By the works of Michiel de Bondt and Arno van den Essen, and Ludwik Drużkowski, it is known that if $F=I+N$, where $I$ is the identity mapping and $N$ is cubic homogeneous polynomials in $n$ complex variables, is invertible, then the Jacobian conjecture is true for $n$ variables.


Consider $J(N)$, the jacobian of the cubic homogeneous polynomials, the works of the above authors shows that $J(N)$ is symmetric and nilpotent in complex variables. In [joint work of Bass, Connell and Wright](https://www.ams.org/journals/bull/1982-07-02/S0273-0979-1982-15032-7), at page 18, the corollary 2.2. It states that for a commutative ring $k$ and $F=I+N$ cubic homogeneous polynomials, if $F$ is invertible, then it implies that for all polynomials of $k^{n}$, if their jacobian is invertible, then they are invertible.


I have many questions about the statement:


1: Does "invertible of $F$" mean having polynomial inverse? And determinant equals to constant?


2: In the assumption, $k$ is just a commutative ring, with no restriction on real number or complex number, does the condition that the $J(N)$ being nilpotent need to hold for the field $\mathbb{R}$?


**Edit**
For the first question, it got resolved by Theorem 2.1 in the above link. For the second question, what I want to know is, can the Jacobian conjecture be solved by proving every cubic homogeneous polynomials with their jacobian nilpotent in $\mathbb{R}$ has polynomial inverse?



Question about Jacobian conjecture on the reals
To (Q1): Yes, $F$ being invertible means in this context exactly that $F$ admits a polynomial *compositional* inverse. If you are working over a general commutative ring $k$, then invertibility of $J\_F$ is equivalent to invertibility of $\det J\_F$ in $k[x\_1,\dots,x\_n]$. In particular, if $k$ is not reduced, then the determinant need not be a constant. Vice versa, even if the determinant is a non-zero constant, this does not guarantee invertibility of $J\_F$, when $k$ is general (it has to be in the group of units of $k$).


To (Q2): Let me state a crucial observation that will hopefully clarify the matter (I believe it is also stated somewhere in BCW's paper.)


*Lemma:* Let $S$ be any $\mathbb{N}\_0$-graded (noncommutative) ring and let $s \in S$ with $s\_0 = 0$ (i.e. $s$ has vanishing degree 0 part). Then $1 + s \in U(S)$ if and only if $s$ is nilpotent, where $U(S)$ denotes the group of units of $S$.


(*Proof*: easy exercise.)


You can now take $S := \operatorname{M}\_n(k[x\_1,\dots,x\_n]) = \operatorname{M}\_n(k)[x\_1,\dots,x\_n]$, graded in the obvious way, for *any commutative ring* $k$. Thus, $J\_F$ is invertible (as a matrix with polynomial entries) if and only if $J\_N$ is nilpotent (as a matrix with polynomial entries).


Let me finally mention that the equivalence between the Jacobian Conjecture and BCW's reduction is a **stable** one: to prove JC, one has to prove JC for **all** $F$ of the form $F = I + N$ with $\deg(N)=3$, i.e. in all dimensions simultaneously. In other words, in order to prove JC in dimension $d$, it does **not** suffice to prove it only for $F: k^d \to k^d$ of the form $F = I + N$.


To address your **Edit**, yes, BCW's reduction theorem applies to any commutative ring $k$ of characteristic 0, in particular to $k = \mathbb{R}$. Let me know if you have any further questions.


