Let $Y \to X$ be a finite branched cover of smooth projective curves over $\mathbb{C}$, so we get a finite extension $K(Y)/K(X)$ where $K(\ )$ is the field of meromorphic functions. Say that $Y \to X$ is solvable if the normal closure of $K(Y)$ over $K(X)$ is a solvable extension of $K(X)$.


I was just wondering this morning: If $Y$ is a generic Riemann surface of some fixed genus $g$, is $Y$ a solvable branched cover of $\mathbb{P}^1$? A generic Riemann surface has gonality $\lceil g/2 \rceil+1$, so we need to go up to at least genus $7$ in order to make sure that there isn't a map to $\mathbb{P}^1$ of degree $\leq 4$, but after that I didn't see any tools to use.



Is a generic genus $g \geq 7$ curve a solvable cover of $\mathbb{P}^1$?
The answer is negative by results of Zariski (weak version) and the stronger result by Guralnick and Neubauer in Theorem B of [Monodromy groups of branched coverings: the generic case](https://doi.org/10.1090/conm/186/02190). They show that for $g\ge7$, the space $\cal M\_g(\text{Sol})$ of all curves of genus $g$ admitting a solvable cover to $\mathbb P^1$ isn't dense in the space $\cal M\_g$ of all genus $g$ curves.


