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I was thinking about this after I read about Jacobian conjecture. But I can't see what I did wrong? Maybe you can help me.

Let $F: \mathbb{C}^n \to \mathbb{C}^n$ be of the form $F(x_1, \dots, x_n)= (F_1(x_1, \dots, x_n), \dots, F_n( \dots , x_1, \dots, x_n) )$ for some $F_1,\dots , F_n \in \mathbb{C}[X_1, \dots, X_n ]$, which I will call polynomial, with constant Jacobian determinant $\det DF$.

Then by Implicit Function Theorem $F$ has a local inverse $F'$ around $0$. Since $F$ is analytical $F'$ is also analytical. We also know that $ DF' = (DF)^{-1}= \frac{1}{\det DF} \text{adj}(DF) $ but the adjugate is computed in terms of the minors of $DF$, $\det DF$ is constant and thus $DF'$ is given by polynomial functions in its components, in particular $D^k F =0$ for some $k\in \mathbb{N}$. Thus $F'$ is polynomial because of its Taylor expansion. Since $F'$ is polynomial it is defined on the whole space $\mathbb{C}^n$.

Now we have $F' \circ F\mid_U= \text{id}$ for some open neighborhood $U$ of $0$, but this implies $F' \circ F= \text{id}$ since $F' \circ F$ is a polynomial.

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Hi user60589. There were some ‘$’ signs missing, hence the edit. – Haskell Curry Feb 1 '13 at 20:13
Why do you presume the adjugate is polynomial? Computing cofactors involves determinants of sub-matrices. – copper.hat Feb 1 '13 at 20:27
They are sub-matrices of $DF$ which consist of the derivatives of $F_i$, so the determinant of it is again a polynomial? – user60589 Feb 1 '13 at 20:31
I think copper.hat is right. The minors involve divisions, and it could happen that you were dividing polynomials with no common factors in the process, which would not yield a polynomial in the end. – busman Feb 2 '13 at 13:17
A minor is just a determinant? Why should they involve divisions? Or am I wrong? – user60589 Feb 2 '13 at 13:28
up vote 1 down vote accepted

You have used the wrong formula $$df^{-1}(y)=(df)^{-1}(y)$$ instead of the correct formula $$df^{-1}(y)=(df)^{-1}\bigl(f^{-1}(y)\bigr)\ .$$

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Look what you did wrong: $$Id=D(F\circ F^{-1})=DF\circ F^{-1} \cdot DF^{-1}\Rightarrow DF^{-1}=(DF\circ F^{-1})^{-1},$$ thus, $DF^{-1} \neq (DF)^{-1}$, which seems to be your crucial assumption.

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