The complete level modular curve $X(p)$ does not have potentially good reduction at $p$ for any $p \neq 2,3,5,7,13$ because then there are cusp forms on $X\_0(p)$ showing up in the cohomology of $X(p)$, whose associated Galois representations have unipotent local monodromy at $p$, which contradicts potentially good reduction.


$X(p)$ clearly does have good reduction at $p$ for $p=2,3,5$ because it has genus $0$ and a rational point.


$X(7)$ nonobviously has potential good reduction at $p=7$ where $X(7)$, the Klein quartic curve in $\mathbb P^2$, degenerates to a smooth hyperelliptic curve. See, for instance, [Elkies's paper on the Klein quartic](http://library.msri.org/books/Book35/files/elkies.pdf).


So $p=13$ is the only remaining case. What's the answer there?


A good first step might be to try to answer the question for $X\_1(13)$, which only has genus $2$.



Does X(13) have potentially good reduction at 13?
Regarding Will's question on $X(13)$, it follows from Michael Stoll's answer and the following lemma that $X(13)$ does not have potentially good reduction.


Lemma. *Let $X\to Y$ be a finite morphism of smooth projective geometrically connected curves over a number field $K$. Suppose that $Y$ has non-zero genus. If $X$ has good reduction over $O\_K$, then $Y$ has good reduction over $O\_K$.*


**Proof.** This follows from the existence of Neron models for hyperbolic curves (not only abelian varieties); see Corollary 4.7 in Liu-Tong <http://arxiv.org/abs/1312.4822> . QED


One now concludes as follows. Suppose that $X(13)$ has good reduction everywhere over some number field $K$. Replacing $K$ by a finite field extension if necessary, there is a (natural) finite morphism $X(13)\to X\_1(13)$. Since $X\_1(13)$ has genus two, it follows from the above lemma that $X\_1(13)$ has good reduction over $O\_K$. This contradicts Stoll's answer.


