Brian, I'm just going to think about the case when J is purely toric because it suffices for my application and I can use "Tate curves". I'm not sure what is going on in the general case. CLAIM: Let 0 --> A --> J --> B --> 0 be an exact sequence of abelian varieties over K=Qp^ur whose Neron models have purely toric reduction at p. Then J(K)-->B(K) is surjective. Proof. This is an application of Hilbert's theorem 90 to the Tate uniformization of A, J, and B. Consider the Tate uniformizations: 0 0 0 | | | X_A --> X_J --> X_B | | | T_A --> T_J --> T_B | | | 0 --> A --> J --> B --> 0 | | | 0 0 0 The maps T_A --> T_J, etc. are induced by the maps A-->J-->B. The kernel of X_A --> X_J is trivial, because X_A and X_J are free abelian groups and the rank of the image of X_A is dim(A). Thus ker(T_A-->T_J) = 1. By the snake lemma, the discrete cokernel of X_J-->X_B is the image of the connected abelian variety A; hence it is trivial. Consequently T_J-->T_B is surjective (this is also "clear" without the snake lemma, by a dimension count). We thus have an exact sequence of tori: 0 --> T_A --> T_J --> T_B --> 0 and a corresponding exact sequence of Kbar-modules: 0 --> T_A(Kbar) --> T_J(Kbar) --> T_B(Kbar) --> 0. The corresponding long exact sequence of Gal(Kbar/K)-cohomology yields the exact sequence 0 --> T_A(K) --> T_J(K) --> T_B(K) --> H^1(Kbar/K, T_A(Kbar)). But T_A(Kbar) is a product of copies of Kbar^*, so by Hilbert's Theorem 90, we have H^1(Kbar/K, T_A(Kbar)) = 0. We now have a square T_J(K) --> T_B(K) | | \|/ \|/ J(K) --> B(K) where the top map is surjective, as are the two vertical maps. It follows that the bottom map is surjective, as was claimed. ------- What do you think?