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



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


