The well known Jacobian conjecture is
Jacobian Conjecture. If $F:\mathbb C^n\rightarrow\mathbb C^n$ is a polynomial map and the Jacobian matrix $J(F)$ is invertible, then $F$ is an invertible polynomial map.
I read in some notes that it is a not very difficult technical result that if the Jacobian Conjecture over $\mathbb C$ is true, then it is true over any field of characteristic zero.
For me it is difficult. Can anyone help me?
There is a problem in the conjecture. Over $\mathbb C$ it is really easy to prove. I believe you mean $\mathbb{A}^n(\mathbb{C})$.
Here is a way of proving it that I like and that can work for proving many things over rings whence you know it over $\mathbb{C}$
First notice that for countability reasons, $\mathbb{C}$ has an infinite transcendance basis. This implies that you can embed $R=\mathbb{Q}[T_i, i\in \mathbb{N}]$ in $\mathbb{C}$. Take $k$ a field that is a finitely generated $\mathbb{Q}$-algebra of characteristic $0$.
Then surely you have a surjective map $p:R\to k$, and take $F:\mathbb{A}^n(k)\to \mathbb{A}^n(k)$ to be polynomial. It is easy to check that if you take arbitrary liftings of the coefficients of $k$ to $R$, then the jacobian of the lifted map will map to the jacobian of your map $F$ through $p$. Of course the Jacobian of your lift is not zero in $\mathbb{C}$, therefore the polynomial is invertible in $\mathbb{C}[X_i]$. But it means it is also invertible in $frac(R)[X_i]$ and thus in $k[X_i]$.
Now if you take an arbitrary field of characteristic zero, then surely the polynomial $F$ will have finitely many coefficients, and working in the sub-$\mathbb{Q}$-algebra generated by the coefficient, you can reduce to the same kind argument.