# Commutative Algebra/Valuation rings

## Augmented ordered Abelian groups

In this section, for reasons that will become apparent soon, we write Abelian groups multiplicatively.

Definition 18.1:

An ordered Abelian group is a group ${\displaystyle G}$ together with a subset ${\displaystyle T\subset G}$ such that:

1. ${\displaystyle T}$ is closed under multiplication (that is, ${\displaystyle a,b\in T\Rightarrow ab\in T}$).
2. If ${\displaystyle a\in T}$, then ${\displaystyle a^{-1}\notin T}$. (This implies in particular that ${\displaystyle 1\notin T}$.)
3. ${\displaystyle G=T\cup \{1\}\cup T^{-1}}$.

We write ordered Abelian groups as pair ${\displaystyle (G,T)}$.

The last two conditions may be summarized as: ${\displaystyle G}$ is the disjoint union of ${\displaystyle T}$, ${\displaystyle \{1\}}$ and ${\displaystyle T^{-1}}$.

Theorem 18.2:

Let an ordered group ${\displaystyle (G,T)}$ be given. Define an order on ${\displaystyle G}$ by

${\displaystyle a, ${\displaystyle a\leq b:\Leftrightarrow a\leq b\vee a=b}$.

Then ${\displaystyle <}$ has the following properties:

1. ${\displaystyle <}$ is a total order of ${\displaystyle G}$.
2. ${\displaystyle <}$ is compatible with multiplication of ${\displaystyle G}$ (that is, ${\displaystyle c\in G}$ and ${\displaystyle a\leq b}$ implies ${\displaystyle ca\leq cb}$).

Proof:

We first prove the first assertion.

${\displaystyle \leq }$ is reflexive by definition. It is also transitive: Let ${\displaystyle a\leq b}$ and ${\displaystyle b\leq c}$. When ${\displaystyle a=b}$ or ${\displaystyle b=c}$, the claim ${\displaystyle a\leq c}$ follows trivially by replacing ${\displaystyle b}$ in either of the given equations. Thus assume ${\displaystyle ab^{-1}\in T}$ and ${\displaystyle bc^{-1}\in T}$. Then ${\displaystyle (ab^{-1})(bc^{-1})=ac^{-1}\in T}$ and hence ${\displaystyle a\leq c}$ (even ${\displaystyle a).

Let ${\displaystyle a\leq b}$ and ${\displaystyle b\leq a}$. Assume ${\displaystyle a\neq b}$ for a contradiction. Then ${\displaystyle ab^{-1}\in T}$ and ${\displaystyle ba^{-1}\in T}$, and since ${\displaystyle T}$ is closed under multiplication, ${\displaystyle 1\in T}$, contradiction. Hence ${\displaystyle a=b}$.

Let ${\displaystyle a,b\in G}$ such that ${\displaystyle a\neq b}$. Since ${\displaystyle G=T\cup \{1\}\cup T^{-1}}$, ${\displaystyle ab^{-1}}$ (which is not equal ${\displaystyle 1}$) is either in ${\displaystyle T}$ or in ${\displaystyle T^{-1}}$ (but not in both, since otherwise ${\displaystyle ab^{-1}\in T^{-1}\Rightarrow (ab^{-1})^{-1}\in T}$ and since ${\displaystyle ab^{-1}\in T}$, ${\displaystyle 1\in T}$, contradiction). Thus either ${\displaystyle a or ${\displaystyle b.

Then we proceed to the second assertion.

Let ${\displaystyle c\in G}$. If ${\displaystyle a=b}$, the claim is trivial. If ${\displaystyle a, then ${\displaystyle ab^{-1}\in T}$, but ${\displaystyle ab^{-1}=acc^{-1}b^{-1}=ac(bc)^{-1}}$. Hence ${\displaystyle ac.${\displaystyle \Box }$

Definition 18.3:

Let ${\displaystyle (G,T)}$ be an ordered Abelian group. An augmented ordered Abelian group is ${\displaystyle (G,T)}$ together with an element ${\displaystyle 0}$ (zero) such that the following rules hold:

${\displaystyle 00=0}$, ${\displaystyle \forall g\in G:0g=0}$.

We write an augmented ordered Abelian group as triple ${\displaystyle (G,T,0)}$.

## Valuations and valuation rings

Definition 18.4:

Let ${\displaystyle \mathbb {F} }$ be a field, and let ${\displaystyle (G,T,0)}$ be an augmented ordered Abelian group. A valuation of the field ${\displaystyle \mathbb {F} }$ is a mapping ${\displaystyle \varphi :\mathbb {F} \to G\cup \{0\}}$ such that:

1. ${\displaystyle \varphi (x)=0\Leftrightarrow x=0}$.
2. ${\displaystyle \forall x,y\in \mathbb {F} :\varphi (xy)=\varphi (x)\varphi (y)}$.
3. ${\displaystyle \forall x,y\in \mathbb {F} :\varphi (x+y)\leq \max\{\varphi (x),\varphi (y)\}}$.

Definition 18.5:

A valuation ring is an integral domain ${\displaystyle D}$, such that there exists an augmented ordered Abelian group ${\displaystyle (G,T,0)}$ and a valuation ${\displaystyle \varphi :\operatorname {Quot} (D)\to G\cup \{0\}}$ with ${\displaystyle D=\{c\in \operatorname {Quot} (D)|\varphi (c)\leq 1\}}$.

Theorem 18.6:

Let ${\displaystyle R}$ be a valuation ring, and let ${\displaystyle \operatorname {Quot} (R)}$ be its field of fractions. Then the following are equivalent:

1. ${\displaystyle R}$ is a valuation ring.
2. ${\displaystyle R}$ is an integral domain and the ideals of ${\displaystyle R}$ are linearly ordered with respect to set inclusion.
3. ${\displaystyle R}$ is an integral domain and for each ${\displaystyle c\in \operatorname {Quot} (R)}$, either ${\displaystyle c\in R}$ or ${\displaystyle {\frac {1}{c}}\in R}$.

Proof:

We begin with 3. ${\displaystyle \Rightarrow }$ 1.; assume that ${\displaystyle R}$

1. ${\displaystyle \Rightarrow }$ 2.: Let ${\displaystyle I,J\leq R}$ any two ideals. Assume there exists ${\displaystyle a\in J\setminus I}$. Let any element ${\displaystyle b\in I}$ be given.

## Properties of valuation rings

Theorem 18.8:

A valuation ring is a local ring.

Proof:

The ideals of a valuation ring ${\displaystyle R}$ are ordered by inclusion. Set ${\displaystyle m:=\bigcup _{I\leq R \atop I\neq R}I}$. We claim that ${\displaystyle m}$ is a proper ideal of ${\displaystyle R}$. Certainly ${\displaystyle 1\notin m}$ for otherwise ${\displaystyle 1\in I}$ for some proper ideal ${\displaystyle I}$ of ${\displaystyle R}$. Furthermore, .

Theorem 18.9:

Let ${\displaystyle R}$ be a Noetherian ring and a valuation ring. Then ${\displaystyle R}$ is a principal ideal domain.

Proof:

For, let ${\displaystyle I\leq R}$ be an ideal; in any Noetherian ring, the ideals are finitely generated. Hence let ${\displaystyle I=\langle a_{1},\ldots ,a_{n}\rangle }$. Consider the ideals of ${\displaystyle R}$ ${\displaystyle \langle a_{1}\rangle ,\ldots ,\langle a_{n}\rangle }$. In a valuation rings, the ideals are totally ordered, so we may renumber the ${\displaystyle a_{j}}$ such that ${\displaystyle \langle a_{1}\rangle \subseteq \langle a_{2}\rangle \subseteq \cdots \subseteq \langle a_{n}\rangle }$. Then ${\displaystyle I=\langle a_{1},\ldots ,a_{n}\rangle =\langle a_{n}\rangle }$.${\displaystyle \Box }$