Proposition 5

Two absolute values $|x|_1, |x|_2$ on a division ring $K$, neither of which is the improper absolute value, are equivalent if and only if the relation $|x|_1 < 1$ implies $|x|_2 < 1$. There exists then a real number $\rho > 0$ such that $|x|_2 = |x|_1^\rho$ for all $x \in K$.

The condition is necessary, for the set of all $x \in K$ such that $|x|_1 < 1$ is the same as the set of all $x$ such that, with respect to the topology defined by the absolute value $|x|_1$, $\lim_{n \to \infty} x^n = 0$.

Suppose conversely that $|x|_1 < 1 \Longrightarrow |x|_2 < 1$. Then $|x|_1 > 1 \Longrightarrow |x|_2 > 1$, because $|x^{-1}|_1 < 1$ and therefore $|x^{-1}|_2 < 1$. Since by hypothesis the absolute value $|x|_1$ is not improper, there exists $x_0 \in K$ such that $|x_0|_1 > 1$. Let $a = |x_0|_1, b = |x_0|_2$ and let $\rho = \log b / \log a > 0$. Let $x \in K^*$ and put $|x|_1 = |x_0|_1^\gamma$. If $m$ and $n$ are integers such that $n > 0$ and $m/n > \gamma$, then $|x|_1 < |x_0|_1^{m/n}$, and therefore $|x^n x_0^{-m}|_1 < 1$; hence $|x^n x_0^{-m}|_2 < 1, |x|_2 < |x_0|_2^{m/n}$. Similarly, if $m/n < \gamma$, we see that $|x|_2 > |x_0|_2^{m/n}$; it follows therefore that $|x|_2 = |x_0|_2^\gamma$; in other words

i.e., $|x|_2 = |x|^p_1$. It is now clear that the neighbourhoods of zero for the topologies defined on $K$ by $|x|_1$ and $|x|_2$ are identical.

Conversely, if $|x|$ is any absolute value on $K$, the function $|x|^p$ is an absolute value on $K$ (equivalent to $|x|$) for all $p$ such that $0 < p \leq 1$. We have only to verify the inequality $|x + y|^p \leq |x|^p + |y|^p$; and since $|x + y|^p \leq (|x| + |y|)^p$ it is enough to show that, if $a > 0$ and $b > 0$, we have $(a + b)^p \leq a^p + b^p$ for any $p$ such that $0 < p \leq 1$. If we put $c = a/(a + b)$ and $d = b/(a + b)$, we have $c + d = 1$, and the inequality to be proved is $c^p + d^p \geq 1$; but this follows immediately from the relations $c^p \geq c$ and $d^p \geq d$, which are valid since $0 < c \leq 1$, $0 < d \leq 1$, $0 < p \leq 1$.

Hence the set of values of $r > 0$ such that $|x|^r$ is an absolute value is a finite or infinite interval of $\mathbf{R}$ with left-hand end-point $0$; if it is finite, it is evidently closed; for if we have $|x + y|^r \leq |x|^r + |y|^r$ for any $x, y$ in $K$ and all $r$ such that $0 < r < r_0$, then by continuity the inequality is still valid for $r = r_0$. If $|x|^r$ is an absolute value for all $r > 0$, then we have

for all $x$ and $y$ in $K$ and all $r > 0$. Now, if $a, b$ are two real numbers $\geq 0$, we have $\lim_{r \to \infty} (a^r + b^r)^{1/c} = \sup (a, b)$; for, supposing for example that $a \geq b$, we have $a \leq (a^r + b^r)^{1/r} \leq 2^{1/r} a$, and the result follows by letting $r \to + \infty$.

Thus, if $|x|^r$ is an absolute value for all $r > 0$, we have

which can be expressed by saying that $v(x) = -\log |x| \ (x \neq 0)$ is a valuation on $K$.