Given a field $k$, consider the polynomial ring $k[x_1,x_2,\dots,x_n]$. Is it possible to find all the automorphisms of this ring that fix the field $k$?
Automorphisms of $k[x_1,x_2,\dots,x_n]$ that fix $k$
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0Conjecturally — 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 ? – 2012-04-15
1 Answers
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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1When talking about the automorphism of $k[x,...]$, it almost always the automorphism of $k$-algebra, i.e. automorphisms which fix (point-wise) $k$. – 2012-04-15