about valuation

Question

Definition. Let $K$ be a field and let $G$ be a totally ordered abelian group. A valuation of $K$ with values in $G$ is a map $v: K\backslash \{0\} \rightarrow G$ such that for all $x, y \in K, x, y \neq 0,$ we have $$ \begin{array}{l} \text { (1) } v(x y)=v(x)+v(y) \\ \text { (2) } v(x+y) \geqslant \min (v(x), v(y)) . \end{array} $$

1. why $G$ must be abelian group in above definition ? if $G$ be any arbitrary group then definition is false ?

2. why $G$ must be totally ordered group in above definition ?

for 2) i think $G$ must be totally ordered group because we can understand any element is "positive" or "negative".(an element $g \in G$ is positive if $0<g$ and negative if $0>g$ )

3. is there a totally ordered abelian group $G$ such that every element $g \in G-\{0\}$ be negative ? or in every non trivial group we have at least one positive element ?

also i want to find simplest example of valuation.

Answer 1

1. The first valuation axiom says that $v$ is a group homomorphism from $K^\times$ (the group of nonzero elements of $K$ with multiplication) to $G$. Since $K$ is a field and one of the field axioms says that multiplication is commutative, that means that the image of $v$ is a commutative subgroup of $G$, regardless of whether $G$ is assumed to be commutative. So there's no point in allowing $G$ to be noncommutative, because we don't actually gain any extra generality that way.
2. Conceptually, the point of a valuation is that it gives a way of comparing the "size" or "multiplicity" of elements of a field. In order to do that, we have to be able to compare any two elements of the value group $G$. You could, of course, generalize this by allowing $G$ to only be partially ordered or something like that, but most of valuation theory probably wouldn't work with this generalization. (For example, you need the value group to be totally ordered for the valuation to induce a metric on $K$ and thus turn $K$ into a metric space.)
3. The definition of an ordered group requires the order relation to be translation-invariant, so if $g < 0$, then $0 = -g + g < -g + 0 = -g$. Thus the inverse operation in the group gives a bijective correspondence between negative elements and positive elements.

As for examples of valuations, here are the simplest ones that come to mind:

1. Let $p$ be a prime number. Define a valuation $v_p$ on the field of rational numbers $\mathbb{Q}$ as follows: for any nonzero $m \in \mathbb{Z}$, define $v_p(m) = k$, where $m = p^k d$ with $\gcd(d, p) = 1$. For any nonzero $m, n \in \mathbb{Z}$, define $v_p(m/n) = v_p(m) - v_p(n)$. So, for example, $v_3(18/5) = 2$, while $v_5(18/5) = -1$.
2. Let $K$ be a field, let $K(x)$ be the field of rational functions in one variable over $K$, and let $c \in K$ be arbitrary. Define a valuation $v_c$ on $K(x)$ as follows: for a polynomial $f \in K[x]$, define $v_c(f)$ to be the order of vanishing of $f$ at $c$, i.e., the multiplicity with which $(x - c)$ occurs in the factorization of $f$. For $f/g \in K(x)$, define $v_c(f/g) = v_c(f) - v_c(g)$; you can think of this as the "order of zeroes or poles". For example, $v_0(x^2/(x - 3) = 2$, while $v_{3}(x^2/(x - 3)) = -1$, and $v_2(x^2/(x - 3)) = 0$.