necessary condition for Archimedean valuation

Question

To fix the terminology, a valuation of a field F will be a map from F to the real numbers such that, on writing $|x|$ for the image of $x$, $|x|\geq 0$ for all $x$, $|x|=0$ iff $x=0$, $|xy|=|x||y|$ for all $x$,$y$ and $|x+y|\leq|x|+|y|$ for all $x,y$. This last inequality is called the triangle inequality. If it is replaced by the stronger inequality $|x+y|\leq max(|x|,|y|)$ for all $x,y$, which is called the ultrametric inequality, the valuation is called non-Archimedean. An Archimedean valuation is a valuation which is not non-Archimedean.

In Ribenboim’s book The Theory of Classical Valuations, he gives an easy proof on page 11 that a valuation is non-Archimedean iff $|n.1|\leq 1$ for all natural numbers $n$ where $n.1$ is the sum of $n$ copies of the identity element $1$ of F. On page 473 of Artin and Whaples’ paper “Axiomatic Characterization of Fields By the Product Formula for Valuations” Bull AMS 51 (1945) pp. 469-492 , they say that $|1+1|>1$ for every Archimedean valuation of F.

It follows from Ribenboim’s theorem that $|1+1|>1$ is sufficient for a valuation to be Archimedean, but why is it necessary?

I tried showing the contrapositive, that a valuation such that $|1+1|≤1$ must be non-Archimedean, and using the fact that a valuation is non-Archimedean iff the set $$S=\{|n.1| : n \text{ is a natural number}\}$$ is bounded (p. 118 of Algebraic Number Theory by Jurgen Neukirch) but could only obtain bounds on particular $|n.1|$, not on the whole set $S$.

Answer 1

Theorem: A valuation on a field $F$ is non-Archimedean if and only if $|1_F+1_F|\leq1$.

Proof: See Lemma 8.2 (p. 18) of "Ultrametric Calculus: An Introduction to p-adic Analysis" - [W.H. Schikhof] - 2006

Answer 2

Assume that $|2\cdot1_F|\leq1$. Then for each $n\geq1$ and its binary expansion $n=\sum_{k=0}^{m}a_k2^k$, we have

$$|n\cdot1_F| \leq\sum_{k=0}^{m} \left| a_k 2^k\cdot1_F \right| \leq\sum_{k=0}^{m} \left| a_k\cdot1_F \right| \leq m+1 \leq\lceil\log_2 n\rceil+1.$$

Now utilize the tensor power trick:

$$|n\cdot1_F|=|n^k\cdot1_F|^{1/k}\leq(1+\lceil k\log_2 n\rceil)^{1/k} \xrightarrow[k\to\infty]{} 1. $$

In view of Ribenboim's theorem, this is enough to guarantee that $|\cdot|$ is non-Archimedean.

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In fact, Ribenboim's theorem can also be proved by the tensor power trick. Assume that $|n\cdot1_F|\leq1$ for all $n\in\mathbb{N}$. Then for each $x, y \in F$,

$$ |x+y| = |(x+y)^n|^{1/n} \leq \left( \sum_{k=0}^{n} \left|\binom{n}{k}x^k y^{n-k}\right| \right)^{1/n} \leq (n+1)^{1/n}\max\{|x|,|y|\}. $$

Letting $n\to\infty$ shows that the ultrametric inequality holds.

Answer 3

$\newcommand\N{\mathbb N}$Suppose $|2|\leq 1$ then in one answer Sangchul has shown that for all $k\in\N$ we have $|k|\leq \lceil \log_2 k \rceil + 1$. Because $\binom{n}{i} \leq 2^n$ for any $0\leq i\leq n$, we have $$\sum_{i=0}^n \left|\binom{n}{i}\right| \leq \sum_{i=0}^n \left(\left\lceil \log_2 \binom{n}{i} \right\rceil + 1\right) \leq \sum_{i=0}^n (n+1) =n^2+n$$

We then simply use the binomial expansion and triangle inequality to get for any $n\in\N$

$$|x+y|^n = |(x+y)^n| = \left|\sum_{i=0}^n \binom{n}{i}x^iy^{n-i}\right| \leq \max(|x|,|y|)^n\sum_{i=0}^n \left|\binom{n}{i}\right|\leq \max(|x|,|y|)^n(n^2+n)$$

So we get for all $n\in \N$ $$ |x+y| \leq (n^2+n)^{1/n} \max(|x|,|y|)$$ take the limit of the above inequalities of non-negative real numbers as $n\rightarrow \infty$ to get what you want.