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



About isogenies of abelian varieties
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. 


