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Jan 22, 2015 at 12:38 comment added jmc It might be worthwile to note that if the field has characteristic $0$, then being foolish does no harm (all group schemes are reduced).
Sep 12, 2010 at 14:07 comment added André Henriques Tank you t3suji. I now understand my mistake, and edited my answer accordingly.
Sep 12, 2010 at 13:23 comment added t3suji @Andre Henriques: It is surjective, so the cokernel is trivial. So it identifies the quotient of the additive group by the Frobenius kernel with the additive group.
Sep 11, 2010 at 13:28 comment added André Henriques @Milne: Concerning your last sentence: Yes, it has a kernel. But what is its cokernel?
Sep 10, 2010 at 12:26 history edited JS Milne CC BY-SA 2.5
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Sep 9, 2010 at 14:42 history answered JS Milne CC BY-SA 2.5