Non-Trivial Commutative Division Ring is Field

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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.

$\blacksquare$


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