It turns out that for my problem it suffices to compute the polynomials 1, f, f^2, ..., f^K
modulo N for some large but known number N, so I suspect that it may be faster to work modulo p for various primes p and then use the Chinese Remainder Theorem. How large, roughly speaking, can I take my prime p to be without slowing down the multiplication in (ZZ/pZZ)[x,y,z] ? Best, Tom -- To post to this group, send an email to sage-devel@googlegroups.com To unsubscribe from this group, send an email to sage-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
