Direct exact certificate

Direct exact certificate#

The most literal certificate differentiates the three displayed polynomials and expands a \(3\times3\) determinant in \(\mathbb Q[x,y,z]\). SymPy, SageMath, and Magma all reduce the result to \(-2\).

The complete executable versions are:

A minimal SymPy fragment is:

import sympy as s

x, y, z = s.symbols("x y z")
F = s.Matrix([
    (1+x*y)**3*z + y**2*(1+x*y)*(4+3*x*y),
    y + 3*x*(1+x*y)**2*z + 3*x*y**2*(4+3*x*y),
    2*x - 3*x**2*y - x**3*z,
])

assert s.expand(F.jacobian((x, y, z)).det()) == -2

Exact substitutions#

At \(P_0=(0,0,-1/4)\), one immediately gets

\[ F(P_0)=(-1/4,0,0). \]

At \(P_+=(1,-3/2,13/2)\), the useful value is \(1+xy=-1/2\). Then

\[\begin{split} \begin{aligned} F_1(P_+)&=(-1/2)^3(13/2)+(9/4)(-1/2)(-1/2)=-1/4,\\ F_2(P_+)&=-3/2+3(1)(1/4)(13/2)+3(1)(9/4)(-1/2)=0,\\ F_3(P_+)&=2+9/2-13/2=0. \end{aligned} \end{split}\]

The involution

\[ (x,y,z)\longmapsto(-x,-y,z) \]

leaves \(F_1\) fixed and changes the signs of \(F_2,F_3\). Since the latter two values are zero at \(P_+\), it sends the calculation to \(P_-=(-1,3/2,13/2)\) automatically.

Why this is not the preferred proof#

A full determinant expansion is decisive but opaque: it confirms cancellation without explaining it. The next chapter compresses the determinant to a two-by-two calculation and exposes the hidden cubic geometry.