# 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 $G$ together with a subset $T\subset G$ such that:
* $T$ is closed under multiplication (that is, $a,b\in T\Rightarrow ab\in T$).
* If $a\in T$, then $a^{-1}\notin T$. (This implies in particular that $1\notin T$.)
* $G=T\cup \{1\}\cup T^{-1}$.
We write ordered Abelian groups as pair $(G,T)$.
The last two conditions may be summarized as: $G$ is the disjoint union of $T$, $\{1\}$ and $T^{-1}$.
Theorem 18.2:
Let an ordered group $(G,T)$ be given. Define an order on $G$ by
    $a<b:\Leftrightarrow ab^{-1}\in T$, $a\leq b:\Leftrightarrow a\leq b\vee a=b$.
Then $<$ has the following properties:
* $<$ is a total order of $G$.
* $<$ is compatible with multiplication of $G$ (that is, $c\in G$ and $a\leq b$ implies $ca\leq cb$).
Proof:
We first prove the first assertion.
$\leq$ is reflexive by definition. It is also transitive: Let $a\leq b$ and $b\leq c$. When $a=b$ or $b=c$, the claim $a\leq c$ follows trivially by replacing $b$ in either of the given equations. Thus assume $ab^{-1}\in T$ and $bc^{-1}\in T$. Then $(ab^{-1})(bc^{-1})=ac^{-1}\in T$ and hence $a\leq c$ (even $a<c$).
Let $a\leq b$ and $b\leq a$. Assume $a\neq b$ for a contradiction. Then $ab^{-1}\in T$ and $ba^{-1}\in T$, and since $T$ is closed under multiplication, $1\in T$, contradiction. Hence $a=b$.
Let $a,b\in G$ such that $a\neq b$. Since $G=T\cup \{1\}\cup T^{-1}$, $ab^{-1}$ (which is not equal $1$) is either in $T$ or in $T^{-1}$ (but not in both, since otherwise $ab^{-1}\in T^{-1}\Rightarrow (ab^{-1})^{-1}\in T$ and since $ab^{-1}\in T$, $1\in T$, contradiction). Thus either $a<b$ or $b<a$.
Then we proceed to the second assertion.
Let $c\in G$. If $a=b$, the claim is trivial. If $a<b$, then $ab^{-1}\in T$, but $ab^{-1}=acc^{-1}b^{-1}=ac(bc)^{-1}$. Hence $ac<bc$.$\Box$
Definition 18.3:
Let $(G,T)$ be an ordered Abelian group. An augmented ordered Abelian group is $(G,T)$ together with an element $0$ (zero) such that the following rules hold:
    $00=0$, $\forall g\in G:0g=0$.
We write an augmented ordered Abelian group as triple $(G,T,0)$.

## Valuations and valuation rings

Definition 18.4:
Let $\mathbb {F}$ be a field, and let $(G,T,0)$ be an augmented ordered Abelian group. A valuation of the field $\mathbb {F}$ is a mapping $\varphi :\mathbb {F} \to G\cup \{0\}$ such that:
* $\varphi (x)=0\Leftrightarrow x=0$.
* $\forall x,y\in \mathbb {F} :\varphi (xy)=\varphi (x)\varphi (y)$.
* $\forall x,y\in \mathbb {F} :\varphi (x+y)\leq \max\{\varphi (x),\varphi (y)\}$.
Definition 18.5:
A valuation ring is an integral domain $D$, such that there exists an augmented ordered Abelian group $(G,T,0)$ and a valuation $\varphi :\operatorname {Quot} (D)\to G\cup \{0\}$ with $D=\{c\in \operatorname {Quot} (D)|\varphi (c)\leq 1\}$.
Theorem 18.6:
Let $R$ be a valuation ring, and let $\operatorname {Quot} (R)$ be its field of fractions. Then the following are equivalent:
* $R$ is a valuation ring.
* $R$ is an integral domain and the ideals of $R$ are linearly ordered with respect to set inclusion.
* $R$ is an integral domain and for each $c\in \operatorname {Quot} (R)$, either $c\in R$ or ${\frac {1}{c}}\in R$.
Proof:
We begin with 3. $\Rightarrow$ 1.; assume that $R$
1. $\Rightarrow$ 2.: Let $I,J\leq R$ any two ideals. Assume there exists $a\in J\setminus I$. Let any element $b\in I$ be given.

## Properties of valuation rings

Theorem 18.8:
A valuation ring is a local ring.
Proof:
The ideals of a valuation ring $R$ are ordered by inclusion. Set $m:=\bigcup _{I\leq R \atop I\neq R}I$. We claim that $m$ is a proper ideal of $R$. Certainly $1\notin m$ for otherwise $1\in I$ for some proper ideal $I$ of $R$. Furthermore, $$.
Theorem 18.9:
Let $R$ be a Noetherian ring and a valuation ring. Then $R$ is a principal ideal domain.
Proof:
For, let $I\leq R$ be an ideal; in any Noetherian ring, the ideals are finitely generated. Hence let $I=\langle a_{1},\ldots ,a_{n}\rangle$. Consider the ideals of $R$ $\langle a_{1}\rangle ,\ldots ,\langle a_{n}\rangle$. In a valuation rings, the ideals are totally ordered, so we may renumber the $a_{j}$ such that $\langle a_{1}\rangle \subseteq \langle a_{2}\rangle \subseteq \cdots \subseteq \langle a_{n}\rangle$. Then $I=\langle a_{1},\ldots ,a_{n}\rangle =\langle a_{n}\rangle$.$\Box$