The following is a known result in algebraic geometry:


Let $k$ be an algebraically closed field of characteristic zero (for example, $k=\mathbb{C}$).
Let $L$ be a field such that $k \subset L \subset k(x,y)$ 
and $L$ is of transcendence degree two over $k$. 
Then there exist $h\_1,h\_2 \in k(x,y)$ such that $L=k(h\_1,h\_2)$.



> 
> Is it possible to find $g\_1,g\_2 \in k[x,y]$ such that $L=k(g\_1,g\_2)$? 
> 
> 
> 


The motivation is the following result: If $k \subset L \subset k(x,y)$ is of transcendence
degree one over $k$, then $L=k(h)$, where $h \in k[x,y]$; see [this answer](https://mathoverflow.net/questions/190920/any-two-bivariate-algebraically-dependent-polynomials-are-always-in-the-same-rin/191058#191058). Perhaps the arguments in that answer are also applicable here?


I have asked the above question in [MSE](https://math.stackexchange.com/questions/3567214/concerning-k-subset-l-subset-kx-y), with no comments. See also [this](https://math.stackexchange.com/questions/2540914/a-pure-algebraic-proof-for-any-unirational-surface-over-mathbbc-is-rationa) question. 


Thank you very much!



Concerning $k \subset L \subset k(x,y)$
No. $M\mathrel{:=}k(x,y)$ has a $k$-automorphism $\sigma:x\mapsto 1/x,\,y\mapsto 1/y$, of order 2. Let $G\mathrel{:=}\langle\sigma\rangle$, and put $L\mathrel{:=}M^{G}$, the fixed field. The elements $x+1/x$ and $y+1/y$ of $L$ are algebraically independent over $k$, hence $L$ has transcendency degree 2. However, no non-constant polynomial $g\in k[x,y]$ can be in $L$ (that is, be invariant under $\sigma$).


By the same token, the automorphism $\tau:x\mapsto 1/y,\,y\mapsto 1/x$, fixes the field $k(x/y)$, which has transcendency degree 1 over $k$, but it cannot fix any non-constant polynomial. So $k(x/y)\neq k(h)$ for every $h\in k[x,y]$. (Here, $M$ is again quadratic over the fixed field of $\tau$, so that the latter is of tr. deg. 2 over $k$ again. A transcendental element independent of $x/y$ is $x+1/y$, for instance).


