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Nov 29, 2012 at 16:40 vote accept user29513
Nov 29, 2012 at 16:37 comment added Lubin The old-fashioned way I see this, and Felipe can slap me down if I’m wrong, is that the transcendence-degree-one field of functions on $A$ has only one purely inseparable subfield of each possible degree $p^r$.
Nov 29, 2012 at 16:29 history answered Felipe Voloch CC BY-SA 3.0