Let us work over mathbbC.
The inclusion ucolonBtoA induces a surjection hatucolonAveetoBvee.
By general facts on Abelian varieties, the kernels of u and hatu have the same number of connected components. Since u is injective, its kernel is trivial, so it follows kerhatu=(kerhatu)0; in other words kerhatu is an Abelian subvariety of Avee.
Therefore we have an exact sequence of Abelian varieties
0tokerhatutoAveetoBveeto0.
By dualizing it, we obtain 0toBtoAto(kerhatu)veeto0,
that is C=(kerhatu)vee.
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