Embedding Division Ring into Quotient Ring of Cauchy Sequences

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Theorem

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

Let $\CC$ be the ring of Cauchy sequences over $R$

Let $\NN = \set {\sequence {x_n}: \ds \lim_{n \mathop \to \infty} x_n = 0}$

Let $\norm {\, \cdot \,}: \CC \, \big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \, \big / \NN$ defined by:

$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN} = \lim_{n \mathop \to \infty} \norm {x_n}$


Let $\phi: R \to \CC \, \big / \NN$ be the mapping from $R$ to the quotient ring $\CC \, \big / \NN$ defined by:

$\forall a \in R: \map \phi a = \sequence {a, a, a, \dotsc} + \NN$

where $\sequence {a, a, a, \dotsc} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {a, a, a, \dotsc}$.

Then:

$\phi$ is a distance-preserving ring monomorphism.


Proof

By the definition of a distance-preserving mapping and a ring monomorphism it has to be shown that:

$(1): \quad \phi$ is a homomorphism.
$(2): \quad \phi$ is an injection.
$(3): \quad \phi$ is distance-preserving.


$(1): \quad \phi$ is a homomorphism

By definition, $\phi$ is the composition of two mappings:

$\phi = q \circ \phi'$

where:

$\text{(a)}: \quad \phi': R \to \CC$, defined by: $\forall a \in R, \map {\phi'} a = \sequence {a, a, a, \dotsc}$
$\text{(b)}: \quad q$ is the quotient mapping $q: \CC \to \CC \, \big / \NN$ defined by: $\map q {\sequence {x_n} } = \sequence {x_n} + \NN$

By Embedding Normed Division Ring into Ring of Cauchy Sequences, $\phi'$ is a ring monomorphism.

By Quotient Ring Epimorphism is Epimorphism, then $q$ is a ring epimorphism.

By Composition of Ring Homomorphisms is Ring Homomorphism then the composition $\phi = q \circ \phi'$ is a ring homomorphism

$\Box$


$(2): \quad \phi$ is an injection

Let $a, b \in R$.

Suppose $\map \phi a = \map \phi b$.

Then:

\(\ds \sequence {a, a, a, \dotsc} + \NN\) \(=\) \(\ds \sequence {b, b, b, \dotsc} + \NN\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds \sequence {a, a, a, \dotsc} - \sequence {b, b, b, \dotsc}\) \(\in\) \(\ds \NN\) Left Cosets are Equal iff Product with Inverse in Subgroup
\(\ds \leadsto \ \ \) \(\ds \sequence {a - b, a - b, a - b, \dotsc}\) \(\in\) \(\ds \NN\) Ring operations on Ring of Cauchy Sequences
\(\ds \leadsto \ \ \) \(\ds \lim_{n \mathop \to \infty} {a - b}\) \(=\) \(\ds 0\) Definition of $\NN$

By Constant Sequence Converges to Constant in Normed Division Ring then:

$\ds \lim_{n \mathop \to \infty} {a - b} = a - b$

Hence $a-b = 0$.

The result follows.

$\Box$


$(3): \quad \phi$ is distance-preserving

Let $a, b \in R$.

Then:

\(\ds \norm {\map \phi a - \map \phi b}\) \(=\) \(\ds \norm {\map \phi {a - b} }\) $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds \norm {\sequence {a - b, a - b, a - b, \dots} + \NN}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty } \norm {a - b}\) Definition of $\norm {\, \cdot \,}$
\(\ds \) \(=\) \(\ds \norm {a - b}\) Constant Sequence Converges to Constant in Normed Division Ring

$\blacksquare$


Sources

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