Let $X_p$ be a projective curve over the finite field $\mathbf{F}_p$ (i.e. a projective $\mathbf{F}_p$-scheme pure of dimension 1) for every prime number $p$. Let $X_\mathbf{Q}$ be a projective curve over $\mathbf{Q}$. When does there exist a flat projective integral normal scheme $X$ over Spec $\mathbf{Z}$ such that the fibre above $p$ is $X_p$ for every prime p and the generic fibre is $X_\mathbf{Q}$?

Suppose that $X$ exists. Then the generic fibre is a smooth projective connected curve over $\mathbf{Q}$. Furthermore, $X_p$ is almost always smooth over $\mathbf{F}_p$. Also, the arithmetic genus is constant in the fibres of $X$. Moreover, the Hilbert polynomial of $X_p$ is independent of $p$.

Example. Suppose that $X_p$ is a supersingular elliptic curve for all $p$. Then there does not exist such an $X$.

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    $\begingroup$ In your example, I think you mean to say that $X_p$ is supersingular for all primes $p$. $\endgroup$ Commented Feb 4, 2011 at 15:20
  • $\begingroup$ You're right. I changed it. $\endgroup$ Commented Feb 5, 2011 at 9:27
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    $\begingroup$ This reminds me of classifying Chebotarev sets, and the answer there is similar to the answer of this question in the case of elliptic curves - if the family comes from an automorphic form (well, conjectured in the number fields case). $\endgroup$ Commented Feb 5, 2011 at 13:01

2 Answers 2


I suspect that even if you had a single curve over $\mathbb{F}_p$, you might not find a lift to $\mathbb{Q}$. Below I sketch an argument that works under the assumption that $\mathcal{M}_g$ does not have Zariski dense set of points. If you believe the conjectures of Lang on rational points, this assumption should be satisfied as soon $\mathcal{M_g}$ is of general type, e.g. if $g \geq 24$.

Lemma. Suppose that $g$ is an integer such that $\mathcal{M}_g(\mathbb{Q})$ is not Zariski dense in $\mathcal{M}_g$. Then for all sufficiently large primes $p$, there are smooth curves $C_p$ of genus $g$ defined over $\mathbb{F}_p$ that are not reductions of curves defined over $\mathbb{Q}$.

Proof. By assumption, the set $Z:=\overline{\mathcal{M}_g(\mathbb{Q})}$ is a proper closed subset of $\mathcal{M}_g$. In particular, the dimension of $Z$ is smaller than the dimension of $\mathcal{M}_g$. By the Lang-Weil estimates, the number of points of $\mathcal{M}_g$ modulo $p$ grows as a polynomial in $p$ of degree $\dim \mathcal{M}_g$, since $\mathcal{M}_g$ is irreducible modulo $p$. Similarly, the number of points of $Z$ modulo $p$ grows as a polynomial in $p$ of degree at most $\dim Z$ (the "at most" comes from the fact that $Z$ need not be geometrically integral). Thus, for sufficiently large $p$, there will be points of $\mathcal{M}_g(\mathbb{F}_p)$ that are not contained in $Z(\mathbb{F}_p)$. These points correspond to smooth curves of genus $g$ defined over $\mathbb{F}_p$ that are not reductions of curves defined over $\mathbb{Q}$, as required.


If every $X_p$ is smooth of non-zero genus $g$, then Fontaine showed ("Il n'y a pas de schema abelien sur $\mathbb Z$") that the answer is "never". From another perspective, the generic fiber $X_{\mathbb Q}$ determines the minimal regular model over $\mathbb Z$ uniquely (again if $g>0$), and then determines $X_p$ whenever $X_p$ is assumed smooth.

  • $\begingroup$ inkspot, I don't know this area as well as I should, could you give a quick reference? $\endgroup$ Commented Feb 5, 2011 at 15:43
  • $\begingroup$ Nevermind, Bhargav Bhatt pointed me in the right direction. I found the reference, "Il n'y a pas de variété abélienne sur Z". Thanks! $\endgroup$ Commented Feb 6, 2011 at 3:16
  • $\begingroup$ It does mean that there is no other X over Z with the desired fibers, assuming that those fibers are smooth of genus at least 1. $\endgroup$
    – inkspot
    Commented Feb 10, 2011 at 17:45

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