An example of a variety of an ideal of a variety...?

Question

I am asked to show an example of some $S \subset k^n$ such that $V(I(S)) \ne S$.

So basically, $S$ is some set of points in $k^n$ a field of $n$ dimensions, and an affine variety, I guess. I've been thinking of some simple ones say in $k[x]$ or $k[x,y]$ but it's making me start to think there's no such thing(while that's obviously wrong)

Can someone give me an example...? I just cannot imagine how an affine variety's ideal's variety can be different from the original variety and why...

Answer 1

For example with $n=1$, take $S$ to be the set of zeroes of some polynomial $P(x)$ with $P(0)\neq 0$, for any such $P$. Then you get everything except the origin in your set. $I(S)=0$ and thus $V(I(S))=\mathbb{A}^1\neq S$.

Answer 2

Since $V(I(S))=\bar S$, the Zariski closure of $S$, any non Zariski closed subset of $k^n$ will yield a counterexample.
[Note carefully that algebraic closedness of $k$ is not assumed]
Explicitly, if you take $k=\mathbb C$ and $S=\mathbb Z\subset k^1=\mathbb C^1$, a non Zariski closed subset, you will get $V(I(\mathbb Z))=V(0)=\mathbb C^1$.