Prove that if $\| n \| \leq 1$ for all integers $n$, then $\| \cdot \|$ is a non-Archimedean norm.

Question

I'm reading the book $p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions by Neal Koblitz, and in Exercise 3 of Chapter 1 we are asked to show that if $\| \cdot \|$ is a norm on a field $F$ such that $\| n \| \leq 1$ for every integer $n$, then $\| \cdot \|$ is non-Archimedean. Here, $n$ denotes $n \cdot 1 = 1 + 1 + \dots + 1$ taken $n$ times.

To show that $\| \cdot \|$ is non-Archimedean, I need to show that it satisfies $\| x + y \| \leq \max\{ \| x \|, \| y \| \}$ for all $x,y \in F$. I'm not able to figure out how to begin solving this problem. Any hints or suggestions would be appreciated.

EDIT: A norm on a field $F$ is a function $\| \cdot \| : F \to \mathbb{R}^{+} \cup \{ 0 \}$ satisfying:

1. $\| x \| = 0 \Leftrightarrow x = 0$,
2. $\| xy \| = \|x \| \cdot \| y \|$,
3. $\| x + y \| \leq \| x \| + \| y \|$

for all $x,y \in F$. A non-Archimedean norm satisfies the stronger condition $\| x + y \| \leq \max\{ \|x \|, \| y \| \}$ instead of 3. above.

A norm will be called Archimedean if it is not non-Archimedean.

Answer 1

Here is a standard trick (which may be considered as an example of the tensor power trick).

Let $n \geq 1$ be an arbitrary positive integer. Then for any $x, y \in F$, we have

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

Now using the assumption, we find that

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

Finally, take $n$-th root both sides and let $n \to \infty$.

Answer 2

First, let’s note that if $u\in\Bbb R$ and $u>1$, then there is $N$ such that $u^N>N+1$.

Now, let’s suppose that $\mid\bullet\mid$ is an absolute value on $F$ for which $\mid n\mid\le1$ for all integers $n$, but that there are elements $x$, $y$ of $F$ with $|x+y|>\max(|x|,|y|)$. Without loss of generality, we can suppose that $|y|\le|x|$, and multiply our offending inequality by $|1/x|$ to get $|1+\frac yx|>\max\bigl(1,|\frac yx|\bigr)$.

But since $|\frac yx|\le1$, the last “max” is $1$, so we set $z=y/x\in F$ and get $|z|\le1$ and $u=|1+z|>1$. Now we get $$ u^N=\bigl|(1+z)^n\bigr|=\bigl|1+Nz+\frac{N(N-1)}2z^2+\cdots+z^N\bigr|\,, $$ where in the binomial sum on the right, every coefficient is an integer, thus having absolute value $\le1$, and every power of $z$ also has absolute value $\le1$, since $|z|\le1$. And there are $N+1$ terms. By the triangle inequality, we have \begin{align} u^N&\le1+\cdots+1\quad\text{(with $N+1$ terms)}\\ &=N+1\,, \end{align} a contradiction. So there are no such $x$ and $y$, and the absolute value is nonarchimedean.