Compute the Zariski clousures $\overline{S} \subset \mathbb{A}^2(\mathbb{Q})$ of the following subsets:
(a) $S=\{(n^2,n^3):n \in \mathbb{N}\}\subset \mathbb{A}^2(\mathbb{Q})$;
(b) $S=\{(x,y): x^2+y^2<1\}\subset \mathbb{A}^2(\mathbb{Q})$;
(c) $S=\{(x,y): x+y \in \mathbb{Z}\}\subset \mathbb{A}^2(\mathbb{Q})$.
I've write (a) and I don't know if it's right. I don't know how to make b and c
(a) Let $P(X,Y)=X^3-Y^2 \in \mathbb{Q}[X,Y]$. $P$ is $0$ in every element of $S$, because $P(n^2,n^3)=(n^2)^3-(n^3)^2=0, \forall n \in \mathbb{N}.$ Fixed $x_0=n^2 \in \mathbb{N}$, $ P(x_0,Y)$ is $0$, at least, in $y_0=n^3$, so $Y-y_0|P(x_0,Y)$. Hence, the number of roots of $P(x,y)$ it's the same than $Card(\mathbb{N})$. $\mathbb{Q}$ is a field, so every polinomial with degree $d$ has $d$ roots. So neccesarily $P(x,y)$ is $0$, so $\overline{S}=\{(a_1,a_2)\in \mathbb{A}^2(\mathbb{Q}): f(a_1,a_2)=0, \forall f \in I(S) \}=\mathbb{Q}^2.$ $$ $$