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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 (*).

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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
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@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
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@ 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

1 Answer 1

up vote 2 down vote accepted

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 $\overline{T}=S$.

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Thanks! This looks correct to me. I upvoted your answer, and I'd like to accept it, but I don't know how to do that. Can anybody help me? [Minor typo in your answer: you wrote "hacve".] –  Pierre-Yves Gaillard Jun 24 '11 at 6:56
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@Pierre: Thanks for pointing out the typo, I've fixed it. Also, this link is the canonical explanation of how to accept an answer; there is a little check mark below the upvote/downvote buttons that turns green when you move your cursor over it; clicking it will accept the answer. –  Zev Chonoles Jun 24 '11 at 7:07

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