If $\mathbb{C}(u,v,x^n)=\mathbb{C}(x,y)$, for every $n \geq 1$, then already $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

Question

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 $C(n)$, is satisfied for every $n \geq 1$: $\mathbb{C}(u,v,x^n)=\mathbb{C}(x,y)$.

Question 1: Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

Remarks:

1. I excluded $C(0)$, since for $n=0$ we trivially have $\mathbb{C}(u,v,x^0)=\mathbb{C}(u,v,1)=\mathbb{C}(u,v)$, so the claim is true.

2. $u=xy, v=x$ is not a counterexample, since $\mathbb{C}(xy,x)=\mathbb{C}(x,y)$ with or without adding $x^n$, $n \geq 1$.

3. $u=x^2, v=y$ is not a counterexample, since $\mathbb{C}(x^2,y,x^2)=\mathbb{C}(x^2,y) \subsetneq \mathbb{C}(x,y)$, so the condition does not hold for $n=2$ (and also does not hold for any even $n$).

4. Maybe using some Galois theory may help, if we further assume something on the degree of the field extension $\mathbb{C}(x,y) : \mathbb{C}(u,v)$. For example, further assume that $d=[\mathbb{C}(x,y) : \mathbb{C}(u,v)]=p$, $p \geq 2$ a prime number. We can think about $u=x^2,v=y$. In this case $d=2$. But, as we mentioned, $C(2m)$ does not hold for every $m \geq 1$, so those $u,v$ are not relevant for this question.

5. If we look for a counterexample, maybe we should try to find $u,v$ with $d=p \geq 2$ prime, $y \in \mathbb{C}(u,v)$ and $x^n \notin \mathbb{C}(u,v)$, for every $n \geq 1$. We can take $v=y$, but I have not yet found an appropriate $u$, namely, $u \in \mathbb{C}[x,y]$ such that $[\mathbb{C}(x,y) : \mathbb{C}(u,y)]=2$ (or any other prime number) with $x^n \notin \mathbb{C}[u,y]$, for every $n \geq 1$, and since the extension is of prime degree, then necessarily $\mathbb{C}(u,y,x^n)=\mathbb{C}(x,y)$, for every $n \geq 1$.

6. Another direction is considering notions like "factorially closed subring".

7. Perhaps one can find a proof by induction on the $y$-degree of $v$, with arbitrary $u$?

8. Perhaps the Cremona group may help.

Thank you very much!

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In view of the counterexample, I would like to change my question to the following one, which I also posted here, in MO:

Let $u,v \in \mathbb{C}(x,y)$.

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

Notice that the above counterexample does not work now, since $u=x^2+x, v=y$ does not satisfy $D(x^2+x)$.

Question 2 has an answer in MO, a counterexample with $u=x+y, v=xy$.

Answer 1

A counterexample: $u=x^2+x, v=y$. The explanation is a as follows:

1. It is clear that $\mathbb{C}(x^2+x,y) \subsetneq \mathbb{C}(x,y)$, so we wish to show that $C(n)$ is satisfied for every $n \geq 1$.
2. $C(n)$ says: $\mathbb{C}(x^2+x,y,x^n)=\mathbb{C}(x,y)$. Fix any $n \geq 1$.

Since $y$ is already in $\mathbb{C}(x^2+x,y,x^n)$, it is left to show that $x \in \mathbb{C}(x^2+x,y,x^n)$ (we can forget about $y$).

Claim: $\mathbb(x^2+x,x^n)=\mathbb{C}(x)$.

Proof of Claim: Apply this answer which says that for $f,g \in \mathbb{C}[x]$, if there exist $a,b,c \in \mathbb{C}$ such that $\gcd(f-a,g-b)=x-c$, then $\mathbb{C}(f,g)=\mathbb{C}(x)$.

Here, it is clear that $\gcd(x^2+x,x^n)=\gcd(x(x+1),x^n)=x$, therefore $\mathbb{C}(x^2+x,x^n)=\mathbb{C}(x)$.

Concluding that $\mathbb{C}(x^2+x,y,x^n)=\mathbb{C}(x^2+x,x^n)(y)=\mathbb{C}(x)(y)=\mathbb{C}(x,y)$, so $C(n)$ is satisfied for every $n \geq 1$.