Let $k$ be an algebraically closed field and $C$ be a smooth projective curve over $k$. Let $p$ be a prime number. Does there exists a finite morphism $f : C \to \mathbb{P}^1$ such that the degree of the Galois closure of the field extension $k(\mathbb{P}^1) \to k(C)$ is coprime to $p$ ?

3$\begingroup$ This is just my gut feeling, but I suspect that the answer is "no". I suspect that for a very general curve $C$ of genus $2$, there is no finite morphism $f:C\to \mathbb{P}^1$ such that the Galois closure has odd order ($p=2$). Every finite, odd order group is solvable. So this begins to sound like that old chestnut about whether the maximal solvable extension is PAC. $\endgroup$ – Jason Starr Apr 6 '16 at 14:31

$\begingroup$ @JasonStarr The old chestnut involves curves defined over number fields, which are much harder, right? If I understand my argument correctly it shows the locus of solvable covers of $\mathbb P^1$ in the moduli space is contained in the union of (1) the locus of curves with nonsimple Jacobians, (2) the locus of curves with gonality lower than the generic value, and (3) countably many subvarieties of dimension at most 6. However (3) might contain all the number field points. $\endgroup$ – Will Sawin Apr 6 '16 at 16:37

$\begingroup$ @JasonStarr Actually not quite 6 but something small. $\endgroup$ – Will Sawin Apr 6 '16 at 16:43

$\begingroup$ @WillSawin. You are correct: the open problem is about number fields, not function fields. By deformationspecialization arguments, this question would follow from the (FALSE!) guess that every geom. integral scheme over $k(\mathbb{P}^1)$ has a solvable point (first disproved by Ambrus Pal, if memory serves). However suggestive it may be, that implication goes the wrong way to solve this problem. Your solution is a true solution. $\endgroup$ – Jason Starr Apr 7 '16 at 11:28

$\begingroup$ @JasonStarr Well there is another subtlety, which is that showing a curve cannot be a solvable cover of $\mathbb P^1$ does not imply it cannot be covered by a curve with a solvable cover of $\mathbb P^1$. I know think it's possible to prove that there exists a curve over $\mathbb Q$ that is not a cover of $\mathbb P^1$ with no solvable Galois group, but it still won't answer that question. $\endgroup$ – Will Sawin Apr 7 '16 at 12:41
Following Jason Starr's suggestion, let $C$ be a very general genus $2$ curve, forming a covering of degree $n$ of $\mathbb P^1$. Without loss of generality there is no intermediate curve between $C$ or $\mathbb P^1$ (it would have to have genus $0$ or $1$. In the first case, we may simplify by replacing $\mathbb P^1$ with that curve, and in the second case we get an elliptic factor in the Jacobian, contradicting "very general".)
Let $G \subseteq S_n$ be the Galois group of the covering. As Jason pointed out, it is solvable. It is also transitive and primitive. Because $G$ is solvable, it has "soluble socle", hence $n=p^d$ is a prime power and $G$ acts as a subgroup of the group of affine transformations of $\mathbb F_{p^d}$ by a fragment of the O'NanScott theorem (The argument goes that because $G$ is solvable, it has some normal subgroup $J$ isomorphic to $\mathbb F_p^d$ for some $d$. $J$ must act transitively, or else the action would not be primitive  the cover would factor through the cover defined y the orbits of $J$. Hence the $n$ points acted on are isomorphic to $\mathbb F_p^d$, and $G$ acts by transformations that normalize the group of translations, i.e. affine transformations).
Now the local monodromy around each ramification point is an affine transformation, hence has at most $p^{d1}$ fixed points. Because the ramification at the nonfixed points has order at least $3$, the local contribution to RiemannHurwitz is at least $(2/3) (p^{d}  p^{d1})$. Because the curve is very general, the number of ramification points is at least $6$, so that after moving the first three to $0,1,\infty$ we still have three parameters. So the total contribution to RiemannHurwitz is at least $4 (p^{d} p^{d1})$ and we obtain the inequality
$$2 = 22g \leq 2 p^d  4 (p^{d}  p^{d1}) = 4 p^{d1} 2 p^d $$
$$\left(1\frac{2}{p} \right) p^d \leq 1 $$
$p$ is at least $3$ so $1\frac{2}{p}$ is at least $1/3$ so $p^d$ is at most $3$ which is an obvious contradiction as for $3$coverings every point has full ramification and then we obtain $22g \leq 6  6 \cdot 2 =  6$.
Because this is a contradiction, a very general genus $2$ curve cannot be so represented.

$\begingroup$ Thank you for the very interesting answer. I don't understand how you showed that the local contribution to RiemannHurwitz is at least $(2/3)(p^d  p^{d1})$. $\endgroup$ – Steve Jenkins Apr 6 '16 at 16:33

$\begingroup$ @JohnWelliaveetil Each ramified point contributes $e1$ to RiemannHurwitz and $e$ to the number of sheets not fixed by RiemannHurwitz. Since $e$ is at least $3$, the former is at least $2/3$ times the latter. $\endgroup$ – Will Sawin Apr 6 '16 at 19:59

$\begingroup$ Thanks for the clarification ! I was reading through your solution again and realized that you only show that no "very general" curve of genus 2 can be a odd number degree Galois cover of $\mathbb{P}^1$. Does this imply that the degree of the Galois closure of any finite morphism from a very general genus 2 curve to $\mathbb{P}^1$ is divisible by 2 ? $\endgroup$ – Steve Jenkins Apr 7 '16 at 12:00

$\begingroup$ @JohnWelliaveetil It's not actually about Galois covers. My argument is about the Galois groups of general covers (i.e. finite morphisms). That's why I must consider arbitrary group actions on finite sets. $\endgroup$ – Will Sawin Apr 7 '16 at 12:42