How to prove a commutative, with unit, Noetherian ring $A$ only has finitely many minimal prime ideals via the following step?

I have proved:

Step. All radical ideals of Noetherian ring $A$ can be expressed as an intersection of finitely many prime ideals.

How to continue?

  • 1
    $\begingroup$ Then how to prove? $\endgroup$
    – Shiquan
    Oct 14, 2013 at 11:15

2 Answers 2


If you know that radical ideals are finite intersections of primes, in particular $\sqrt{(0)}=\mathfrak p_1\cap\cdots\cap\mathfrak p_n$. Let $\mathfrak p$ be a minimal prime. Since $(0)\subseteq\mathfrak p$ we get $\sqrt{(0)}\subseteq\sqrt{\mathfrak p}=\mathfrak p$, that is, $\mathfrak p_1\cap\cdots\cap\mathfrak p_n\subseteq\mathfrak p$. It follows that there exists $\mathfrak p_i\subseteq\mathfrak p$ and since $\mathfrak p$ is minimal we must have $\mathfrak p_i=\mathfrak p$. (In other words, the minimal primes of $A$ are among the primes $\mathfrak p_1,,\dots,\mathfrak p_n$.)

  • $\begingroup$ "radical ideals are finite intersections of primes"? This is not true in general, is it? $\endgroup$
    – metalder9
    Jan 14 at 20:54
  • 1
    $\begingroup$ I believe you need the Noetherian condition for finiteness, hence the statement can be confusing but this is referring to radical ideals of Noetherian rings. $\endgroup$
    – metalder9
    Jan 16 at 3:34

Not sure about the approach you're taking but here is how one usually proves this fact.

Exercise 1: Show that in a ring $A$, the irreducible components of $\operatorname{Spec} A$ are in bijection with the minimal primes $\mathfrak{p}$. The bijection is given by $\mathfrak{p} \mapsto V(\mathfrak{p})$.

Exercise 2: Show if $A$ is Noetherian that $\operatorname{Spec} A$ is Noetherian. That is, it satisfies the DCC on closed subsets.

Exercise 3: Show that $\operatorname{Spec} A$ is the finite union of its irreducible components. (Hint: use the fact that every non-empty collection of closed sets ordered by inclusion in a Noetherian space has a minimal element).

Alternative approach: In a Noetherian ring $A$, $\text{Ass}(A)$ is a finite set. In a Noetherian ring every minimal prime is associated from which it follows that the set of minimal primes is finite too.

  • 3
    $\begingroup$ Maybe you want to say "how geometers usually proves this fact". In CA the simplest way is to show that in a noetherian ring $(0)$ is a product of finitely many primes which follows, in this case, from Step 2. $\endgroup$
    – user26857
    Oct 14, 2013 at 9:26
  • $\begingroup$ @YACP Or an alternative approach: In a Noetherian ring, the associated primes form a finite set. Every minimal prime is associated. Q.E.D. $\endgroup$
    – user38268
    Oct 14, 2013 at 13:00
  • $\begingroup$ @user38268 You don't even need to bring in associated points. In any Noetherian space, the number of irreducible components is finite. But, if $A$ is Noetherian, then $\text{Spec}(A)$ is a Noetherian space, and thus it has finitely many irreducible components. But, since minimal primes in $\text{Spec}(A)$ correspond to irreducible components, it follows that there are finitely many minimal primes. $\endgroup$ Oct 15, 2013 at 9:44
  • $\begingroup$ @AlexYoucis Those are the exercises (1) - (3) in my answer above. $\endgroup$
    – user38268
    Oct 15, 2013 at 11:39
  • $\begingroup$ @user38268 Ah, yes, I'm sorry--I wrote that late last night :) (+1) But, moving to associated points is an over complication, your original idea is much cleaner. $\endgroup$ Oct 16, 2013 at 4:18

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