Let $k$ be a field of characteristic zero and let $R_{-1}:=k[x,x^{-1},y]$ be the $k$-algebra of polynomials in $x,y$ containing the inverse of $x$, denoted by $x^{-1}$.

Let $f: k[x,y] \to R_{-1}$ be an *injective* $k$-algebra homomorphism,
and denote $u:=f(x)$, $v:=g(y)$.

$u,v$ are algebraically independent over $k$, by the injectivity of $f$, if I am not wrong.

(1)Is it true that the Jacobian of $u,v$, $\operatorname{Jac}(u,v):=u_xv_y-u_yv_x$, belongs to $k^{\times}=k-\{0\}$?

For example: $u=x, v=y+x^{-1}$ 'seems' an injective homomorphism.

In addition assume that:

**(i)** $u=x+\tilde{u}$ and $v=y+\tilde{v}$,
where $v_{1,-1}(\tilde{u}) < 1$ and $v_{1,-1}(\tilde{v}) < -1$;
in other words, $l_{1,-1}(u)=x$ and $l_{1,-1}(v)=y$.

**(ii)** The Jacobian of $u$ and $v$ equals $1$;
equivalently, $\operatorname{Jac}(u,v) \in k^{\times}$.
(In case injectivity does not imply this).

(2)Is it possible to say something interesting about that $f$?

A simple example of such $f$ is as follows: $u=x+y+x^{-2}$ and $v=y+x^{-2}$. (The Jacobian here equals $1$).

What I had in mind is that such $f$ must be of a 'similar' form to the form of the above simple example, namely, a 'triangular' map, with no monomials of the form $x^iy^j$, $ij \neq 0$. I have tried to prove this by considerations of $(1,-1)$-degrees, but things are getting complicated, so I wonder if one can find a counterexample or a trickier proof.

**Remark:** I have already asked this qustion in separate MSE questions 1 2 3, but I have not got any comments.

Possible variations of the above questions are:

**(I)** Replacing $R_{-1}$ by $R_{\frac{1}{m}}:=k[x,x^{-\frac{1}{m}},y]$.

**(II)** Replacing $R_{-1}$ by $\tilde{A_1(k)}$ the first Weyl algebra $A_1(k)$ with a third generator $x^{-1}$ and the obvious relations.