Determinant-one integral normalization

Some formulations normalize the constant Jacobian to \(1\) and the linear part to the identity. Both conditions can be imposed without losing integer coefficients.

The linear part of \(F\) is

\[ L(x,y,z)=(z,y,2x). \]

Let \(D=\operatorname{diag}(2,1,1)\), and define

\[ H=D^{-1}\circ L^{-1}\circ F\circ D. \]

Writing \(T=1+2XY\), this becomes

\[\begin{split} \boxed{ \begin{aligned} H_1&=X-3X^2Y-2X^3Z,\\ H_2&=Y+6XT^2Z+6XY^2(4+6XY),\\ H_3&=T^3Z+Y^2T(4+6XY). \end{aligned}} \end{split}\]

It follows either by conjugation or exact differentiation that

\[ H(0)=0,\qquad dH_0=I_3,\qquad \det JH=1. \]

Nevertheless,

\[\begin{split} \begin{aligned} H(0,0,-1/4) &=H(1/2,-3/2,13/2)\\ &=H(-1/2,3/2,13/2)=(0,0,-1/4). \end{aligned} \end{split}\]

Thus the counterexample survives the strongest elementary affine normalization. Its coordinate degrees, in the displayed output order, are \((4,6,7)\).

Why conjugation preserves the conclusion

Both \(L\) and \(D\) are invertible linear maps. Pre- and postcomposition by such maps preserves injectivity and polynomial invertibility. The determinant changes by the product of the constant linear determinants; here the choice of \(L^{-1}\) changes \(-2\) to \(1\), while conjugation by \(D\) changes nothing.