Let $k[x_1,\ldots,x_n]$ be a polynomial ring over a field $k$ of characteristic zero.

When $n=2$, it is known that every automorphism of $k[x_1,x_2]$ is tame, namely, a finite product of elementary automorphisms.

For $n=3$, in their paper, Shestakov and Umirbaev showed that the Nagata map is wild (=non-tame, not belongs to the subgroup generated by elementary automorphisms).

My question: For $n \geq 3$, if we know that a given automorphism $g$ is of finite order (namely $g^m=1$ for some $m$), must it be tame?

Or is it possible to have a wild automorphism of finite order?

Sorry if my question is trivial.

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    $\begingroup$ This is wide open as far as I am aware. On page xvi of the introduction of van den Essen's book Polynomial Automorphisms and the Jacobian Conjecture it is written that it is not known whether finite order automorphisms are linearizable (an affirmative answer to which will imply an affirmative answer to the Cancellation problem: does $V \times \mathbb{C} \cong \mathbb{C}^{n+1}$ imply $V \cong \mathbb{C}^n$?). It seems nonetheless that most believe the answer to be negative: there should be wild polynomial involutions. $\endgroup$ – Vesselin Dimitrov Jan 17 '16 at 21:25
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    $\begingroup$ If you take an automorphism of finite order, and conjugate it by a wild automorphism, you'll get an automorphism of finite order. Any reason to believe that this won't be wild? $\endgroup$ – André Henriques Jan 18 '16 at 0:51
  • $\begingroup$ @AndréHenriques This is right, and I was thinking of something else (a finite order automorphism that cannot be conjugated to an affine one). $\endgroup$ – Vesselin Dimitrov Jan 18 '16 at 10:09
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    $\begingroup$ @AndréHenriques Thanks for the clarification. As you remark, for example, for $n=3$, if we conjugate the involution $x \mapsto y$, $y \mapsto x$, $z \mapsto z$ by Nagata's wild automorphism, we get an involution which is (probably?) wild. But what about a wild involution not conjugate to an involution from the tame subgroup, as I guess Vesselin Dimitrov meant? $\endgroup$ – user237522 Jan 18 '16 at 17:40

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