Leigh.Samphier/Sandbox/Distance-Preserving Homomorphism Preserves Norm

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Theorem

Let $\struct {R_1, \norm {\, \cdot \,}_1 }, \struct {R_2, \norm {\, \cdot \,}_2 }$ be normed division rings.

Let $\phi: R_1 \to R_2$ be a distance-preserving ring homomorphism.


Then:

$\forall x \in R_1 : \norm{\map \phi x}_2 = \norm x_1$

Proof

\(\displaystyle \norm x_1\) \(=\) \(\displaystyle \norm {x - 0}_1\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {\map \phi x - \map \phi 0}_2\) As $\phi$ is distance-preserving
\(\displaystyle \) \(=\) \(\displaystyle \norm {\map \phi {x - 0} }_2\) As $\phi$ is a ring homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \norm {\map \phi x}_2\)

$\blacksquare$