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After asking this MO question, I wish to ask about the following special case:

Let $f$ be a $\mathbb{C}$-algebra automorphism of $\mathbb{C}(x,y)$ and denote $u:=f(x),v:=f(y)$.

Is it possible to find 'a general form' of $(u,v)$, in the special case where $u,v \in \mathbb{C}[x,y]$ and $\operatorname{Jac}(u,v) \in \mathbb{C}[x,y]-\mathbb{C}$?

For example: $(u,v)=(x,xy)$; in this example $(u,v)$ is not a Jacobian pair.

Please also see this MSE question (in which Mohan suggests $u=x+y^2, v=y(x+y^2)=yu$),

Thank you very much!

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  • $\begingroup$ What do you mean by "a general form"? $\endgroup$
    – YCor
    Commented Nov 24, 2019 at 12:59
  • $\begingroup$ @YCor, thanks, good question.. Perhaps something similar to 'a general form' of a $\mathbb{C}$-algebra automorphism of $\mathbb{C}[x,y]$ (affine or triangular). But other types of answers are welcome too. (Actually, I wished to restrict to the case where $\operatorname{Jac}(u,v) \in \mathbb{C}[x,y]-\mathbb{C}$ to exclude, by Keller's theorem, automorphisms of $\mathbb{C}[x,y]$; I will add this). $\endgroup$
    – user237522
    Commented Nov 24, 2019 at 13:51
  • $\begingroup$ But an automorphism of $C[x,y]$ is not always affine or triangular... $\endgroup$
    – YCor
    Commented Nov 24, 2019 at 13:58
  • $\begingroup$ @YCor, you are right... but at least it is a finite product of such, though writing a general pair (as a pair of elements of $\mathbb{C}[x,y]$) is impossible... Any other suggestions? (Or a similar result for a pair of polynomials not being a Jacobian pair?). $\endgroup$
    – user237522
    Commented Nov 24, 2019 at 14:26
  • $\begingroup$ @YCor, please, could you elaborate on one of your previous comments: "But in Cremona 'describe the group' can have a totally different meaning. E.g., it can consist in describing the set of pairs of rational functions that indeed define a element of the Cremona group..." $\endgroup$
    – user237522
    Commented Nov 24, 2019 at 14:29

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The monoid that you are looking for is the set of birational endomorphisms of the affine plane. It is of course closed under compositions and the invertible elements are the automorphisms. You would like to study the non-trivial elements, i.e. birational endomorphisms whose inverse is not an (auto)-morphism.

The simplest example is $\pi\colon (x,y)\mapsto (xy,y)$, which contracts exactly one curve and thus cannot be a composition of two (or more) non-trivial birational endomorphisms.

The elements of the form $\alpha \circ \pi \circ \beta$, where $\alpha,\beta$ are automorphisms, are usually called simple affine contractions. In the seventies, the natural question of knowing if every birational endomorphism was a composition of such ones, or equivalently if $\pi$ generates, together with automorphisms, all the monoid of birational endomorphisms, was asked.

The answer is no and there are many counterexamples. You can for instance have a look at the article "Birational endomorphisms of the affine plane " of Daniel Daigle: https://projecteuclid.org/download/pdf_1/euclid.kjm/1250519792

The whole monoid of birational endomorphisms is quite complicated and still mysterious now, even after a lot of nice results (the article above cites some of them, but you can find a lot more by looking on internet).

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  • $\begingroup$ Thank you very much! Your answer is very helpful. Please, what if we further assume that $u(x,0) v(x,0) \neq 0$? You have mentioned that the monoid of birational endomorphisms is quite complicated, so I guess that this further assumption would not help. Perhaps further assuming the contrary, namely, that $u(x,0)=0$ or $v(x,0)=0$ may yield something interesting? In the 2 examples in my question $\mathbb{C}[u(x,0),v(x,0)]=\mathbb{C}[x]$. In your example, $\mathbb{C}[u(x,0),v(x,0)]=\mathbb{C}$. (The non-zero assumption is probably 'not solvable'). $\endgroup$
    – user237522
    Commented Nov 27, 2019 at 17:29

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