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representation obtained by reduction?

representation obtained by reduction?

John Perry · Fri Oct 17, 2008 11:34 pm

Suppose that B is an ideal, f is a polynomial, and r is obtained via
r = reduce(f,B);
Then there exists a list H of polynomials such that
f == H1*B1 + ... + Hm*Bm + r

Does Singular have a function/option to determine H, after computing r?

representation obtained by reduction?

bricken · Thu Oct 30, 2008 9:52 am

Hi John!
Lift might do the job.
http://www.singular.uni-kl.de/Manual/3-0-4/sing_235.htm
Michael

Lift yes, but...

John Perry · Thu Oct 30, 2008 11:13 pm

bricken wrote:
Hi John!
Lift might do the job.
http://www.singular.uni-kl.de/Manual/3-0-4/sing_235.htm
Michael
Thanks, that helps a great deal. I don't know that it's as efficient as I'd like, maybe you can say. To use lift to "recover" the H's, I understand that I need to type
Code:
r = reduce(f,B);
lift(B,f-r);
When Singular performs the reduction, it computes H once.

Then, when Singular computes the lift, it computes H a second time.

I'd prefer to compute H only the first time, and recover it without having Singular go through it again.

representation obtained by reduction?

greuel · Sat Nov 15, 2008 3:12 am

Hi,
I guess 'division' is what you want.

representation obtained by reduction?

Guest · Sat Nov 15, 2008 5:59 am

greuel wrote:
Hi,
I guess 'division' is what you want.

Thanks. Unfortunately, B is not necessarily a standard basis, and division appears to compute a standard basis of B before doing the division. For example,
Code:
> ring R=0,(x,y),dp;
> ideal l = xy+1,y2+1;
> poly f = x-y;
> division(f,l);
[1]:
   _[1,1]=-y
   _[2,1]=x
[2]:
   _[1]=0
[3]:
   _[1,1]=1
The answer I want is
Code:
[1]:
    _[1,1]=0
    _[2,1]=0
[2]:
    _[1]=x-y
[3]:
    _[1,1]=1
If I read the documentation correctly, division computes a standard basis, then lifts to l; I want the result from the non-standard basis. Is there an option to turn that behavior off?