Construction of Inverse Completion/Quotient Structure

Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $C \subseteq S$ be the set of cancellable elements of $S$.

Let $\left({S \times C, \oplus}\right)$ be the external direct product of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$, where:

• $\circ \restriction_C$ is the restriction of $\circ$ to $C \times C$, and
• $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.

Let $\mathcal R$ be the congruence relation $\mathcal R$ defined on $\left({S \times C, \oplus}\right)$ by:

$\left({x_1, y_1}\right) \ \mathcal R \ \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Let the quotient structure defined by $\mathcal R$ be:

$\displaystyle \left({T\,', \oplus'}\right) := \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$

where $\oplus_\mathcal R$ is the operation induced on $\displaystyle \frac {S \times C} {\mathcal R}$ by $\oplus$.

Quotient Structure is Commutative Semigroup

$\left({T\,', \oplus'}\right)$ is a commutative semigroup.

Quotient Mapping is Injective

Let the mapping $\psi: S \to T\,'$ be defined as:

$\forall x \in S: \psi \left({x}\right) = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\mathcal R$

Then $\psi: S \to T\,'$ is an injection, and does not depend on the particular element $a$ chosen.

Quotient Mapping is Monomorphism

The mapping $\psi: S \to T\,'$ is a monomorphism.

Image of Quotient Mapping is Subsemigroup

Let $S\,'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\left({S\,', \oplus'}\right)$ is a subsemigroup of $\left({T\,', \oplus'}\right)$.

Quotient Mapping to Image is Isomorphism

Let $S\,'$ be the image $\psi \left({S}\right)$ of $S$.

Then $\psi$ is an isomorphism from $S$ onto $S\,'$.

Image of Cancellable Elements in Quotient Mapping

The set $C\,'$ of cancellable elements of the semigroup $S\,'$ is $\psi \left({C}\right)$.

Proof

From the defined equivalence relation, we have that:

$\left({x_1, y_1}\right) \ \mathcal R \ \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

is a congruence relation on $\left({S \times C, \oplus}\right)$.

From the definition of the members of the equivalence classes:

$(1) \quad \forall x, y \in S, a, b \in C: \left({x \circ a, a}\right) \ \mathcal R \ \left({y \circ b, b}\right) \iff x = y$

$(2) \quad \forall x, y \in S, a, b \in C: \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\mathcal R = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$

From the definition of the equivalence class of equal elements:

$(3) \quad \forall c, d \in C: \left({c, c}\right) \ \mathcal R \ \left({d, d}\right)$

where $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$ is the equivalence class of $\left({x, y}\right)$ under $\mathcal R$.

Hence we are justified in asserting the existence of the quotient structure:

$\displaystyle \left({T\,', \oplus'}\right) = \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 20$