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](http://www.ams.org/journals/jams/2004-17-01/S0894-0347-03-00440-5/), 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.



Is a wild automorphism of $k[x_1,\ldots,x_n]$, $n \geq 3$, necessarily of infinite order?
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.


