The Archimedean property occurs in various areas of mathematics; for instance it is defined for ordered groups, ordered fields, partially ordered vector spaces and normed fields. In each of these contexts it is roughly the following property:
Archimedean property. For any two (strictly) positive elements $x$ and $y$ there is some $n\in\mathbb{N}$ such that $n \cdot x$ exceeds $y$.
This definition might not be adequate in each of the mentioned contexts, but at least it conveys the general idea. Indeed, in the context of normed fields we have the following definition (paraphrasing the definition given on Wikipedia):
Definition. Let $F$ be a field with an absolute value $\left|\:\cdot\:\right|$, that is, a function $\left|\:\cdot\:\right| : F \to \mathbb{R}_{\geq 0}$ satisfying the following properties:
- $|x| = 0$ if and only if $x = 0$;
- For all $x,y\in F$ we have $|xy| = |x|\cdot |y|$;
- For all $x,y\in F$ we have $|x + y| \leq |x| + |y|$.
Then $F$ is said to be Archimedean if for any non-zero $x\in F$ there exists some $n\in\mathbb{N}$ such that $$ \big|\:\underbrace{x + \cdots + x}_{n\ \text{times}}\:\big| > 1. $$ An absolute value that does not satisfy this property is called non-Archimedean.
However, in the literature the term non-Archimedean absolute value is usually used as a synonym for an absolute value which satisfies the ultrametric inequality:
- For any $x,y\in F$ we have $|x + y| \leq \max(|x|,|y|)$.
It is not so hard to see that an ultrametric absolute value can never be Archimedean: one easily proves that $|1| = 1$ holds, and then we find $|1 + 1| \leq 1$, followed by $|1 + 1 + 1| \leq 1$ and so on (repeatedly using the ultrametric inequality).
It is however not so clear to me that any non-Archimedean absolute value must necessarily satisfy the ultrametric inequaltiy. Is this always true? Or is it only true for certain fields, say $\mathbb{Q}$, that happen to be the most common fields in the study of absolute values on fields?