How do you show that an absolute value is non-archimedean if $\sup\{|n|: n \text{ is an integer}\}= C<\infty$ and how to show that $C=1$?

Question

I've been trying to work this out but I've been stuck and can't seem to see how to work through a proof of this. It's part of a lemma I'm trying to study in p-adic analysis but the textbook doesn't give a proof for me to follow.

Answer 1

$R$ is an integral domain with an absolute value such that $\sup_{n\in \Bbb{Z}} |n|<\infty$.

With $|n|$ I mean $|n 1_R|$.

If $|n|> 1$ then $\lim_{k\to \infty} |n^k| = \infty$. So $\sup_{n\in \Bbb{Z}} |n|<\infty$ implies $\sup_{n\in \Bbb{Z}} |n|\le 1$.

Assume that for some $a,b\in R, |a+b|>|a|\ge |b|$. Then $$|(a+b)^k|\le \sum_{m=0}^k | {k \choose m}a^m b^{k-m}| \le (k+1) |a|^k$$ But $\lim_{k\to \infty} \frac{|a+b|^k}{(k+1)|a|^k}= \infty$ so this is a contradiction.

Therefore $\forall a,b\in R, |a+b|\le \max(|a|,|b|)$ ie. $|.|$ is non-archimedian.