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

This is to fill some prerequisites to BCnrd's commentanswer. First of all, there are several definitions of a polarization on an abelian variety, and the most "coordinatefree" one is that it is a homomorphism $\lambda:A\to A^t = Pic^0(A)$ given by some (nonunique) 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:
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 skewsymmetric, 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 commentanswer. 


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. 

