My question is the following:$\newcommand{\Q}{\Bbb Q} \newcommand{\Z}{\Bbb Z}$

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 $\varepsilon=\frac{5+\sqrt{29}}{2}$. – GreginGre Dec 5 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 $K = \Q(i, \sqrt[5]{2}, \sqrt{1 - \zeta_5})$. – Watson Dec 5 at 12:15
  • 2
    According to this document, Fontaine's proof also shows works «for “small” fields $K$, e.g. $\Q(\zeta_n)$ for $n \leq 7$» – Watson Dec 5 at 12:42

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. – Watson Dec 5 at 9:50

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.