Non-Archimedean Division Ring Iff Non-Archimedean Completion

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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:

$\norm {\, \cdot \,}$ is non-archimedean if and only if $\norm {\, \cdot \,}'$ is non-archimedean.

Proof

By the definition of a normed division ring completion then:

  1. there exists a distance-preserving ring monomorphism $\phi: R \to R'$.
  2. $\struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
  3. $\phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.

By Normed Division Ring is Dense Subring of Completion then $\phi \sqbrk R$ is a dense normed division subring of $R'$ and $\phi: R \to \map \phi R$ is an isometric isomorphism.

By Isometrically Isomorphic Non-Archimedean Division Rings then $\struct {R, \norm {\, \cdot \,} }$ is non-archimedean if and only if $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is.

So it remains to show that $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ non-archimedean if and only if $\struct {R', \norm {\, \cdot \,}' }$ is.


Necessary Condition

Let $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ be non-archimedean.

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

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

By Closure of Subset of Metric Space by Convergent Sequence then:

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

By sum rule for convergent sequences then:

$\displaystyle \lim_{n \to \infty} x_n' + y_n' = x' + y'$

By modulus of limit then:

$\displaystyle \lim_{n \mathop \to \infty} \norm{x_n'}' = \norm{x'}'$
$\displaystyle \lim_{n \mathop \to \infty} \norm{y_n'}' = \norm{y'}'$
$\displaystyle \lim_{n \mathop \to \infty} \norm{x_n' + y_n'}' = \norm{x' + y'}'$

By Maximum Rule for Real Sequences then:

$\displaystyle \lim_{n \mathop \to \infty} \max \set {\norm{x_n'}', \norm{y_n'}' } = \max \set {\norm{x'}', \norm{y'}'}$

Since $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is non-archimedean then:

$\forall n: \norm{x_n' + y_n'}' \le \max \set {\norm{x_n'}', \norm{y_n'}' }$

By Inequality Rule for Real Sequences then:

$\displaystyle \norm{x' + y'}' = \lim_{n \mathop \to \infty} \norm{x_n' + y_n'}' \le \lim_{n \mathop \to \infty} \max \set {\norm{x_n'}', \norm{y_n'}' } = \max \set {\norm{x'}', \norm{y'}'}$

The result follows.

$\Box$


Sufficient Condition

Let $\struct {R', \norm {\, \cdot \,}' }$ be non-archimedean.

By Subring of Non-Archimedean Division Ring then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is non-archimedean.

$\blacksquare$


Sources