Closure of quasi-projective variety: if $X=V(I)\setminus V(J)$, must $\overline{X}=V(I)$?

Question

Let $X=V(I)\setminus V(J)$ in a complex projective space $\Bbb P^n$, where $I,J$ are ideals of complex polynomials in $n+1$ variables and $V(\ldots)$ their common zeros. I mean $X$ is locally closed set in the Zariski topology. I was told: the closure of X is $V(I)$ and $V(J)$ is equal to that closure ‘less’ $X$. For me it seems false, I mean I can only say that closure is inside $V(I)$.

Answer 1

You're correct that if one write $X=V(I)\setminus V(J)$, it is not necessarily true that $\overline{X}=V(I)$, but given a locally closed subset $X$ one can always choose $I,J$ so that $\overline{X}=V(I)$: since $\overline{X}$ is closed, it is of the form $V(I')$ for some homogeneous ideal $I'$, and then you can write $\overline{X}\setminus X = V(I')\setminus V(J+I')$, as $V(J)\cap V(I') = V(J+I')$.

Here's a counterexample: let $X=V(xy)\setminus V(y)\subset\Bbb P^2$. Then $\overline{X}=V(x)\neq V(xy)$, so you can write $X=V(x)\setminus V(x,y)$.