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Let $k$ be an arbitrary closed field (of arbitrary characteristic). Assume that we have a short exact sequence of k-algebraic abelian connected groups $$ 1\rightarrow K\rightarrow G \rightarrow H\rightarrow 1 $$ where $H$ is an abelian variety and $K$ is a (commutative) linear algebaic group. Assume moreover (if it helps) that $$ 1\rightarrow Lie(K)\rightarrow Lie(G) \rightarrow Lie(H)\rightarrow 1 $$ is exact. Let $R$ be a ring and assume that we have an injection $\rho:R\rightarrow End_k(G)$.

Q: Does $\rho$ necessarily restrict to $K$?

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  • $\begingroup$ If I don't miss anything, the question is just whether every element of $End_k(G)$ stabilizes $K$. I guess $K$ is the intersection of all homomorphisms from $G$ to abelian varieties, which suggests a positive answer. $\endgroup$
    – YCor
    Commented May 27, 2014 at 13:08
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    $\begingroup$ Indeed Chevalley structure theorem tells you that the group $K$ is the unique normal linear subgroup of $G$ such that $G/K$ is an abelian variety, so it is preserved by any endomorphism of $G$. $\endgroup$
    – abx
    Commented May 27, 2014 at 13:36
  • $\begingroup$ Nice, so this answers completely my question. $\endgroup$
    – Hugo Chapdelaine
    Commented May 27, 2014 at 14:23
  • $\begingroup$ But doesn't it follow from the simpler observation that if $\theta\in End_k(G)$ and $\pi:G\rightarrow A$ then the map $\pi\circ\theta:K\rightarrow A$ is constant since $K$ is an affine variety and $A$ is projective? $\endgroup$
    – Hugo Chapdelaine
    Commented May 27, 2014 at 14:32
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    $\begingroup$ There can be nontrivial maps from an affine variety into a projective one, think of $\mathbb{A}^n\subset \mathbb{P}^n$. However a linear group is rational (over an algebraically closed field), and there is indeed no nontrivial map from a rational variety into an abelian variety. $\endgroup$
    – abx
    Commented May 27, 2014 at 15:07

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