[sage-devel] Re: f in ZZ[x] mod p gives ZZ[x]
On Fri, Apr 24, 2009 at 9:03 AM, Robert Miller rlmills...@gmail.com wrote: sage: x = polygen(ZZ) sage: f = 2*x^2 sage: f.mod(2)==0 False You should do f.mod? and read the docstring, which says: Return a representative for self modulo the ideal I (or the ideal generated by the elements of I if I is not an ideal.) I believe f itself is a representative for f mod the ideal 2. :-) You're assuming that the mod function does something interesting, but it is in this case just some generic code which does what its definition says, which in this case happens to be nothing. So make it better! :-) William sage: type(f.mod(2)) type 'sage.rings.polynomial.polynomial_integer_dense_flint.Polynomial_integer_dense_flint' Even this doesn't work: sage: R.x = ZZ[] sage: f.mod(2*R)==0 False But last I checked, 2 | 2x^2. -- William Stein Associate Professor of Mathematics University of Washington http://wstein.org --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to sage-devel-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: f in ZZ[x] mod p gives ZZ[x]
Worse still: sage: x = polygen(QQ) sage: h = 4*x sage: h%3 0 --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to sage-devel-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: f in ZZ[x] mod p gives ZZ[x]
Worse still: sage: x = polygen(QQ) sage: h = 4*x sage: h%3 0 Over QQ[x], isn't 4*x = 3 * (4/3*x) ? Over ZZ, it's fine: sage: x = polygen(ZZ) sage: h = 4*x sage: h%3 x -cc --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to sage-devel-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: f in ZZ[x] mod p gives ZZ[x]
Yeah, I should have mentioned that my point was that maybe h%3 should raise an error over QQ. On Apr 24, 11:00 am, Craig Citro craigci...@gmail.com wrote: Worse still: sage: x = polygen(QQ) sage: h = 4*x sage: h%3 0 Over QQ[x], isn't 4*x = 3 * (4/3*x) ? Over ZZ, it's fine: sage: x = polygen(ZZ) sage: h = 4*x sage: h%3 x -cc --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to sage-devel-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
[sage-devel] Re: f in ZZ[x] mod p gives ZZ[x]
On Fri, Apr 24, 2009 at 11:53 AM, Robert Miller rlmills...@gmail.com wrote: Yeah, I should have mentioned that my point was that maybe h%3 should raise an error over QQ. Over QQ, the number 3 generates the unit ideal, so everything is 0 modulo it :-). William On Apr 24, 11:00 am, Craig Citro craigci...@gmail.com wrote: Worse still: sage: x = polygen(QQ) sage: h = 4*x sage: h%3 0 Over QQ[x], isn't 4*x = 3 * (4/3*x) ? Over ZZ, it's fine: sage: x = polygen(ZZ) sage: h = 4*x sage: h%3 x -cc -- William Stein Associate Professor of Mathematics University of Washington http://wstein.org --~--~-~--~~~---~--~~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to sage-devel-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
