Let $I, J \subseteq K[x_0, \dotsc, x_n ]$ with $I \subseteq J$, then I have $V(J) \subseteq V(I)$ (looking at it from the affine space).
Now I was wondering whether also the converse direction holds, i.e. for $V(J) \subseteq V(I)$, I can deduce that $I \subseteq J$?
So far, I've always only seen it stated in one direction.
It would be great if anybody could tell whether or under which circumstances also the other directions holds. Thank you!
The converse is not necessarily true. Consider the ideals $$J=((x-1)(x-2)^2),~I=((x-1)(x-2)(x-3))\subseteq K[x]$$
Then $$V(J)=\{1,2\}\subset V(I)=\{1,2,3\}$$ but $$I\not\subset J$$
I think you can also see why this is a problem, it's because of $(x-2)^2$, the square. It doesn't change the set of roots, but it makes the ideal smaller.
What is true, is that if $K$ is algrebraically closed, then $$V(J)\subseteq V(I)\rightarrow \sqrt I\subseteq\sqrt J$$ because then we have no square factors. This can be proven using Hilbert's Nullstellensatz.
Indeed, $\sqrt J=((x-1)(x-2))$ and $\sqrt I=((x-1)(x-2)(x-3))$, so $$\sqrt I\subseteq \sqrt J$$
An even simpler counter-example:
Consider the special case in which $V(I)=V(J)$ (that is $V(I)\subseteq V(J)$ and $V(J)\subseteq V(I)$), if the conserve was true this would imply $J\subseteq I$ and $I\subseteq J$, that is, $I=J$. Now take $$I=(x-1)^2,~J=(x-1)\subseteq K[x]$$
$V(I)=V(J)$, but $I\neq J$.