Let $G$ be a connected algebraic group in positive characteristic $p$. If the Frobenius kernel $G_{(p)}=ker (F:G\to G^{(p)})$ is unipotent, do we have $G$ also unipotent?
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1$\begingroup$ What about if $G$ is etale of order prime to $p$? $\endgroup$– user30035Feb 15, 2013 at 23:08
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$\begingroup$ I forgot to say $G$ is connected. $\endgroup$– XingtingFeb 16, 2013 at 2:28
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2$\begingroup$ I suppose $G$ is meant to be affine (otherwise a supersingular elliptic curve would be a counterexample). We can assume the ground field is algebraically closed, and then the smooth connected subgroup $G_{\rm{red}}$ inherits the hypothesis, so it contains no nontrivial torus. Consequently, $G_{\rm{red}}$ must be unipotent, by the structure theory of smooth connected affine groups over alg. closed fields. Are you interested in non-smooth $G$? If so then 4.3.1 in Exp. XVII of SGA3 gives the affirmative answer (absence of $\mu_p$ forces unipotence as a group scheme, assuming connectedness). $\endgroup$– user30379Feb 16, 2013 at 8:49
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$\begingroup$ Yes, I'm interested in non-smooth ones. Thanks for the reference you gave. $\endgroup$– XingtingFeb 16, 2013 at 17:33
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