Definition of abelian variety

Question

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

Answer 1

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 $f=g$.

Answer 2

There are varieties which are isomorphic to abelian varieties over the algebraic closure but are not themselves abelian varieties (this answers both your questions). They are principal homogeneous spaces for abelian varieties. To ensure that such a beast is an abelian variety, it turns out you only need to require the existence of a rational point. So to give a concrete example, just use a cubic with no rational points, e.g. $x^3+2y^3+4z^3=0$ over $\mathbb{Q}$.