Hilbert's Nullstellensatz asserts, that if $k$ is an algebraically closed fields, $I \subset k[x_1,\dots,x_n]$ is an ideal, then $I(V(I)) = \sqrt{I}$. So every polynomial that vanishes on the variety defined by $I$ is actually in the radical of $I$.
Obviously we always have the inclusion $\sqrt{I} \subset I(V(I))$.
Of course this is not true, if $k$ is not algebraically closed, the standard example is $f = x^2 +1 \in \mathbb{R}$. Then $V(f) = \emptyset$, i.e. $I(V(f)) = k[x]$.
My question is: Are there any other examples where the Nullstellensatz does not hold, but $I$ actually has zeroes? Or is this the only class of counterexamples?
