Hi,


I have the following definition of what an Abelian Variety $A$ over an arbitrary field $k$ is:


it is a geometrically integral and proper group scheme over $Spec(k)$.


By a group scheme I understand a scheme over $k$ which has $k-$ morphisms


$m: A\times A \rightarrow A$,


$i: A \rightarrow A$,


$e: Spec(k) \rightarrow A$,


which formally fulfill the axioms of a group, i.e. one has the corresponding commutative diagrams.


By Yoneda this is equivalent to the datum of: for all $k-schemes$ $T$ the structure of a group on $A(T)$ which is functorial in $T$, i.e. for a $k-$morphism $S \rightarrow T$ one has a homomorphism of groups $A(T) \rightarrow A(S)$.


In particular this holds for the algebraic closure $\bar k$ of $k$.


Now my question: is it already enough to have the structure of a group on $A(\bar k)$ in order to regain the whole data described above?


And similarly for morphisms of abelian varieties: does a homomorphism between $A(\bar k)$ and $B(\bar k)$ already induce a scheme morphism $A \rightarrow B$?


Thanks



Definition of abelian variety
You cannot expect any group structure on $A(\bar k)$ or any homomorphism $A(\bar k)\to B(\bar k)$ to come from morphisms, just consider an elliptic curve over $k=\bar k=\mathbb C$ where $A(\bar k)\cong\mathbb R^2/\mathbb Z^2$ as groups (for *any* group scheme structure on $A$).


However, two morphisms $f,g\colon A\to B$ inducing the same map $A(\bar k)\to B(\bar k)$ must be equal (so in particular, there is at most one group scheme structure on $A$ inducing a given group structure on $A(\bar k)$). To see this, consider the equalizer of $f$ and $g$. Since $B$ is separated, it is a closed subscheme of $A$. It contains all closed points, so its underlying set is all of $A$, and since $A$ is reduced, it must equal $A$, so $f=g$.


