The map and the claim

Let \(k\) be a field of characteristic different from \(2\). For the classical conjecture take \(k=\mathbb C\). Define \(F=(F_1,F_2,F_3):\mathbb A_k^3\to\mathbb A_k^3\) by

\[\begin{split} \begin{aligned} F_1&=(1+xy)^3z+y^2(1+xy)(4+3xy),\\ F_2&=y+3x(1+xy)^2z+3xy^2(4+3xy),\\ F_3&=2x-3x^2y-x^3z. \end{aligned} \end{split}\]

We use

\[ A=F_1,\qquad B=F_2,\qquad C=F_3 \]

when the outputs themselves are being manipulated. Lower-case \((a,b,c)\) denotes a fixed target point.

The coordinate degrees are

\[ (\deg F_1,\deg F_2,\deg F_3)=(7,6,4). \]

Main theorem

Theorem — Keller counterexample in dimension three

The map \(F\) satisfies

\[ \det\left(\frac{\partial(F_1,F_2,F_3)} {\partial(x,y,z)}\right)=-2. \]

Moreover,

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

The determinant is a nonzero constant, so \(F\) is a Keller map and an étale morphism. The three inputs are distinct, so \(F\) is not injective and cannot be a polynomial automorphism. This is exactly a counterexample to the usual Jacobian conjecture.

No theorem equating injectivity and polynomial invertibility is needed for the negative conclusion: a map with an inverse of any kind is necessarily injective.

Coefficient field

The determinant identity lies in \(\mathbb Z[x,y,z]\), and the collision lies in \(\mathbb Z[1/2]\). The same example therefore works over every field of odd characteristic. Characteristic zero is the historically meaningful case; positive-characteristic analogues were already known to fail for simpler reasons.