# 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 $K$?

• "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). Commented Oct 1, 2013 at 3:21
• Commented Dec 2, 2018 at 21:57
• Possibly relevant: "GENUS TWO CURVES WITH EVERYWHERE TWISTED GOOD REDUCTION" available at projecteuclid.org/download/pdf_1/euclid.rmjm/1370267178. According to this Master thesis, « A possibly easier, but to the author’s knowledge still open, problem is the following: fix an integer $g >2$ and find an explicit example of a number field $K$ and a curve of genus g over K with good reduction at all places of $K$ » Commented Dec 3, 2018 at 12:55
• (The MathOverflow question I quoted above only deals with potential good reduction, while in our case, $K$ is fixed at the beginning ; so it doesn't address your question. However, this MO question seems to be related to the open problem quoted above). Commented Dec 3, 2018 at 13:23
• Commented Dec 4, 2018 at 16:59

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.

• 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 $2^{12}$. Commented Oct 1, 2013 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, Commented Oct 1, 2013 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). Commented Oct 1, 2013 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! Commented Oct 1, 2013 at 3:30