# Homomorphisms from $k[x,y]$ to $k[x,x^{-1},y]$

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.

(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.

• For $f(x) = x^2$, and $f(y) = y^2$ the Jacobian is $4xy$ vanishing at 0 – InfiniteLooper Jun 2 at 15:41
• @InfiniteLooper, thank you. So injectivity does not imply that the Jacobian is a non-zero scalar, but probably it must be at least a non-zero element of $k[x,x^{-1},y]$. (Please see math.stackexchange.com/questions/3147174/…). – user237522 Jun 3 at 8:36