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.

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 $2$.

  • $\begingroup$ The Jacobian $J(13)$ admits as a factor $J_0(169)^+$, which is a 3-dimensional variety isogenous to $A_f$, where $f \in S_2(\Gamma_0(169))^+$ is a newform with coefficients in $\mathbf{Q}(\zeta_7)^+$. Does this $A_f$ have potential good reduction at $13$? $\endgroup$ Mar 16, 2016 at 18:48
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    $\begingroup$ The stable reduction of all $X(p)$ has been worked out by Bouw and Wewers in their paper "Stable reduction of modular curves". In particular, it also follows from their results that $X(13)$ does not have potentially good reduction. $\endgroup$
    – naf
    Mar 19, 2016 at 6:54

3 Answers 3


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.


Regarding Will's question on $X_1(13)$:

An equation of this genus 2 curve is $$y^2 = x^6 - 2 x^5 + x^4 - 2 x^3 + 6 x^2 - 4 x + 1.$$ The polynomial on the right, when reduced mod 13, has a triple root at $-3$ and three simple roots. Over ${\mathbb Q}_{13}$, the roots reducing to $-3$ are of the form $-3+\theta_j$ with $v_{13}(\theta_j) = \frac{1}{3}$ and $v_{13}(\theta_i - \theta_j) = \frac{1}{3}$ when $i \neq j$. This implies that over a sufficiently ramified finite extension of ${\mathbb Q}_{13}$ (ramification index divisible by 6 is sufficient), the stable model of $X_1(13)$ will have special fiber consisting of two elliptic curves meeting transversally in one point (if my computation is correct, then both curves have $j$-invariant zero). This in turn means that the curve $X_1(13)$ has bad reduction over any extension of ${\mathbb Q}_{13}$, but its Jacobian $J_1(13)$ has good reduction over sufficiently ramified extensions (its reduction is the product of the two elliptic curves and hence an abelian variety). So this is a case of (potentially) "mildly bad reduction".

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    $\begingroup$ Indeed $J_1(13)$ has good reduction over $\mathbf{Q}(\zeta_{13})$ (see my comment to @znt's answer). Actually over that number field $J_1(13)$ is isogenous to the product of two conjugate elliptic curves defined over $\mathbf{Q}(\sqrt{13})$. These elliptic curves have been determined in the article "Q-curves and their Manin ideals" by Josep González and Joan-C. Lario. $\endgroup$ Mar 18, 2016 at 12:08

If there were a newform which didn't have potentially good reduction and whose associated abelian variety were isogenous to a subvariety of $Jac(X(p))$ then the associated local automorphic representation would have to be a twist of the Steinberg. The character that one is twisting by would have to have conductor 1 or $p$ (otherwise the conductor of the local automorphic rep at $p$ would be bigger than $p^2$) and there's a global character which looks like this locally on inertia; untwisting by this global char shows that the twisted modular form contributes to the cohomology of $X_0(p)$. So if $X_0(p)$ has genus 0 then $Jac(X(p))$ has potentially good reduction at $p$. I don't think this quite proves that $X(p)$ has good reduction at $p$ but it answers François' question at least. The implicit claim in the question that $X_0(p)$ has good reduction at $p$ because it has genus 0 isn't quite a complete argument: twists of the projective line can have bad reduction at some primes, but Will is of course OK because $X_0(p)$ has cusps and hence points.

  • $\begingroup$ PS I'm new here -- how do I get the cedilla on Francois using a US keyboard? $\endgroup$
    – znt
    Mar 16, 2016 at 19:05
  • $\begingroup$ Thanks for answering my question. For the cedilla the simplest is copy&paste, but I'm not really sensible to it $\endgroup$ Mar 16, 2016 at 21:44
  • $\begingroup$ On a mac I believe option+c does the trick. $\endgroup$ Mar 16, 2016 at 21:54
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    $\begingroup$ @znt on a mac it actually works the same as with an ipad/iphone: press and hold a key and all the possible modifications of the letter will eventually be displayed for you to choose from $\endgroup$ Mar 17, 2016 at 8:42
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    $\begingroup$ According to Katz-Mazur, Deligne and Rapoport showed that for $p$ prime, the abelian variety $J_1(p)/J_0(p)$ has good reduction over $\mathbf{Q}(\zeta_p)$. I wonder what can be said about $J(p)/J_0(p)$. $\endgroup$ Mar 18, 2016 at 12:01

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