On Wednesday, April 29, 2015 at 3:15:44 AM UTC-7, Jeroen Demeyer wrote: > > No, it's not, the trailing coefficient is O(5^20): > > sage: R.<x> = Qp(5)[] > sage: f = x^2 > sage: parent(f[0]) > 5-adic Field with capped relative precision 20 > Note the capped *relative* precision. These are "floats*. That's to say that elements are represented as p^v*u where u is a unit represented by an element in Z/p^rZ with r<=20 ... except that 0 admits representations like that for any v as long as r=0. Apparently we ended up having an "exact" zero in Qp(5).
So it is correct *in this case* that you can actually positively conclude that f is a square, since it's a monomial of degree 2. This is such a special case, however, that it doesn't make sense to support it in factorization, so we don't. I suspect that after removing content from the polynomial, factorization first moves to a (capped?) absolute precision representation, i.e., effectively we probably work with factorization of polynomials in (Z/p^nZ)[x], where f cannot be distinguished from polynomials that come from the square-free locus. -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
