On Aug 29, 11:47 am, "John Cremona" <[EMAIL PROTECTED]> wrote: > If f is a real polynomial then f.roots(RR) gives the real roots, and > f.roots(CC) gives the complex (including real) roots. Is there a > fool-proof way of getting at the non-real roots? (You may assume that > the coefficients of f are exact, perhaps they are rationals, so the > questions is certainly well-defined.)
You can do f.roots(QQbar), and then go through the roots and check r.imag()!=0 (this is an exact computation). > But now I need it again in the context of K.embeddings(L) for K a > number field. With L=RR it gives the real embeddings, and with K=CC > it gives the real-and-complex embeddings, all of which have codomain > CC even if they are actually real. But in my code I need to treat > real embeddings differently from complex ones! I cannot be the first > person to need this. Again, you can do K.embeddings(QQbar) and then check the imaginary part of the image of the generator. Unfortunately, this may be much slower than the corresponding call to f.roots(QQbar), because of technical details in the implementation of .embeddings() (it sorts the roots of the polynomial; over QQbar, this sorting involves a lot of exact computation to prove that the real parts of conjugate pairs are actually equal). Carl --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-devel@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-devel URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
