The following question is a direct continuation of [this](https://math.stackexchange.com/questions/4758087/if-mathbbcu-v-xn-mathbbcx-y-for-every-n-geq-1-then-already-m) elaborate question; it is mentioned there at the end:


Let $u,v \in \mathbb{C}(x,y)$ or $u,v \in \mathbb{C}[x,y]$, if it is easier to answer in this case.


Assume that the following condition, call it $D(f)$, is satisfied for every $f \in \mathbb{C}[x]-\mathbb{C}$:


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



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


The notation is of fields of fractions.


Remarks:


1. It seems that $y \in \mathbb{C}(u,v)$, but actually I am not sure about this, since I only have the following argument:


We know that $y \in \mathbb{C}(u,v,f)$, so there exist $F,G \in \mathbb{C}[r,s,t]$ such that $y=\frac{F(u,v,f)}{G(u,v,f)}$.


For some $\alpha \in \mathbb{C}$, $f(\alpha)=0$, and then
$y=\frac{F(u(\alpha,y),v(\alpha,y),f(\alpha))}{G(u(\alpha,y),v(\alpha,y),f(\alpha))}=
\frac{F(u(\alpha,y),v(\alpha,y))}{G(u(\alpha,y),v(\alpha,y))}$,
which just shows that $y \in \mathbb{C}(u(\alpha,y),v(\alpha,y))$.


2. I do not know how to show that $x \in \mathbb{C}(u,v)$, if this is true.
3. I guess such $u,v$ must be algebraically independent.


Thank you very much!



If $\mathbb{C}(u(x,y),v(x,y),f(x))=\mathbb{C}(x,y)$, for every $f(x) \in \mathbb{C}[x]-\mathbb{C}$, then already $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?
The answer to the Question is ``no".
As an example, let $u=xy$, $v=x+y$ be the elementary symmetric functions in $x$ and $y$. It is well-known that $[\mathbb{C}(x,y): \mathbb{C}(u,v)]=2$, so those two fields are not equal.
On the other hand, consider $K\_f:=\mathbb{C}(u,v,f(x))$ for any nonconstant polynomial $f(x)\in \mathbb{C}[x]$. Note that $\mathbb{C}(u,v)$ already contains the functions $f(x)+f(y)$ and $\frac{f(x)-f(y)}{x-y}$, since indeed those are symmetric functions in $x$ and $y$. So successively we find that $K\_f$ contains the elements $(f(x)+f(y))-f(x)=f(y)$, thus also $f(x)-f(y)$, and thus also $\frac{f(x)-f(y)}{(f(x)-f(y))/(x-y)}=x-y$.
Finally, it also contains $x+y$ and thus $x$ and $y$, showing $K\_f=\mathbb{C}(x,y)$.


(**Edit**: One could have of course argued non-constructively even quicker: Since the field of symmetric functions is of index $2$ in $\mathbb{C}(x,y)$ and doesn't contain any nonconstant function $f(x)$, adjoining any such function must give the whole thing by lack of intermediate fields)


