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| linear part of an ideal https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1902 |
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| Author: | aniakedz [ Mon Jan 10, 2011 2:59 pm ] |
| Post subject: | linear part of an ideal |
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 |
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| Author: | vinay [ Tue Jan 11, 2011 1:06 pm ] |
| Post subject: | Re: linear part of an ideal |
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 |
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| Author: | gorzel [ Wed Jan 12, 2011 9:22 pm ] |
| Post subject: | Re: linear part of an ideal |
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 |
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