Structure of Inverse Completion of Commutative Semigroup
Theorem
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
- $T = S \circ' C^{-1}$
where:
- $C^{-1}$ is the inverse of $C$ in $T$
- $S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.
Proof
Let $a \in C$.
\(\ds x\) | \(\in\) | \(\ds S\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds x \circ \paren {a \circ' a^{-1} }\) | Definition of Invertible Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \paren {x \circ a} \circ' a^{-1}\) | Definition of Associative Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds S \circ' C^{-1}\) | Definition of Subset Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds S \circ' C^{-1}\) | Definition of Subset |
Then:
\(\ds y\) | \(\in\) | \(\ds C\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^{-1}\) | \(\in\) | \(\ds C^{-1}\) | Definition of Inverse of Subset of Monoid | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^{-1}\) | \(=\) | \(\ds a \circ' a^{-1} \circ' y^{-1}\) | Definition of Invertible Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^{-1}\) | \(=\) | \(\ds a \circ' \paren {y \circ' a}^{-1}\) | Inverse of Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^{-1}\) | \(=\) | \(\ds a \circ' \paren {y \circ a}^{-1}\) | Definition of Extension of Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^{-1}\) | \(\in\) | \(\ds S \circ' C^{-1}\) | Definition of Subset Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds C^{-1}\) | \(\subseteq\) | \(\ds S \circ' C^{-1}\) | Definition of Subset |
Thus, as:
- $C^{-1} \subseteq S \circ' C^{-1}$
and:
- $S \subseteq S \circ' C^{-1}$
by Union is Smallest Superset it follows that:
- $S \cup C^{-1} \subseteq S \circ' C^{-1}$
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
- $S \circ' C^{-1}$ is a commutative semigroup.
So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$.
By definition of generator of semigroup, it follows that:
- $\gen {S \cup C^{-1} } \subseteq S \circ' C^{-1}$
Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.
Then by definition:
- $x \in S$
and:
- $y^{-1} \in C^{-1}$
and so by definition of generator of semigroup:
- $x \circ' y^{-1} \in \gen {S \cup C^{-1} }$
Thus: $S \circ' C^{-1} \subseteq \gen {S \cup C^{-1} }$
By definition of set equality:
- $S \circ' C^{-1} = \gen {S \cup C^{-1} }$
and so by definition of the inverse completion:
- $T = S \circ' C^{-1}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 1^\circ$