Radical of a sum of ideals ($\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{I+J}$)

In a commutative ring, the radical of an ideal is monotone and idempotent: inclusion $I\subseteq J$ gives $\sqrt{I}\subseteq \sqrt{J}$, and $a^n\in \sqrt{I}$ means $a^{nm}\in I$ for some $m$, giving $\sqrt{\sqrt{I}}=\sqrt{I}$. Since $\sqrt{I+J}$ contains both $\sqrt{I}$ and $\sqrt{J}$, it contains their sum of ideals. Taking radicals of ideals gives $\sqrt{\sqrt{I}+\sqrt{J}}\subseteq \sqrt{I+J}$. The reverse inclusion follows from $I+J\subseteq \sqrt{I}+\sqrt{J}$.

In particular, $\sqrt{I}+\sqrt{J}=R$ implies $I+J=R$. One can also see this directly: write $1=a+b$ with $a^m\in I$ and $b^n\in J$, then expand $(a+b{)}^{m+n-1}$ by the binomial theorem. Each term belongs to $I$ or $J$, so $1\in I+J$.