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.
Asked
Active
Viewed 200 times
1
Torsten Schoeneberg
- 29,325
-
It might be a good idea to tell us the textbook you're following. – Christian E. Ramirez Jan 25 '23 at 00:45
-
2I'm using P-adic numbers by Gouvea 3rd edition. He gives a problem after a lemma but doesn't provide a proof. It's in the second chapter – Jan 25 '23 at 00:47
-
Related: https://math.stackexchange.com/q/1956792/96384, https://math.stackexchange.com/q/1748861/96384. – Torsten Schoeneberg Jan 25 '23 at 05:24
-
Please do not change the title to something that makes the question incomprehensible. I have reverted that. – Torsten Schoeneberg Feb 09 '23 at 21:21
-
Do the close voters understand the question (and the answer)?? – reuns Feb 09 '23 at 21:27
1 Answers
2
$R$ 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.
reuns
- 79,880