## Tag `00I8`

## 10.49. Valuation rings

Here are some definitions.

Definition 10.49.1. Valuation rings.

- 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 $\mathfrak m_A = A \cap \mathfrak m_B$.- Let $A$ be a ring. We say $A$ is a
valuation ringif $A$ is a local domain and if $A$ is maximal for the relation of domination among local rings contained in the fraction field of $A$.- Let $A$ be a valuation ring with fraction field $K$. If $R \subset K$ is a subring of $K$, then we say $A$ is
centeredon $R$ 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 the fraction field $F$ of $B$ and $K$ by $K \cap F$. 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 groupis 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.

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

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:

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

Proof.Omitted. $\square$Lemma 10.49.15. Let $A$ be a ring. The following are equivalent

- $A$ is a valuation ring,
- $A$ 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 isprimeif $\gamma + \gamma' \in I, \gamma, \gamma' \geq 0 \Rightarrow \gamma \in I \text{ or } \gamma' \in I$.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 11138–11542 (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 the fraction field $F$ of $B$ and $K$ by
$K \cap F$. 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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