I have to prove that:
The closed sets in Zariski topology of $\mathbb{A}_\mathbb{C}^2$ are the sets of form $Z(g)\cup F$, where $g\in \mathbb{C}[x,y]$ and $F$ is a finite set.
So, I thinked the following:
We have $Z(\langle g,f\rangle)=Z(\langle g\rangle +\langle f\rangle )=Z(g)\cap Z(f)$.
Note that we have two options:
1) There's a polynomial $m_{(g,f)}$ with degree $>0$ s.t. this is a "maximal divisor common" of $g$ and $f$. So, $g=m_{(g,f)}g_1$ and $f=m_{(g,f)}f_1$, where there's not a polynomial $m_{(g_1,f_1)}$ with degree $>0$ s.t. this is a "maximal divisor common" of $g_1$ and $f_1$ (ie, they are relatively irreducible).So $Z(g)\cap Z(f)=Z(m_{(g,f)}\langle g_1,f_1\rangle)=Z(m_{(g,f)})\cup (Z(g_1)\cap Z(f_1))$.
2) There's not a polynomial $m_{(g,f)}$ with degree $>0$ s.t. this is a "maximal divisor common" of $g$ and $f$ (ie, they are relatively irreducible). So, the zeros of the two are at most the product of the degrees (intersections of curves), so finite.
In any case, $Z(\langle g,f\rangle)=Z(h)+F$. The completely answer is by induction.
Is it correct?
I have doubts at language and also at this point: Is it possible determine (or at least know the existence) of some set $S$ of polynomials s.t. $Z(S)=F$ for given $F$?
Many thanks!