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Let $k$ be a field of characteristic zero.

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

It is known that the generators of the group of $k$-automorphisms of $k[x,y]$ are the affines $(x,y) \mapsto (ax+by+u,cx+dy+v)$, $ad-bc \in k-\{0\}$, and the triangulars $(x,y) \mapsto (ax,by+H(x))$, $ab \neq 0$, $H(x) \in k[x]$.

My question: Is it true that the generators of the group of $k$-automorphisms of $k[[x,y]]$ are $(x,y) \mapsto (ax+by,cx+dy)$, $ad-bc \in k-\{0\}$, and $(x,y) \mapsto (ax,by+H(x))$, $ab \neq 0$, $H(x) \in k[[x]]$ with no constant term?

Perhaps this is too good to be true? Can one find a $k$-automorphism which is not a finite composition of the above suggested generators?

Edit: It seems that the suggested generators do not generate all automorphisms of $k[[x,y]]$. Therefore, I change my question to: Is it possible to find additional "nice" generators that together indeed generate all automorphisms of $k[[x,y]]$? (I do not define 'nice', but I hope that it is understandable that I do not consider all automorphisms as a nice family of generators).

user237522
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    I doubt that your conjecture is true: how about the automorphism ${x\mapsto x+x^2,y\mapsto y+y^3},$? Can you exhibit it as composition of automorphisms of the limited form you specify? – Lubin Apr 16 '17 at 04:04
  • Probably not (I mentioned that it is perhaps too good to be true). But how to prove this? And, more importantly, can one find additional "nice" generators which generate the group of automorphisms of $k[[x,y]]$? – user237522 Apr 16 '17 at 04:09
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    Obviously the above example is NOT an automorphism of $k[x,y]$, since its Jacobian is not in $k-{0}$, so it is not a finite product of the suggested generators; I guess this argument is enough? – user237522 Apr 16 '17 at 04:11
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    Since automorphisms are easy to describe as you have, what would you consider nice? $A=x+P(x,y),B=y+Q(x,y)$ gives an automorphism for ANY $P,Q$ of order at least 2. This forms a normal subgroup with the quotient $GL_2(k)$. – Mohan Apr 16 '17 at 23:10
  • My "definition" of nice is probably something like (using your notation): $P \in k[[x]]$ and $Q \in k[[y]]$ (both with no constant term, of course). – user237522 Apr 17 '17 at 19:07
  • Perhaps the "new triangular" should look like $(x,y) \mapsto (ax+H_1,by+H_2)$, with $a,b \in k-{0}$, and $H_i \in k[[x]]$ or $H_i \in k[[y]]$ (four options with no constant term). – user237522 Apr 17 '17 at 19:13

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