Radical of an ideal equals the intersection of all prime ideals containing it.

I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $\sqrt{I}\subset N$. Proving the reverse containinment is not as simple. Given $x\notin \sqrt{I}$, consider the collection $\Omega$ of all ideals $J \supset I$ such that for all $n\in \Bbb N$, $x^n \notin J$. Partially order $\Omega$ by inclusion, and show by Zorn's lemma that $\Omega$ has a maximal element $\frak{p}$. Since $\mathfrak{p}\supset I$, if you can show that $\frak{p}$ is prime, then you can claim $x\notin N$ (since $x\notin \frak{p}$). Take $a,b\notin \frak{p}$, and using maximality of $\mathfrak{p}$, show that $\mathfrak{p} + (ab) \notin \Omega$; this will imply $ab\notin \mathfrak{p}$ and thus $\mathfrak{p}$ is prime.