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.