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The following theorem is a slightly modified version of Theorem 1, p.6 of Chevalley's Introduction to the theory of algebraic functions of one variable. He proved it using Zorn's lemma. However, Weil wrote, in his review of the book, that this can be proved without it. I wonder how.

Theorem Let $k$ be a field. Let $K$ be a finitely generated extension field of $k$ of transcendence degree one. Let $A$ be a subring of $K$ containing $k$. Let $P$ be a prime ideal of $A$. Then there exists a valuation ring $R$ of $K$ dominating $A_P$.

EDIT Weil wrote:

One might observe here that, in a function-field of dimension 1, every valuation-ring is finitely generated over the field of constants, and therefore , if a slightly different arrangement had been adopted, the use of Zorn's lemma (or of Zermelo's axiom) could have been avoided altogether; since Theorem 1 is formulated only for such fields, this treatment would have been more consistent, and the distinct features of dimension 1 would have appeared more clearly.

EDIT We can assume that $A$ contains a transcendental element $x$ over $k$(otherwise the theorem would be trivial). If $A$ is finitely generated over $k$, we can prove the theorem without using Zorn's lemma; It is well known that the integral closure $B$ of $A$ in $K$ is finitely generated as an $A$-module. Hence $B$ is Noetherian and integrally closed. We can assume $P ≠ 0$. Since $B$ is integral over $A$, there exists a prime ideal ideal $M$ of $B$ which lies over $P$. Since $B$ is Noetherian, this can be proved without Zorn's lemma. Since dim $B$ = 1, $B_M$ is a discrete valuation ring dominating $A_P$. I wonder if Weil was talking about this case.

EDIT[July 10, 2012] As the answers to this question show, the following line in the above is wrong: "Since $B$ is Noetherian, this can be proved without Zorn's lemma." Surely this can be proved without Zorn's lemma, but it's not because $B$ is Noetherian, but because $B$ is finitely generated as an $A$-module.

EDIT[July 13, 2012] I think I solved the problem. However, I don't think this is how Weil did. I guess his proof was simpler.

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    Dear Makoto, Perhaps I'm confused, but I don't understand the statement that the valuation rings will be f.g. over $k$. Rather, I think they will be *essentially of finite type over $k$*, i.e. localizations of f.g. $k$-algebras. Regards,2012-07-10
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    Dear Matt, I think you are right. I think he meant essentially of finite type over k. Weil wrote that in 1951. At that time, commutative algebra was not developed well. So I guess there was no such terminology.2012-07-10

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