Equivalent Norms are both Non-Archimedean or both Archimedean

Theorem

Let $R$ be a division ring with unity $1_R$.

Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.

Then $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are either both non-Archimedean or both Archimedean.

Proof

By Characterisation of Non-Archimedean Division Ring Norms then:

$\norm {\,\cdot\,}_1$ is non-Archimedean $\iff \forall n \in \N_{>0}: \norm{n \cdot 1_R}_1 \le 1$.

By the definition of norm equivalence then:

$\forall n \in \N: \norm {n \cdot 1_R}_1 \le 1 \iff \norm {n \cdot 1_R}_2 \le 1$

Similarly, by Characterisation of Non-Archimedean Division Ring Norms then:

$\forall n \in \N_{>0}: \norm {n \cdot 1_R}_2 \le 1 \iff \norm {\,\cdot\,}_2$ is non-Archimedean.

The result follows.

$\blacksquare$

Sources

• 2007: Svetlana Katok: p-adic Analysis Compared with Real ... (previous) ... (next): $\S 1.2$ Normed fields: Proposition $1.18$