# Which number fields allow higher genus curves with everywhere good reduction

The field of rational numbers is not such a number field. That is, there does not exist a smooth projective morphism $X\to\text{Spec } \mathbf{Z}$ such that the generic fibre is a curve of genus $\geq 1$.

Which number fields allow (or do not allow) the existence of such curves?

For any number field $K/\mathbf{Q}$ of degree $>1$, does there exist a smooth projective geometrically connected curve $X$ over $K$ with good reduction over $K$?

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"For any number $K/\bf Q$ of positive degree" should be "For any number field $K/\bf Q$ of degree $>1$" (number field, not number; and the degree $[K:{\bf Q}]$ is automatically positive). – Noam D. Elkies Oct 1 '13 at 3:21

This is an interesting question.

Here is, I think, a partial result for genus $1$. Take everything I write with a grain of salt (edit: make that a whole pinch. See the discussion in the comments.).

Let $X$ be the projective curve over $\mathbf Q$ given by the homogeneous equation of degree 12 $$-16(4a^3+27b^2) = d^{12}$$ in the three variables $(a,b,d)$, where $a$ has graded degree $4$, $b$ has graded degree $6$, and $d$ has graded degree $1$.

Proposition: Let $P=(a_P, b_P, d_P)$ be a $\overline{\mathbf Q}$-point of $X$, with field of definition $K_P = \mathbf Q(a_P, b_P, d_P)$. If $d_P \neq 0$, then the elliptic curve

$$E_P \: : \: y^2 = x^3 +a_Px + b_P$$

defined over $K_P$ has good reduction everywhere.

Indeed, the condition for $E_P/K_P$ to have good reduction at some prime $\mathfrak p$ of $\mathcal O_K$ is that the minimal model of $E_{P, \mathfrak p} : =E_P\times_{K}K_{\mathfrak p}$ over $K_\mathfrak p$ have unit discriminant; this is so if and only if $\mathcal v(\Delta_{E_P}) \equiv 0 \mod 12$, i.e. $\Delta_{E_P}$ is a twelvth-power in $K^\times$ (by Tate's algorithm) (Edit: the "only if" part isn't right. See the comments.). This is precisely what the equation defining $X_\Delta$ expresses.

What this shows in fact is that $X_\Delta$ is a $12$-sheeted (ramified, singular) cover of the projective $j$-line $\mathbf P^1_j$, which pulls apart into twelve the fibres of the fake universal elliptic curve over $\mathbf P^1_j$.

A setback of the Proposition is that its converse is false, i.e. not all elliptic curves over $K$ with everywhere good reduction correspond to a $K$-point of $X$. Indeed, what if the discriminant $\Delta(E) \in K^\times$ satisfies $\nu(\Delta(E)) \equiv 0 \mod 12$ for each place $\nu$ of $K$, but is not a global $12$-th power in $K$? This obstruction is measured by the finite Galois-cohomological object $$K(S,m) := \{ x \in K^\times/(K^\times)^m : \nu(x) \equiv 0 \mod 12 \: \forall \: \nu\}$$ which we could call the $\mathbf Z/m\mathbf Z$-module of 'Weil periods'. To get the right correspondence, I believe we should extend the scalars of $X_\Delta$ from $\mathbf Q$ to the 12-th cyclotomic field $\mathbf Q(\zeta_{12})$.

Remark: I'm not quite sure what the genus of (the normalization of) $X_\Delta$ is, but it is certainly greater than $1$. Together with Falting's theorem, this seems to imply:

Claim: Let $K$ be a number field. Then, up to $K$-isomorphism, there are finitely many elliptic curves over $K$ with good reduction everywhere.

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Careful: the condition that $\Delta$ be a 12th power is necessary but not sufficient for $E$ to have good reduction. For example, the elliptic curve $y^2 = x^3 - 4x$ has discriminant $2^{12}$. – Noam D. Elkies Oct 1 '13 at 3:03
@NoamD.Elkies Hmmm, you are right! I guess funny things happen at $2$ and $3$. Can it be salvaged? By the way, it is an honor for me! Regards, – Bruno Joyal Oct 1 '13 at 3:11
The finiteness result is true but I don't think the geometrical picture is right. At any rate, this isn't just a small-primes phenomenon: for any prime $p$ there are curves with bad reduction at $p$ whose discriminant has $p$-valuation $12m$ (e.g. with a I$_{12m}$ fiber). – Noam D. Elkies Oct 1 '13 at 3:18
@NoamD.Elkies I see! I think that I misunderstood the criterion. Oh well, I will leave it up so others can learn from my mistake. At any rate, thank you for reading it! – Bruno Joyal Oct 1 '13 at 3:30