Isometric Isomorphism is Norm-Preserving

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Theorem

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

Let $\phi: R \to S$ be a ring isomorphism.


Then $\phi: R \to S$ is an isometric isomorphism if and only if $\phi$ satisfies:

$\forall x \in R: \norm {\map \phi x}_S = \norm x_R $


Proof

Let $d_R$ and $d_S$ be the metric induced by the norms $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively.


Necessary Condition

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


Then for $x \in R$:

\(\ds \norm {\map \phi x}_S\) \(=\) \(\ds \norm {\map \phi x - 0_S}_S\) $0_S$ is the zero of $S$
\(\ds \) \(=\) \(\ds \norm {\map \phi x - \map \phi {0_R} }_S\) Ring Homomorphism Preserves Zero
\(\ds \) \(=\) \(\ds \map {d_S} {\map \phi x, \map \phi {0_R} }\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(=\) \(\ds \map {d_R} {x, {0_R} }\) Definition of Isometry (Metric Spaces)
\(\ds \) \(=\) \(\ds \norm {x - {0_R} }_R\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(=\) \(\ds \norm x_R\) $0_R$ is the zero of $R$

The result follows.

$\Box$


Sufficient Condition

Let $\phi: R \to S$ satisfy:

$\forall x \in R: \norm {\map \phi x}_S = \norm x_R$


Then for $x, y \in R$:

\(\ds \map {d_S} {\map \phi x, \map \phi y}\) \(=\) \(\ds \norm {\map \phi x - \map \phi y}_S\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(=\) \(\ds \norm {\map \phi {x - y} }_S\) $\phi$ is a Ring Isomorphism
\(\ds \) \(=\) \(\ds \norm {x - y}_R\) by hypothesis
\(\ds \) \(=\) \(\ds \map {d_R} {x, y}\) Definition of Metric Induced by Norm on Division Ring

The result follows.

$\blacksquare$