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Tag 00I8

10.49. Valuation rings

Here are some definitions.

Definition 10.49.1. Valuation rings.

  1. Let $K$ be a field. Let $A$, $B$ be local rings contained in $K$. We say that $B$ dominates $A$ if $A \subset B$ and $\mathfrak m_A = A \cap \mathfrak m_B$.
  2. Let $A$ be a ring. We say $A$ is a valuation ring if $A$ is a local domain and if $A$ is maximal for the relation of domination among local rings contained in the fraction field of $A$.
  3. Let $A$ be a valuation ring with fraction field $K$. If $R \subset K$ is a subring of $K$, then we say $A$ is centered on $R$ if $R \subset A$.

With this definition a field is a valuation ring.

Lemma 10.49.2. Let $K$ be a field. Let $A \subset K$ be a local subring. Then there exists a valuation ring with fraction field $K$ dominating $A$.

Proof. We consider the collection of local subrings of $K$ as a partially ordered set using the relation of domination. Suppose that $\{A_i\}_{i \in I}$ is a totally ordered collection of local subrings of $K$. Then $B = \bigcup A_i$ is a local subring which dominates all of the $A_i$. Hence by Zorn's Lemma, it suffices to show that if $A \subset K$ is a local ring whose fraction field is not $K$, then there exists a local ring $B \subset K$, $B \not = A$ dominating $A$.

Pick $t \in K$ which is not in the fraction field of $A$. If $t$ is transcendental over $A$, then $A[t] \subset K$ and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$. Then for some $a \in A$ the element $at$ is integral over $A$. In this case the subring $A' \subset K$ generated by $A$ and $ta$ is finite over $A$. By Lemma 10.35.17 there exists a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. If $A = A'_{\mathfrak m'}$, then $t$ is in the fraction field of $A$ which we assumed not to be the case. Thus $A \not = A'_{\mathfrak m'}$ as desired. $\square$

Lemma 10.49.3. Let $A$ be a valuation ring with maximal ideal $\mathfrak m$ and fraction field $K$. Let $x \in K$. Then either $x \in A$ or $x^{-1} \in A$ or both.

Proof. Assume that $x$ is not in $A$. Let $A'$ denote the subring of $K$ generated by $A$ and $x$. Since $A$ is a valuation ring we see that there is no prime of $A'$ lying over $\mathfrak m$. Since $\mathfrak m$ is maximal we see that $V(\mathfrak m A') = \emptyset$. Then $\mathfrak m A' = A'$ by Lemma 10.16.2. Hence we can write $1 = \sum_{i = 0}^d t_i x^i$ with $t_i \in \mathfrak m$. This implies that $(1 - t_0) (x^{-1})^d - \sum t_i (x^{-1})^{d - i} = 0$. In particular we see that $x^{-1}$ is integral over $A$. Thus the subring $A''$ of $K$ generated by $A$ and $x^{-1}$ is finite over $A$ and we see there exists a prime ideal $\mathfrak m'' \subset A''$ lying over $\mathfrak m$ by Lemma 10.35.17. Since $A$ is a valuation ring we conclude that $A = (A'')_{\mathfrak m''}$ and hence $x^{-1} \in A$. $\square$

Lemma 10.49.4. Let $A \subset K$ be a subring of a field $K$ such that for all $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both. Then $A$ is a valuation ring with fraction field $K$.

Proof. If $A$ is not $K$, then $A$ is not a field and there is a nonzero maximal ideal $\mathfrak m$. If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$ with $x \in \mathfrak m$, $y \not \in \mathfrak m$, $x \not \in \mathfrak m'$, and $y \in \mathfrak m'$ (see Lemma 10.14.2). Then neither $x/y \in A$ nor $y/x \in A$ contradicting the assumption of the lemma. Thus we see that $A$ is a local ring. Suppose that $A'$ is a local ring contained in $K$ which dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_A$. But then $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$. $\square$

Lemma 10.49.5. Let $I$ be a directed set. Let $(A_i, \varphi_{ij})$ be a system of valuation rings over $I$. Then $A = \mathop{\rm colim}\nolimits A_i$ is a valuation ring.

Proof. It is clear that $A$ is a domain. Let $a, b \in A$. Lemma 10.49.4 tells us we have to show that either $a | b$ or $b | a$ in $A$. Choose $i$ so large that there exist $a_i, b_i \in A_i$ mapping to $a, b$. Then Lemma 10.49.3 applied to $a_i, b_i$ in $A_i$ implies the result for $a, b$ in $A$. $\square$

Lemma 10.49.6. Let $K \subset L$ be an extension of fields. If $B \subset L$ is a valuation ring, then $A = K \cap B$ is a valuation ring.

Proof. We can replace $L$ by $f.f.(B)$ and $K$ by $K \cap f.f.(B)$. Then the lemma follows from a combination of Lemmas 10.49.3 and 10.49.4. $\square$

Lemma 10.49.7. Let $K \subset L$ be an algebraic extension of fields. If $B \subset L$ is a valuation ring with fraction field $L$ and not a field, then $A = K \cap B$ is a valuation ring and not a field.

Proof. By Lemma 10.49.6 the ring $A$ is a valuation ring. If $A$ is a field, then $A = K$. Then $A = K \subset B$ is an integral extension, hence there are no proper inclusions among the primes of $B$ (Lemma 10.35.20). This contradicts the assumption that $B$ is a local domain and not a field. $\square$

Lemma 10.49.8. Let $A$ be a valuation ring. For any prime ideal $\mathfrak p \subset A$ the quotient $A/\mathfrak p$ is a valuation ring. The same is true for the localization $A_\mathfrak p$ and in fact any localization of $A$.

Proof. Use the characterization of valuation rings given in Lemma 10.49.4. $\square$

Lemma 10.49.9. Let $A'$ be a valuation ring with residue field $K$. Let $A$ be a valuation ring with fraction field $K$. Then $C = \{\lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A\}$ is a valuation ring.

Proof. Note that $\mathfrak m_{A'} \subset C$ and $C/\mathfrak m_{A'} = A$. In particular, the fraction field of $C$ is equal to the fraction field of $A'$. We will use the criterion of Lemma 10.49.4 to prove the lemma. Let $x$ be an element of the fraction field of $C$. By the lemma we may assume $x \in A'$. If $x \in \mathfrak m_{A'}$, then we see $x \in C$. If not, then $x$ is a unit of $A'$ and we also have $x^{-1} \in A'$. Hence either $x$ or $x^{-1}$ maps to an element of $A$ by the lemma again. $\square$

Lemma 10.49.10. Let $A$ be a valuation ring. Then $A$ is a normal domain.

Proof. Suppose $x$ is in the field of fractions of $A$ and integral over $A$, say $x^{d + 1} + \sum_{i \leq d} a_i x^i = 0$. By Lemma 10.49.4 either $x \in A$ (and we're done) or $x^{-1} \in A$. In the second case we see that $x = - \sum a_i x^{i - d} \in A$ as well. $\square$

Lemma 10.49.11. Let $A$ be a normal domain with fraction field $K$. For every $x \in K$, $x \not \in A$ there exists a valuation ring $A \subset V \subset K$ with fraction field $K$ such that $x \not \in V$. In other words, $A$ is the intersection of all valuation rings in $K$ containing $A$.

Proof. Suppose $x \in K$, $x \not \in A$. Consider $B = A[x^{-1}]$. Then $x \not \in B$. Namely, if $x = a_0 + a_1x^{-1} + \ldots + a_d x^{-d}$ then $x^{d + 1} - a_0x^d - \ldots - a_d = 0$ and $x$ is integral over $A$ in contradiction with the fact that $A$ is normal. Thus $x^{-1}$ is not a unit in $B$. Thus $V(x^{-1}) \subset \mathop{\rm Spec}(B)$ is not empty (Lemma 10.16.2), and we can choose a prime $\mathfrak p \subset B$ with $x^{-1} \in \mathfrak p$. Choose a valuation ring $V \subset K$ dominating $B_\mathfrak p$ (Lemma 10.49.2). Then $x \not \in V$ as $x^{-1} \in \mathfrak m_V$. $\square$

An totally ordered abelian group is a pair $(\Gamma, \geq)$ consisting of an abelian group $\Gamma$ endowed with a total ordering $\geq$ such that $\gamma \geq \gamma' \Rightarrow \gamma + \gamma'' \geq \gamma' + \gamma''$ for all $\gamma, \gamma', \gamma'' \in \Gamma$.

Lemma 10.49.12. Let $A$ be a valuation ring with field of fractions $K$. Set $\Gamma = K^*/A^*$ (with group law written additively). For $\gamma, \gamma' \in \Gamma$ define $\gamma \geq \gamma'$ if and only if $\gamma - \gamma'$ is in the image of $A - \{0\} \to \Gamma$. Then $(\Gamma, \geq)$ is a totally ordered abelian group.

Proof. Omitted, but follows easily from Lemma 10.49.3. Note that in case $A = K$ we obtain the zero group $\Gamma = \{0\}$ endowed with its unique total ordering. $\square$

Definition 10.49.13. Let $A$ be a valuation ring.

  1. The totally ordered abelian group $(\Gamma, \geq)$ of Lemma 10.49.12 is called the value group of the valuation ring $A$.
  2. The map $v : A - \{0\} \to \Gamma$ and also $v : K^* \to \Gamma$ is called the valuation associated to $A$.
  3. The valuation ring $A$ is called a discrete valuation ring if $\Gamma \cong \mathbf{Z}$.

Note that if $\Gamma \cong \mathbf{Z}$ then there is a unique such isomorphism such that $1 \geq 0$. If the isomorphism is chosen in this way, then the ordering becomes the usual ordering of the integers.

Lemma 10.49.14. Let $A$ be a valuation ring. The valuation $v : A -\{0\} \to \Gamma_{\geq 0}$ has the following properties:

  1. $v(a) = 0 \Leftrightarrow a \in A^*$,
  2. $v(ab) = v(a) + v(b)$,
  3. $v(a + b) \geq \min(v(a), v(b))$.

Proof. Omitted. $\square$

Lemma 10.49.15. Let $A$ be a ring. The following are equivalent

  1. $A$ is a valuation ring,
  2. $A$ is a local domain and every finitely generated ideal of $A$ is principal.

Proof. Assume $A$ is a valuation ring and let $f_1, \ldots, f_n \in A$. Choose $i$ such that $v(f_i)$ is minimal among $v(f_j)$. Then $(f_i) = (f_1, \ldots, f_n)$. Conversely, assume $A$ is a local domain and every finitely generated ideal of $A$ is principal. Pick $f, g \in A$ and write $(f, g) = (h)$. Then $f = ah$ and $g = bh$ and $h = cf + dg$ for some $a, b, c, d \in A$. Thus $ac + bd = 1$ and we see that either $a$ or $b$ is a unit, i.e., either $g/f$ or $f/g$ is an element of $A$. This shows $A$ is a valuation ring by Lemma 10.49.4. $\square$

Lemma 10.49.16. Let $(\Gamma, \geq)$ be a totally ordered abelian group. Let $K$ be a field. Let $v : K^* \to \Gamma$ be a homomorphism of abelian groups such that $v(a + b) \geq \min(v(a), v(b))$ for $a, b \in K$ with $a, b, a + b$ not zero. Then $$ A = \{ x \in K \mid x = 0 \text{ or } v(x) \geq 0 \} $$ is a valuation ring with value group $\mathop{\rm Im}(v) \subset \Gamma$, with maximal ideal $$ \mathfrak m = \{ x \in K \mid x = 0 \text{ or } v(x) > 0 \} $$ and with group of units $$ A^* = \{ x \in K^* \mid v(x) = 0 \}. $$

Proof. Omitted. $\square$

Let $(\Gamma, \geq)$ be a totally ordered abelian group. An ideal of $\Gamma$ is a subset $I \subset \Gamma$ such that all elements of $I$ are $\geq 0$ and $\gamma \in I$, $\gamma' \geq \gamma$ implies $\gamma' \in I$. We say that such an ideal is prime if $\gamma + \gamma' \in I, \gamma, \gamma' \geq 0 \Rightarrow \gamma \in I \text{ or } \gamma' \in I$.

Lemma 10.49.17. Let $A$ be a valuation ring. Ideals in $A$ correspond $1 - 1$ with ideals of $\Gamma$. This bijection is inclusion preserving, and maps prime ideals to prime ideals.

Proof. Omitted. $\square$

Lemma 10.49.18. A valuation ring is Noetherian if and only if it is a discrete valuation ring or a field.

Proof. Suppose $A$ is a discrete valuation ring with valuation $v : A \setminus \{0\} \to \mathbf{Z}$ normalized so that $\mathop{\rm Im}(v) \subset \mathbf{Z}_{\geq 0}$. By Lemma 10.49.17 the ideals of $A$ are the subsets $I_n = \{0\} \cup v^{-1}(\mathbf{Z}_{\geq n})$. It is clear that any element $x \in A$ with $v(x) = n$ generates $I_n$. Hence $A$ is a PID so certainly Noetherian.

Suppose $A$ is a Noetherian valuation ring with value group $\Gamma$. By Lemma 10.49.17 we see the ascending chain condition holds for ideals in $\Gamma$. We may assume $A$ is not a field, i.e., there is a $\gamma \in \Gamma$ with $\gamma > 0$. Applying the ascending chain condition to the subsets $\gamma + \Gamma_{\geq 0}$ with $\gamma > 0$ we see there exists a smallest element $\gamma_0$ which is bigger than $0$. Let $\gamma \in \Gamma$ be an element $\gamma > 0$. Consider the sequence of elements $\gamma$, $\gamma - \gamma_0$, $\gamma - 2\gamma_0$, etc. By the ascending chain condition these cannot all be $> 0$. Let $\gamma - n \gamma_0$ be the last one $\geq 0$. By minimality of $\gamma_0$ we see that $0 = \gamma - n \gamma_0$. Hence $\Gamma$ is a cyclic group as desired. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 11105–11509 (see updates for more information).

    \section{Valuation rings}
    \label{section-valuation-rings}
    
    \noindent
    Here are some definitions.
    
    \begin{definition}
    \label{definition-valuation-ring}
    Valuation rings.
    \begin{enumerate}
    \item Let $K$ be a field. Let $A$, $B$ be local rings contained
    in $K$. We say that $B$ {\it dominates} $A$ if $A \subset B$
    and $\mathfrak m_A = A \cap \mathfrak m_B$.
    \item Let $A$ be a ring. We say $A$ is a {\it valuation ring}
    if $A$ is a local domain and if $A$ is maximal
    for the relation of domination among local rings contained in
    the fraction field of $A$.
    \item Let $A$ be a valuation ring with fraction field $K$.
    If $R \subset K$ is a subring of $K$, then we say $A$
    is {\it centered} on $R$ if $R \subset A$.
    \end{enumerate}
    \end{definition}
    
    \noindent
    With this definition a field is a valuation ring.
    
    \begin{lemma}
    \label{lemma-dominate}
    Let $K$ be a field. Let $A \subset K$ be a local subring.
    Then there exists a valuation ring with fraction field $K$
    dominating $A$.
    \end{lemma}
    
    \begin{proof}
    We consider the collection of local subrings
    of $K$ as a partially ordered set using the relation of domination.
    Suppose that $\{A_i\}_{i \in I}$ is a totally ordered
    collection of local subrings of $K$. Then $B = \bigcup A_i$
    is a local subring which dominates all of the $A_i$.
    Hence by Zorn's Lemma, it suffices to show that if $A \subset K$
    is a local ring whose fraction field is not $K$, then there
    exists a local ring $B \subset K$, $B \not = A$ dominating $A$.
    
    \medskip\noindent
    Pick $t \in K$ which is not in the fraction field of $A$.
    If $t$ is transcendental over $A$, then $A[t] \subset K$
    and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring
    distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$.
    Then for some $a \in A$ the element $at$ is integral over $A$.
    In this case the subring $A' \subset K$ generated by $A$ and
    $ta$ is finite over $A$.
    By Lemma \ref{lemma-integral-overring-surjective} there exists
    a prime ideal $\mathfrak m' \subset A'$ lying over
    $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates
    $A$. If $A = A'_{\mathfrak m'}$, then $t$
    is in the fraction field of $A$ which we assumed not to be the case.
    Thus $A \not = A'_{\mathfrak m'}$ as desired.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-valuation-ring-x-or-x-inverse}
    Let $A$ be a valuation ring with maximal ideal $\mathfrak m$ and
    fraction field $K$.
    Let $x \in K$. Then either $x \in A$ or $x^{-1} \in A$ or both.
    \end{lemma}
    
    \begin{proof}
    Assume that $x$ is not in $A$.
    Let $A'$ denote the subring of $K$ generated by $A$ and $x$.
    Since $A$ is a valuation ring we see that there is no prime
    of $A'$ lying over $\mathfrak m$. Since $\mathfrak m$ is maximal
    we see that $V(\mathfrak m A') = \emptyset$. Then $\mathfrak m A' = A'$
    by Lemma \ref{lemma-Zariski-topology}.
    Hence we can write
    $1 = \sum_{i = 0}^d t_i x^i$ with $t_i \in \mathfrak m$.
    This implies that $(1 - t_0) (x^{-1})^d - \sum t_i (x^{-1})^{d - i} = 0$.
    In particular we see that $x^{-1}$ is integral over $A$.
    Thus the subring $A''$ of $K$ generated by $A$ and $x^{-1}$ is
    finite over $A$ and we see there exists a prime ideal
    $\mathfrak m'' \subset A''$ lying over $\mathfrak m$ by
    Lemma \ref{lemma-integral-overring-surjective}. Since $A$
    is a valuation ring we conclude that $A = (A'')_{\mathfrak m''}$
    and hence $x^{-1} \in A$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-x-or-x-inverse-valuation-ring}
    Let $A \subset K$ be a subring of a field $K$ such that for all
    $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both.
    Then $A$ is a valuation ring with fraction field $K$.
    \end{lemma}
    
    \begin{proof}
    If $A$ is not $K$, then $A$ is not a field and there is a nonzero
    maximal ideal $\mathfrak m$.
    If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$
    with $x \in \mathfrak m$, $y \not \in \mathfrak m$,
    $x \not \in \mathfrak m'$, and $y \in \mathfrak m'$ (see
    Lemma \ref{lemma-silly}). Then neither $x/y \in A$ nor $y/x \in A$
    contradicting the assumption of the lemma. Thus we see that $A$ is
    a local ring. Suppose that $A'$ is a local ring contained in $K$ which
    dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then
    $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_A$. But then
    $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-colimit-valuation-rings}
    Let $I$ be a directed set. Let $(A_i, \varphi_{ij})$
    be a system of valuation rings over $I$.
    Then $A = \colim A_i$ is a valuation ring.
    \end{lemma}
    
    \begin{proof}
    It is clear that $A$ is a domain. Let $a, b \in A$.
    Lemma \ref{lemma-x-or-x-inverse-valuation-ring} tells us we have
    to show that either $a | b$ or $b | a$ in $A$. Choose $i$
    so large that there exist $a_i, b_i \in A_i$ mapping to $a, b$.
    Then Lemma \ref{lemma-valuation-ring-x-or-x-inverse}
    applied to $a_i, b_i$ in $A_i$ implies the result for $a, b$ in $A$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-valuation-ring-cap-field}
    Let $K \subset L$ be an extension of fields. If $B \subset L$
    is a valuation ring, then $A = K \cap B$ is a valuation ring.
    \end{lemma}
    
    \begin{proof}
    We can replace $L$ by $f.f.(B)$ and $K$ by $K \cap f.f.(B)$.
    Then the lemma follows from a combination of
    Lemmas \ref{lemma-valuation-ring-x-or-x-inverse} and
    \ref{lemma-x-or-x-inverse-valuation-ring}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-valuation-ring-cap-field-finite}
    Let $K \subset L$ be an algebraic extension of fields. If $B \subset L$
    is a valuation ring with fraction field $L$ and not a field, then
    $A = K \cap B$ is a valuation ring and not a field.
    \end{lemma}
    
    \begin{proof}
    By Lemma \ref{lemma-valuation-ring-cap-field} the ring $A$ is a valuation
    ring. If $A$ is a field, then $A = K$. Then $A = K \subset B$ is an integral
    extension, hence there are no proper inclusions among the primes of $B$
    (Lemma \ref{lemma-integral-no-inclusion}).
    This contradicts the assumption that $B$ is a local domain and not a field.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-make-valuation-rings}
    Let $A$ be a valuation ring. For any prime ideal $\mathfrak p \subset A$ the
    quotient $A/\mathfrak p$ is a valuation ring. The same is true for the
    localization $A_\mathfrak p$ and in fact any localization of $A$.
    \end{lemma}
    
    \begin{proof}
    Use the characterization of valuation rings given
    in Lemma \ref{lemma-x-or-x-inverse-valuation-ring}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-stack-valuation-rings}
    Let $A'$ be a valuation ring with residue field $K$.
    Let $A$ be a valuation ring with fraction field $K$.
    Then
    $C = \{\lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A\}$
    is a valuation ring.
    \end{lemma}
    
    \begin{proof}
    Note that $\mathfrak m_{A'} \subset C$ and $C/\mathfrak m_{A'} = A$.
    In particular, the fraction field of $C$ is equal to the fraction field
    of $A'$. We will use the criterion of
    Lemma \ref{lemma-x-or-x-inverse-valuation-ring} to prove the lemma.
    Let $x$ be an element of the fraction field of $C$.
    By the lemma we may assume $x \in A'$. If $x \in \mathfrak m_{A'}$,
    then we see $x \in C$. If not, then $x$ is a unit of $A'$ and we
    also have $x^{-1} \in A'$. Hence either $x$ or $x^{-1}$ maps to
    an element of $A$ by the lemma again.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-valuation-ring-normal}
    Let $A$ be a valuation ring.
    Then $A$ is a normal domain.
    \end{lemma}
    
    \begin{proof}
    Suppose $x$ is in the field of fractions of $A$ and integral over $A$,
    say $x^{d + 1} + \sum_{i \leq d} a_i x^i = 0$. By
    Lemma \ref{lemma-x-or-x-inverse-valuation-ring}
    either $x \in A$ (and we're done) or $x^{-1} \in A$. In the second case
    we see that $x = - \sum a_i x^{i - d} \in A$ as well.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-find-valuation-rings}
    Let $A$ be a normal domain with fraction field $K$.
    For every $x \in K$, $x \not \in A$ there exists a
    valuation ring $A \subset V \subset K$ with fraction field $K$
    such that $x \not \in V$. In other words, $A$ is the intersection
    of all valuation rings in $K$ containing $A$.
    \end{lemma}
    
    \begin{proof}
    Suppose $x \in K$, $x \not \in A$. Consider $B = A[x^{-1}]$.
    Then $x \not \in B$. Namely, if $x = a_0 + a_1x^{-1} + \ldots + a_d x^{-d}$
    then $x^{d + 1} - a_0x^d - \ldots - a_d = 0$ and $x$ is integral
    over $A$ in contradiction with the fact that $A$ is normal.
    Thus $x^{-1}$ is not a unit in $B$. Thus $V(x^{-1}) \subset \Spec(B)$
    is not empty (Lemma \ref{lemma-Zariski-topology}), and we can choose a prime
    $\mathfrak p \subset B$ with $x^{-1} \in \mathfrak p$.
    Choose a valuation ring $V \subset K$ dominating $B_\mathfrak p$
    (Lemma \ref{lemma-dominate}).
    Then $x \not \in V$ as $x^{-1} \in \mathfrak m_V$.
    \end{proof}
    
    \noindent
    An {\it totally ordered abelian group} is a pair $(\Gamma, \geq)$
    consisting of an abelian group $\Gamma$ endowed with a total
    ordering $\geq$ such that $\gamma \geq \gamma' \Rightarrow
    \gamma + \gamma'' \geq \gamma' + \gamma''$ for all
    $\gamma, \gamma', \gamma'' \in \Gamma$.
    
    \begin{lemma}
    \label{lemma-valuation-group}
    Let $A$ be a valuation ring with field of fractions $K$.
    Set $\Gamma = K^*/A^*$ (with group law written additively).
    For $\gamma, \gamma' \in \Gamma$
    define $\gamma \geq \gamma'$ if and only if
    $\gamma - \gamma'$ is in the image of $A - \{0\} \to \Gamma$.
    Then $(\Gamma, \geq)$ is a totally ordered abelian group.
    \end{lemma}
    
    \begin{proof}
    Omitted, but follows easily from
    Lemma \ref{lemma-valuation-ring-x-or-x-inverse}.
    Note that in case $A = K$ we obtain the zero group $\Gamma = \{0\}$
    endowed with its unique total ordering.
    \end{proof}
    
    \begin{definition}
    \label{definition-value-group}
    Let $A$ be a valuation ring.
    \begin{enumerate}
    \item The totally ordered abelian group $(\Gamma, \geq)$ of
    Lemma \ref{lemma-valuation-group} is called the
    {\it value group} of the valuation ring $A$.
    \item The map $v : A - \{0\} \to \Gamma$ and also $v : K^* \to \Gamma$ is
    called the {\it valuation} associated to $A$.
    \item The valuation ring $A$ is called a {\it discrete valuation ring}
    if $\Gamma \cong \mathbf{Z}$.
    \end{enumerate}
    \end{definition}
    
    \noindent
    Note that if $\Gamma \cong \mathbf{Z}$ then there is a unique such
    isomorphism such that $1 \geq 0$. If the isomorphism is chosen in this
    way, then the ordering becomes the usual ordering of the integers.
    
    \begin{lemma}
    \label{lemma-properties-valuation}
    Let $A$ be a valuation ring. The valuation $v : A -\{0\} \to \Gamma_{\geq 0}$
    has the following properties:
    \begin{enumerate}
    \item $v(a) = 0 \Leftrightarrow a \in A^*$,
    \item $v(ab) = v(a) + v(b)$,
    \item $v(a + b) \geq \min(v(a), v(b))$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-characterize-valuation-ring}
    Let $A$ be a ring. The following are equivalent
    \begin{enumerate}
    \item $A$ is a valuation ring,
    \item $A$ is a local domain and every finitely generated
    ideal of $A$ is principal.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Assume $A$ is a valuation ring and let $f_1, \ldots, f_n \in A$.
    Choose $i$ such that $v(f_i)$ is minimal among $v(f_j)$.
    Then $(f_i) = (f_1, \ldots, f_n)$. Conversely, assume $A$ is
    a local domain and every finitely generated ideal of $A$ is principal.
    Pick $f, g \in A$ and write $(f, g) = (h)$. Then $f = ah$ and $g = bh$
    and $h = cf + dg$ for some $a, b, c, d \in A$. Thus $ac + bd = 1$
    and we see that either $a$ or $b$ is a unit, i.e., either
    $g/f$ or $f/g$ is an element of $A$. This shows $A$ is a valuation ring
    by Lemma \ref{lemma-x-or-x-inverse-valuation-ring}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-valuation-valuation-ring}
    Let $(\Gamma, \geq)$ be a totally ordered abelian group.
    Let $K$ be a field. Let $v : K^* \to \Gamma$ be a homomorphism
    of abelian groups such that $v(a + b) \geq \min(v(a), v(b))$ for
    $a, b \in K$ with $a, b, a + b$ not zero. Then
    $$
    A =
    \{
    x \in K \mid x = 0 \text{ or } v(x) \geq 0
    \}
    $$
    is a valuation ring with value group $\Im(v) \subset \Gamma$,
    with maximal ideal
    $$
    \mathfrak m =
    \{
    x \in K \mid x = 0 \text{ or } v(x) > 0
    \}
    $$
    and with group of units
    $$
    A^* =
    \{
    x \in K^* \mid v(x) = 0
    \}.
    $$
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \noindent
    Let $(\Gamma, \geq)$ be a totally ordered abelian group.
    An {\it ideal of $\Gamma$} is a subset $I \subset \Gamma$ such
    that all elements of $I$ are $\geq 0$ and $\gamma \in I$,
    $\gamma' \geq \gamma$ implies $\gamma' \in I$. We say that such
    an ideal is {\it prime} if $\gamma + \gamma' \in I, \gamma, \gamma' \geq 0
    \Rightarrow \gamma \in I \text{ or } \gamma' \in I$.
    
    \begin{lemma}
    \label{lemma-ideals-valuation-ring}
    Let $A$ be a valuation ring.
    Ideals in $A$ correspond $1 - 1$ with ideals of $\Gamma$.
    This bijection is inclusion preserving, and maps prime
    ideals to prime ideals.
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-valuation-ring-Noetherian-discrete}
    A valuation ring is Noetherian if and only if it is
    a discrete valuation ring or a field.
    \end{lemma}
    
    \begin{proof}
    Suppose $A$ is a discrete valuation ring
    with valuation $v : A \setminus \{0\} \to \mathbf{Z}$
    normalized so that $\Im(v) \subset \mathbf{Z}_{\geq 0}$.
    By Lemma \ref{lemma-ideals-valuation-ring} the ideals of $A$ are the subsets
    $I_n = \{0\} \cup v^{-1}(\mathbf{Z}_{\geq n})$. It is clear
    that any element $x \in A$ with $v(x) = n$ generates $I_n$.
    Hence $A$ is a PID so certainly Noetherian.
    
    \medskip\noindent
    Suppose $A$ is a Noetherian valuation ring with value group $\Gamma$.
    By Lemma \ref{lemma-ideals-valuation-ring} we see the ascending chain
    condition holds for ideals in $\Gamma$.
    We may assume $A$ is not a field, i.e., there is a $\gamma \in \Gamma$
    with $\gamma > 0$. Applying the ascending chain condition to the subsets
    $\gamma + \Gamma_{\geq 0}$ with $\gamma > 0$ we see
    there exists a smallest element $\gamma_0$ which is bigger than $0$.
    Let $\gamma \in \Gamma$ be an element $\gamma > 0$. Consider the sequence
    of elements $\gamma$, $\gamma - \gamma_0$, $\gamma - 2\gamma_0$,
    etc. By the ascending chain condition these cannot all be $> 0$.
    Let $\gamma - n \gamma_0$ be the last one $\geq 0$. By minimality
    of $\gamma_0$ we see that $0 = \gamma - n \gamma_0$. Hence $\Gamma$
    is a cyclic group as desired.
    \end{proof}

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