For $r \geq 2$, let $A_r=\mathbb{C}[x_1,\ldots,x_r]$, $F_r=\mathbb{C}(x_1,\ldots,x_r)$ the field of fractions of $A_r$, and $E_r \subseteq F_r$ an arbitrary subfield of $F_r$ with $[F_r:E_r] < \infty$ (this happens if the transcendence degree of $E_r$ over $\mathbb{C}$ is two).
Here $\mathbb{N}$ includes zero.
For $(i_1,\ldots,i_r) \in \mathbb{N}^r$ denote by $C_{E_r,i_1,\ldots,i_r}$ the following condition: $E_r(x_1^{i_1},\ldots,x_r^{i_r})=F_r$, namely, $x_1^{i_1},\ldots,x_r^{i_r}$ is a primitive element for the extension $E_r \subseteq F_r$.
Claim: Fix $E_r$ and assume that for every $(i_1,\ldots,i_r) \in \mathbb{N}^r-\{(0,\ldots,0)\}$, $C_{E_r,i_1,\ldots,i_r}$ holds. (Of course, we removed $(0,\ldots,0)$ since $C_{E_r,0,\ldots,0}$ says that $E_r(1)=F_r$, so $E_r=F_r$). Then $[F_r:E_r] \leq 2$.
It seems $E_r$ of the claim should be 'large', since it has 'small' generators, for example, for $r=3$, among other conditions, $F_3=E_3(x_1)=E_3(x_2)=E_3(x_3)$. But maybe I am wrong.
Remarks:
(1) I have excluded $r=1$ since there are counterexamples in the one-dimensional case, for example: $E_1=\mathbb{C}(x_1^m-x_1^{m-1})$; $C_{E_1,i}$ holds for every $i \geq 1$, but $[F_1:E_1] = m$, so for $m \geq 3$, this is a counterexample. (Where $C_{E_1,i}$ says that $\mathbb{C}(x_1^m-x_1^{m-1})(x_1^i)=E_1(x_1^i)=F_1=\mathbb{C}(x_1)$).
(2) It seems that we can replace $\mathbb{C}$ by any field $k$ of characteristic zero.
(3) In the above question I have tried to generalize this question.
Thank you very much!
