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Prove that if $I$ is a radical ideal and $ab\in I$, then $I=\operatorname{rad}(I+(a))\cap \operatorname{rad}(I+(b))$.

Deduce that every radical ideal in a Noetherian ring is a finite intersection of primes.

I've done the first part (showing the sets are equal), but don't know how to do the second.

user26857
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user137090
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    As an additional note, every radical ideal is the intersection of the primes minimal over it. But in a Noetherian ring, there are always only finitely many primes minimal over a given ideal. – Siddharth Venkatesh Apr 26 '14 at 17:21
  • Is Noetherian needed? Seems like one can use Zorn's lemma to get that a radical ideal is a finite intersection of primes. –  Sep 29 '15 at 09:29

1 Answers1

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If $I$ is a radical ideal which is not prime, then there are elements $a,b$ with $ab \in I$ but $a \notin I$ and $b \notin I$. By the first part you can write $I$ as an intersection of two radical ideals $I_1, I_2$ containing $a$ and $b$ respectively. The same procedure can be applied for $I_1$ and $I_2$, and since the ring is Noetherian you will reach prime ideals after finitely many steps (otherwise there would be an infinite ascending chain of (radical) ideals).

Dune
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    I'm curious if others agree with me, but I don't think the parenthetical part is that obvious. Just because each chain of ideals is finite, how do you know this entire procedure terminates after finitely many steps? I think it's equivalent to saying that in a binary tree, if each branch is finite, then the tree is finite. I agree it's true but I think at least in a formal class setting it requires careful proof. – CJD Jul 17 '18 at 12:03
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    For others who may be wondering how to get around the snag @CJD brought up, the argument can be completed as follows: Since the ring is Noetherian, there would necessarily exist a radical ideal maximal with respect to the property of not being equal to the intersection of finitely many primes. Then $\sqrt{I+(a)}$ and $\sqrt{I+(b)}$ would both be the intersection of finitely many primes, which is a contradiction. – BHT May 13 '19 at 02:21
  • Proposition $4$ in here. – Bowei Tang Jun 17 '25 at 13:35