Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

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 $f=g$.

share|improve this answer
    
The group axioms all have the form $f=g$ as in the second paragraph, so can be checked on $\bar k$-valued points. –  user2035 Sep 10 '11 at 10:39
    
The preceding comment was an answer to a comment which has now disappeared. –  user2035 Sep 10 '11 at 10:48
    
Sorry, the question in the comment became clear to me somewhat afterwards. But just one last thing: a closed subscheme which contains all closed points is the whole space? And why is the equalizer reduced? Thank you! –  Descartes Sep 10 '11 at 11:22
    
In a scheme which is locally of finite type over a field, the set of closed points is dense. In the reasoning above, we only know that the equalizer is reduced after we have concluded that it equals $A$. –  user2035 Sep 10 '11 at 11:40
add comment

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 $\mathbb{Q}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.