Valuation Ideal is Maximal Ideal of Induced Valuation Ring/Corollary 1

Theorem
Let $\struct {R, \norm{\,\cdot\,}}$ be a non-Archimedean normed division ring with zero $0_R$ and unity $1_R$.

Let $\mathcal O$ be the valuation ring induced by the non-Archimedean norm $\norm{\,\cdot\,}$, that is:
 * $\mathcal O = \set{x \in R : \norm{x} \le 1}$

Then:
 * $\mathcal O$ is a local ring.

Proof
Let $\mathcal P$ be the valuation ideal induced by the non-Archimedean norm $\norm{\,\cdot\,}$, that is:
 * $\mathcal P = \set{x \in R : \norm{x} \lt 1}$

By Valuation Ideal is Maximal Ideal of Induced Valuation Ring then:
 * $\mathcal P$ is a maximal left ideal of $\mathcal O$.

Let $J \subsetneq \mathcal O$ be any maximal left ideal of $\mathcal O$.

Let $x \in \mathcal O \setminus \mathcal P$.

Aiming for a contradiction, suppose $x \in J$.

By Norm of Inverse then:
 * $\norm {x^{-1}} = 1 / \norm x = 1 / 1 = 1$

Hence:
 * $x^{-1} \in \mathcal O$

Since $J$ is a left ideal then:
 * $x^{-1} x = 1_R \in J$

Thus:
 * $\forall y \in \mathcal O: y \cdot 1_R = y \in J$

That is:
 * $J = \mathcal O$

This contradicts the assumption that $J \neq \mathcal O$.

So:
 * $x \notin J$

Hence:
 * $\paren {\mathcal O \setminus \mathcal P } \cap J = \empty$

That is:
 * $J \subseteq \mathcal P$

Since $J$ and $\mathcal P$ are both maximal left ideals then:
 * $J = \mathcal P$

The result follows.