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
Best Answer
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.