at 5:27
  • $\begingroup$ @jgon: okay ... added now $\endgroup$ – user521337 Dec 31 '18 at 5:34
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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.

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  • $\begingroup$ amazing ... thanks ... as a side question, may I ask if d.c.c. on radical ideals would imply a.c.c. on radical ideals with only finite dimension assumption (with say also given the ring is local )? $\endgroup$ – user521337 Dec 31 '18 at 5:53
  • $\begingroup$ I don't know. ${}$ $\endgroup$ – Eric Wofsey Dec 31 '18 at 6:12

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