Identity of Inverse Completion of Commutative Monoid
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Theorem
Let $\struct {S, \circ}$ be a commutative monoid whose identity is $e$.
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 $e \in T$ is the identity for $\circ'$.
Proof
Let $e$ be the identity for $\circ$.
Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$.
Then:
\(\ds y\) | \(=\) | \(\ds e \circ y\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ' y^{-1} } \circ' y\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ' \paren {y^{-1} \circ' y}\) | $\circ'$ is associative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ' e\) | Definition of Inverse Element |
Thus $e = y^{-1} \circ' y$, and $y^{-1} \circ' y$ is the identity for $\circ'$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 4^\circ$