Let $R\_n$ be the integral polynomial ring $\mathbb{Z}[x\_1,x\_2,...,x\_n]$, let $A\_n$ be the group of ring automorphisms $\mathrm{Aut}(R\_n)$, and for $f\in R\_n$ let $\mathrm{Aut}(f)=\{\alpha\in A\_n\ |\ \alpha(f)=f\}$.


Define a polynomial $f\in R\_n$ to be *interesting* if $\deg\_{x\_i}(f)\geq 1$ for $1\leq i\leq n$ and there is an infinite order element in $\mathrm{Aut}(f)$. 


Since $A\_1=\{n\pm x\ |\ n\in \mathbb{Z}\}$ there do not appear to be any interesting members of $R\_1$.


On the other hand, $\kappa=x^2+y^2+z^2-xyz-2$ is interesting since Horowitz showed in *Induced automorphisms on Fricke characters of free groups* that $\mathrm{Aut}(\kappa)\cong \mathrm{PGL}(2, \mathbb{Z}) \rtimes (\mathbb{Z}/2 \oplus \mathbb{Z}/2)$.


Hence, there are interesting members of $R\_n$ for all $n\geq 3$.


Here is my question:



> 
> Are there any interesting planar polynomials?
> 
> 
> 


Precisely, do there exist $f\in \mathbb{Z}[x,y]$ with degree in $x$ and $y$ at least 1 so that $\mathrm{Aut}(f)$ contains an infinite order element?


I imagine that the structure of the affine Cremona group will be relevant here. See *Two-dimensional Cremona groups acting on simplicial complexes* by Wright for a structure theorem relevant to $A\_2$ (Theorem 2.4 with $k=\mathbb{Q}$).


**Update**: Given the helpful comments by Yves, which give an answer to the original post (and even the slightly edited version), I have a second question to ask that is too related to post separately. 


We now say $f\in R\_n$ is *very interesting* if it is interesting and $f\not=\alpha(g)$ where $\alpha\in A\_n$ and $g$ has degree 0 in some variable. I believe $\kappa$ remains an example (although it does not seem obvious to me).


Here is the second question:



> 
> Are there any very interesting planar polynomials?
> 
> 
> 



Infinite order automorphisms of planar polynomials
First, YCor answered the first question in the comments. Here is a summary:


Take any polynomial $f(x)$ in only one variable $x$ and apply any automorphism $\alpha$ to get $\alpha(f)$. Then $\alpha\circ T\circ \alpha^{-1}$ is an automorphism that fixes $\alpha(f)$ whenever $T$ is an automorphism that fixes $x$. In particular, taking $T$ to be $T(x,y)=(x,y+P(x))$ for any non-zero $P(x)$ gives an infinite order element that fixes $\alpha(f)$. Note that $T$ is invertible since $T^{-1}(x,y)=(x,y-P(x))$.


As to the second question, the paper *[Automorphisms of the plane preserving a curve](http://content.algebraicgeometry.nl/2015-2/2015-2-009.pdf)* by Jérémy Blanc and Immanuel Stampfli describes similar automorphism groups quite explicitly over fields (arbitrary characteristic).


In particular, their Theorem 1 gives evidence that the answer to the second question is no. It says that the automorphism group (in the context of the paper) is algebraic (rigid) if and only if there is no automorphism that sends $f$ to a one-variable polynomial (a "fence" in the language of the paper).


Here is an illustrative concrete example from their paper:


If $f(x,y)=xy-1$, which does not appear to be equivalent to a one variable polynomial, then over $\mathbb{k}$ the automorphism group that fixes $f$ is $\mathbb{k}^\*\rtimes \mathbb{Z}/2\mathbb{Z}$ generated by $(x,y)\mapsto (y,x)$ and $t\mapsto (tx,t^{-1}y)$. If $|t|\not=1$, then there is an infinite order element. However, over $\mathbb{Z}$ the scalar $t=\pm 1$ and so there would appear to be no such infinite order element. 


I suspect this example is similar to what happens in general.


