Yes. Use "places": sage: K.<a>=NumberField(x^2-3) sage: K.places()
[Ring morphism: From: Number Field in a with defining polynomial x^2 - 3 To: Real Field with 106 bits of precision Defn: a |--> -1.732050807568877293527446341506, Ring morphism: From: Number Field in a with defining polynomial x^2 - 3 To: Real Field with 106 bits of precision Defn: a |--> 1.732050807568877293527446341506] See the docstring. You can set all_complex=True to get both conjugates if needed. John Cremona On Jul 24, 10:41 am, mac8090 <bonzerpot...@hotmail.com> wrote: > For a number field X, the complex embeddings of X can be given by > > X.complex_embeddings() > > However, is there then a way to give embConj, the set of all > embeddings without conjugates? > > ie for X=QQ(d) with conjugate pair of embeddings: > > a: d --> x + I*y > b: d --> x - I*y > > then if embConj contains a, it does not contain b - so len(emb) = > 2*len(embConj) > > any ideas? --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
