The radical of the product of two ideals is the intersection of the radicals of the ideals

Question

Let $R$ be a Noetherian ring (not sure if "Noetherianness" is needed). I'm trying to show that $r(IJ)=r(I)\cap r(J)$ ($r$ for radical).

One direction looks easy. If $x\in r(IJ)$, then for some $k$, $x^k\in IJ$. Since $IJ\subset I\cap J$, $x^k\in I\cap J$, so $x\in r(I)\cap r(J)$.

The other direction is not clear to me. Let $x\in r(I)\cap r(J)$. Then $x^n\in I, x^m\in J$. So $x^{\max(n,m)}\in I\cap J$. If $I+J=R$, then $I\cap J\subset IJ$, and I would be able to conclude that $x\in r(IJ)$. But it need not be the casse that $I+J=R$. How do I proceed then?

Answer 1

Hint: What can you say about $x^m\times x^n$?