Bill Allombert on Thu, 14 Nov 2013 14:33:27 +0100 |
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
Re: factorpadic() for non-squarefree polynomials |
On Thu, Nov 14, 2013 at 01:30:08PM +0100, Jeroen Demeyer wrote: > Hello pari-users, > > The documentation for ??factorpadic states that the discriminant of > the polynomial much have p-adic valuation less than r. While I > understand where the condition comes from, does that mean that the > following is undefined behaviour: I think the documentation is inaccurate. This function should return the Q_p factorization reduced to the precision p^r, which is very different from factoring over Z/p^rZ. What matter is the precision of the polynomial itself. If the polynomial has inexact entries and its discriminant is 0 to the precision, then the Q_p factorization is not well determined (it depend on a lifting). On the other hand, if the polynomial has exact entries, there is no issue. > gp> factorpadic(t^2, 3, 5) > %13 = > [(1 + O(3^5))*t + O(3^5) 2] t^2 is exact, so there is no ambiguity. > Indeed, the following example should be equivalent but it's not: > > gp> factorpadic(t^2 + 3^5, 3, 5) > %14 = > [(1 + O(3^5))*t^2 + O(3^5)*t + O(3^0) 1] No, it is not equivalent. t^2 + 3^5 is exact and is different from t^2. t^2 + 3^5 is irreducible in Q_p[t]. I do not know why it returns + O(3^0). Cheers, Bill.