= Solution
{c}
If $\sqrt I+\sqrt J=R$, write $1=a+b$ with $a\in\sqrt I$ and $b\in\sqrt J$. Choose $m,n\geq1$ such that $a^m\in I$ and $b^n\in J$. In the <binomial theorem> expansion
$$
1=(a+b)^{m+n-1},
$$
each term has either at least $m$ factors $a$, or at least $n$ factors $b$. Hence every term belongs to the <sum of ideals> $I+J$, and
$$
\boxed{\sqrt I+\sqrt J=R\ \Longrightarrow\ I+J=R.}
$$
Thus <comaximal ideals> can be recognized from their <radicals of ideals>.
For the general identity, $I+J\subseteq\sqrt I+\sqrt J$ gives
$$
\sqrt{I+J}\subseteq\sqrt{\sqrt I+\sqrt J}.
$$
On 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 <2/a> yields
$$
\sqrt{\sqrt I+\sqrt J}\subseteq\sqrt{\sqrt{I+J}}=\sqrt{I+J}.
$$
Therefore the <radical of a sum of ideals> satisfies
$$
\boxed{\operatorname{rad}(\operatorname{rad}(I)+\operatorname{rad}(J))=\operatorname{rad}(I+J).}
$$
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