Equivalence of Definitions of Division Ring

Theorem

The following definitions of the concept of Division Ring are equivalent:

A division ring is a ring with unity $\struct {R, +, \circ}$ with the following properties:

Definition 1

$\forall x \in R_{\ne 0_R}: \exists! x^{-1} \in R_{\ne 0_R}: x^{-1} \circ x = x \circ x^{-1} = 1_R$

where $R^*$ denotes the set of elements of $R$ without the ring zero $0_R$:

$R_{\ne 0_R} = R \setminus \set {0_R}$

That is, every non-zero element of $R$ has a (unique) non-zero product inverse.

Definition 2

Every non-zero element of $R$ is a unit.

Definition 3

$R$ has no proper elements.

Proof

In the following, let:

$0_R$ denote the zero of $R$

$1_R$ denote the unity of $R$

$R^*$ denote the set of elements of $R$ without $0_R$.

$(1)$ implies $(2)$

Let $\struct {R, +, \circ}$ be a division ring by definition 1.

Then by definition:

$\forall x \in R^*: \exists! x^{-1} \in R^*: x^{-1} \circ x = x \circ x^{-1} = 1_R$

That is, $x^{-1}$ is the (unique) product inverse of $x$.

Thus, by definition, $x$ is a unit of $R$.

Thus $\struct {R, +, \circ}$ is a division ring by definition 2.

$\Box$

$(2)$ implies $(1)$

Let $\struct {R, +, \circ}$ be a division ring by definition 2.

Then by definition:

$\forall x \in \R^*: x$ is a unit.

Thus, by definition, $x$ has a product inverse $x^{-1}$.

From Product Inverse in Ring is Unique it follows that:

$\forall x \in R^*: \exists! x^{-1} \in R^*: x^{-1} \circ x = x \circ x^{-1} = 1_R$

Thus $\struct {R, +, \circ}$ is a division ring by definition 1.

$\Box$

$(2)$ is equivalent to $(3)$

By definition, a unit of $R$ is an element of $R$ which has a product inverse.

Also by definition, a proper element of $R$ is a non-zero element of $R$ which does not have a product inverse.

Hence:

a ring with unity whose non-zero elements are all units

and:

a ring with unity whose non-zero elements are all not proper elements

are the same thing.

$\blacksquare$