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Aug 13, 2019 at 6:45 comment added Joakim Arnlind 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!
Aug 13, 2019 at 6:38 history edited Joakim Arnlind CC BY-SA 4.0
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Aug 12, 2019 at 22:44 comment added YCor 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.
Aug 12, 2019 at 22:40 comment added YCor 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 $p\in k[x]$.
Aug 12, 2019 at 20:49 comment added Jesse Elliott 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 $p(x)$?
Aug 12, 2019 at 17:50 review Close votes
Aug 19, 2019 at 6:47
Aug 12, 2019 at 15:45 comment added abx What are "automorphisms of rational functions"?
Aug 12, 2019 at 14:47 history asked Joakim Arnlind CC BY-SA 4.0