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Consider the collection $C$ of all ideals $I$ of $R$ such that $\sqrt{I}$ is not a finite intersection of prime ideals. Now using Zorn's lemma $C$ has a maximal element and that won't be prime. Continuing it seems we can get that radical of any ideal $I$ is a finite intersection of prime ideals. Where is my mistake?

I have seen this result stated with Noetherian hypothesis, but the above indicated proof does not use that hypothesis.

user26857
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    How did you prove that $C$ is inductively ordered? And what do you mean by "continuing"? – user26857 Sep 29 '15 at 12:33
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    Ok, I assumed the existence of maximal element $I$ by Zorn's and then observe that it is not prime and choose $x,y\notin I$ with $xy\in I$. Then just show that $\sqrt{I}=\sqrt{I+x}\cap \sqrt{I+y}$ is a finite intersection of prime ideals to get a contradiction. But I guess I cannot apply Zorn's lemma. –  Sep 29 '15 at 13:06

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Yes, it is.

Let $R=K^{\mathbb N}$ be the ring of sequences with elements in a field $K$. Then $R$ is reduced, and every prime ideal of $R$ is maximal. Now suppose that $(0)=\bigcap_{i=1}^nP_i$. Then, by CRT we get that $R\simeq\prod_{i=1}^nR/P_i$, that is, $R$ is isomorphic to a finite direct product of fields, so $R$ is noetherian, a contradiction.

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