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. $\endgroup$ Jan 17, 2016 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? $\endgroup$ Jan 18, 2016 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). $\endgroup$ Jan 18, 2016 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, 2016 at 17:40
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    $\begingroup$ @VesselinDimitrov it is true that we still do not have any example of an involution (or simpy an element of finite order) not conjugate to a linear map and if no such exists, then the cancellation problem holds. However, this is not directly related to the fact of being tame or not, as the result of Edo and Poloni cited below shows. $\endgroup$ Nov 27, 2019 at 7:26

1 Answer 1


The answer is no.

Let $K$ be a field of characteristic zero and let us take the Nagata automorphism of $\mathbb{A}^3_K$ given by

$$N\colon (x,y,z)\mapsto (x+(x^2-yz)z,y+2(x^2-yz)x+(x^2-yz)^2z,z)$$ It is part of an action of the additive group given by $$f_t\colon (x,y,z)\mapsto (x+t(x^2-yz)z,y+2t(x^2-yz)x+t^2(x^2-yz)^2z,z)$$ which satisfies $f_t\circ f_s=f_{s+t}$ for each $s,t\in K$. As $N=f_1$ we get $N^{-1}=f_{-1}$. Choosing $\alpha=(x,y,z)\mapsto (-x,y,z)$ which is of order $2$, you obtain $\alpha \circ N\circ \alpha=N^{-1}$, so $\alpha\circ N$ is of order $2$. As $\alpha$ is tame and $N$ is not tame, the involution $\alpha\circ N$ is not tame.

In the article "The Nagata automorphism is shifted linearizable" of Eric Edo and Pierre-Marie Poloni, you can find that composing the Nagata automorphism with a linear map can give an element which is conjugate to a linear map even if it is not tame. So conjugation indeed changes the fact to be tame or not.

  • $\begingroup$ Thank you very much! Interesting. $\endgroup$
    – user237522
    Nov 27, 2019 at 7:37
  • $\begingroup$ You are welcome. $\endgroup$ Nov 27, 2019 at 8:17

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