It is a well known fact that for a noetherian ring $A$ there are finitely many minimal primes. Now I'm wondering if this is true for every subset in the primes of $A$. My question, specifically, would be if for every non empty $B\subseteq \operatorname{Spec} A $ the set $X=\{P\in B\mid P\text{ is minimal in } B\} $ is finite.
I came up with this question when trying to prove something about $\operatorname{Supp}M$, where $M $ is an $A$-module. Perhaps if the answer to the first question is no, it turn out to be yes when $B=\operatorname{Supp}M $.
I couldn't find nothing about it on internet. Any ideas?
Let $A$ be a noetherian ring with infinitely many maximal ideals (e.g., $A=K[x]$, where $K$ is an infinite field).
Now let $B$ be the set of maximal ideals of $A$.
Clearly all elements of $B$ are minimal in $B$.
As to your question "Ok, but what happen when it is closed?", suppose $B$ is a closed subset of ${\text{Spec}}\; A$.
Then $B=V(I)=\{P\in {\text{Spec}}\; A\mid P\supseteq I\}$ for some ideal $I$ of $A$.
Note that the minimal elements of $B$ correspond to the minimal prime ideals of $A/I$, hence there are at most fininitely many, since $A$ noetherian implies $A/I$ is noetherian.