# automorphism of $k[x,y]$ that fix k.

I was doing an exercise, and to finish it, I need to find something, and I conclude that is just this group of automorphism. The questions is, I can say more about it? ( There has some form, etc..) ? So the question is: Given a field k , Consider the polynomial ring $k[x_1,x_2,...x_n]$. It´s possible to find all the automorphism of this ring, that fix the field k?

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 Conjecturally — jacobian conjecture —, it is given by polynomial maps $k^2 \to k^2$ with constant jacobian, if $k$ has zero characteristic. However, some weaker results are known and they might be enough for your needs. Could you detail ? – Lierre Apr 15 '12 at 17:06

Let $X = (x_1, x_2, \ldots , x_n)$. Automorphisms are polynomial substitutions $f(X) \to f(P(X))$ where $P$ has a polynomial inverse.

Structure theory of the automorphism group is complicated for $n \geq 3$. A recent breakthrough was the proof that not all of the group is generated by substitutions $x_i \to ax_i +b$ where $a$ is a constant and $b$ is a polynomial in the other variables.

http://www.pnas.org/content/100/22/12561.full

For $n=2$ these substitutions (called tame automorphisms) generate the full automorphism group and the structure of the group is explained in Cohn's book on free rings. The structure of the tame subgroup for $n=3$ is determined in

http://arxiv.org/abs/math/0607028

This is all very closely connected to the Jacobian conjecture.

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 You are talking about automorphism in general of $k[x_1,..x_n]$ or considering the automorphism that fixes k? – Arkj Apr 15 '12 at 17:31 When talking about the automorphism of $k[x,...]$, it almost always the automorphism of $k$-algebra, i.e. automorphisms which fix (point-wise) $k$. – Lierre Apr 15 '12 at 17:58