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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?

I have asked the above question in MSE, with no comments. See also this question.

Thank you very much!

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  • $\begingroup$ @YCor, thank you very much for adding the highly relevant tags. $\endgroup$ – user237522 Mar 3 at 19:02
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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).

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  • $\begingroup$ Thank you for the nice answer! $\endgroup$ – user237522 Apr 11 at 21:27

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