Question

Why is the Jacobian Conjecture so hard to prove when only a little knowledge beyond Calculus is needed to understand it?

Answer 1

F has a polynomial inverse function G: k^N → k^N, then J_F has a polynomial reciprocal, so is a nonzero constant. The Jacobian conjecture is the following partial converse:

Firstly, this question does not have any meaningful answer. I don't know any satisfactory way of describing why a problem is hard.

So, I shall rather explain what the Jacobian Conjecture is. More details to be found on Ludwik Druzkowskis "Effective approach to Keller's Jacobian Conjecture".

Let k have characteristic 0. If J_F is a non-zero constant, then F has an inverse function G: k^N → k^N which is regular, meaning its components are polynomials.

According to van den Essen (1997), the problem was first conjectured by Keller in 1939 for the limited case of two variables and integer coefficients (which has been proved — see § Results).

The obvious analogue of the Jacobian conjecture fails if k has characteristic p > 0 even for 1 variable. The characteristic of a field must be prime, so it is at least 2. The polynomial x − x^p has derivative 1 − px x^p−2 which is 1 (because px is 0) but it has no inverse function. However, Adjamagbo (1995) suggested extending the Jacobian conjecture to characteristic p > 0 by adding the hypothesis that p does not divide the degree of the field extension k(X) / k(F).

The condition J_F ≠ 0 is related to the inverse function theorem in multivariable calculus. In fact for smooth functions (and so in particular for polynomials) a smooth local inverse function to F exists at every point where J_F is non-zero. For example, the map x → x + x³ has a smooth global inverse, but the inverse is not polynomial.