Jacobian Conjecture

Importance: High ✭✭✭
Recomm. for undergrads: no
Posted by: Charles
on: July 6th, 2008

\begin{conjecture} Let $k$ be a field of characteristic zero. A collection $f_1,\ldots,f_n$ of polynomials in variables $x_1,\ldots,x_n$ defines an automorphism of $k^n$ if and only if the Jacobian matrix is a nonzero constant. \end{conjecture}

The Jacobian determinant is the determinant of the matrix $A$ with $a_{ij}=\frac{\partial f_i}{\partial x_j}$. It is elementary to show that if the map $F:k^n\to k^n$ is an automorphism, then the Jacobian determinant is a nonzero constant, by using the inverse map. The other direction has turned out to be rather difficult.

It is known that the Conjecture holds for polynomials of degree 2, and that the general case follows from a special case in degree 3.


% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

* indicates original appearance(s) of problem.