Equivalence of Definitions of Division Ring

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

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.

$(2)$ 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.

$(2)$ is equivalent to $(3)$
By definition, a unit is an element of $R$ which is not a proper element.

Hence:
 * a ring with unity whose non-zero elements are all unit

is also:
 * a ring with unity whose non-zero are all not proper elements.

$(2)$ is equivalent to $(4)$
By definition, a proper zero divisor of $R$ is an element $x \in R^*$ such that:


 * $\exists y \in R^*: x \circ y = 0_R$

Thus if all the elements of $R^*$ are all units, none of them can be proper zero divisors.