On minimal elements, w.r.t. inclusion, of non-empty subset of prime ideals of commutative rings

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 ?

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.