 Radical of an ideal is the intersection of the prime ideals containing it

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 $T=\{1,a,a^2,\dots \}$ 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.