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Let $A$ be an abelian variety over a field of characteristic $0$ and let $R \subset \mathrm{End}(A)$ be a (commutative, if that matters) subring. Suppose that $B \subset A$ is an $R$-stable abelian subvariety. Does $B$ have an $R$-stable complement? More precisely, does there exist an $R$-stable abelian subvariety $C \subset A$ such that $B + C = A$ and $B \cap C$ is finite?

I am guessing that a $C$ does not exist in general, but I do not have an example. In the positive direction, when $R = \mathbb{Z}[G]$ is a group ring, I think $C$ should exist by building a $G$-equivariant polarization and running the usual argument with the induced polarization on $B$ (Proposition 13.5.1 in the book "Complex abelian varieties" of Birkenhake and Lange is a result of this flavor).

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    $\begingroup$ Consider an elliptic curve $E$ and let $R$ be the (commutative) subring generated by the endomorphism $(x,y)\mapsto (0,x)$. At the level of the tangent space at 0, the only invariant subspace is that of $\{0\}\times E$. It's easy to deduce that $\{0\}\times E$ is the only 1-dimensional abelian subvariety of $E\times E$ stable under $R$. In particular it has no stable complement. $\endgroup$
    – YCor
    Nov 3, 2016 at 19:07
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    $\begingroup$ (the previous example is essentially a group ring, with $G$ cyclic infinite. In your fact about group rings, you probably mean $G$ to be finite.) $\endgroup$
    – YCor
    Nov 3, 2016 at 19:09
  • $\begingroup$ @YCor: Thank you. Yes, in my remark I implicitly assumed that $G$ is finite. $\endgroup$
    – Lisa S.
    Nov 3, 2016 at 19:55

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