Hi Dusan, On 1 Aug., 11:35, Dušan Orlović <[email protected]> wrote: > we get f=y*I.0 -2*y^3*I.1 -x*I.2 = x*y^2 > We CAN reduce f on [x*y^3, 2*x^2 + x*y, 3*x*y, 2*y^2] to zero because > f = y * (3*x*y) - x * (2*y^2) .
No, we can't. You have shown that f belongs to the ideal generated by [x*y^3, 2*x^2 + x*y, 3*x*y, 2*y^2]. But the reduction process relies on the question whether the leading term (i.e., including the coefficient) of f is divisible by the leading term of any of the elements of [x*y^3, 2*x^2 + x*y, 3*x*y, 2*y^2]. The leading term of f is x*y^2, and this is *not* divisible by either of x*y^3, 2*x^2, 3*x*y or 2*y^2 (note that the coefficients are in ZZ, not in QQ). Best regards, Simon -- To post to this group, send an email to [email protected] To unsubscribe from this group, send an email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
