Re: [sage-support] Re: How to find one element of residue field as a vector over base field

On Wednesday, May 15, 2019 at 10:08:05 PM UTC+9, Santanu wrote: > > > What is the value of $\frac{xy}{(x^2 + x + 1) } + >>> >>> \frac{1}{x^2 + x + 1}+$ Place $(x^2 + x + 1, x y + 1)$? >>> >>> >> You cannot add an element of the function field with a place. >> > Actually by this we meant the

Re: [sage-support] Re: How to find one element of residue field as a vector over base field

On Wednesday, May 15, 2019 at 10:08:05 PM UTC+9, Santanu wrote: > > > > On Wed, 15 May 2019 at 17:03, Kwankyu > > wrote: > >> Hi Chandra, >> >> What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by >>> >>> (x^2 + x + 1, x*y + 1). >>> >>> >> No. Place (x^2 + x + 1, x*y + 1) is the

Re: [sage-support] Re: How to find one element of residue field as a vector over base field

On Wed, 15 May 2019 at 17:03, Kwankyu wrote: > Hi Chandra, > > What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by >> >> (x^2 + x + 1, x*y + 1). >> >> > No. Place (x^2 + x + 1, x*y + 1) is the unique place of the function field > > at which both functions x^2 + x +1, x*y + 1 vanish. >

[sage-support] Re: How to find one element of residue field as a vector over base field

Hi Chandra, What is Place (x^2 + x + 1, x*y + 1)? Is it ideal generated by > > (x^2 + x + 1, x*y + 1). > > No. Place (x^2 + x + 1, x*y + 1) is the unique place of the function field at which both functions x^2 + x +1, x*y + 1 vanish. > What is the value of $\frac{xy}{(x^2 + x + 1) } + > >