About Zariski closure Let $A=\left\{\left(a^{2}+1, a^{3}+a\right) \mid a \in \mathbb{R}\right\} \subset \mathbb{R}^{2}$.

Find Zariski closure of $A$ .

I think Zariski closure of $A$ is $\mathbb{R}^{2}$ because let $ u=a^{2}+1$   and $ v= a^{3}+a  $ and let $ f(x,y) \in \mathbb{R}[x,y]$ then $f(u,v) =0$ now we must show $f$ is zero polynomial .
 A: Note that for any $a \in \Bbb R$:
$$(a^2 + 1)^3 - (a^2+1)^2 - (a^3+a)^2 = \\
a^6 + 3a^4 + 3a^2 + 1 - (a^4 + 2a^2 +1) - (a^6+ 2a^4 + a^2) = 0$$ so $A \subseteq \{(x,y)\mid x^3 - x^2 - y^2 = 0\}$ which is a Zariski closed subset by definition.
So the closure is at most $Z(x^3 - x^2- y^2)$, maybe even equal to it, thought I cannot show it yet. It probably is, though.
A: Not sure about my solution, but here's what I thought:
As others already mentioned, if we set $A(k) = \{(a^{2} + 1, a^{3} + a) | a \in k\}$ and $B(k) = \{(x, y)| y^{2} - x^{3} -x^{2} =0, (x, y) \in k^{2}\}$ for a field $k$, then we have $A(\mathbb{R}) \subset B(\mathbb{R})$. Unfortunately, two sets aren't the same since $(0, 0) \in B(\mathbb{R}) \backslash A(\mathbb{R})$. However, we can show $A(\mathbb{C}) = B(\mathbb{C})$: if $(x, y) \in B(\mathbb{C})$, then for $x\neq 0$, we have $(x, y) = (t^{2} + 1, t^{3}+ t)$ when $t = y/x$. For $x =0$, we can treat some cases by hands. What this shows is that the Zariski closure of $A(\mathbb{C})$ in $\mathbb{A}_{\mathbb{C}}^{2}$ is indeed $Z(y^{2} - x^{3} - x^{2})$. From this, we obtain that (which is the part that I'm not convinced enough) any polynomial $f(x, y) \in \mathbb{C}[x, y]$ with $f(t^{2} +1, t^{3} + t)=0$ for all $t\in \mathbb{C}$ can be divided by $y^{2} - x^{3} - x^{2}$, i.e. $f(x, y) = (y^{2}- x^{3} - x^{2})g(x, y)$ for some $g \in \mathbb{C}[x, y]$. *Now if $f(x, y) \in \mathbb{R}[x, y]$, then $f = \bar{f}$ (take conjugation of all coefficients) gives $g = \bar{g}$ and so $g \in \mathbb{R}[x, y]$, which concludes that the similar statement for real polynomial is also true.
