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This is a problem from an algebra textbook.

Let $R$ be a ring, and $I$ be an ideal. If the radical of $I$ is $I$ itself, i.e. $\operatorname{rad}(I) = I$, then $I$ is an intersection of prime ideals.

My friend and I cannot figure out how to prove it.

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What are you taking to be the definition of $rad(I)$? There are a few equivalent ways to define it. –  Matt Jul 4 '11 at 1:22
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2 Answers

up vote 8 down vote accepted

Hint: Show that the radical of the zero ideal (the nilradical) is the intersection of all prime ideals of $R.$ Then apply this to $R/I$ and use the fourth isomorphism theorem to obtain your desired result.

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Here's my attempt. If you didn't want an actual solution then don't read this. Please correct me if I made any mistakes! I'm using the definition $\mbox{rad}(I) = \{ r \in R \,|\,\,\, r^k \in I \text{ for some } k \in \mathbb N \}$.

Let $P$ be a prime ideal containing $I$. If $r \in R$ is such that $r^k \in I$, then $r^k \in P$, so $r \in P$ since $P$ is prime. Thus $\mbox{rad}(I) \subset \bigcap_{P \supset I} P$.

Conversely, if $r \notin \mbox{rad}(I)$, then $r^k \notin I$ for any $k$, so $S = \{1, r, r^2, \ldots \}$ is a multiplicatively closed set disjoint from $I$. By a basic theorem on prime ideals, we have that $R \smallsetminus S$ contains a prime ideal $P_r$ containing I. Since $r \notin P_r$, we have $r \notin \bigcap_{P \supset I} P$. Therefore $\mbox{rad}(I) = \bigcap_{P \supset I} P$.

The problem posted is the special case where $I = \mbox{rad}(I)$.

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Thinking about it, I think this might only work for commutative rings.. –  J.D. Mohr Jul 4 '11 at 3:32
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