Construction of Inverse Completion

Theorem

This page consists of a series of linked theorems, each of which builds towards one result.

To access the proofs for the individual theorems, click on the links which form the titles of each major section.

Initial Definitions

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

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

Commutative Semigroup by its Cancellable Elements is a Commutative Semigroup

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 $C$.

That is:

$\forall \left({x, y}\right), \left({u, v}\right) \in S \times C: \left({x, y}\right) \oplus \left({u, v}\right) = \left({x \circ u, y \circ \restriction_C v}\right)$

Then $\left({S \times C, \oplus}\right)$ is a commutative semigroup.

Equivalence Relation on Semigroup Product with Cancellable Elements

The relation $\mathcal R$ defined on $S \times C$ by:

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

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

Members of Equivalence Classes

We have:

$\forall x, y \in S, a, b \in C:$

• $\left({x \circ a, a}\right) \mathcal R \left({y \circ b, b}\right) \iff x = y$

• $\left[\!\left[{x \circ a, y \circ a}\right]\!\right]_\mathcal R = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$

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

Equivalence Class of Equal Elements

We also have:

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

Comment

In the context of the naturally ordered semigroup, the Unique Minus is defined:

$n \ominus m = p \iff m \circ p = n$

from which it can be seen that the above congruence can be understood as:

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

Thus this congruence defines an equivalence between pairs of elements which have the same Unique Minus (or, informally at this stage, difference).

Quotient Structure

Let the quotient structure defined by $\mathcal R$ be $\displaystyle \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$

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

Let us use $T'$ to denote the quotient set $\displaystyle \frac {S \times C} {\mathcal R}$.

Let us use $\oplus'$ to denote the operation $\oplus_\mathcal R$.

Thus $\displaystyle \left({\frac {S \times C} {\mathcal R}, \oplus_\mathcal R}\right)$ is now denoted $\left({T', \oplus'}\right)$.

Quotient Mapping 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.

Quotient Mapping to Image is Isomorphism

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

Then:

• $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$;
• $\psi$ 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 $\psi \left({C}\right)$.

Properties of Quotient Structure

Identity of Quotient Structure

We have that:

$\forall c \in C: \left[\!\left[{\left({c, c}\right)}\right]\!\right]_\mathcal R$

is the identity of $T'$.

We denote the identity of $T'$ as $e_{T'}$, as usual.

Invertible Elements in Quotient Structure

Every cancellable element of $S'$ is invertible in $T'$.

Generator for Quotient Structure

$T' = S' \cup \left({C'}\right)^{-1}$ is a generator for the semigroup $T'$.

Quotient Structure is Inverse Completion

$T'$ is an inverse completion of its subsemigroup $S'$.

Notes

• Elements of $T'$ are equivalence classes of ordered pairs $\left({x, y}\right)$ where $x \in S, y \in C$.

Each of the elements of $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\mathcal R$ are such that $x \circ y^{-1}$ have the same value, where $y^{-1} \in C^{-1}$.

Hence is it a natural progression to define an operation $\odot$, say, such that:

$x \odot y \equiv x \circ y^{-1}$

In the context of the integers, this operation is minus, hence:

$x - y \equiv x + \left({-y}\right)$

In the context of the rational numbers, this operation is divided by, hence:

$\dfrac x y \equiv x \times y^{-1}$

• Each element of $S$, and hence in $C$, is identified (via the isomorphism $\psi$) with one of these equivalence classes.

If $S$ is a monoid, then it has an identity $e_S$, say, which is in $C$.

Hence:

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

In particular:

$\psi \left({e_S}\right) = \left[\!\left[{\left({e_S, e_S}\right)}\right]\!\right]_\mathcal R$

Sources

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