4
$\begingroup$

Let $A$ be an abelian variety over a number field $K$ with semi-stable reduction over $O_K$.

Does the Weil restriction $\textrm{Res}_{K/\mathbf{Q}}A$ of $A$ to $\mathbf{Q}$ have semi-stable reduction over $\mathbf{Z}$?

Note that $\textrm{Res}_{K/\mathbf{Q}}A$ is an abelian variety over $\mathbf{Q}$ of dimension $\dim A \cdot [K:\mathbf{Q}]$.

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ The answer is "not in general". The question is local, and you've got problems at primes which ramify in $K/\mathbb{Q}$. You can read off whether or not you have semi-stable reduction at primes above $p$ by looking at the $\ell$-adic Tate module for some $\ell\not=p$; Weil restriction just corresponds to induction on the Tate module side, and now the full story is just an easy exercise. For example if $A$ has positive dimension and good reduction at a ramified prime, then the Weil restriction will have potentially good, but not good, reduction by Neron-Ogg-Shafarevich, so can't have semistbrdn $\endgroup$
    – Kevin Buzzard
    Commented Aug 5, 2012 at 15:49
  • 2
    $\begingroup$ @Kevin: Your "easy exercise" (which shows the answer is "never", if $A \ne 0$, $K \ne \mathbf{Q}$) uses Grothendieck's inertial semistable reduction criterion, a very deep result. There is an easier (and more hands-on) method. The key point is as follows. If $k$ is field, $R$ is a non-etale finite local $k$-algebra, and $G$ is a nonzero quasi-projective smooth connected $R$-group then the geometric fiber of ${\rm{Res}}_{R/k}(G)$ contains a nonzero smooth connected unipotent group, built with the nonzero maximal ideal of some local factor of the (non-etale!) geometric fiber of $R$ over $k$. $\endgroup$
    – user22479
    Commented Aug 5, 2012 at 18:19
  • $\begingroup$ @Kevin: I should have also noted that the nonzero smooth connected unipotent subgroup can be arranged to be normal (I don't assume $G$ is commutative). Anyway, the point of the alternative method that I am suggesting is that it makes more transparent how the failure of semistable reduction is caused by the presence of ramification, in the sense that one can really "see" the link between non-etaleness in the base extension and a nontrivial unipotent radical on a geometric fiber. Appealing to inertial criteria may leave it a bit mysterious, though is of course insightful in other ways. $\endgroup$
    – user22479
    Commented Aug 5, 2012 at 19:10
  • 1
    $\begingroup$ Note that Weil restriction commutes with an everywhere unramified extension of number fields. In this case, it never commutes because $\bf Z$ is simply connected (i.e. has no non-trivial everywhere unramified finite extension). $\endgroup$
    – Damian Rössler
    Commented Aug 6, 2012 at 12:16
  • $\begingroup$ @quasi-coherent. Sorry, I meant to write "Weil restriction commutes (is compatible with) semistable reduction", ie the answer to your question is affirmative when you consider an everywhere unramified extension of number fields. $\endgroup$
    – Damian Rössler
    Commented Aug 7, 2012 at 11:08

0

Reset to default

You must log in to answer this question.