laurentbartholdi's profile - activity

I was informed by Pierre-Jean Spaenlehauer that the Gröbner basis code in FGb solves the problem in about a minute.

It's been running for a day without success. I remember that I had gotten something similar to work in less than an hour, in 2013; but too many of my brain cells died since then. Note that I.elimination_ideal([c1,c2]) already fails. I suspect that it should be possible to use the symmetry of the polynomials in some way or other.

Hi, I'm trying to eliminate 5 variables from a system of 6 equations in 7 unknowns, to obtain a planar curve. Am I asking for too much?

This is what I tried:

R.<x,v,c1,c2,c3,c4,c5> = PolynomialRing(QQ)
p0 = -16*c1*c2*c3*c4*c5+16*v
p1 = 16*c1*c2*c3*c4-16*(-c1*c2*c3-(c1*c2-(-c1-c2)*c3)*c4)*c5+40*v
p2 = -16*c1*c2*c3-16*(c1*c2-(-c1-c2)*c3)*c4-16*(c1*c2-(-c1-c2)*c3-(-c1-c2-c3)*c4)*c5+25*v
p3 = 16*c1*c2-16*(-c1-c2)*c3-16*(-c1-c2-c3)*c4-16*(-c1-c2-c3-c4)*c5-25;
p4 = -16*c1-16*c2-16*c3-16*c4-16*c5-40
I = ideal((c1-c3)*(c2-c4)-x*(c1-c4)*(c2-c3),p0,p1,p2,p3,p4)
J = I.elimination_ideal([c1,c2,c3,c4,c5])

Note that the system is symmetric in permutations of c1...c5, except for one equation that says that x is the cross-ratio of c1...c4.

I'm all the more embarrassed because I already computed the solution some time ago, but forgot how: it should be a polynomial J of degree 30 in v and 16 in x.

Many thanks in advance! Laurent