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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.

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?

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up vote 10 down vote accepted

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.

This is due to Raynaud, I believe, and is also discussed in SGA 7. For precise references, allow me (out of laziness) to cite one of my papers --- see the discussion at the beginning of Section 3.

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Just a little comment. In fact, the whole semi-abelian variety $G_A$ has good reduction: it extends to a split semi-abelian scheme over the ring of integers. – Keerthi Madapusi Pera Jan 8 '11 at 17:54
Also, the Raynaud reference is: `Variétés abéliennes et géométrie rigide'. Actes du Congrès International des Mathématiciens (Nice, 1970), Tome 1, pp. 473–477. Gauthier-Villars, Paris, 1971 – Keerthi Madapusi Pera Jan 8 '11 at 17:59

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