Abelian Varieties and Neron Models over local fields: Quotient of points of k mod valuation ring is finite I am working on a paper and I am not sure why the following fact is true. Suppose A/k is an abelian variety over a nn-archimedean local field k and $N/\mathcal{O}$ should be the Neron model. Then let $N^0$ be the connected component of the Neron Model, ie the normal open subgroup scheme, which is fibrewise the connected component of the identity. We are in the situation $N^0 \neq N$. Now they state that the quotient
$N^0(k)/N^0(\mathcal{O})$ is finite.
I am wondering, why this is true.
Maybe this has something to do with the reduction map, ie. the map $N(\mathcal{O}) \rightarrow N(\kappa)$, $\kappa$ the residue field, which is surjective?
I would be happy for an answer :)
 A: In case someone needs this fact too, here my suggestion:
For the definition and all the properties I use, see SGA 3, $VI_B$, §3.
First remark that $N^0$ is compatible with base change, hence $(N^0)_k= (N_k)^0=A^0=A$ since $A/k$ is ab abelian variety and hence connected. Using this we obtain $N^0(k)=A(k)=N(\mathcal{O})$ via the Neron mapping property.
So we are reduced to show $N(\mathcal{O})/N^0(\mathcal{O})$ is finite. Now go back to SGA, where they defined $N^0$ and we see that
$N^0(\mathcal{O})=Ker(N(\mathcal{O}) \rightarrow N(\kappa)/N_\kappa^0(\kappa))$.
ie you have to look at the fibers for $s \in Spec(\mathcal{O})$. But being a non-archimedean local field there were just two points, 0 and the unique prime ideal $\mathfrak{p}$. For $s=\mathfrak{p}$ we obtain the reduction map. This quotient is finite, since $\kappa$ is a finite field and we obtain the desired property.
If you have any comments, please feel free to write, I would be happy, if there were some things, which are not clear and we could discuss.
