When do we have $Rad(I)=I$ for an ideal $I$ of a ring $R$? This is kind of a follow-up question about calculating the radical of an ideal. Since

$Rad(I)$ is the intersection of all the prime ideals of $R$ that contain $I$,

which is a property I learned from this article in wikipedia, we have that 
$$
Rad(I)=I
$$
whenever $I$ is a prime ideal. My question is:

Can this be true for some $I$ which is not a prime ideal? [EDIT: And when is this NOT true?] Is there an equivalent easy-to-check conditions for this kind of $I$?

Let $R={\Bbb Z}[x]$, for example. $I=\langle x,2\rangle$ is a prime ideal and thus $Rad(I)=I$. For any ideal $I\unlhd R$, (say $I=\langle x^2+1\rangle$ or $I=\langle x^2+2\rangle$, etc.) the key point is to check
$$
Rad(I)\subset I
$$
since $I\subset Rad(I)$ is always true. But I don't know a quick way to check this relation. 
 A: You can, in fact, have $I=\mathrm{Rad}(I)$ for nonprime $I$. For example, consider the ideal $\langle xy\rangle$ of the ring $\mathbb{Q}[x,y]$. This isn't prime since the generator isn't irreducible, but can easily be seen to be radical.
More generally, if you take a monomial ideal, i.e. an ideal generated by monomials, in a polynomial ring over some field, its radical will be generated by the "roots" of those monomials. E.g. if $I=\langle x^2y^4,z^3\rangle$, then $\sqrt{I}=\langle xy^2,z\rangle$.
A: By definition, every intersection of a set of prime ideals is semiprime (aka a radical ideal).
Let $\cap P_i=I$, where the $P_i$ are all prime. Then the set of all prime ideals containing $I$ contains the $P_i$. Thus, $\cap\{P\mid P\supseteq I\}\subseteq \cap P_i$. The left hand intersection involves "more" prime ideals, and so the intersection of more ideals should be smaller than just the set of $P_i$.
Thus in total:
$$
\cap P_i=I\subseteq\cap\{P\mid P\supseteq I\}\subseteq \cap P_i
$$
So, there is equality all across.
It is relatively easy to find examples where prime ideals do not intersect to a prime ideal. For example, the intersection of two prime ideals, neither of which contains the other, cannot be prime. (Explain why!)
