# Concerning $k \subset L \subset k(x,y)$

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. Perhaps the arguments in that answer are also applicable here?

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).