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

## 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

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

$\blacksquare$