8
$\begingroup$

$\newcommand{\Spec}{\operatorname{Spec}}$ It is a well-known fact that a smooth projective variety over $\mathbb Q$ has good reduction almost everywhere, i.e. everywhere apart from finitely many points. In the language of schemes, this is almost obvious -- given projective equations of the variety $X/\mathbb Q$, for some $n$ we can construct a projective model $\mathcal X\to\Spec\mathbb Z[1/n]$ of $X$. Since a projective morphism is proper and the singular locus of $\mathcal X$ is closed, the set of points of $\Spec\mathbb Z[1/n]$ with singular fibers is closed, hence finite (since the generic fiber is smooth).

When I've first learned this proof, it seemed somewhat clear to me this works for smooth proper varieties over $\mathbb Q$, but my professor has pointed out that while the latter part of the argument goes through, there may be problems in constructing a model over a subscheme of $\Spec\mathbb Z$. He couldn't, off the top of his head, answer whether such varieties could have infinitely many places of bad reduction, but he speculated the answer is yes. This is my question:

Can a smooth proper variety over $\mathbb Q$ have infinitely many places of bad reduction?

I have speculated about a much stronger failure as well:

Can a smooth proper variety over $\mathbb Q$ have bad reduction everywhere (i.e. at every finite prime)?

Note: I have added a reference-request tag because I strongly suspect the answer has been already addressed in the literature, but my searches were unsuccessful, mostly returning results about varieties with everywhere good reduction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.