On Jun 16, 2009, at 11:18 PM, Craig Citro wrote: > >> How about >> >> sage: p = 7 >> sage: K.<q> = QQ[p^(1/p)] >> sage: q^p >> 7 >> sage: F.<qbar> = K.residue_field(q) >> sage: F >> Residue field of Fractional ideal (a) >> >> Of course as p splits completely the residue field is always >> isomorphic to Z/pZ (with the obvious reduction map, as q * q^(p-1) == >> p). >> > > Actually, p is totally ramified in that extension -- it doesn't split > at all
Sorry, yes, there's only one prime above p, namely q. - Robert --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
