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Let $k$ be a field. In 1941, Jung showed that all polynomial $k$-algebra automorphisms of the rational (polynomial) functions in two variables, denoted by $k(x,y)$ can be written as compositions of the basic automorphisms

(1) $\phi(x)=\alpha_1x+\beta_1x+\gamma_1$ and $\phi(y)=\alpha_2x+\beta_2y+\gamma_2$ (with $\alpha_1\beta_2-\alpha_2\beta_1\neq 0$)

(2) $\phi(x)=x$ and $\phi(y)=y+p(x)$ with $p(x)\in k[x]$

What can one say about all $k$-algebra automorphisms of $k(x,y)$?

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    $\begingroup$ What are "automorphisms of rational functions"? $\endgroup$
    – abx
    Commented Aug 12, 2019 at 15:45
  • $\begingroup$ Do you mean $\phi$ is a $K$-algebra automorphism of $K(x,y)$ with $\phi(x),\phi(y) \in K[x,y]$, where $K$ is a field, and you're wondering what happens if the restriction that $\phi(x),\phi(y) \in K[x,y]$ be dropped? Also, what is $p(x)$? $\endgroup$
    – Jesse Elliott
    Commented Aug 12, 2019 at 20:49
  • $\begingroup$ Jung's 1941 result (which the OP failed to quote correctly) is that the group of $k$-automorphisms of $k[x,y]$, $k$ a field, or equivalently of the affine $k$-plane, is generated by affine automorphisms and by those $(x,y)\mapsto (x,y+p(x))$ for $p\in k[x]$. $\endgroup$
    – YCor
    Commented Aug 12, 2019 at 22:40
  • $\begingroup$ So the question is apparently about what's a generating subset for the Cremona group, i.e., the automorphism group of the $k$-algebra $k(x,y)$. For $k$ algebraically closed (char 0?) there's a similar theorem (attributed to Max Noether). But for $k$ not algebraically closed it's more complicated. $\endgroup$
    – YCor
    Commented Aug 12, 2019 at 22:44
  • $\begingroup$ I have updated the question an added a little bit of notation (and yes, Jung doesn't exactly write the polynomial automorphisms (or, "ganze birationale transformationen") in this form; he splits them into three basic ones). The question is indeed about all $k$-algebra automorphisms of $k(x,y)$. A reference to the statement for algebraically closed (and char 0) fields, indicated by @YCor, would be nice. Thanks! $\endgroup$
    – Joakim Arnlind
    Commented Aug 13, 2019 at 6:45

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