= Solution
{c}
The <radical of an ideal> $I$ is
$$
\boxed{\sqrt I=\operatorname{rad}(I)=\{a\in R:a^n\in I\text{ for some }n\geq1\}.}
$$
We first establish the properties needed below. If $a^m,b^n\in I$, every term of the <binomial theorem> expansion of $(a+b)^{m+n-1}$ contains either $a^m$ or $b^n$, so $a+b\in\sqrt I$. Also $(-a)^m\in I$ and $(ra)^m=r^ma^m\in I$. Thus $\sqrt I$ is an <ideal>. The definition gives $I\subseteq\sqrt I$ and shows that $I\subseteq J$ implies $\sqrt I\subseteq\sqrt J$. Finally, if $a^n\in\sqrt I$, then $(a^n)^m\in I$ for some $m$, so $a\in\sqrt I$. Hence $\sqrt{\sqrt I}=\sqrt I$; in other words, $\sqrt I$ is a <radical ideal>.
We prove the stronger fact that the <radical of an ideal is the intersection of the prime ideals containing it>:
$$
\sqrt I=\bigcap_{\substack{P\supseteq I\\P\text{ prime}}}P.
$$
If $a\in\sqrt I$ and $P\supseteq I$ is a <prime ideal>, then $a^n\in P$ for some $n$, and repeated application of the definition of a <prime ideal> gives $a\in P$. This proves one inclusion.
Conversely, fix $a\notin\sqrt I$. The <multiplicatively closed set> $T=\{1,a,a^2,\ldots\}$ is disjoint from $I$. Consider the <ideals> containing $I$ and disjoint from $T$. This collection is nonempty, and the union of any chain in it is another such <ideal>. By <Zorn's lemma>, there is a maximal member $P$. It is proper because $1\in T$.
To prove that $P$ is a <prime ideal>, suppose $bc\in P$ but $b,c\notin P$. Maximality forces both $P+(b)$ and $P+(c)$ to meet $T$. Thus, for some $u,v\geq0$,
$$
a^u=p+rb,\qquad a^v=q+sc,
\qquad p,q\in P,\quad r,s\in R.
$$
Multiplication gives $a^{u+v}\in P$, since every term lies in $P$, including $rsbc$. This contradicts disjointness from $T$. Therefore $P$ is a <prime ideal> containing $I$ and avoiding $a$. This proves the other inclusion, including the required <prime ideal avoiding a multiplicative subset> argument without using any unproved facts about the <nilradical>.
If $I=\sqrt I$, the displayed intersection represents $I$ as required. Conversely, an <intersection of ideals> that are all <prime ideals> is a <radical ideal>: if $a^n$ belongs to every one of them, then so does $a$. Consequently
$$
\boxed{I=\sqrt I\iff I\text{ is an intersection of prime ideals}.}
$$
The <unit ideal> $I=R$ is included by interpreting the empty intersection as $R$.
Back to article page
