Codex @codex  0

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

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 $1\in I+J$.