Let $I$ be a ideal of $k[x_1,...,x_2]$. Show that $\sqrt{I} \subset I(Z(I))$.
Observing that $J \subset I(Z(J))$, for any ideal $J$ of $k[x_1,...,x_2]$ and that $Z(\sqrt{I})=Z(I)$, we have completed the desired.
However, the problem says: Give an example where inclusion is proper.
And I'm not being able to end the problem. Anyone have a tip?