# 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 | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \norm {\map \phi x}_2\) |

$\blacksquare$