Construction of Inverse Completion/Invertible Elements in Quotient Structure
Theorem
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
Let $\struct {C, \circ {\restriction_C} } \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.
Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.
Let $\boxtimes$ be the cross-relation on $S \times C$, defined as:
- $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
This cross-relation is a congruence relation on $S \times C$.
Let the quotient structure defined by $\boxtimes$ be:
- $\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.
Every cancellable element of $S'$ is invertible in $T'$.
Proof
From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.
Call this identity $e_{T'}$.
Let the mapping $\psi: S \to T'$ be defined as:
- $\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
From Image of Cancellable Elements in Quotient Mapping:
- $C' = \psi \sqbrk C$
So:
\(\ds x'\) | \(\in\) | \(\ds C'\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in C: \, \) | \(\ds x'\) | \(=\) | \(\ds \map \psi x\) | as $\psi$ is a surjection | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall a \in C: \, \) | \(\ds x'\) | \(=\) | \(\ds \eqclass {\tuple {x \circ a, a} } \boxtimes\) | Definition of $\psi$ |
The inverse of $x'$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$, as follows:
\(\ds a \circ x\) | \(\in\) | \(\ds C\) | Cancellable Elements of Semigroup form Subsemigroup | |||||||||||
\(\ds a \circ a \circ x\) | \(\in\) | \(\ds C\) | Cancellable Elements of Semigroup form Subsemigroup | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ x} } \boxtimes\) | \(=\) | \(\ds \eqclass {\tuple {x \circ a \circ a, a \circ a \circ x} } \boxtimes\) | Definition of $\oplus'$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass {\tuple {a \circ a \circ x, a \circ a \circ x} } \boxtimes\) | Commutativity of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e_{T'}\) | Identity of Quotient Structure |
thus showing that the inverse of $\eqclass {\tuple {x \circ a, a} } \boxtimes$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers