# Geometry of abelian varieties

if $X$ and $Y$ are abelian varieties over a field $k$ and

$f:X\rightarrow Y$ is a homomorphism of abelian varieties, are then the following true:

1) $\dim(X)=\dim_0(\operatorname{ker}(f))+\dim(\operatorname{im}(f)),$ where I mean the connected component in the identity resp. the scheme theoretic image, and always the Krull dimension of schemes.

2) If a subvariety of $X$ is zerodimensional, then it is a finite set of points and vice versa.

Thanks

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Yes, your assertions are true. – Matt E Sep 23 '11 at 16:34
I mean, 2) is relatively clear. But I would be interested in a hint how to prove the dimension formula in 1). – Veen Sep 23 '11 at 19:13

As for (1), if $f: X \to Y$ is a dominant (e.g. surjective) morphism of varieties, then the dimension of a generic fibre is equal to $\dim X - \dim Y$. Now in the case when $f$ is a morphism of group varieties, all the fibres are of the same dimension (in fact they are all isomorphic, since they are cosets of the kernel, and thus all translates of one another). Thus all the fibres are of dimension $\dim X - \dim Y$, and in particular the kernel is of this dimension.
[Note that any (not necessarily surjective) homomorphism $f$ of abelian varieties is a morphism between projective schemes, hence is proper, and hence its image is closed, connected (since the source is connected), and thus is itself an abelian variety. Thus the above reasoning holds, when applied to the induced map $X \to im(f)$.]
Thanks, Matt. Just two questions: 1) where do I find a reference for the assertion that the dimension of a generic fibre is $dimX-dimY$? 2) The dimension formula does not hold if one considers just the connected component in zero, as I originally posted the question? – Veen Sep 27 '11 at 8:49