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Let $k[x,y]$ be the polynomial ring in two variables over a field $k$ of characteristic zero. Every $k$-algebra automorphism of $k[x,y]$ is tame (e.g. the paper of McKay and Wang). It was pointed out in an answer to this question that $k[x,y,z]$ has wild automorphisms of finite order.

My question is, is every automorphism of $k[x,y]$ of finite order a composition of elementary automorphisms of type (ii) as in the abstract of McKay and Wang? My rough thinking is that elementary automorphisms of type (i) increase total degree and therefore cannot be finite order, but the example of the wild automorphism of finite order for $k[x,y,z]$ has me questioning this reasoning.

EDIT: I would also be happy with a similar statement about finite order automorphisms of $k[x]$, if $k[x,y]$ is too ambitious.

McKay, James H.; Wang, Stuart Sui-Sheng, An elementay proof of the automorphism theorem for the polynomial ring in two variables, J. Pure Appl. Algebra 52, No. 1-2, 91-102 (1988). ZBL0656.13002.

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    $\begingroup$ Automorphisms of $k[x]$ are contained in automorphisms of $k(x)$ and the latter are well-understood in terms of Móbius transformations. $\endgroup$
    – Kapil
    Commented Mar 29, 2022 at 2:47
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    $\begingroup$ Automorphisms of $k[x]$ have the form $P(x)\mapsto P(ax+b)$ for $(a,b)\in K^*\times K$. In particular, for $k$ of char. zero, a non-identity automorphism has finite order if and only it has such a form with $a\neq 1$ root of unity. $\endgroup$
    – YCor
    Commented Mar 29, 2022 at 15:23

2 Answers 2

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The answer seems to be no. The automorphism group of $k[x,y]$ can be described as the amalgamated product of the affine transformations with the group of maps of type (i) as in the abstract of McKay and Wang. The torsion elements in an amalgamated product are those elements conjugate to a torsion element of one of the factors, so the torsion elements of $\operatorname{Aut}(k[x,y])$ are those conjugate to affine transformations. This includes many automorphisms which are not affine.

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  • $\begingroup$ The automorphisms of finite order are not affine but always conjugate to affine $\endgroup$
    – Jérémy Blanc
    Commented Dec 20, 2023 at 13:01
  • $\begingroup$ @JérémyBlanc Isn't it what the answer says? $\endgroup$
    – YCor
    Commented Dec 20, 2023 at 13:29
  • $\begingroup$ You are absolutely right. Sorry I read it falsely. My answer is then basically the same... $\endgroup$
    – Jérémy Blanc
    Commented Dec 20, 2023 at 14:43
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As explained in the article of McKay and Wang, the result of Jung says that $\operatorname{Aut}_k(k[x,y])$ is an amalgamated product of the affine automorphisms and Jonquières ones, that preserve the fibres of the fibration $(x,y)\mapsto x$. So, as explained by Joshua Ruiter, any group of finite order is conjugate to a subgroup of one of the two factors. In positive characteristic, there are examples of automorphisms of finite order not conjugate to affine automorphisms but in characteristic zero, every finite subgroup is conjugate to a group of affine transformations. This is a very classical result. You can for instance check Theorem 2 of the article Finite groups of polynomial automorphisms in $\mathbf{C}^n$ of Mikio Furushima, together with the references cited there (Kambayashi for instance).

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