Normed Division Ring is Field iff Completion is Field

Theorem
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.

Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$

Then:
 * $R$ is a field $R'$ is a field.

Proof
By the definition of a normed division ring completion then:
 * there exists a distance-preserving ring monomorphism $\phi:R \to R'$.
 * $\struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
 * $\phi \paren{R}$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.

By image of a ring homomorphism is a subring then $\phi \paren{R}$ is a subring of $R'$ and $\phi:R \to \phi \paren{R}$ is an isomorphism.

Necessary Condition
Let $x',y' \in R'$.

By the definition of a dense subset then $cl \paren {\phi \paren{R} } = R'$.

By Closure of Subset of Metric Space by Convergent Sequence then:
 * there exists a sequence $\sequence {x_n'} \subseteq \phi \paren{R}$ that converges to $x'$, that is, $\displaystyle \lim_{n \to \infty} x_n' = x'$
 * there exists a sequence $\sequence {y_n'} \subseteq \phi \paren{R}$ that converges to $y'$, that is, $\displaystyle \lim_{n \to \infty} y_n' = y'$

By product rule for convergent sequences then:
 * $\displaystyle \lim_{n \to \infty} x_n' y_n' = x' y'$
 * $\displaystyle \lim_{n \to \infty} y_n' x_n' = y' x'$

By the definition of a field, then $R$ is commutative.

Since $\phi:R \to \phi \paren{R}$ is an isomorphism, by Isomorphism Preserves Commutativity then $\phi \paren{R}$ is a commutative ring.

Since $\sequence {x_n'} \subseteq \phi \paren{R}$ and $\sequence {y_n'} \subseteq \phi \paren{R}$ then:
 * $\forall n: x_n' y_n' = y_n' x_n'$

Hence:
 * $\displaystyle x' y' = \lim_{n \to \infty} x_n' y_n' = \lim_{n \to \infty} y_n' x_n' = y' x'$

Since $x', y' \in R'$ were arbitrary, then $R'$ is a commutative ring.

The result follows.

Sufficient Condition
By Restriction of Commutative Operation is Commutative then $\phi \paren R$ is a commutative ring.

By Monomorphism Image Isomorphic to Domain then $\phi:R \to \phi \paren R$ is a ring isomorphism.

By Inverse of Algebraic Structure Isomorphism is Isomorphism then $\phi^{-1}:\phi \paren R \to R$ is a ring isomorphism.

By Isomorphism Preserves Commutativity then $R$ is a commutative ring.

The result follows.