Hyperelliptic modular curves in characteristic p Ogg characterized the finitely many N such that $X_0(N)_{\mathbb{Q}}$ is hyperelliptic, and Poonen proved in "Gonality of modular curves in characteristic p" that for large enough N, $X_0(N)_{\mathbb{F}_p}$ is not hyperelliptic.
Question: Are there any N such that $X_0(N)_{\mathbb{Q}}$ is not hyperelliptic but for some p not dividing N $X_0(N)_{\mathbb{F}_p}$ is hyperelliptic? 
I'm also interested in this question for other modular curves of the form $X_H$, where H is a congruence subgroup.
 A: A curve $C$ is hyperelliptic if and only if the canonical map $C \to \mathbb P^*(\Omega^1(C))$, which sends a point $p$ to the codimension 1 subspace $V_p \subset\Omega^1(C)$ of all one forms vanishing at $p$, is a two to one map. In the hyperelliptic case its image will be a $\mathbb P^1$, and the degree two map will be the quotient by the hyperelliptic involution. The map $\mathbb P^1 \to \mathbb P^*(\Omega^1(C))$ will basically be a Veronese embedding and hence its image will lie on a lot of quadrics. One can use this to search for possible candidates. I found none for $\Gamma_0(N)$ with $N \leq 151$. I think my search is provably complete, showing that for $N \leq 151$ and all $p$ the only hyperelliptic $X_0(N)_{\mathbb F_p}$ are the ones who are already hyperelliptic over $\mathbb Q$. Together with the general bound $${\rm gon_{\mathbb F_p^2}} X_0(N) \geq [PSL_2(\mathbb Z) : \Gamma_0(N)]\frac {p-1} {12(p^2+1)}$$ which is mentioned in Poonen's paper this implies that $X_0(N)_{\mathbb F_p}$ will never be hyperelliptic modulo $2,3$ or $5$ unless of course $X_0(N)_{\mathbb Q}$ is.
Now for $N \geq 151$ the genus of $X_0(N)$ already starts to become reasonably big. So I would not expect there to be any examples with $p>5$ and $N>151$ either. But I don't see how to prove it yet since Poonen's bound depends heavily on $p$.
Using this idea for $X_H$ with other congruence subgroups $H$ I found an example though. There might be more but this is just the first one I found. Anyway here is the explicit example:
Consider the modular curve $X_1(37)/\langle 4\rangle$, it is a double cover of $X_0(37)$ and its genus is $4$. Its global one forms are generated by the modular forms with q-expansion:
\begin{gather}
x_0 := q - 2q^{5} + 2q^{6} - 3q^{7} + 2q^{9} + O(q^{10})\\
x_1 := q^{2} - q^{5} - q^{6} + O(q^{10})\\
x_2 := q^{3} - 2q^{7} + O(q^{10})\\
x_3 := q^{4} - q^{5} + q^{6} - 2q^{7} + 2q^{9} + O(q^{10}).\\
\end{gather}
These satisfy the relations:
\begin{gather}
- x_{1}^{2} + x_{0} x_{2} - 2 x_{2} x_{3} = 0 \\
x_{1}^{2} x_{2} - 2 x_{1} x_{2}^{2} + 2 x_{2}^{3} -  x_{0} x_{1} x_{3} +
x_{1}^{2} x_{3} + x_{1} x_{2} x_{3} - 4 x_{2}^{2} x_{3} -  x_{0}
x_{3}^{2} + x_{1} x_{3}^{2} + 4 x_{2} x_{3}^{2} + 2 x_{3}^{3} = 0, \\
\end{gather}
which describe the canonical model $X_1(37)/\langle 4\rangle $ as the intersection of a smooth quadric and cubic in $\mathbb P^3$.
Modulo $2$ however $x_0,x_1,x_2,x_3$ satisfy two additional relations:
$$
x_{1} x_{2} + x_{2}^{2} + x_{0} x_{3} + x_{2} x_{3} =0 \\
x_{2}^{2} + x_{1} x_{3} + x_{2} x_{3} = 0,\\
$$
which shows that in this case the image of the canonical embedding is a $\mathbb P^1$. 2 is the only prime for which it happens for $X_1(37)/\langle 4 \rangle$.
A: This is sort of a cheap answer, but I would look at values of N where X_0(N) has genus 3 and is not hyperelliptic (over Q).  In genus 3, the locus of hyperelliptic curves has codimension 1 in the whole moduli space of curves, so one would expect a non-hyperelliptic curve to reduce to a hyperelliptic curve mod p for some finite (but typically non-empty) set of primes p.
Presumably this could be made explicit:  a non-hyperelliptic genus 3 curve is a plane quartic, and so you've got to figure that there is an invariant F of ternary quartic forms with the property that F(P) is a multiple of precisely those primes where the curve cut out by P reduces to a hyperelliptic curve.  Or so I would guess.
