Show that \(\text{rad}(IJ) = \text{rad}(I) \intersect \text{rad}(J)\)
Solution
1. One has \(\operatorname{rad}(IJ)\subseteq\operatorname{rad}(I)\cap\operatorname{rad}(J)\).
Since \(IJ\subseteq I\) and \(IJ\subseteq J\), monotonicity of radicals gives \[ \operatorname{rad}(IJ)\subseteq\operatorname{rad}(I), \qquad \operatorname{rad}(IJ)\subseteq\operatorname{rad}(J). \] Hence \[ \operatorname{rad}(IJ)\subseteq\operatorname{rad}(I)\cap\operatorname{rad}(J). \]
2. One has \(\operatorname{rad}(I)\cap\operatorname{rad}(J)\subseteq\operatorname{rad}(IJ)\).
Let \(x\in\operatorname{rad}(I)\cap\operatorname{rad}(J)\). Then for some integers \(m,n\ge1\), \[ x^m\in I, \qquad x^n\in J. \] Therefore \[ x^{m+n}=x^m x^n\in IJ, \] so \(x\in\operatorname{rad}(IJ)\).
3. Therefore\[ \operatorname{rad}(IJ)=\operatorname{rad}(I)\cap\operatorname{rad}(J). \]
Combine 1 and 2.