Isometrically Isomorphic Non-Archimedean Division Rings

Theorem

Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be normed division rings.

Let $\phi:R \to S$ be an isometric isomorphism.

Then:

$\norm {\,\cdot\,}_R$ is a non-archimedean norm if and only if $\norm {\,\cdot\,}_S$ is a non-archimedean norm.

Proof

Necessary Condition

Let $\norm {\,\cdot\,}_R$ be a non-archimedean norm.

Then for all $x,y \in R$:

\(\ds \norm {x + y}_S\)

\(=\)

\(\ds \norm {\map \phi {\map {\phi^{-1} } x} + \map \phi {\map {\phi^{-1} } y} }_S\)

$\phi$ is a bijection

\(\ds \)

\(=\)

\(\ds \norm {\map {\phi^{-1} } x + \map {\phi^{-1} } y}_R\)

$\phi$ is an isometry.

\(\ds \)

\(\le\)

\(\ds \max \set {\norm {\map {\phi^{-1} } x}_R, \norm {\map {\phi^{-1} } y}_R}\)

$\norm {\,\cdot\,}_R$ is non-archimedean.

\(\ds \)

\(=\)

\(\ds \max \set {\norm {\map \phi {\map {\phi^{-1} } x} }_S, \norm {\map \phi {\map {\phi^{-1} } y} }_S}\)

$\phi$ is an isometry.

\(\ds \)

\(=\)

\(\ds \max \set {\norm x_S, \norm y_S}\)

$\phi$ is a bijection.

$\Box$

Sufficient Condition

Let $\norm {\, \cdot \,}_S$ be a non-archimedean norm.

By Inverse of Isometric Isomorphism, $\phi^{-1}: S \to R$ is an isometric isomorphism.

By the necessary condition, $\norm {\, \cdot \,}_R$ is non-archimedean.

$\blacksquare$