# Intersection of prime ideals

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.

• 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
• You can check Chapter 2 of the book "A Primer of Commutative Algebra" written by James S. Milne. – chenf Aug 18 at 12:21

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.

• For what special case, every radical ideal can be written as an intersection of maximal ideals? I have just seen the result, that for radical ideals of $k[x_1, x_2, \cdot \cdot x_n]$ where $k$ is an algebraically closed field, every radical ideal is the intersection of maximal ideals containing that ideal. Can you give me a simple argument why the result in this question reduces to maximal ideals for the ring mentioned above. – ramanujan_dirac Feb 18 '15 at 11:47

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 $I = \mbox{rad}(I)$.

• Thinking about it, I think this might only work for commutative rings.. – talkloud Jul 4 '11 at 3:32
• Why's that? I don't see what would fail in the non-commutative case. – MathManiac Aug 15 '16 at 14:16
• In the non-commutative case, $Rad (I)$ is not an ideal . See what happens when you try to prove that $ri \in I$ when $r \in R$ and $i \in I$ – Astrid A. Olave H. Aug 29 '16 at 4:57
• In the noncommutative setting, radical ideals are replaced by semiprime ideals, which are ideals $I$ such that if $J$ is an ideal such that $J^2\subseteq I$ then $J\subseteq I$. Equivalently, $I$ is semiprime if and only if $a\in R$ and $aRa\subseteq I$ imply $a\in I$. It is true that any semiprime ideal of a noncommutative ring is the intersection of the prime ideals which contain it (this is Nagata's lemma for the Baer radical). – Jose Brox Oct 28 '17 at 10:40