The exact three-point fiber

Substitution proves that at least three points lie over \(q=(-1/4,0,0)\). A Gröbner basis proves that there are no others and that no hidden multiplicities occur.

In \(\mathbb Q[x,y,z]\), exact ideal reduction gives

\[ \boxed{ \left(F_1+\frac14,F_2,F_3\right) =\left(z-\frac{27}{4}x^2+\frac14, y+\frac32x, x^3-x\right).} \]

The quotient has \(\mathbb Q\)-basis \(1,x,x^2\), so the fiber scheme has length three. Since

\[ x^3-x=x(x-1)(x+1) \]

has distinct roots in characteristic zero, the quotient is reduced. Solving the triangular basis gives exactly

\[ (0,0,-1/4),\quad(1,-3/2,13/2),\quad(-1,3/2,13/2). \]

A direct proof without Gröbner bases

Put \(u=1+xy\) and

\[ r=2-3xy-x^2z, \]

so \(F_3=xr\). If \(F_3=0\), there are two cases.

If \(x=0\), then \(F_2=y\) and \(F_1=z+4y^2\). The target forces \(y=0,z=-1/4\).

If \(x\ne0\), then \(r=0\). In the weighted form one has

\[ F_2=\frac{4u+2}{x},\qquad F_1=\frac{u(u+1)}{x^2}. \]

The target forces \(u=-1/2\), followed by \(x^2=1\). Then \(y=(u-1)/x\) and \(z=(2-3xy)/x^2\), producing precisely the two remaining points.

The Magma file 03_fiber_certificate.m verifies ideal equality, zero-dimensionality, and radicality directly.