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Topic review - Milnor global number
Author Message
  Post subject:  Re: Milnor global number  Reply with quote
Hello everybody

I found the solution.
Singular command codim(I, J) return the vectorspace dimension of I/J , I and J ideals.
Here We consider the ring S=K[x,y] as ideal, generated by 1.
If J is the ideal J = Jacob (f), the Milnor global number is the codimension
of the ideal generated by the jacobian and f^2 (for 3 variables f^3)
So, codim((1), std(I)) where I=J, f*f;
Example for f= (y^2*(y+1)-x^3)^2-x^3*y^3,
Milnor global number it is indeed 20.

gstic
Post Posted: Thu Aug 25, 2016 2:12 pm
  Post subject:  Re: Milnor global number  Reply with quote
see http://www.singular.uni-kl.de/Manual/4-0-3/sing_418.htm
(vector space dimension of a polynomial ring modulo an ideal)
see http://www.singular.uni-kl.de/Manual/4- ... g_1567.htm
(Milnor number)
Code:
> ring r=0,(x,y> ring r=0,(x,y),dp;
> poly f= (y^2*(y+1)-x^3)^2-x^3*y^3;
// first possibility:
> ideal i=jacob(f);
> ideal j=std(i);
> vdim(j);
25
// or, second possibility:
> LIB "sing.lib";
> milnor(f);
25
// and the vector space basis:
> kbase(j);
....
Post Posted: Thu Aug 25, 2016 10:23 am
  Post subject:  Milnor global number  Reply with quote
Hello,

I want to calculate Milnor global number for a polynomial f in 2 variables in polynomial ring S=K[x,y].
If J is the ideal J = Jacob (f), the Milnor global is the size of S/J as a vector space.
How do I calculate the size of S/J as a vector space over K?
Example for f= (y^2*(y+1)-x^3)^2-x^3*y^3,
Milnor_global need to get = 20.

Thanks in advance
Post Posted: Wed Aug 24, 2016 8:42 pm


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