A subfield $R \subseteq \mathbb{C}(x,y)$ with 'many' generators $w$, $R(w)=\mathbb{C}(x,y)$

Question

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.

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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 $\mathbb{C}(u,v)$.
• $\mathbb{C}(u,v,x^ny^n) \subsetneq \mathbb{C}(x,y)$.
• $x^ny^n \in \mathbb{C}(u,v)$.

Perhaps the third option, $x^ny^n \in \mathbb{C}(u,v)$, has the following solution, based on the answer to the original question:

For a given field $F$ and a given pair $(i,j) \in \mathbb{N} \times \mathbb{N}$, denote by $C_{F,i,j}$ the condition/property that $F(x^iy^j)=\mathbb{C}(x,y)$, in other words, the condition that $x^iy^j$ is a primitive element for the extension $F \subseteq \mathbb{C}(x,y)$.

Then the answer shows the following: If $C_{R,i,j}$ holds for every $(i,j) \in \mathbb{N} \times \mathbb{N} - (0,0)$, then $[\mathbb{C}(x,y) : R]=2$.

(Trivially, if $C_{R,0,0}$ holds, then $R=\mathbb{C}(x,y)$, so in this case $[\mathbb{C}(x,y) : R ]=1$).

Now take $\tilde{R}=\mathbb{C}(\tilde{u},\tilde{v})$, where $\tilde{u}, \tilde{v} \in \mathbb{C}[x,y]$ are algebraically independent over $\mathbb{C}$, and assume that $C_{\tilde{R},i,j}$ holds for every $(i,j) \in \mathbb{N} \times \mathbb{N} - D$, where $D= \{(n,n)\}_{n \in \mathbb{N}}$.

Moreover, assume that for every $n \in \mathbb{N}$, $x^ny^n \in \tilde{R}$.

Claim: $[\mathbb{C}(x,y):\tilde{R}]\leq 2$.

Proof of claim: Same as the answer, except that the last two lines are changed, the map is not surjective, $\tilde{R} \subseteq \mathbb{C}(x,y) \subseteq L$. Therefore, by considerations of degree extensions, $[\mathbb{C}(x,y):\tilde{R}]\leq 2$.

Maybe I am missing something.

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

Answer 1

Start with $R=\mathbb{C}(u,v)$ and consider its quadratic extension $L=R(\sqrt{u})=R(s)=\mathbb{C}(s,v)$ with $s^2=u$.

Take an element $z$ of $L$, viewed as a rational function $z(s,v)$. Then $R(z)=L$ iff $z\notin R$ (because $[L:R]=2$) iff $\sigma(z)\neq z$ where $\sigma(z)$ is the $\mathrm{Gal}(L/R)$-conjugate of $z$ which is $z(-s,v)$.

We can also view $L$ as $\mathbb{C}(x,y)$ in many different ways; let us pick $$x=s+v, \quad y=s+2v.$$ For $i, j\in \mathbb{N}$ we have $x^i y^j= (s+v)^i (s+2v)^j$ and $\sigma(x^i y^j)= (-s+v)^i (-s+2v)^j$. If $(i,j)\neq(0,0)$ these are always different, so $x^i y^j\notin R$ and thus $R(x^i y^j)=L$.