 Solution

If $\sqrt{I}+\sqrt{J}=R$, write $1=a+b$ with $a\in \sqrt{I}$ and $b\in \sqrt{J}$. Choose $m,n\ge 1$ such that $a^m\in I$ and $b^n\in J$. In the binomial theorem expansioneach term has either at least $m$ factors $a$, or at least $n$ factors $b$. Hence every term belongs to the sum of ideals $I+J$, andThus comaximal ideals can be recognized from their radicals of ideals.

For the general identity, $I+J\subseteq \sqrt{I}+\sqrt{J}$ givesOn the other hand, $I,J\subseteq I+J$ implies $\sqrt{I},\sqrt{J}\subseteq \sqrt{I+J}$. Since the latter is an ideal, $\sqrt{I}+\sqrt{J}\subseteq \sqrt{I+J}$. Taking radicals of ideals and using the idempotence proved in a yieldsTherefore the radical of a sum of ideals satisfies