Polynomial mappings with Jacobian determinant of bounded degree (Q1086295)

Let \(\mathbb{C}\) denote the complex numbers and let \(F: \mathbb{C}^n\to \mathbb{C}^n\) be a map whose coordinate functions, \(F_1,\ldots,F_n\), are polynomials. If \(F\) has a globally defined inverse then it is easy to see that \(F\) must have a Jacobian determinant that is identically a non-zero constant. The Jacobian problem asks if the converse is true. The problem is trivially true for \(n=1\) and unsolved otherwise. The most studied special case is the case \(n=2\). In this case, \textit{S. S. Abhyankar} has shown [Lectures on expansion techniques in algebraic geometry. Tata Inst. Fund. Res. 57. Bombay: Tata Institute of Fundamental Research (1977; Zbl 0818.14001)] that the Jacobian problem can be answered in the affirmative if we can show that the \(F_1\) and \(F_2\) must have only one point at infinity when the Jacobian determinant of \(F\) is identically a non-zero constant. Abhyankar has also shown in the book cited above that \(F_1\) and \(F_2\) have at most two points at infinity given the assumption on the Jacobian determinant. The latter result was also proven independently by \textit{L. G. Makar-Limanov} [Funct. Anal. Appl. 4(1970), 262--264 (1971); translation from Funkts. Anal. Prilozh. 4, No. 3, 107--108 (1971; Zbl 0218.13006)]. In this paper we show that if \(F\) has Jacobian determinant that is identically constant, the \(F_1, F_2\) have two zeros at infinity and both of these zeros have the same multiplicity, in a canonical sense, then the Jacobian determinant of \(F\) must be identically zero.