Valuation Ideal is Maximal Ideal of Induced Valuation Ring

Theorem
Let $\struct {R, \norm{\,\cdot\,}}$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$.

Let $0_R$ be the zero of $R$ and $1_R$ be the unity of $R$.

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

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

Then $\mathcal P$ is an ideal of $\mathcal O$:
 * $(a):\quad \mathcal P$ is a maximal left ideal
 * $(b):\quad \mathcal P$ is a maximal right ideal
 * $(c):\quad$ the quotient ring $\mathcal O / \mathcal P$ is a division ring.

Proof
First it is shown that $\mathcal P$ is an ideal of $\mathcal O$ by applying Test for Ideal, that is, it is shown that:
 * $(1): \quad \mathcal P \ne \empty$
 * $(2): \quad \forall x, y \in \mathcal P: x + \paren {-y} \in \mathcal P$
 * $(3): \quad \forall x \in \mathcal P, y \in \mathcal O: x y \in \mathcal P$

(1)

By Norm axiom (N1) (Positive Definiteness),
 * $\norm{0_R} = 0$

Hence:
 * $0_R \in \mathcal P \ne \empty$

(2)

Let $x, y \in \mathcal P$.

Then:

Hence:
 * $x + \paren {-y} \in \mathcal P$

(3)

Let $x \in \mathcal P, y \in \mathcal O$.

Then:

Hence:
 * $x y \in \mathcal P$

By Test for Ideal then $\mathcal P$ is an ideal of $\mathcal O$.

By Maximal Left and Right Ideal iff Quotient Ring is Division Ring the statements (a), (b) and (c) above are equivalent.

It is now shown that statement (a) holds.

Let $J$ be a left ideal of $\mathcal O$:
 * $\mathcal P \subsetneq J \subset \mathcal O$

Let $x \in J \setminus \mathcal P$, then:
 * $\norm x = 1$

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, $J = \mathcal O$

Hence $\mathcal P$ is a maximal left ideal.

The result follows.