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?

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    $\begingroup$ It might be wary to write $\mathrm{Aut}(R_n,f)$ instead of $\mathrm{Aut}(f)$. Indeed, $\mathrm{Aut}(f)$ is not the same is $f$ is viewed as element of $R_m$ for some $m>n$. Also I'd avoid using "interesting" in any formal sense, since it deprives the informal language of a quite useful word. $\endgroup$
    – YCor
    May 25, 2019 at 9:01
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    $\begingroup$ Wright deals with the Cremona group, which is much larger than the automorphism group. The latter (over a field) is a suitable amalgam (Jung / van der Kulk) and is quite different. Some description over $\mathbf{Z}$ should follow, since it's then a subgroup of the corresponding group over $\mathbf{Q}$. $\endgroup$
    – YCor
    May 25, 2019 at 9:07
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    $\begingroup$ By the way any polynomial $P(x)\in\mathbf{Z}[x,y]$ satisfies your property, since $x\mapsto x$, $y\mapsto y+Q(x)$ for any given nonzero polynomial $Q$, has infinite order. The argument immediately extends to $P(ax+by)$. $\endgroup$
    – YCor
    May 25, 2019 at 10:38
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    $\begingroup$ @YCor Thank you for your comments, especially the last two (+1 for both). I made a couple edits in response to those. Of course, "interesting" is not a good word to use in a published paper that introduces a term, but I view this as an informal discussion and a contextual term is sufficient from my point-of-view (but I respect there are differences of opinion here). And of course you are right that $\mathrm{Aut}(f)$ depends on $n$, but that is the notation I have seen used in the literature and in context it is clear enough. $\endgroup$ May 25, 2019 at 13:40
  • $\begingroup$ @YCor What do you mean by "the argument immediately extends...". Are you saying you have an answer to the question? If so, please post it. You certainly do not mean take $P(ax+by)$ and use the same automorphism $\tau_Q$ defined by $x\mapsto x$ and $y\mapsto y+ Q(x)$ for that does not work even for $P(x)=x$. Perhaps you mean take a linear automorphism $L$ so $x\mapsto ax+by$ and $y\mapsto cx+dy$ where $ad-bc=\pm 1$ and then conjugate $\tau_Q$ by $L$? Yeah, that seems to work! Please post so it is answered. $\endgroup$ May 25, 2019 at 14:05

1 Answer 1


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 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.


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