The Jacobian conjecture meets a counterexample

Suppose $F$ has a polynomial inverse $G$, so $G(F(x)) = x$ for all $x$. The chain rule gives $DG(F(x))\, DF(x) = I$, and taking determinants,

So $\det DF$ is a polynomial that never vanishes on $\mathbb{C}^n$, and such a polynomial must be a constant. Keller asked in 1939 whether the converse holds: does a constant nonzero determinant guarantee a polynomial inverse?

By the inverse function theorem, $\det DF(a) \neq 0$ means $F$ is invertible on a small neighbourhood of $a$. The conjecture asks for much more: one inverse that works everywhere, and is itself a polynomial. In one variable the answer is easy, since $F'(x) = c \neq 0$ forces $F(x) = cx + d$. Over the real numbers the answer to a stronger question is no: Pinchuk (1994) found a real polynomial map of the plane whose Jacobian never vanishes but which is not injective. His Jacobian is not constant, so it did not settle Keller's question.

Białynicki-Birula and Rosenlicht proved in 1962 that an injective polynomial map $\mathbb{C}^n \to \mathbb{C}^n$ is automatically bijective, with a polynomial inverse. So a counterexample must fail to be injective: there must be points $a \neq b$ with

That is exactly the form of the new counterexample: a constant determinant, and a collision.

Wang (1980) proved the conjecture for maps of degree at most two. Bass, Connell and Wright (1982) and Yagzhev (1980) showed that it would be enough to prove it for maps $F(x) = x + H(x)$ with $H$ homogeneous of degree three, in any number of variables; for these the condition says that the matrix $DH$ is nilpotent. A counterexample $F$ in three variables gives one in every higher dimension, since $(x, y) \mapsto (F(x), y)$ has the same determinant and the same collision. The case of two variables is untouched by this: it would need a genuinely new example or a proof.