Commutative Algebra/Valuation rings

Augmented ordered Abelian groups

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

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

Proof:

We first prove the first assertion.

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

Let $a\le b$ and $b\le a$. Assume $a\ne b$ for a contradiction. Then $a{b}^{-1}\in T$ and $b{a}^{-1}\in T$, and since $T$ is closed under multiplication, $1\in T$, contradiction. Hence $a=b$.

Let $a,b\in G$ such that $a\ne b$. Since $G=T\cup \{1\}\cup T^{-1}$, $a{b}^{-1}$ (which is not equal $1$) is either in $T$ or in $T^{-1}$ (but not in both, since otherwise $a{b}^{-1}\in T^{-1}\Rightarrow (a{b}^{-1}{)}^{-1}\in T$ and since $a{b}^{-1}\in T$, $1\in T$, 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 $a{b}^{-1}\in T$, but $a{b}^{-1}=ac{c}^{-1}{b}^{-1}=ac(bc{)}^{-1}$. Hence $ac<bc$.$◻$

Valuations and valuation rings

Proof:

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

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

Properties of valuation rings

Proof:

The ideals of a valuation ring $R$ are ordered by inclusion. Set $m:=\bigcup _{\genfrac{}{}{0ex}{}{I\le R}{I\ne R}}I$. 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, ${\displaystyle }$.

Proof:

For, let $I\le R$ be an ideal; in any Noetherian ring, the ideals are finitely generated. Hence let $I=⟨a_1,\dots ,a_n⟩$. Consider the ideals of $R$ $⟨a_1⟩,\dots ,⟨a_n⟩$. In a valuation rings, the ideals are totally ordered, so we may renumber the $a_j$ such that $⟨a_1⟩\subseteq ⟨a_2⟩\subseteq \cdots \subseteq ⟨a_n⟩$. Then $I=⟨a_1,\dots ,a_n⟩=⟨a_n⟩$.$◻$