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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.

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    $\begingroup$ Hey Bryden, non-affine quotient existence results by action of something more than a finite flat group probably cannot be entirely "easy", since one cannot easily see where the coordinate rings on the quotient are to come from. Within the framework of algebraic spaces one can make good quotients in great generality (it is the method to use in general when the base is not an artin ring), and then sometimes prove the alg. space quotient is a scheme (e.g., alg. space group lft over field is scheme...); sometimes qt isn't a scheme! In your case, trick via Poincare red. seems simplest (to me). $\endgroup$
    – BCnrd
    Commented Sep 2, 2010 at 21:13
  • $\begingroup$ Hey Brian, and thanks for your thoughts. I now agree that one has to use some things which are not "trivial", whether it be representability of alg. space quotients or Poincare reducibility or properties of $\mathcal{E}xt(\cdot,\mathbb{G}_m)$, as in my comment to Francesco's answer below. $\endgroup$
    – B. Cais
    Commented Sep 3, 2010 at 13:08

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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 $C = (\ker \hat{u})^{\vee}$.

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    $\begingroup$ Oh yeah, the trick of taking kernel of a well-chosen map between dual abelian varieties (using an ample line bundle) and then dualizing back (or something like that) is what also underlies the proof of Poincare reducibility, isn't it? $\endgroup$
    – BCnrd
    Commented Sep 2, 2010 at 22:29
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    $\begingroup$ Thanks, Francesco. Comments: 1. Are you working over $\mathbb{C}$ to ensure $\mathrm{ker} \hat{u}$ is reduced? For general {\em perfect} $k$, \{($\mathrm{ker}\hat{u}$)_0\}_{red} is an abelian variety, but then one loses exactness. 2. The fact that "dualizing" is exact isn't obvious to me. I had to apply $\mathcal{H}om(\cdot,\mathbb{G}_m)$ and use that $\mathcal{E}xt^i(G,\mathbb{G}_m)$ vanishes for $i\ge 2$ (Oort, Comm gp sch, II 12.3) and is isomorphic to $G^{\vee}$ when $G$ is an ab. var. and $i=1$. 3. Can you explain why the kernels of $u$ and its dual have same no. of conn comps? $\endgroup$
    – B. Cais
    Commented Sep 3, 2010 at 12:29
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    $\begingroup$ 1. I'm working over $\mathbb{C}$ just because I'm more familiar with complex numbers and I learnt the theory of Abelian variety mostly in this setting. Probably some part of the argument can be carried out also in positive characteristic, but I prefer not to write down things if I'm not sure. 2. The fact that "dualizing" is exact comes from [Birkenake-Lange, Complex Abelian Varieties, Proposition 2.4.2]. The proof actually works for any complex torus $X$, and uses only the fact that $X = C^n/ \Gamma$, where $\Gamma$ is a lattice... $\endgroup$
    – Francesco Polizzi
    Commented Sep 3, 2010 at 13:40
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    $\begingroup$ ...3. The fact that $u$ and its dual have the same number of connected components is not difficult. Standard trick about factorization of a morphism of Abelian varieties by using the quotient by the connected components of the kernel containing 0 and an isogeny, plus some easy argument involving duality. A complete proof is in [Debarre, Tores and varietes abeliennes complexes, Proposition 4.6 pag 71]. I do not know whether this results is also true in positive characteristic. $\endgroup$
    – Francesco Polizzi
    Commented Sep 3, 2010 at 13:42
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    $\begingroup$ Bryden, the exactness of duality over any field is proved by using the following criterion: a 3-term complex of abelian varieties is short exact (in the evident concrete sense) if and only if the induced sequence of $n$-torsion is exact for every integer $n \ne 0$. That's a nice exercise. Then use the functorial isomorphism between $n$-torsion of dual abelian variety and Cartier dual of $n$-torsion (see Oda's thesis), coupled with the exactness of Cartier duality on finite flat commutative group schemes. That's all over a field, but works for abelian schemes (once know dual exists). $\endgroup$
    – BCnrd
    Commented Sep 3, 2010 at 15:04

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