Non-Trivial Commutative Division Ring is Field

Theorem
Let $\struct {R, +, \circ}$ be a non-trivial division ring such that $\circ$ is commutative.

Then $\struct {R, +, \circ}$ is a field.

Similarly, let $\struct {F, +, \circ}$ be a field.

Then $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.

Proof
Suppose $\struct {R, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.

By definition $\struct {R, +}$ is an abelian group.

Thus field axioms $(\text A 0)$ to $(\text A 4)$ are satisfied.

Also by definition, $\struct {R, \circ}$ is a semigroup such that $\circ$ is commutative.

Thus field axioms $(\text M 0)$ to $(\text M 2)$ are satisfied.

As $\struct {R, +, \circ}$ is a ring with unity, $(\text M 3)$ is satisfied.

Field axiom $(\text D)$ is satisfied by dint of $\struct {R, +, \circ}$ being a ring.

Finally note that by definition of division ring, field axiom $(\text M 4)$ is satisfied.

Thus all the field axioms are satisfied, and so $\struct {R, +, \circ}$ is a field.

Suppose $\struct {F, +, \circ}$ is a field.

Then by definition $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.