# Curves over number fields with everywhere good reduction


What is known about number fields $$K$$ fulfilling the condition $$C_{g,K}$$ "there is a smooth projective curve of genus $$g$$ over $$K$$, having everywhere good reduction" for some $$g \geq 1$$ ?

By the work of Fontaine and Abrashkin, it is known that for every $$g>0$$, the field $$K = \Q$$ does not satisfy $$C_{g,\Q}$$. Notice that the condition $$C_{g,K}$$ (for some $$g>0$$) implies, by the functorial properties of the Jacobian, the property $$A_K$$ "there is a non-zero abelian variety over $$K$$ having good reduction everywhere". (Using the theory of Néron models, it should be equivalent to state that there is no non-trivial abelian scheme over the ring of integers $$O_K$$ (see here, where it is also explained that a CM abelian variety has potentially "good reduction everywhere")).

Thus a closely related question is:

Do we expect the existence of (quadratic?) number fields $$K \neq \Q$$ such that the assertion $$A_K$$ does not hold (so that in particular, there is no smooth projective curve of genus $$>0$$ over $$K$$ with everywhere good reduction)?

For instance, what happens if $$K = \Bbb Q\left(\sqrt{-2}\right)$$ or $$K = \Bbb Q(\sqrt{2})$$?

It is mentioned here that there is no elliptic curve over $$\Q(\sqrt 2)$$ with everywhere good reduction. The same happens with $$\Q(i)$$, see this answer. Examples of abelian surfaces with everywhere good reduction (i.e. the opposite of what I'm looking for) are mentioned here.

By some analogy discussed here, it may be useful to note that $$\Q(i)$$ and $$\Q(\sqrt 2)$$ have no non-trivial unramified extension (see here).

It was asked here whether every number field $$K \neq \Q$$ satisfies $$C_{g,K}$$ for some $$g>0$$, but the answer is only very partial. Furtherfore, it is explained here that for every $$g \geq 0$$, there is some number field $$K$$ such that $$C_{g,K}$$ holds — this is different from my question, where I want $$K$$ to be fixed at the beginning.

• According to comments to the accepted answer in mathoverflow.net/questions/16600/…, there is an example of elliptic curve over $\mathbb{Q}(\sqrt{29})$ which has good reduction everywhere. It is $E:y^2+xy+\varepsilon^2y=x^3$, where $\varepsilon=\frac{5+\sqrt{29}}{2}$. Dec 5, 2018 at 10:49
• I am aware of this (I already linked this question in my post). Moreover, for $g=2$, it is claimed here that the curve $y^2=x^5-1$ has good reduction everywhere over $K = \Q(i, \sqrt[5]{2}, \sqrt{1 - \zeta_5})$. Dec 5, 2018 at 12:15
• According to this document, Fontaine's proof also shows works «for “small” fields $K$, e.g. $\Q(\zeta_n)$ for $n \leq 7$» Dec 5, 2018 at 12:42
• See also the following papers by R. Schoof: mat.uniroma2.it/~schoof/zeta20.pdf, mat.uniroma2.it/~schoof/sqrt6.pdf, mat.uniroma2.it/~schoof/abcyc.pdf. Apr 14, 2019 at 12:00
• According to Mestre's Formules explicites et minorations de conducteurs de variétés algébriques, remarque 3, there are examples of curves with good reduction everywhere over $\Bbb Q(\sqrt D)$ where $D=6, 28, 29, 41$ Dec 5, 2019 at 18:01

Do we expect the existence of (quadratic?) number fields $$K \neq \Q$$ such that the assertion $$A_K$$ does not hold (so that in particular, there is no smooth projective curve of genus $$>0$$ over $$K$$ with everywhere good reduction)?
Yes, we do. Moreover, we do know that such number fields exist. Look at the original Fontaine's paper "Il n’y a pas de vari´et´e abelienne sur $$\mathbf Z$$" where he proves that there are no abelian varieties with everywhere good reduction over the following number fields: $$\mathbf Q, \mathbf Q(i), \mathbf Q(\sqrt{-3})$$ and $$\mathbf Q(\sqrt{5})$$.
• Thank you! I shouldn't have missed this corollary in Fontaine's paper. Still my first question remains. Moreover, one could wonder about the assertion $A_K$ for $K=\Q(\sqrt 2)$ or $K = \Bbb Q(\sqrt{-2})$ (notice that those fields have no non-trivial unramified extension, as the four number fields given in Fontaine's result). At least, it is known that there is no elliptic curve over $\Bbb Q(\sqrt 2)$ (or over $\Bbb Q(\sqrt{-2})$) with everywhere good reduction. Dec 5, 2018 at 9:50