Suppose some algebraic curve is given by $\{F=0\}$, $F(X,Y)\in \mathbb{R}[X,Y]$. I am wondering whether we can use the existence of a solution of $F = 0$ when viewed as an element of $(\mathbb{R}[X])[Y]$ to determine whether or not $\{F=0\}$ is not closed. (When I say closed I do not mean in the topological sense, but rather that the curve 'ends at the same point it starts')
Specifically, since $F\in \mathbb{R}[X,Y]$, and $\mathbb{R}[X,Y] \simeq (\mathbb{R}[X])[Y]$, $F$ may be viewed as a polynomial over the indeterminate $Y$, with coefficients that are polynomials over $X$. Suppose I identify $F\in \mathbb{R}[X,Y]$ with $\bar{F} \in (\mathbb{R}[X])[Y]$ (in the obvious way). I think that if $\bar{F}(a(x)) = 0$ for any polynomial $a(x) \in \mathbb{R}[x]$, then $\{F=0\}$ is not closed.
My justification for this is that $$\{F=0\} = \{(x,y)\in \mathbb{R}^2: F(x,y) = 0\}$$ But if $\bar{F}(a_0(x)) = 0 \in \mathbb{R}[X]$, then for every real number $x$, the tuple $(x,(a_0(x))\in \{F=0\}$ since $\bar{F}(a_0(x))$ is the zero polynomial. But this means that the curve must extend off in the "x-axis" to infinity as the curve is defined "for every point $x$".
Is this reasoning correct? I am not sure if this is meant to be trivial, or if I am missing something and the statement is completely wrong.