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:** 

**(1)** If [this][1] question has a positive answer, then we can apply it here, since the conditon here includes $\mathbb{C}(u,v,y^n)=\mathbb{C}(x,y)$, and conclude that $R=\mathbb{C}(x,y)$.

**(2)** If I am not wrong, the answer below shows that $[\mathbb{C}(x,y): \mathbb{C}(u,v)] \leq 2$.


  [1]: https://math.stackexchange.com/questions/4942550/if-for-every-n-geq-1-kft-gt-tn-kt-then-kft-gt-kt