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Feb 21, 2016 at 1:46 vote accept Drike
Feb 14, 2016 at 6:19 comment added nfdc23 @zeno: The proof in B.1 seems elementary and self-contained, especially if one runs it over a perfect field. Is that more complicated than the approach via k[F]-modules when one includes the work needed to set up the relationship to k[F]-modules? (Perhaps this is just a matter of taste.)
Feb 13, 2016 at 18:40 comment added zeno Over a perfect field, the statement follows from the description of commutative algebraic groups with zero Verschiebung in terms of modules over k[F] (cf. Demazure Gabriel p.523). The treatment in CGP is more complicated because they are interested in imperfect base fields.
Feb 13, 2016 at 16:00 history answered nfdc23 CC BY-SA 3.0