On Tue, Nov 18, 2008 at 7:43 PM, John H Palmieri <[EMAIL PROTECTED]> wrote: > > Thanks for the answers. By the way: > >> Infinite fields of characteristic p aren't perfect, > > Isn't the algebraic closure of F_p perfect? For fields of > characteristic p, perfect should mean that every element has a pth > root. (I agree that some infinite fields of characteristic p aren't > perfect, but this is not true for all of them.)
Sorry, Fpbar is indeed an exception. I was thinking of "global fields", which are fields that are finitely generated over their prime subfield (in this case, F_p). Any algebraically closed field is perfect, since there are no nontrivial finite extensions. If a field F is infinite and finitely generated over its prime subfield, it has to have transcendence degree at least 1, hence contain k(t), where k is a finite field and t is an indeterminate. If F were perfect, it would contain all p-power roots of t, hence it wouldn't be finitely generated over F_p. -- William --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
