= Radical of an ideal is the intersection of the prime ideals containing it
{title2=$\sqrt I=\bigcap_{P\supseteq I,\ P\text{ prime}}P$}
For any <ideal> $I$ in a <commutative ring>, every <prime ideal> containing $I$ contains $\sqrt I$, because $a^n\in P$ implies $a\in P$. Conversely, if $a\notin\sqrt I$, the <multiplicatively closed set> $T=\{1,a,a^2,\ldots\}$ is disjoint from $I$. By <Zorn's lemma>, there is an <ideal> $P\supseteq I$ maximal among those disjoint from $T$.
This $P$ is a <prime ideal>: if $bc\in P$ but $b,c\notin P$, maximality makes both $P+(b)$ and $P+(c)$ meet $T$. Multiplication of representatives from these two intersections puts an element of $T$ in $P$, a contradiction. Thus there is a <prime ideal> containing $I$ but avoiding $a$, establishing the identity. For $I=R$, the intersection is empty and means $R$. It follows that an <ideal> is a <radical ideal> exactly when it is an <intersection of ideals> that are all <prime ideals>.
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