Non-Trivial Commutative Division Ring is Field

Theorem
Let $\left({R, +, \circ}\right)$ be a division ring such that $\circ$ is commutative.

Then $\left({R, +, \circ}\right)$ is a field.

Similarly, let $\left({F, +, \circ}\right)$ be a field.

Then $\left({F, +, \circ}\right)$ is a division ring such that $\circ$ is commutative.

Proof
Suppose $\left({R, +, \circ}\right)$ is a division ring such that $\circ$ is commutative.

Then by definition $\left({R, +}\right)$ is an abelian group.

Thus field axioms $(A0)$ to $(A4)$ are satisfied.

Also by definition, $\left({R, \circ}\right)$ is a semigroup such that $\circ$ is commutative.

Thus field axioms $(M0)$ to $(M2)$ are satisfied.

As $\left({R, +, \circ}\right)$ is a ring with unity, $(M3)$ is satisfied.

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

Finally note that by definition of division ring, field axiom $(M4)$ is satisfied.

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

Suppose $\left({F, +, \circ}\right)$ is a field.

Then by definition $\left({F, +, \circ}\right)$ is a division ring such that $\circ$ is commutative.