Characterisation of Non-Archimedean Absolute Values

Theorem

Let $|\cdot|$ be an absolute value on a field $k$ with unity $1_k$.

Then $|\cdot |$ is non-archimedean if and only if $ | n \cdot 1_k | \leq 1$ for all $n \geq 1$.

Corollary

If $k$ has characteristic $p > 0$, then there are no archimedean absolute values on $k$.

Proof of Theorem

Suppose that $|\cdot|$ is non-archimedean.

Then by the definition of a non-archimedean absolute value, for $n \in \N$,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert n \cdot 1_k \right\vert\)	\(=\)	\(\displaystyle \left\vert 1_k + \cdots + 1_k \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	($n$ summands)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \max \left\{ \left\vert 1_k \right\vert, \ldots, \left\vert 1_k \right\vert \right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because $\left\vert 1_k \right\vert = 1$

Conversely, suppose that $ | n \cdot 1_k | \leq 1$ for all $n \geq 1$. We compute

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert x + y \right\vert^n\)	\(=\)	\(\displaystyle \left\vert \sum_{i = 0} ^ n { n \choose i } x^{n-i}y^i \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the Binomial Theorem
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \sum_{i = 0} ^ n \left\vert { n \choose i } \right\vert \cdot \left\vert x \right\vert^{n-i} \left\vert y \right\vert^i\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \left\vert B(n + 1) \right\vert \max\{ \left\vert x \right\vert, \left\vert y \right\vert \}^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	where $B$ is a bound for $n \choose i$, $i = 0, \ldots, n$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \max\{ \left\vert x \right\vert, \left\vert y \right\vert \}^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Taking $n^\text{th}$ roots yields $|x + y| \leq \max\{ |x|, |y| \}$.

$\blacksquare$

Proof of Corollary

Since $k$ has characteristic $p > 0$, the set

$ \{ n \cdot 1_k : n \in \Z \}$

has cardinality $p-1$, so any absolute value must be bounded by

$ 1 = \max \left\{ | 1_k |, \cdots, |1_k| \right\}$

$\blacksquare$