Norm of Unity of Division Ring

Theorem
Let $\left({R,+,\circ}\right)$ be a division ring, and denote its unity by $1_R$.

Let $\left\Vert{\cdot}\right\Vert$ be a norm on $R$.

Then $\left \Vert {1_R} \right \Vert = 1$.

Proof
By definition:
 * $\forall x, y \in R: \left \Vert{x \circ y}\right \Vert = \left \Vert{x}\right \Vert \left \Vert{y}\right \Vert$

So:


 * $\left \Vert{1_R}\right \Vert = \left \Vert{1_R \circ 1_R}\right \Vert = \left \Vert{1_R}\right \Vert \left \Vert{1_R}\right \Vert$

By the definition of norms, $\left\Vert{1_R}\right\Vert\ne 0$, so $\left\Vert{1_R}\right\Vert$ has an inverse in $\R$.

By multiplying with this inverse, we get:


 * $ \left \Vert{1_R}\right \Vert \left \Vert{1_R}\right \Vert=\left \Vert{1_R}\right \Vert\iff\left \Vert{1_R}\right \Vert=1$.