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.
$\blacksquare$