Real algebraic curves of $\mathbb R^2\sim\mathbb C$ containing all zeroes of infinitely many rational polynomials  Is there a nice description of all real algebraic curves of $\mathbb R^2$ 
which have the property that every such curve contains all zeroes (after the obvious identification of $\mathbb C$ with $\mathbb R^2$) of infinitely many monic rational polynomials of $\mathbb Q[z]$ having only simple roots.
Examples are the real line, the unit circle and there are many hyperelliptic curves
(obtained by taking preimages of such curves with respect to suitable rational functions).
(It is of course easy to construct also non-simple examples of such curves, eg. by considering 
all real lines determined by sets of zeroes of cyclotomic polynomials.)
I guess that there exists a curve $\mathcal C\subset \mathbb R^2\sim\mathbb C$ 
defined by a polynomial in $\mathbb Z[x,y]$
which contains the rootsets of only finitely many rational monic polynomials without multiple roots. Can someone exhibit such a curve?
 A: Yes. There exist many such curves:
For any rational monic polynomial without multiple roots if $x+iy$ is a root, then $x-iy$ is a root as well. So if your curve $C$ satisfies $y\not =0,\\ (x,y)\in C \rightarrow (x,-y)\not\in C$ then it cannot contain the rootset of any polynomial with complex roots. To finish, make the curve have finitely many real roots. For example: $$C: x-y=0$$

As for your first question, it seems hard to characterize all such curves. There two types of families I have thought of. Of course you can multiply curves that satisfy the property, so let us only talk about minimal curves with respect to the property.


*

*As you mention, the real line.

*Genus 0 or 1 curves with a rational point of infinite order and the required property (mentioned in comments) that $y$ appears only with even powers. This uses the simple fact that conjugate numbers in $\mathbb{Q}(\sqrt{-1})$ have minimal polynomial over the rationals. If we want to extend this to higher genus, there is the following question: Given a rational class in the jacobian of a curve, does it have infinitely many multiples such that $a(x)$, in its representation as a reduced divisor $[(a(x), b(x,y)]$, has only real roots? If so, then the property holds for all curves with a rational point on the jacobian of infinite order.

*The unit circle fits into the above. But as you mention, there is another way, which we can generalize: curves of the form $||f(z)||=1, z=x+iy$. The question is now: how many points in the union of all CM fields does this equation have? If I had to guess, infinitely many, so I think these probably satisfy the property.
