Equivalent Absolute Values

Theorem

Let $k$ be a field.

Let $|\cdot|_1$, $|\cdot|_2$ be absolute values on $k$.

Then the following are equivalent:

$(1): \quad |\cdot|_1$ and $|\cdot|_2$ induce the same topology on $k$

$(2): \quad |x|_1 < 1 \iff |x|_2 < 1$ for all $x \in k$

$(3): \quad$ There exists a fixed $c > 0$ such that $|x|_1 = |x|_2 ^ c$ for all $x \in k$

Condition $(1)$ is known as Cauchy equivalence.

Proof

$(1) \implies (2)$

If $|\cdot|_1$ and $|\cdot|_2$ induce the same topology on $k$, then a sequence $y_n$ converges to a limit $Y$ with respect to $|\cdot|_1$ if and only if the same is true for $|\cdot|_2$.

Let $x \in k$ be such that $|x|_1 < 1$.

Then

$|x^n|_1 = |x|^n_1 \to 0$

as $n \to \infty$. Therefore also

$|x|_2^n = |x^n|_2 \to 0$

as $n \to \infty$. Thus $|x|_2 < 1$.

$(2) \implies (3)$

First note that $(2)$ implies that $|x|_1 > 1 \iff |x|_2 > 1$ for all $x \in k$, for if $|x|_1 > 1$ then $|x^{-1}|_1 <1$.

Then $|x^{-1}|_2 <1$, so also $|x|_2 > 1$.

Thus

$|x|_1 \wedge 1 \iff |x|_2 \wedge 1 \qquad (1)$

whenever $\wedge$ is one of the symbols $=$, $<$ or $>$.

Now fix some $a \in k$, $|a|_1 \neq 0, 1$.

By $(1)$, there is some $c > 0$ such that $|a|_1 = |a|_2 ^ c$.

Now let $x \in k$ be arbitrary, subject to $|x|_1 \neq 0, 1$ (if $|x|_1 = 1$ or $|x|_1 = 0$ the result is trivial).

There exists some $t \in \R$ such that $|x|_1 = |a|_1 ^ t$.

Suppose that $m/n \in \Q$ with $m/n < t$. Then

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert a \right\vert_1 ^{m/n}\)	\(<\)	\(\displaystyle \left\vert x \right\vert _1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left\vert a^m \right\vert_1\)	\(<\)	\(\displaystyle \left\vert x^n \right\vert_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the properties of absolute values
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left\vert \frac{a^m}{x^n} \right\vert_1\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the properties of absolute values
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left\vert \frac{a^m}{x^n} \right\vert_2\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert a \right\vert_2 ^{m/n}\)	\(<\)	\(\displaystyle \left\vert x \right\vert _2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	performing the same manipulations in reverse

Similarly one shows that if $m/n \in \Q$ with $m/n > t$ then $\left\vert a \right\vert_2 ^{m/n} > |x|_2$.

Also, by Continuous Functions on Valued Fields, any absolute value $|\cdot| : k \to \R$ is continuous.

So

$ \left\vert a \right\vert_2 ^{m/n} < \left\vert x \right\vert _2 < \left\vert a \right\vert_2 ^{m/n} $

and by the Squeeze Theorem $|x|_2 = |a|_2^t$. Now

$ |x|_1 = |a|_1 ^ c = |a|_2^{ct} = |x|_2^c $

$(3) \implies (1)$

For $i \in \{1,2\}$, let $B_{r,i}(x) = \left\{ y \in k : |y - x|_i < r \right\}$ be the open ball of radius $r$ in $k$ with respect to the $i^\text{th}$ metric. We compute

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle B_{r,2}(x)\)	\(=\)	\(\displaystyle \left\{ y \in k : \left\vert y - x \right\vert_2 < r \right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ y \in k : \left\vert y - x \right\vert_2^{c^{-1} } < r \right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\{ y \in k : \left\vert y - x \right\vert_2 < r^c \right\}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle B_{r^c,1}(x)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Let $X \subseteq (k, |\cdot|_1)$ be open, and $x \in X$.

Then for some $r > 0$, $B_{r,1}(x) \subseteq X$.

By the above computation $B_{r^{c^{-1}},2}(x) \subseteq X$.

Therefore $X$ is open in $(k,|\cdot |_2)$, and the metric spaces $(k,|\cdot|_1)$, $(k,|\cdot|_2)$ are identical.

$\blacksquare$