Let $R \subseteq \mathbb{C}(x,y)$ and assume that $R=\mathbb{C}(u,v)$, where $u,v \in \mathbb{C}[x,y]$ are algebraically independent over $\mathbb{C}$.

Here $\mathbb{N}$ includes $0$.

Assume that $R$ satisfies the following 'rare' property: For every monomial $x^iy^j$, $i \in \mathbb{N}$, $j \in \mathbb{N}$ (except the case $i=j=0$, for which we assume nothing), we have $\mathbb{C}(u,v,x^iy^j)=\mathbb{C}(x,y)$.

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

I am not able to find a counterexample, but perhaps there is such.

I do not mind to further assume that $x+y$ also satisfies $\mathbb{C}(u,v,x+y)=\mathbb{C}(x,y)$.

Any help is welcome! Thank you very much.

Edit: Please, what about the following version of the above question:

Same as the above question, except that now the property does not include the diagonal, namely, $\mathbb{C}(u,v,x^ny^n)=\mathbb{C}(x,y)$, $n \geq 1$, may not hold.

I am not sure which of the following three options is better to assume for every $n \geq 1$:

• There is no information about adding $x^ny^n$ to $\mathbb{C}(u,v)$.
• $\mathbb{C}(u,v,x^ny^n) \subsetneq \mathbb{C}(x,y)$.
• $x^ny^n \in \mathbb{C}(u,v)$.

Should I ask this as a new question, or is it ok to ask it here?