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May 27, 2014 at 18:01 comment added YCor PS In my previous comment I meant "the intersection of all kernels of homomorphisms", of course, sorry for the typo.
May 27, 2014 at 16:10 comment added Hugo Chapdelaine Yes you're right. So I guess that the key point is that a subgroup of an abelian variety is necessarily an abelian variety and a quotient of a linear group is linear. Thus the only linear group which is an abelian variety is the trivial group.
May 27, 2014 at 15:07 comment added abx 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.
May 27, 2014 at 14:34 history undeleted Hugo Chapdelaine
May 27, 2014 at 14:34 history deleted Hugo Chapdelaine via Vote
May 27, 2014 at 14:32 comment added Hugo Chapdelaine 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?
May 27, 2014 at 14:23 comment added Hugo Chapdelaine Nice, so this answers completely my question.
May 27, 2014 at 13:36 comment added abx 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 $G$.
May 27, 2014 at 13:08 comment added YCor 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.
May 27, 2014 at 12:54 history edited Hugo Chapdelaine CC BY-SA 3.0
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May 27, 2014 at 12:48 history asked Hugo Chapdelaine CC BY-SA 3.0