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Milnor global number http://www.singular.unikl.de/forum/viewtopic.php?f=10&t=2553 
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Author:  gstic [ Wed Aug 24, 2016 8:42 pm ] 
Post subject:  Milnor global number 
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)^2x^3*y^3, Milnor_global need to get = 20. Thanks in advance 
Author:  hannes [ Thu Aug 25, 2016 10:23 am ] 
Post subject:  Re: Milnor global number 
see http://www.singular.unikl.de/Manual/403/sing_418.htm (vector space dimension of a polynomial ring modulo an ideal) see http://www.singular.unikl.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)^2x^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); .... 
Author:  gstic [ Thu Aug 25, 2016 2:12 pm ] 
Post subject:  Re: Milnor global number 
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)^2x^3*y^3, Milnor global number it is indeed 20. gstic 
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