Proposition

For an arbitrary field, show that $\sqrt{IJ} = \sqrt{I \cap J}$. Give an example to show that the product of radical ideals need not be radical. Also give an example to show that $\sqrt{IJ}$ can differ from $\sqrt{I}\sqrt{J}$.

Solution

Since $IJ \subset I \cap J$, $\sqrt{IJ} \subset \sqrt{I \cap J}$.

Let $f \in \sqrt{I \cap J}$. Then $f^m \in I \cap J$ for some $m \geq 1$. This implies $f^m \in I$ and $f^m \in J$. Therefore, $(f^m)(f^m) \in IJ$, so $f^{2m} \in IJ$. Hence, $f \in \sqrt{IJ}$.

Let $I = J = \ev{x}$. Then $IJ = \ev{x^2}$, and $IJ$ is clearly not radical. Similarly, $\sqrt{IJ} = \ev{x} \ne \ev{x^2} = \sqrt{I}\sqrt{J}$.