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 if and only if $R'$ is a field.

Proof

By the definition of a normed division ring completion then:

$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$.

$(2): \quad$ $\struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.

$(3): \quad$ The image $\phi \sqbrk R$ of $\phi$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.

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

Necessary Condition

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

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

From Closure of Subset of Metric Space by Convergent Sequence:

there exists a sequence $\sequence {x_n'} \subseteq \phi \sqbrk R$ that converges to $x'$, that is, $\ds \lim_{n \mathop \to \infty} x_n' = x'$

there exists a sequence $\sequence {y_n'} \subseteq \phi \sqbrk R$ that converges to $y'$, that is, $\ds \lim_{n \mathop \to \infty} y_n' = y'$

From Product Rule for Sequences in Normed Division Ring:

$\ds \lim_{n \mathop \to \infty} x_n' y_n' = x' y'$

$\ds \lim_{n \mathop \to \infty} y_n' x_n' = y' x'$

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

Because $\phi: R \to \phi \sqbrk R$ is an isomorphism, by Isomorphism Preserves Commutativity, $\phi \sqbrk R$ is a commutative ring.

Because $\sequence {x_n'} \subseteq \phi \sqbrk R$ and $\sequence {y_n'} \subseteq \phi \sqbrk R$:

$\forall n: x_n' y_n' = y_n' x_n'$

Hence:

$\ds x' y' = \lim_{n \mathop \to \infty} x_n' y_n' = \lim_{n \mathop \to \infty} y_n' x_n' = y' x'$

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

The result follows.

$\Box$

Sufficient Condition

By Restriction of Commutative Operation is Commutative, $\phi \sqbrk R$ is a commutative ring.

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

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

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

The result follows.

$\blacksquare$

Sources

• 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.2$: Completions