= Radical of a sum of ideals
{title2=$\sqrt{\sqrt I+\sqrt J}=\sqrt{I+J}$}
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 $1\in I+J$.
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