Isometric Isomorphism is Norm-Preserving

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 $\phi$ 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 $x \in R$:

The result follows.

Sufficient Condition
Let $\phi:R \to S$ satisfy:
 * $\forall x \in R: \norm { \map \phi x }_S = \norm { x }_R $

Then for $x, y \in R$:

The result follows.