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Let $A$ and $B$ be two isogenous abelian varieties over a number field $K$ and let $\mathcal{A}$ and $\mathcal{B}$ denote their Neron models over $\mathcal{O}_K$. Let $v \in M_K^0$ denote a finite prime of $K$, $k_v$ its residue field, $\mathcal{A}_v = \mathcal{A} \times _{\mathcal{O}_K} k_v$ the special fiber of the reduction of $A$ at $v$, and $\mathcal{A}_v^0$ the connected component of the identity section in the special fiber $\mathcal{A}_v$.

Is it true that the cardinalities of the $k_v$-rational points of $\mathcal{A}_v^0$ and $\mathcal{B}_v^0$ are the same, i.e.

$|\mathcal{A}_v^0(k_v)| = |\mathcal{B}_v^0(k_v)| ,\ \forall v \in M_K^0$?

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up vote 6 down vote accepted

I think the answer is yes. This can be deduced for example from the results of SGA7, Expose IX (p.14-15) as follows:

Firstly, the dimension of the unipotent part, the toric part and the abelian part of the connected component of the special fibre are the same for $\mathcal{A}_v^0$ and $\mathcal{B}_v^0$. The toric and abelian parts of $\mathcal{A}_v^0$ and $\mathcal{B}_v^0$ are moreover isogenous and so have the same number of points in the residue field. The unipotent parts also have the same number of points since that only depends on the dimension. Since the connected component of the special fibre is an iterated extension of these group schemes the claim follows follows from Lang's theorem i.e., that torsors for connected groups over finite fields are trivial.

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Thanks a lot for the fast answer! –  Stefan Keil Jun 17 '11 at 13:55

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