Let $R$ be a commutative ring with unity. Let $\operatorname{Spec}R$ denote the set of prime ideals of $R$. For a non-empty subset $\mathcal A \subseteq \operatorname{Spec}R$, let us say that $P \in \mathcal A$ is a minimal element of $\mathcal A$ if $Q \in \mathcal A$ and $Q \subseteq P \implies Q=P$. Let $\min \mathcal A$ denote the set of all minimal elements in $\mathcal A$. Now consider the following two statements:
(1) $\min \mathcal A \ne \emptyset$
(2) Prime ideals in $\mathcal A$ satisfies d.c.c. (i.e. every descending chain of prime ideals in $\mathcal A$ terminates).
Now I can easily see that (2) $\implies $ (1).
My question is: Is it true that (1) $\implies$ (2) ?
If this is not true in general, is it at least true when $\mathcal A=\operatorname{Supp}M=\{P \in \operatorname{Spec}R : M_P\ne 0\}$ ($=V(\operatorname{Ann}_R M)$), where $M$ is a finitely generated, non-zero $R$-module ?
Best Answer
It is not true that $(1)\implies(2)$; consider $\mathcal{A}=\operatorname{Spec}R$ where $R=A[X_1,X_2,\ldots]$ is a polynomial ring in countably infinitely many variables over an integral domain. Here the zero ideal is minimal because $R$ is an integral domain, so $\min\mathcal{A}\neq\varnothing$. But the prime ideals $I_k:=(X_k,X_{k+1},\ldots)$ form an infinite strictly descending chain, so prime ideals in $\mathcal{A}$ do not satisfy d.c.c.