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Let $k$ be a field of characteristic zero (I do not mind to assume that $k=\mathbb{C}$ if this helps).

A $k$-automorphism of $k[[x,y]]$ was described in the answer to this question: $f: (x,y) \mapsto (A,B)$ is a $k$-automorphism of $k[[x,y]]$ if and only if $A,B$ are power series with no constant term whose linear parts are linearly independent.

If $f$ is an automorphism of $k[[x,y]]$ of finite order and $A,B \in k[x,y]$, is it true that such $f$ is also an automorphism of $k[x,y]$?

Remarks: (1) $(x,y) \mapsto (x+x^2,y+y^2)$ is an automorphism of $k[[x,y]]$ of infinite order, and it is not an automorphism of $k[x,y]$ (because its Jacobian is not invertible). So the condition that $f$ is of finite order is necessary. (2) Perhaps this question is relevant (but I do not wish to make any change of coordinates).

user237522
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  • Can you rewrite your question in a more coherent way? Right now, you ask a question, you provide a negative answer to the question (the example in (1)), then you deduce that "the condition that $f$ is of finite order is necessary": no it's not necessary: it's a restriction for which maybe you expect a positive answer. – YCor Apr 18 '17 at 15:52
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    The answer is trivially yes if $f$ has finite order: indeed if $f^n$ is the identity and $f$ maps $x,y$ to polynomials, then so does $f^{-1}=f^{n-1}$. – YCor Apr 18 '17 at 15:52
  • But why should those polynomials generate all $k[x,y]$? – user237522 Apr 18 '17 at 15:54
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    Both $f$ and $f^{-1}$ map $k[x,y]$ into $k[x,y]$. It's an exercise to conclude. – YCor Apr 18 '17 at 15:55
  • Oh, of course...so trivial, sorry... – user237522 Apr 18 '17 at 15:56
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    Yes. Call this automorphism $\sigma$. Your assumption implies $\sigma$ gives an endomorphism of $k[x,y]$. But, $\sigma^n$ is identity and then it follows that $\sigma$ is both injective and surjective. – Mohan Apr 18 '17 at 15:58
  • Thank you very much YCor and Mohan! I really apologize for asking this question here (I felt it should have a positive answer, but was too tired to see the easy solution). Is it possible to move it to MathSE? (I have just flagged it, asking to move it to MathSE). – user237522 Apr 18 '17 at 16:04

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