MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

share|cite|improve this question
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.

share|cite|improve this answer
Thanks a lot for the fast answer! – Stefan Keil Jun 17 '11 at 13:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.