Are there non-Archimedean fields without associated valuation or being a non-archimedan field implies it is a valuation field?
I understand that a non-Archimedean field is a field which does not satisfy the Archimedean property.
[Math] Non-Archimedean Fields
field-theoryordered-fieldsvaluation-theory
Best Answer
"Non-Archimedean Field" is a term used to refer to a non-Archimedean valued field or a non-Archimedean ordered field. Of course, this term is used within a context that allows you to decide whether we are treating with the former or latter case. Now I am going to define these two concepts.
Valued field: It is a field $K$ equipped with a map $|\cdot|:K\to\mathbb{R}$ that satisfies: for all $x,y\in K$,
This map is called valuation. A valuation is said to be non-Archimedean if it satisfies $$|x+y|\leq\max\{|x|,|y|\}, \mbox{ for all }x,y\in K.$$ A valuation is said to be Archimedean if it is not non-Archimedean. Finally, a non-Archimedean valued field is a valued field with non-Archimedean valuation.
A non-Archimedean valuation can be generalized (to something usually called Krull valuation or general valuation, or just valuation) by allowing to the codomain to be more general, not just the real numbers. This generalization looks like follows:
Take a map $|\cdot|:K\mapsto G\cup\{0\}$ satisfying
where $G$ is an arbitrary multiplicative ordered group and $0$ is an element such that $0<g$ for all $g\in G$. If the group $G$ is isomorphic to a subgroup of the open interval of real numbers $(0,\infty)$ equipped with the multiplication, then we say that the valuation has rank 1, and if this is not the case, (e.g. $G=\mathbb{Z}^5$) then we say that the valuation is of higher rank.
Ordered field: for the definition of ordered field see this. An ordered field $K$ is said to be Archimedean if and only if for all $a\in K$, there exists some $n\in\mathbb{N}$ such that $a<n1_K$. Here, $1_K$ stands for the multiplicative identity of the field $K$. On the other side, $K$ is a non-Archimedean ordered field if and only if there exists some $a\in K$ such that $a>n1_K$ for all $n\in\mathbb{N}$.
Example: Consider $\mathbb{R}[x]$, the ring of polynomials with coefficients in $\mathbb{R}$. Consider the field of rational functions $\mathbb{R}(x)=\{p/q:p,q\in\mathbb{R}[x], q\neq0\}$.
$\mathbb{R}(x)$ as non-Archimedean valued field: For any $p\in\mathbb{R}[x]$, $p\neq0$ let $deg(p)$ be the degree of the polynomial $p$ and put $|p|:=e^{deg(p)}$ and $|0|:=0$. Then on $\mathbb{R}(x)$ the map $|\cdot|$ defined by $|p/q|:=|p|/|q|$ is a non-Archimedean valuation.
$\mathbb{R}(x)$ as non-Archimedean ordered field: Define an order on $\mathbb{R}[x]$ as follows: for constant polynomials consider the order of $\mathbb{R}$. For $p=a_0+a_1x+\cdots+a_nx^n\in\mathbb{R}[x]$, with $a_n\neq0$, put $p>0$ whenever $a_n>0$. Now, in the field of rational functions $\mathbb{R}(x)$ consider the order defined as follows: $p/q>0$ whenever $pq>0$ in $\mathbb{R}[x]$. Here, for all $f,g\in\mathbb{R}(x)$, $f<g$ whenever $0<g-f$.
This field is an extension of $\mathbb{R}$ as ordered field where $a<x^2$ for all $a\in\mathbb{R}$. Therefore $\mathbb{R}(x)$ is a non-Archimedean ordered field.
Epic facts: (a) The topology on $\mathbb{R}(x)$ induced by the metric $D(f,g)=|f-g|$ coincides with the order topology induced by the order in $\mathbb{R}(x)$. Let's denote this topology by $\tau$.
(b) The topology on $\mathbb{R}$ as a topological subspace of $(\mathbb{R}(x),\tau)$ is the discrete topology.
This is just another example of the following:
Remark: Let $(A,<)$ be an ordered set equipped with the corresponding order topology $\tau_A$ and let $B$ be a subset of $A$. The order in $A$ induces an order in $B$ which induces a topology $\tau_<$ on B. In general, the subspace topology $\tau_A\cap B$ may be different than the topology $\tau_<$. For other examples of this situation, see page 90 of the book: Topology, James R Munkres, Prentice Hall, 2000.
Proof of epic facts: The proof of the statement (a) is consequence of the following inclusions which can be verified straightforward for all $n\in\mathbb{N}$:
$$\{f\in\mathbb{R}(x):|f|<e^n\}\subset(-x^n,x^n)\subset \{f\in\mathbb{R}(x):|f|<e^{n+1}\}$$ $$\{f\in\mathbb{R}(x):|f|<e^{-n-1}\}\subset(-x^{-n},x^{-n})\subset \{f\in\mathbb{R}(x):|f|<e^{-n}\}.$$
The proof of the statement (b) follows from (a) and the fact that the valuation $|\cdot|$ is the trivial valuation when it is restricted to $\mathbb{R}$. An alternative proof is that for all $a\in\mathbb{R}$ and for all $n\in\mathbb{N}$, $$\{a\}=(a-x^{-n},a+x^{-n})\cap\mathbb{R}.$$ Notice that $\{x^{-n}:n\in\mathbb{N}\}$ is a set of infinitesimals.