Hi Simon, please change upstream bug that you have reported, because I found that the problem is in reduce command. Please read this book<http://books.google.com/books?id=Caoxi78WaIAC&pg=PA201&dq=adams+loustaunau+introduction+to+grobner+bases+chapter+4&hl=sr&ei=CwdYTJzPHcGe4AaZsKmhBw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q=adams%20loustaunau%20introduction%20to%20grobner%20bases%20chapter%204&f=false>(just first five pages in chapter 4). I think the problem is in reduce command. Thanks,
On Sun, Aug 1, 2010 at 1:30 PM, Simon King <[email protected]> wrote: > 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]<sage-devel%[email protected]> > For more options, visit this group at > http://groups.google.com/group/sage-devel > URL: http://www.sagemath.org > -- 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
