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May 17, 2016 at 2:28 comment added nfdc23 Curiously, the proof becomes a bit easier if one doesn't pass to the quasi-simple case, as one can formulate the result with a strong uniqueness aspect (pinning down $(k'/k, G')$ uniquely up to unique isomorphism, with $k'$ a nonzero finite etale $k$-algebra and $G'$ a smooth affine $k'$-group whose fibers are connected semisimple, absolutely simple, and simply connected -- or adjoint); that handles the Galois descent by pure thought. See Proposition A.5.14 in the book "Pseudo-reductive groups" for such a proof, the version over fields for the reference given by Cesnavicius over rings.
May 16, 2016 at 20:36 comment added zeno The second sentence of the proof: in general $G$ will only be isogenous to such a product.
May 16, 2016 at 20:34 history edited zeno CC BY-SA 3.0
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May 16, 2016 at 19:27 comment added Daniel Loughran Thanks. Though can you please emphasise at which point in the argument you use that $G$ simply connected or adjoint?
May 16, 2016 at 18:51 history answered zeno CC BY-SA 3.0