Does every closed set of prime ideals of a noetherian (commutative) ring contain a finite dense subset?
EDIT 1. Here is the motivation. In books like Mumford’s red book or Eisenbud-Harris, the authors describe the topology of the spectrum of some interesting noetherian rings by describing only the closures of points. Zev’s answer shows that they are right after all: this tells the whole story, because any closed set is a finite union of such closures.
EDIT 2. For the sake of completeness here is a proof of the following statement:
(*) A noetherian ring contains only finitely many minimal prime ideals.
Let $A$ be our noetherian ring, and let $R$ be the set of those ideals $\mathfrak a$ of $A$ which are equal to their radical $r(\mathfrak a)$. We'll use the following facts:
(a) $r(\mathfrak a\mathfrak b)=r(\mathfrak a\cap\mathfrak b)=r(\mathfrak a)\cap r(\mathfrak b)$ for all ideals $\mathfrak a,\mathfrak b$; in particular $\mathfrak a,\mathfrak b\in R\Rightarrow$ $\mathfrak a\cap\mathfrak b\in R$,
(b) if $\mathfrak p\in R$ and if the conditions $\mathfrak a,\mathfrak b\in R$ and $\mathfrak p=\mathfrak a\cap\mathfrak b$ imply $\mathfrak p=\mathfrak a$ or $\mathfrak p=\mathfrak b$, then $\mathfrak p$ is prime.
Claim (a) is clear. We prove (b). Let $\mathfrak p$ satisfy the assumptions of (b) and suppose by contradiction that $\mathfrak p$ is not prime. By (a) there are ideals $\mathfrak c,\mathfrak d$ such that $\mathfrak p\supset\mathfrak c\cap\mathfrak d$, $\mathfrak p\not\supset\mathfrak c$, $\mathfrak p\not\supset\mathfrak d$.
Indeed, as $\mathfrak p$ is not prime, there are ideals $\mathfrak c',\mathfrak d'$ such that $\mathfrak p\supset\mathfrak c'\mathfrak d'$, $\mathfrak p\not\supset\mathfrak c'$, $\mathfrak p\not\supset\mathfrak d'$, and it suffices to set $\mathfrak c:=r(\mathfrak c')$, $\mathfrak d:=r(\mathfrak d')$. (Note $\mathfrak c\cap\mathfrak d=r(\mathfrak c'\mathfrak d')\subset\mathfrak p$ by (a).)
In particular, $\mathfrak p$ contains neither $\mathfrak a:=r(\mathfrak c+\mathfrak p)$ nor $\mathfrak b:=r(\mathfrak d+\mathfrak p)$, and we have by (a) $$ \mathfrak a\cap\mathfrak b=r((\mathfrak c+\mathfrak p)(\mathfrak d+\mathfrak p))\subset\mathfrak p\subset\mathfrak a\cap\mathfrak b, $$ and thus $\mathfrak p=\mathfrak a\cap\mathfrak b$. But the choice of $\mathfrak p$ implies $\mathfrak p=\mathfrak a$ of $\mathfrak p=\mathfrak b$, a contradiction. This proves (b).
To prove (*), let $I\subset R$ be the set of finite intersections of primes. We claim $I=R$.
To prove this equality, we argue again by contradiction, assuming $I\neq R$. Let $\mathfrak p$ be a maximal element of $R\setminus I$. In particular $\mathfrak p$ is not prime, and (b) implies that there are $\mathfrak a,\mathfrak b\in R$ such that $\mathfrak p=\mathfrak a\cap\mathfrak b$ and $\mathfrak a\neq \mathfrak p\neq\mathfrak b$. By maximality of $\mathfrak p$ we have $\mathfrak a,\mathfrak b\in I$, and thus $\mathfrak p\in I$, contradiction. This shows $I=R$.
In particular, the nilradical $\mathfrak n$, being in $R=I$, is the intersection of the primes $\mathfrak p_1,\dots,\mathfrak p_n$. Let $\mathfrak p$ be an arbitrary prime. As $\mathfrak p$ contains $\mathfrak n$, it contains one the $\mathfrak p_i$. This proves (*).
