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Topic review - linear part of an ideal
Author Message
  Post subject:  Re: linear part of an ideal  Reply with quote
If your input is homogneous then you may set the variable degBound
http://www.singular.uni-kl.de/Manual/la ... htm#SEC385

Code:
>   ring r = 0,(x,y,z),dp;
>  ideal J = x2+xy, y2-xz, x2+z2;
>  std(J);
_[1]=y2-xz
_[2]=xy-z2
_[3]=x2+z2
_[4]=yz2+z3
_[5]=xz2+yz2
_[6]=z4
> degBound;  // default value
0
>  degBound = 3;  // computation up to degree 3
>  std(J);
_[1]=y2-xz
_[2]=xy-z2
_[3]=x2+z2
_[4]=yz2+z3
_[5]=xz2+yz2

> option(prot);    // enable prot, to see that the computation is indeed shorter
>  degBound = 2;
>  std(J);
[1023:1]2(2)sss
product criterion:1 chain criterion:0
_[1]=y2-xz
_[2]=xy-z2
_[3]=x2+z2

> degBound =0;  // reset
>  std(J);               // the std computation is longer
[1023:1]2(2)sss3ss4(3)-s-5--
product criterion:6 chain criterion:1
_[1]=y2-xz
_[2]=xy-z2
_[3]=x2+z2
_[4]=yz2+z3
_[5]=xz2+yz2
_[6]=z4
Post Posted: Wed Jan 12, 2011 9:22 pm
  Post subject:  Re: linear part of an ideal  Reply with quote
Here is a trick you can try:
Step 1: Define an ideal I = maxideal(2), i.e. 2nd power of the generic maximal ideal.
Step 2: Define quotient ring with this ideal. (i.e. kill higher degree terms)
Step 3: Consider your original ideal with the new ring. (The only survivors will be the linear terms!!!)

I hope this is what you wanted...

-- VInay
Post Posted: Tue Jan 11, 2011 1:06 pm
  Post subject:  linear part of an ideal  Reply with quote
hello!
I would like to calculate a basis of a rather large ideal- it is impossible to calculate the full basis, but I'm interested only in its linear part.

Is there a function that calculates directly a linear part of an ideal?
Alternatively, is there a function that calculates Groebner basis up a given degree? (1 in this case)?
I searched but to no avail. If anyone is knows how to go about this problem, I'd highly appreciate some advice!
Best,
Anna
Post Posted: Mon Jan 10, 2011 2:59 pm


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