I am studying ring theory and trying to solve this problem:
Let $R$ be a commutative ring. For an ideal $I$ of $R$, let \begin{equation*} \sqrt{I} = \{x\in R\mid x^n \in I \text{ for some } n\geq 1\}. \end{equation*}
Show that $\sqrt{IJ} = \sqrt{I\cap J}$ for any two ideals $I$ and $J$.
We need to show both sides of inclusion. I first let $x\in\sqrt{IJ}$. Then there exists $n_1\geq 1$ such that $x^{n_1}\in IJ$. Now I need to show that there exists $n_2$ such that $x^{n_2}\in I\cap J$ and I am stuck at here. Can anyone help me proceed the proof?
By definition of ideals, and product of ideals, $IJ\subseteq I\cap J$. So if $x^{n}\in IJ$ for some integer $n$, then certainly $x^{n}$ is in $I\cap J$ too.
The other direction isn't as immediate. However, note that $(I\cap J)(I\cap J)\subseteq IJ$, which means that if $x^{n}\in I\cap J$ for some integer $n$, then $x^{2n}\in IJ$.