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| Dimension of a finite-dimensional quotient ring https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=2832 |
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| Author: | guest12 [ Wed Mar 06, 2019 6:30 pm ] |
| Post subject: | Dimension of a finite-dimensional quotient ring |
I would like to compute the dimension of a coinvariant algebra. I'm not familiar with Singular so I have a few naive questions, listed after my code : Code: LIB "finvar.lib"; ring R = 0, (a,b,c,d), dp; matrix A[4][4] = -1,3,0,0,0,1,0,0,0,0,-1,0,0,0,3,1; matrix B[4][4] = 1,0,0,0,1,-1,0,0,0,0,1,1,0,0,0,-1; matrix P,S,IS = invariant_ring(A,B,intvec(0,0,1)); 
ideal I = S[0],S[1],S[2],S[3],S[4],S[5],S[6],S[7],S[8],S[9],S[10],S[11],S[12],P[1],P[2],P[3],P[4]; qring Q = groebner(I); 1) How can I compute the dimension of Q as a vector space ? 2) When I ask explicitly the generators of I, I obtained a lot of "gen(1)" as a variable, what does it mean ? 3) If I want to repeat the process with a bigger vector space and a bigger group, how can I directly get the ideal generated by all the components of P and S without copying everything ? |
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