Let $k$ be a field and $A$ an abelian variety over $k$. Suppose that $B$ is an abelian subvariety of $A$. Consider the following fact:
There exists an abelian variety $C$ over $k$ and a surjective morphism $A\twoheadrightarrow C$ with kernel exactly $B$.
This is proved in section 9.5 of the book "Abelian varieties, theta functions and the Fourier transform" By Alexander Polishchuk on the way to proving Poincare reducibility. The proof there seems (to me) to be a bit complicated, so I'm wondering if anyone knows of a "simple" proof. I think I could probably devise a proof of the above fact using Poincare reducibility (employing the proof of the latter result in Milne's chapter of Cornell-Silverman, Proposition 12.1 to avoid circular logic), but somehow I'm not so satisfied by this as it seems like it ought to be an "easy" fact.
Best Answer
Let us work over $\mathbb{C}$.
The inclusion $u \colon B \to A$ induces a surjection $\hat{u} \colon A^{\vee} \to B^{\vee}$. By general facts on Abelian varieties, the kernels of $u$ and $\hat{u}$ have the same number of connected components. Since $u$ is injective, its kernel is trivial, so it follows $\ker \hat{u}=(\ker \hat{u})_0$; in other words $\ker \hat{u}$ is an Abelian subvariety of $A^{\vee}$.
Therefore we have an exact sequence of Abelian varieties $$0 \to \ker \hat{u} \to A^{\vee} \to B^{\vee} \to 0.$$ By dualizing it, we obtain $$0 \to B \to A \to (\ker \hat{u})^{\vee} \to 0,$$ that is $C = (\ker \hat{u})^{\vee}$.