About isogenies of abelian varieties

Question

Why it is true that, over an algebraically closed field, any abelian variety is isogenous to a principally polarized abelian variety?

Answer 1

This is to fill some prerequisites to BCnrd's comment-answer.

First of all, there are several definitions of a polarization on an abelian variety, and the most "coordinate-free" one is that it is a homomorphism $\lambda:A\to A^t = Pic^0(A)$ given by some (non-unique) ample divisor $D$, so that $\lambda(a) = \mathcal O_A( T^*_a D - D)$, where $T_a:A\to A$ is the translation by $a\in A$. A polarization is principal if $\lambda$ is an isomorphism.

Then the basic steps, all requiring proof, are:

1. Every abelian variety (over a field) has a polarization.

2. $\lambda$ is surjective, and $K(\lambda)=\ker\lambda$ is a finite group subscheme of $A$ of length $d^2$. The degree of $\lambda$ is defined to be $d$. Thus, a polarization is principal iff $d=1$.

3. $K(\lambda)$ comes with a perfect skew symmetric Weil pairing $b: K(\lambda)\times K(\lambda) \to \mathbb G_m$ with the values in the multiplicative group. "Perfect pairing" means that it defines an iso from $K(\lambda)$ to its Cartier dual.

4. If $H\subset K(\lambda)$ is a subgroup which is isotropic w.r.t. this pairing (i.e. $b$ restricted to $H\times H$ is trivial), then $\lambda$ descends to a polarization on $B=A/H$ of degree $d/|H|$.

So then it is enough to find a nontrivial isotropic subgroup $H$, replace $A$ by $A/H$, and continue by induction, until you reach $d=1$.

In char 0, this is trivial: just pick any cyclic subgroup in $K(\lambda)$. Since $b$ is skew-symmetric, it is automatically isotropic.

In char p, you need to get into the classification of finite group schemes, and learn the difference between $\mathbb Z/p\mathbb Z$, $\mu_p$, and $\alpha_p$. Once you learn that (Introduction to affine group schemes by Waterhouse is a good source), then you are perhaps ready to read BCnrd's comment-answer.

Answer 2

Over $\mathbb{C}$ the proof of this fact is very simple.

In fact, given any complex Abelian variety $X:= V/\Gamma$ of dimension $g$, one can find strictly positive integers $d_1, \ldots ,d_g$ and a basis $\gamma_1, \ldots ,\gamma_{2g}$ of $\Gamma$ such that the matrix of the Kähler form in this basis is

$\left[\begin{matrix}0 & \Delta \cr - \Delta & 0 \end{matrix}\right],$

where $\Delta$ is the diagonal matrix with diagonal coefficients $d_1, \ldots, d_g$.

Now let $\Gamma'$ be the lattice generated by

$\gamma_1/d_1, \ldots, \gamma_g/d_g, \gamma_{g+1}, \ldots, \gamma_{2g}$

and set $Y:=V/ \Gamma'$.

The Kähler form is integer and unimodular when restricted on $\Gamma'$, hence $Y$ is a principally polarized Abelian variety. Moreover, the natural surjection $X \to Y$ is an isogeny.