Does every non-Archimedean absolute value satisfy the ultrametric inequality?

Question

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?

Answer 1

Indeed, a non-Archimedean absolute value automatically satisfies the ultrametric inequality (as pointed out by Robert Israel). In my original question, I used a slightly different formulation of the Archimedean property (and the referenced lecture notes might not be online forever), so here is a full proof.

Proposition. Let $F$ be a field and let $|\cdot|$ be a non-Archimedean absolute value. Then $|\cdot|$ satisfies the ultrametric inequality.

Proof. Since $|\cdot|$ is non-Archimedean, we may choose some non-zero $x\in F$ such that $$ \big|\:\underbrace{x + \cdots + x}_{n\ \text{times}}\:\big| \leq 1,\tag*{for all $n\in\mathbb{N}$.} $$ We may interpret any element of $\mathbb{N}$ (or $\mathbb{Z}$, for that matter) as an element of $F$ by identifying it with its image under the natural ring homomorphism $\mathbb{Z} \to F$. Then the above becomes $$ |n|\cdot |x| = |n\cdot x| \leq 1,\tag*{for all $n\in\mathbb{N}$.} $$ Since $x$ is non-zero by assumption, we have $|x| \neq 0$, hence $$ |n| \leq \frac{1}{|x|},\tag*{for all $n\in\mathbb{N}$.} $$ Now let $y,z\in F$ be given. By the binomial theorem, for all $k\in\mathbb{N}$ we have $$ (y + z)^k \: = \: \sum_{j=0}^k \binom{k}{j} y^j z^{k-j}, $$ hence $$ |y+z|^k \: = \: |(y+z)^k| \: = \: \left|\sum_{j=0}^k \binom{k}{j} y^j z^{k-j}\right| \: \leq \: \sum_{j=0}^k \frac{|y|^j\cdot |z|^{k-j}}{|x|} \: \leq \: \frac{k+1}{|x|}\cdot \max(|y|,|z|)^k. $$ Equivalently: for all $k\in\mathbb{Z}_{> 0}$ we have $$ |y+z| \: \leq \: \sqrt[k]{\frac{k+1}{|x|}}\cdot \max(|y|,|z|). $$ As $k$ increases, this factor $\sqrt[k]{\frac{k+1}{|x|}}$ converges (decreasingly) to one, so we have $$ |y + z| \: \leq \: \inf_{k\to\infty} \sqrt[k]{\frac{k+1}{|x|}}\cdot \max(|y|,|z|) \: = \: \max(|y|,|z|).\tag*{$\Box$} $$

This peculiar little trick is now standard in the literature. It is also used in many textbooks, for instance:

• W. Schikhof, Ultrametric Calculus: An Introduction to p-Adic Analysis, Cambridge Studies in Advanced Mathematics. Cambridge: Cambridge University Press. doi:10.1017/CBO9780511623844 (Lemma 8.2)

• Paulo Ribenboim, The Theory of Classical Valuations, Springer Monographs in Mathematics (section 1.2, fact E).

• Antonio J. Engler & Alexander Prestel, Valued Fields, Springer Monographs in Mathematics (proposition 1.1.1).

• Pierre Antoine Grillet, Abstract Algebra, Second Edition, Springer Graduate Texts in Mathematics 242 (chapter VI, proposition 3.2).

• Alain M. Robert, A Course in p-adic Analysis, Springer Graduate Texts in Mathematics 198 (chapter 2, section 1.6, first theorem).

Answer 2

Yes, a non-Archimedean absolute value must be ultrametric. See e.g. Theorem 2.1 of these notes.