Does every closed set of prime ideals of a noetherian ring contain a finite dense subset?

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 fact that a noetherian ring $A$ contains only finitely many minimal prime ideals. 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) $r(\mathfrak a\mathfrak b)=r(\mathfrak a\cap\mathfrak b=r(\mathfrak a)\cap r(\mathfrak b)$; 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. Let $\mathfrak p$ satisfy the assumptions of (b) and suppose that $\mathfrak p$ is not prime. By (a) there are ideals $\mathfrak c,\mathfrak d$ such that $\mathfrak p\supset\mathfrak c\cap\mathfrak d$, $\mathfrak p\not\supset\mathfrak c$, $\mathfrak p\not\supset\mathfrak d$. Then $\mathfrak p$ contains neither $\mathfrak a:=r(\mathfrak c+\mathfrak p)$ nor $\mathfrak b:=r(\mathfrak d+\mathfrak p)$. To get a contradiction it suffices to prove $\mathfrak p=\mathfrak a\cap\mathfrak b$. The inclusion $\mathfrak p\subset\mathfrak a\cap\mathfrak b$ is obvious, and the reverse inclusion follows from (a).

Let $I\subset R$ be the set of finite intersections of primes. As the existence of a maximal element of $R\backslash I$ would contradict (b), we have $I=R$. In particular, the nilradical $\mathfrak n$ is the intersection of the primes $\mathfrak p_1,\dots,\mathfrak p_n$. Let $\mathfrak p$ be a prime. As $\mathfrak p$ contains $\mathfrak n$, it contains one the $\mathfrak p_i$.

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For any closed $S\subseteq\text{Spec}(R)$, the set of minimal elements of $S$ with respect to inclusion is certainly dense in $S$, but I am not sure how to prove it is always finite. – Zev Chonoles Jun 24 '11 at 6:23
@Zev Chonoles - Thank you very much for the comment you just made, and also for the one you deleted. (To the other readers: I had forgotten the closedness assumption, without which the question is silly. Zev very kindly pointed that out to me in a comment that he deleted after my editing the question.) – Pierre-Yves Gaillard Jun 24 '11 at 6:30
@Pierre: It's no problem. I think I have got the answer below, the noetherian hypothesis looks like it guarantees the finiteness. – Zev Chonoles Jun 24 '11 at 6:36
@ Zev. Yes this should follow from primary decomposition. – jspecter Jun 24 '11 at 6:39
@jspecter: Ah yes. I should really review that section of Atiyah-Macdonald more, I am very rusty... – Zev Chonoles Jun 24 '11 at 6:43
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The answer is yes. Given a closed set $S\subseteq\text{Spec}(R)$, there is an ideal $I\subseteq R$ for which $S=V(I)$ (this is just the definition of the Zariski topology). Because $R$ is noetherian, there are only finitely many primes minimal over $I$; let the set of minimal primes of $I$ be $T$. We have that $T\subseteq S$, and that every element of $S$ contains some element of $T$, so that by the definition of the Zariski topology, we have that $\overline{T}\supseteq S$; but because $S$ is closed, we have $\overline{T}=S$.