Construction of Inverse Completion/Identity of 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 $\oplus$.
Let $c \in C$ be arbitrary.
Then:
- $\eqclass {\tuple {c, c} } \boxtimes$
is the identity of $T'$.
Proof
\(\ds \paren {x \circ c} \circ y\) | \(=\) | \(\ds x \circ \paren {c \circ y}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {y \circ c}\) | Definition of Commutative Semigroup | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \eqclass {\tuple {x, y} } \boxtimes \oplus' \eqclass {\tuple {c, c} } \boxtimes\) | \(=\) | \(\ds \eqclass {\tuple {x \circ c, y \circ c} } \boxtimes\) | Definition of $\oplus'$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass {\tuple {x, y} } \boxtimes\) | Cancellability of elements of $C$ |
Hence the result, by definition of identity element.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers