The integral closure of ideals (trying to use reesclos.lib)
Dmitry · Thu Dec 19, 2013 11:05 am
I need to compute the integral closures of ideals (ie.g. in k[[x,y]]). So, I'm playing with the following:
LIB "reesclos.lib";
ring r=0,(x,y),dp;
poly f=.....
ideal I=x*diff(f,x),y*diff(f,y);
list JJ=normalI(I);
for weighted homogeneous cases (e.g. f=x^p+y^q) the answer comes quickly. For non-weighted homogeneous, but Newton-non-degenerate (e.g. f=x^5+x^2*y^2+y^5) I could not get the answer in char=0, but at least the computation is successful for : >>>ring r=11,(x,y),dp;
In the simplest degenerate case, f=(x^2-y^3)^2+x*y^4;, I could not get the answer even in char=11. (After several minutes I just stopped Singular.)
Am I doing smth wrong/stupid?
Are there any faster ways to compute the integral closures of ideals?
Thanks!
LIB "reesclos.lib";
ring r=0,(x,y),dp;
poly f=.....
ideal I=x*diff(f,x),y*diff(f,y);
list JJ=normalI(I);
for weighted homogeneous cases (e.g. f=x^p+y^q) the answer comes quickly. For non-weighted homogeneous, but Newton-non-degenerate (e.g. f=x^5+x^2*y^2+y^5) I could not get the answer in char=0, but at least the computation is successful for : >>>ring r=11,(x,y),dp;
In the simplest degenerate case, f=(x^2-y^3)^2+x*y^4;, I could not get the answer even in char=11. (After several minutes I just stopped Singular.)
Am I doing smth wrong/stupid?
Are there any faster ways to compute the integral closures of ideals?
Thanks!