Any finite number of curves over $\mathbb{Q}$ have a common cover

Question

Given a finite number of algebraic curves over $\mathbb{Q}$ is there a curve that covers all of them?

Answer 1

If we are talking about branched covers, yes: Let the curves be $C_1$, $C_2$, ..., $C_r$. Take the product $\prod C_i$, embed it into projective space, and intersect with a general codimension $r-1$ plane $H$. Then (by Bertini) $H \cap \prod C_i$ will be a smooth curve and, for $G$ chosen generically, the projection onto each of the $C_i$ will be nonzero.

If we are talking about unbranched covers, then the literal answer is no: Curves of genus zero have no non-trivial unbranched covers, so we'd need to exclude those. Also, for curves of genus one, all unbranched covers are also of genus one, so two curves of genus one have a common branched cover if and only if they are isogneous.

A harder question is if, given curves over $\mathbb{Q}$ of genuses $\geq 2$, they have a common unbranched cover.

I strongly expect the answer to be "no", and I can prove it for curves over $\mathbb{C}$. Fix integers $(d_1, d_2, g_1, g_2)$ with $d_1 (g_1-1) = d_2 (g_2-1)$ and let $X(d_1, d_2, g_1, g_2) \subset M_{g_1} \times M_{g_2}$ be the moduli space of pairs of curves $(C_1, C_2)$ with a common cover $C$, such that $C \to C_i$ is degree $d_i$. Given a curve $C_i$, it has only finitely many covers of given degree $d_i$, so the projection of $X(d_1, d_2, g_1, g_2)$ to either factor is finite to one, and thus the $\dim X(d_1, d_2, g_1, g_2) \leq \max \left( \dim M_{g_1}, \dim M_{g_2} \right) < \dim \left( M_{g_1} \times M_{g_2} \right)$.

So each $X(d_1, d_2, g_1, g_2)$ has positive codimension in $M_{g_1} \times M_{g_2}$, and a complex variety can't be the union of countably many positive dimensional subvarieties.

Over $\mathbb{Q}$, or even $\overline{\mathbb{Q}}$, this argument fails, but I would bet the conclusion is still right.