> 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 ... but the residue field is still Z/pZ. In the above example: sage: K.factor(p) (Fractional ideal (-a))^7 sage: P = K.factor(p)[0][0] sage: P.ramification_index() 7 sage: K.residue_field(P) Residue field of Fractional ideal (-a) sage: K.residue_field(P).order() 7 -cc --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
