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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 $L=k(h_1,h_2)$.

This seems a known result in algebraic geometry, according to the comments in this question (especially the last one).

Please:

(1) Is there a pure algebraic proof for this result?

(2) Is it possible to find $h_1,h_2 \in k[x,y]$? 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 the answer to this question.

Thank you very much!

Edit: (1) Wikipedia only brings the algebraic geometry terminology. Also, this notes talk in algebraic geometry terminology (except for the first chapter). (2) This question is relevant.

user237522
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1 Answers1

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This is a classical result in algebraic geometry. It is a direct consequence of the Castelnuovo criterion of rationality.

You can find a proof of it for instance in

Beauville: " Complex algebraic surfaces" Chapter V

Or

Badescu : "Algebraic surfaces" Chapter 13

Carletto
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  • Thank you for trying to help me. But I wished to see a pure algebraic proof, without using algebraic geometry. – user237522 Nov 28 '17 at 17:28