Hi,
I'm trying to obtain the planar projection of a curve by eliminating 5 variables in a system with 7 unknowns and 6 equations:
Code:
ring r=0,(x,v,c1,c2,c3,c4,c5),lp;
ideal I=(-16*c1*c2*c3*c4*c5+16*v,16*c1*c2*c3*c4-16*(-c1*c2*c3-(c1*c2-(-c1-c2)*c3)*c4)*c5+40*v,-16*c1*c2*c3-16*(c1*c2-(-c1-c2)*c3)*c4-16*(c1*c2-(-c1-c2)*c3-(-c1-c2-c3)*c4)*c5+25*v,16*c1*c2-16*(-c1-c2)*c3-16*(-c1-c2-c3)*c4-16*(-c1-c2-c3-c4)*c5-25,-16*c1-16*c2-16*c3-16*c4-16*c5-40,(c1-c3)*(c2-c4)-x*(c1-c4)*(c2-c3));
eliminate(I,c1*c2*c3*c4*c5);
ideal I=(-16*c1*c2*c3*c4*c5+16*v,16*c1*c2*c3*c4-16*(-c1*c2*c3-(c1*c2-(-c1-c2)*c3)*c4)*c5+40*v,-16*c1*c2*c3-16*(c1*c2-(-c1-c2)*c3)*c4-16*(c1*c2-(-c1-c2)*c3-(-c1-c2-c3)*c4)*c5+25*v,16*c1*c2-16*(-c1-c2)*c3-16*(-c1-c2-c3)*c4-16*(-c1-c2-c3-c4)*c5-25,-16*c1-16*c2-16*c3-16*c4-16*c5-40,(c1-c3)*(c2-c4)-x*(c1-c4)*(c2-c3));
eliminate(I,c1*c2*c3*c4*c5);
but it seems that I'm being far too naive. I know that the answer should have degree 30 in v and 16 in x, so it should be manageable; and I also remember that I had computed it in 2013, I just forgot how
Note that all equations are symmetric in c1...c5, except for the last one which says that x is the cross ratio of c1...c4.
Many thanks in advance! Laurent