Tate models for semistable algebraic varieties with mixed reduction over a local field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T01:34:17Z http://mathoverflow.net/feeds/question/51464 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51464/tate-models-for-semistable-algebraic-varieties-with-mixed-reduction-over-a-local Tate models for semistable algebraic varieties with mixed reduction over a local field Dmitry Vaintrob 2011-01-08T05:28:45Z 2011-01-08T07:41:47Z <p>It's known that if $A$ is an abelian variety of totally multiplicative reduction over a p-adic field K, then, after taking a finite field extension, it becomes isomorphic, as a rigid analytic group, to $G_m^g/\prod_{i=1}^g q_i^{\mathbb{Z}}$ where $q_i$ are points (after a finite field extension) of $G_m^g$ which generate a discrete subgroup. </p> <p>My question is, what can be said when $A$ is not totally multiplicative or good reduction, but is some semistable abelian variety of mixed multiplicative-good reduction. I would guess that instead of being a quotient of $G_m$ by a discrete subgroup, $A$ will be (as a rigid analytic variety) a quotient of a good-reduction semi-abelian variety (i.e. a variety which is an extension by $G_m^m$ of an abelian variety of good reduction) by a free discrete subgroup. Does anyone know whether this is true, or whether there's something else replacing the Tate uniformization in this case?</p> http://mathoverflow.net/questions/51464/tate-models-for-semistable-algebraic-varieties-with-mixed-reduction-over-a-local/51465#51465 Answer by Emerton for Tate models for semistable algebraic varieties with mixed reduction over a local field Emerton 2011-01-08T05:42:41Z 2011-01-08T05:42:41Z <p>If $A$ has semi-abelian reduction, then $A$ is uniformized by a semi-abelian variety $G_A$, namely there is an exact sequence $$0 \to \Gamma_A \to G_A \to A \to 0,$$ where $\Gamma_A$ is free of finite rank, $G_A$ is semi-abelian, and the maximal abelian variety quotient of $G_A$ has good reduction. </p> <p>This is due to Raynaud, I believe, and is also discussed in SGA 7. For precise references, allow me (out of laziness) to cite <a href="http://www.math.northwestern.edu/~emerton/pdffiles/optimal.pdf" rel="nofollow">one of my papers</a> --- see the discussion at the beginning of Section 3.</p>