The following question is a direct continuation of [this](https://mathoverflow.net/questions/453393/if-mathbbcux-y-vx-y-fx-mathbbcx-y-for-every-fx-in-mathbb/453411?noredirect=1#comment1173100_453411) question:


Let $u,v \in \mathbb{C}[x,y]$.


Assume that for every $f \in \mathbb{C}[x]$ and every $g \in \mathbb{C}[y]$ (excluding the cases where $f+g \in \mathbb{C}$), the following condition, call it $E(f,g)$, is satisfied:


$\mathbb{C}(u,v,f+g)=\mathbb{C}(x,y)$.


The notation is of fields of fractions.


Then we can ask the following question:



> 
> **Question:** Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?
> 
> 
> 


Notice that the counterexample for the quoted question, $u=x+y,v=xy$, is not a counterexample here, since condition $E(x,y)$ does not hold,
since $\mathbb{C}(x+y,xy,x+y)=\mathbb{C}(x+y,xy) \subsetneq \mathbb{C}(x,y)$ (Also, conditions $E(x^n,y^n)$ do not hold, but it is enough for us that one required condition does not hold).


**Remarks:**


**(1)** I also thought about the following condition:
Assume that for every $f \in \mathbb{C}[x]$ and every $g \in \mathbb{C}[y]$ (excluding the cases where $fg \in \mathbb{C}$), the following condition, call it $M(f,g)$, is satisfied:


$\mathbb{C}(u,v,fg)=\mathbb{C}(x,y)$.


However, it seems that Joachim König also gave a counterexample for conditions $M(f,g)$:
$u=x+y^3,v=y+x^3$, although he referred to adding $(ax+by+c)^n$ , $n \geq 1$, $(ax+by+c)^n \notin \mathbb{C}$.


I think I can explain why his counterexample also applies for adding $fg$, $f \in \mathbb{C}[x], g \in \mathbb{C}[y]$, $fg \notin \mathbb{C}$.


The argument would be (I do not yet have a proof for the argument, see [this](https://math.stackexchange.com/questions/4723640/generalizing-a-result-in-mathbbcx-to-a-result-in-mathbbcx-y?noredirect=1&lq=1)): Given $u,v,w \in \mathbb{C}[x,y]$, if there exist some $a,b,c \in \mathbb{C}$ such that the ideal generated by $u-a,v-b,w-c$ is maximal, then $\mathbb{C}(u,v,w)=\mathbb{C}(x,y)$.


**(2)** Notice that $u=x+y^3, v=y+x^3$ is not a counterexample for the current question, since, for example, conditions $E(x,y^3)$ and $E(x^3,y)$ do not hold.


Thank you very much!



$\mathbb{C}(u(x,y),v(x,y),f(x)+g(y))=\mathbb{C}(x,y)$ implies $\mathbb{C}(u(x,y),v(x,y))=\mathbb{C}(x,y)$?
Actually, there are still counterexamples of essentially the same nature as the one given in the previous (linked) thread: Take $u=x+xy$ and $v=x^2y$, so that $\mathbb{C}(u,v)$ is the field of functions symmetric in $x$ and $xy$ (i.e., the fixed field of the involution $\sigma:x\mapsto xy$ and $y = \frac{xy}{x}\mapsto \frac{x}{xy}=\frac{1}{y}$). This is of index $2$ in $\mathbb{C}(x,xy)=\mathbb{C}(x,y)$ (but of course not identical to the field of symmetric functions in $x$ and $y$). If $f(x),g(y)$ are polynomials, not both constant, then $\sigma(f(x)+g(y))=f(xy)+g(\frac{1}{y})$, and this can never equal $f(x)+g(y)$ (indeed, if $g$ is nonconstant, then $f(xy)+g(\frac{1}{y})$ is not even in $\mathbb{C}[x,y]$, whereas if it is constant, then $f(x)+g\in \mathbb{C}[x]$, but $f(xy)+g\notin \mathbb{C}[x]$. Thus, $f(x)+g(y)\notin Fix(\sigma)=\mathbb{C}(u,v)$, and hence $\mathbb{C}(u,v,f(x)+g(y))=\mathbb{C}(x,y)$.


