Valuation Ring of Non-Archimedean Division Ring is Subring

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:
 * $\mathcal O = {B_1}^- \paren {0_R} = \set{x \in R : \norm{x} \le 1}$

where ${B_1}^- \paren {0_R}$ denotes the closed ball with center $0_R$ and radius $1$

Then $\mathcal O$ is a subring of $R$:
 * with a unity: $1_R$
 * in which there are no (proper) zero divisors, that is:
 * $\forall x, y \in \mathcal O: x \circ y = 0_R \implies x = 0_R \text{ or } y = 0_R$

Proof
To show that $\mathcal O$ is a subring the Subring Test is used by showing:
 * $(1): \quad \mathcal O \ne \empty$
 * $(2): \quad \forall x, y \in \mathcal O: x + \paren {-y} \in \mathcal O$
 * $(3): \quad \forall x, y \in \mathcal O: x y \in \mathcal O$

(1)

By Norm of Unity,
 * $\norm{1_R} = 1$

Hence:
 * $1_R \in \mathcal O \ne \empty$

(2)

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

Then:

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

(3)

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

Then:

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

By Subring Test it follows that $\mathcal O$ is a subring of $R$.

Since $1_R \in S$ and $1_R$ is the unity of $R$ then $1_R$ is the unity of $\mathcal O$.

By Division Ring has No Proper Zero Divisors then $R$ has no proper zero divisors.

Hence $\mathcal O$ has no proper zero divisors.