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Let $X$ be a hyperelliptic curve of genus at least two.

Let $Y\to X$ be a finite etale morphism with $Y$ connected.Then $Y$ is a smooth projective connected curve.

Is $Y$ hyperelliptic?

More generally, is $Y$ a Galois cover of the projective line?

I expect the answer to be no in general (to both questions). On the other hand, double etale covers of curves of genus two are curves of genus three which (I think) should be hyperelliptic.

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As pointed out by Stankewicz, the answer to tour first question is no.

Let me give an explicit example of a curve $F$ of genus $4$ which is an $\acute{\text{e}}$tale triple cover of a curve of genus $2$ but which is not hyperelliptic. The details of this construction can be found in my paper

Surfaces of general type with $p_g=q=1, K^2=8$ and bicanonical map of degree 2, Trans. Amer. Math. Soc. 358 (2006), no. 2, 759-798,

see in particular Section 4.

Let us consider the symmetric group $S_3$, presented as $$S_3 = \langle r, \, s \;| \; r^3=s^2=1, \, sr = r^2s \rangle$$ and the fuchsian group $\Gamma$ of genus zero and of presentation $$ \Gamma = \langle x_1, \ldots, x_6 \; | \; x_i^2=1, \; x_1x_2 \cdots x_6=1 \rangle.$$ Then the epimorphism $$x_1, \, x_2 \mapsto s, \quad x_3, \,x_4 \mapsto rs, \quad x_5, \,x_6 \mapsto r^2s$$ defines a Galois cover $F \longrightarrow \mathbb{P}^1$, with Galois group $S_3$, branched at $6$ points, such that the $3$-cycles $r$ and $r^2$ act on $F$ without fixed points. Then the Hurwitz formula gives $g(F)=4$.

Moreover:

$(i)$ we have $g(F/\langle r \rangle)=2$, i.e. $F$ is an $\acute{\text{e}}$tale triple cover of a curve of genus $2$;

$(ii)$ we have $g(F/\langle sr \rangle)=1$, i.e. $F$ is a double cover of an elliptic curve (branched at $6$ points).

Now, it is well-known that a bielliptic curve of genus at least $4$ cannot be hyperelliptic (the references are in my paper, see Remark 3.5). Hence $F$ is not hyperelliptic and we are done.

In this example $F$ is Galois cover of the projective line, but perhaps the construction can be modified in order to find also counterexamples to your second question.

Remark. Your guess about the genus $3$ case is correct. In fact, a curve of genus $3$ is an $\acute{\text{e}}$tale double cover of a curve of genus $2$ if and only if it is both hyperelliptic and bielliptic, see Theorem 3.4 in the paper I quoted.

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Certainly not. Ogg proved in "Real points on Shimura curves" that there are only finitely many Shimura curves (of the form $X^D_0(N)$) which are hyperelliptic. If there is a prime divisor $q$ of $D$ such that $q\equiv 1\bmod 12$ then for all $N$ coprime to $D$, $X^D_0(N) \to X^D_0(1)$ is a connected finite \'etale cover.

Now take $D = 26$. $X^{26}_0(1)$ is of genus 2 and hence is hyperelliptic. Take $N$ large enough and you're done.

I also guess there are easier examples but this was the first one to come to me.

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  • $\begingroup$ Nice example. I don't think this is an answer to the second question, though. In fact, the OP seems to ask whether $Y$ is a Galois cover of the projective line, as opposed to the composite cover $Y\to X\to \mathbf P^1$ being Galois. These are two different things, unless I'm misunderstanding... $\endgroup$ – Ariyan Javanpeykar Apr 1 '14 at 12:35
  • $\begingroup$ You're right. I still think there's a way to say that if $Y$ was also Galois then there's a way to break it up into a composite cover with $X$ in the middle but I don't really have any evidence for that. $\endgroup$ – stankewicz Apr 1 '14 at 16:15

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