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 Axiom $\text A$: Addition forms an Abelian Group are satisfied.
Also by definition, $\struct {R, \circ}$ is a semigroup such that $\circ$ is commutative.
Thus Field Axiom $\text M0$: Closure under Product to Field Axiom $\text M2$: Commutativity of Product are satisfied.
As $\struct {R, +, \circ}$ is a ring with unity, Field Axiom $\text M3$: Identity for Product is satisfied.
Field Axiom $\text D$: Distributivity of Product over Addition is satisfied by dint of $\struct {R, +, \circ}$ being a ring.
Finally note that by definition of division ring, Field Axiom $\text M4$: Inverses for Product is satisfied.
Thus all the field axioms are satisfied, and so $\struct {R, +, \circ}$ is a field.
$\Box$
Suppose $\struct {F, +, \circ}$ is a field.
Then by definition $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers