Let $f=f(x,y),g=g(x,y) \in \mathbb{C}[x,y]$, each of degree $\geq 1$, and $f,g$ are algebraically independent over $\mathbb{C}$ (= their Jacobian $\in \mathbb{C}[x,y]-\{0\}$).



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> **(1)** Is there a sufficient condition that will guarantee that $\mathbb{C}(f,g)=\mathbb{C}(x,y)$?
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Perhaps it would help if we will consider $f$ and $g$ as polynomials in one variable $y$ over $\mathbb{C}(x) \subset \bar{\mathbb{C}(x)}$?


A related question I have asked is [this question](https://math.stackexchange.com/questions/2772853/two-polynomials-generating-mathbbcx-y2), but I have not got an answer for it, so I decided to ask the above related question here, hoping to get some additional help.


**A remark:** I have just found three relevant papers: [1](https://www.worldscientific.com/doi/abs/10.1142/S0219498815400058), [2](https://arxiv.org/abs/1308.0154) and
[3](https://arxiv.org/pdf/1307.3752.pdf), where the last one seems to answer my question (Theorem 1.10, or more accurately, since here the base field is of characteristic zero, Theorem 1.10 in characteristic zero is a result of K.P. Russell "Field generators in two variables", J. Math. Kyoto Univ., 1.



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> **(2)** Does it make a difference if we allow $f,g \in \mathbb{C}(x,y)$? 
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For example, $f=x$ and $g=\frac{1}{y}$ (what is the analog notion of being algebraically independent?).


Any comments are welcome! Thank you.



Two bivariate polynomials (or rational functions) that generate $\mathbb{C}(x,y)$
The two polynomials that you are giving provide a morphism $\tau\colon\mathbb{C}^2\to \mathbb{C}^2$, given by $(x,y)\mapsto (f(x,y),g(x,y))$. 


This map $\tau$ is dominant if and only if $f$ and $g$ are algebraically independent (which seems what you already ask).


This map $\tau$ is birational if and only if $\mathbb{C}(f,g)=\mathbb{C}(x,y)$. This is also equivalent to ask that is is generically injective, which means that the preimage of a general point consists of one single point.


In practical, you can extend your morphism to a birational map from $\mathbb{P}^2$ to $\mathbb{P}^2$ by homogenising, and use Bézout Theorem: you find the common roots of the two homogenised polynomials in $\mathbb{P}^2$ (all these are on the line at infinity because the map is a morphism on $\mathbb{C}^2$). You then blow-up these and count the multiplicities of all base-points, including infinitely near ones. Then the map is birational if and only if $d^2-\sum m\_i^2=1$, where $d$ is the degree (max of the degree of $f$ and $g$) and the $m\_i$ are the multiplicities.


The article you mention are dealing with a related question which takes only one polynomial $f$ and try to find if it is a variable, i.e. if there is $g$ such that $(f,g)$ is birational.


