Existence of valuation rings in an algebraic function field of one variable

Question

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.

Answer 1

I borrowed the idea of the Bourbaki's proof of Krull-Akizuki theorem.

Definition Let $A$ be a not-necessarily commutative ring. Let $M$ be a left $A$-module. Suppose $M$ has a composition series, the lengths of each series are the same by Jordan-Hoelder theorem. We denote it by $leng_A M$. If $M$ does not have a composition series, we define $leng_A M = \infty$.

Lemma 1 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $f$ be a non-zero element of $A$. Then $A/fA$ is a finite $k$-module.

Proof: Clear.

Lemma 2 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $M$ be a torsion $A$-module of finite type. Then $M$ is a finite $k$-module.

Proof: Let $x_1, ..., x_n$ be generating elements of $M$. There exists a non-zero element $f$ of $A$ such that $fx_i = 0$, $i = 1, ..., n$. Let $\psi:A^n \rightarrow M$ be the morphism defined by $\psi(e_i) = x_i$, $i = 1, ..., n$, where $e_1, ..., e_n$ is the canonical basis of $A^n$. By Lemma 1, $A^n/fA^n$ is a finite $k$-module. Since $\psi$ induces a surjective mophism $A^n/fA^n \rightarrow M$, $M$ is a finite $k$-module. QED

Lemma 3 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $M$ be an $A$-module. Then $length_A M < \infty$ if and only if $M$ is a finite $k$-module.

Proof: Suppose $length_A M < \infty$. Let $M = M_0 \supset M_1 \supset ... \supset M_n = 0$ be a composition series. Each $M_i/M_{i+1}$ is isomorphic to $A/f_iA$, where $f_i$ is an irreducible polynomial in $A$. Since $dim_k A/f_iA$ is finite by Lemma 1, $dim_k M$ is finite.

The converse is clear. QED

Lemma 4 Let $A$ be a not necessarily commutative ring. Let $M$ be a left $A$-module. Let $(M_i)_I$ be a family of $A$-submodules of $M$ indexed be a set $I$. Suppose $(M_i)_I$ satisfies the following condition.

$M = \cup_i M_i$, and for any $i, j \in I$, there exists $k \in I$ such that $M_i \subset M_k$ and $M_j \subset M_k$.

Then $leng_A M = sup_i leng_A M_i$.

Proof: Suppose $sup_i leng_A M_i = \infty$. Since $sup_i leng_A M_i \leq leng_A M$, $leng_A M = \infty$. Hence we can assume that $sup_i leng_A M_i = n < \infty$. Let $n = leng_A M_{i_0}$. For each $i \in I$, there exists $k \in I$ such that $M_{i_0} \subset M_k$ and $M_i \subset M_k$. Since $leng_A M_k = n$, $M_{i_0} = M_k$, $M_i \subset M_{i_0}$. Since $M = \cup_i M_i$, $M = M_{i_0}$. Hence $leng_A M = n$. QED

Lemma 5 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $K$ be the field of fractions of $A$. Let $M$ be a torsion-free $A$-module of finite type. Let $r = dim_K M \otimes_A K$ Let $f$ be a non-zero element of $A$. Then $leng_A M/fM \leq r(leng_A A/fA)$

Proof: There exists a $A$-submodule $L$ of $M$ such that $L$ is isomorphic to $A^r$ and $Q = M/L$ is a torsion module of finite type over $A$. Hence, by Lemma 2, $Q$ is a finite $k$-module. The kernel of $M/f^nM \rightarrow Q/f^nQ$ is $(L + f^nM)/f^nM$ which is isomorphic to $L/(f^nM \cap L)$. Since $f^nL \subset f^nM \cap L$, $leng_A M/f^nM \leq leng_A L/f^nL + leng_A Q/f^nQ \leq leng_A L/f^nL + leng_A Q$. Since $M$ is torsion-free, $f$ induces isomorphism $M/fM \rightarrow fM/f^2M$. Hence $leng_A M/f^nM = n(leng_A M/fM)$. Similarly $leng_A L/f^nL = n(leng_A L/fL)$. Hence $leng_A M/fM \leq leng_A L/fL + (1/n) leng_A Q$. Since $L$ is isomorphic to $A^r$, $leng_A L/fL = r(leng_A A/fA)$. Hence $leng_A M/fM \leq r(Leng_A A/fA)$. QED

Lemma 6 Let $A = k[X]$ be the polynomial ring of one variable over a field $k$. Let $K$ be the field of fractions of $A$. Let $M$ be a torsion-free $A$-module. Suppose $r = dim_K M \otimes_A K$ is finite. Let $f$ be a non-zero element of $A$. Then $leng_A M/fM \leq r(Leng_A A/fA)$

Proof: Let $(M_i)_I$ be the family of finitely generated $A$-submodules of $M$. $M/fM = \cup_i (M_i + fM)/fM =\cup_i M_i/(M_i \cap fM)$. Since $fM_i \subset M_i \cap fM$, $M_i/(M_i \cap fM)$ is isomorphic to a quotient of $M_i/fM_i$. Hence, by Lemma 5, $leng_A M_i/(M_i \cap fM) \leq r(leng_A A/fA)$. Hence, by Lemma 4, $leng_A M/fM \leq r(leng_A A/fA)$ QED

Lemma 7 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $K$ be the field of fractions of $A$. Let $L$ be a finite extension field of $K$. Let $B$ be a subring of $L$ containing $A$. Then $B/fB$ is a finite $k$-module for every non-zero element $f \in B$.

Proof: Since $L$ is a finite extension of $K$, $a_rf^r + ... + a_1f + a_0 = 0$, where $a_i \in A, a_0 \neq 0$. Then $a_0 \in fB$. Since $B \otimes_A K \subset L$, $dim_K B \otimes_A K \leq [L : K]$. Hence, by Lemma 6, $leng_A B/a_0B$ is finite. Hence $leng_A B/fB$ is finite. Hence, by Lemma 3, the assertion follows. QED

Lemma 8 Let $A$ be an integrally closed domain containing a field $k$ as a subring. Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$. Let $S$ be a multiplicative subset of $A$. Let $A_S$ be the localization with respect to $S$. Then $A_S$ is an integrally closed domain containing a field $k$ as a subring and $A_S/fA_S$ is a finite $k$-module for every non-zero element $f \in A_S$.

Proof: Let $K$ be the field of fractions of $A$. Suppose that $x \in K$ is integral over $A_S$. $x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0$, where $a_i \in A_S$. Hence there exists $s \in S$ such that $sx$ is integral over $A$. Since $A$ is integrally closed, $sx \in A$. Hence $x \in A_S$. Hence $A_S$ is integrally closed.

Let $f$ be a non-zero element of $A_S$. $f = a/s$, where $a \in A, s \in S$. Then $fA_S = aA_S$. By this, $aA$ is a product of prime ideals of $A$. Let $P$ be a non-zero prime ideal $P$ of $A$. Since $P$ is maximal, $A_S/P^nA_S$ is isomorphic to $A/P^n$ or $0$. Hence $A_S/aA_S$ is a finite $k$-module. QED

Lemma 9 Let $A$ be an integrally closed domain containing a field $k$ as a subring. Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$. Let $P$ be a non-zero prime ideal of $A$. Then $A_P$ is a discrete valuation ring.

Proof: By Lemma 8 and this, every non-zero ideal of $A_P$ has a unique factorization as a product of prime ideals. Hence $PA_P \neq P^2A_p$. Let $x \in PA_P - P^2A_P$. Since $A_P$ is the only non-zero prime ideal of $A_P$, $xA = PA_P$. Since every non-zero ideal of $A_P$ can be written $P^nA_P$, $A_P$ is a principal ideal domain. Hence $A_P$ is a discrete valuation ring. QED

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 $A_P$.

Proof: We can assume that $A$ contains a transcendental element $x$ over $k$(otherwise the theorem would be trivial). We can also assume that $P \neq 0$.

Let $B$ be the integral closure of $A$ in $K$. By Lemma 7, $B/fB$ is a finite $k$-module for every non-zero element $f \in B$. Let $S = A - P$. Let $B_P$ and $A_P$ be the localizations of $B$ and $A$ with rspect to $S$ respectively. Let $y \in P$ be a non-zer element. By Lemma 8, $B_P/yB_P$ is a finite k-module. Since $yB_P \subset PB_P$ and $PB_P \neq B_P$, $yB_P \neq B_P$. Hence there exists a maximal ideal $Q$ of $B_P$ containing $y$. Since $B_P$ is integral over $A_P$ and $PA_P$ is a unique maximal ideal of $A_P$, $P = Q \cap A_P$. Let $Q' = Q \cap B$. Then $Q'$ is a prime ideal of $B$ lying over $P$. By Lemma 9, $B_Q'$ is a discrete valuation ring and it dominates $A_P$. QED