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 $$ab^{-1} \in T$$ and $$bc^{-1} \in T$$. Then $$(ab^{-1})(bc^{-1}) = ac^{-1} \in T$$ and hence $$a \le c$$ (even $$a < c$$).

Let $$a \le b$$ and $$b \le a$$. 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 $$ac < bc$$.

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 $$m := \bigcup_{I \le R \atop I \neq R} I$$. 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, $$$$.

Proof:

For, let $$I \le R$$ 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 $$I = \langle a_1, \ldots, a_n \rangle = \langle a_n \rangle$$.