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Is there exist a similar conjecture to the famous Jacobian Conjecture with $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$ instead of $\mathbb{C}[x_1,\ldots,x_n]$?

Namely, let $f$ be $\mathbb{C}$-algebra endomorphism of $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, denote $f_i:= f(x_i)$, and further assume that the Jacobian of $\{f_1,\ldots,f_n\}$ is in $\mathbb{C}-\{0\}$. Is such $f$ an automorphism?

(I guess that first one should be familiar with the group of automorphisms of $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, see this question).

Edit: I also wonder if there exists any nice connection between the Jacobian Conjecture and my above conjecture (which is not exactly phrased yet); for example, are the two conjectures equivalent?

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Counterexample: the endomorphism of the product of two punctured lines (complement of the curve of equation $xy=0$ in the plane) given by $$(x,y)\mapsto f(x,y)=\left(\frac{x}{y},y^2\right)$$

We have $$\begin{pmatrix}\partial_1f_1 & \partial_2f_1\\ \partial_1f_2 & \partial_2f_2\end{pmatrix}=\begin{pmatrix}\frac{1}{y} & -\frac{x}{y^2}\\ 0 & 2y\end{pmatrix},$$ so the Jacobian is constant equal to 1.

It's not an automorphism, since both $(1,1)$ and $(-1,-1)$ map to $(1,1)$.

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    $\begingroup$ About injectivity, see: en.wikipedia.org/wiki/Ax%E2%80%93Grothendieck_theorem (I haven't checked it works in this context, although it sounds likely at first sight). But it's clearly of a different spirit, because the Jacobian condition is some immediately checkable condition, while the injectivity is not. $\endgroup$
    – YCor
    Commented Apr 13, 2017 at 22:44
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    $\begingroup$ Note that constancy of Jacobian is not necessary, even in 1-d: take $x\mapsto 1/x$, for example. A necessary condition is that the Jacobian be a monomial. $\endgroup$
    – ACL
    Commented Apr 13, 2017 at 22:45
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    $\begingroup$ Well, it is invertible. In the sense that it is a non vanishing function on the torus. $\endgroup$
    – ACL
    Commented Apr 13, 2017 at 22:50
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    $\begingroup$ More generally, any isogeny of algebraic tori: $f\colon \mathbf G_m^n\to\mathbf G_m^n$ has an invertible Jacobian. (Proof: either by computation, or by remarking that there is an “inverse” isogeny, whose composition is $(x_1,\dots,x_n)\to (x_1^d,\dots,x_n^d)$, where $d$ is the degree of $f$.) On the other hand, the first proof gives the clue to the initial question. $\endgroup$
    – ACL
    Commented Apr 13, 2017 at 22:57
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    $\begingroup$ @user237522: the point in ACL's last comments is that the endomorphism semigroup of the torus is considerably smaller than that of the affine $n$-space, so checking whether it's an automorphism is immediate (namely, you need that the matrix of exponents, which is in $\mathrm{M}_n(\mathbf{Z})$, belongs to $\mathrm{GL}_n(\mathbf{Z})$) $\endgroup$
    – YCor
    Commented Apr 13, 2017 at 23:07

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