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Any map $A \to B$ of abelian varieties of the same dimension over a global field $K$ induces a map $\mathcal{A} \to \mathcal{B}$ on the corresponding Neron models over $X$ (where $X=Spec{\mathcal{O}_K}$ for any number field $K$ or $X$ is a complete smooth curve over a perfect field with function field $K$) which further restricts to a map $\mathcal{A}^0 \to \mathcal{B}^0$ on the respective identity components. Thus we have a map

$${\textrm{Res}}^0:{\textrm{Hom}}_K(A,B) \cong {\textrm{Hom}}_X(\mathcal{A},\mathcal{B}) \to {\textrm{Hom}}_X(\mathcal{A}^0,\mathcal{B}^0)$$

My question is - are there maps from $\mathcal{A}^0$ to $\mathcal{B}^0$ over $X$ that do not arise from any map over $K$ from $A$ to $B$ (equivalently, from any map over $X$ from $\mathcal{A}$ to $\mathcal{B}$)? In other words, is ${\textrm{Res}}^0$ surjective? I'm particularly interested to know if it is possible to have an isogeny of identity components that does not lie in the image of ${\textrm{Res}}^0$ above.

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up vote 5 down vote accepted

If I'm not mistaken, the map $Res^0$ is a bijection. Indeed, restriction induces maps $$Hom_X(\mathcal A,\mathcal B) \to Hom_X(\mathcal A^0,\mathcal B^0)\to Hom_K(A,B).$$ The composite is a bijection, by the universal mapping property of Neron models, and both maps are injective, since $A$ is Zariski dense in $\mathcal A^0$, which in turn is Zariksi dense in $\mathcal A$. Thus both maps are bijections, and hence $Res^0$ is a bijection.

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Thanks! That was very helpful. –  Saikat Biswas Mar 30 '11 at 21:16

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