Inverse Completion of Integral Domain Exists

Theorem
Let $\left({D, +, \circ}\right)$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.

Then an inverse completion of $\left({D, \circ}\right)$ can be constructed.

Proof
From the definition of an integral domain:
 * All elements of $D^* = D \setminus \left\{{0_D}\right\}$ are cancellable;
 * $\left({D^*, \circ}\right)$ is a commutative semigroup.

So by the Inverse Completion Theorem, there exists an inverse completion of $\left({D, \circ}\right)$.

From Construction of Inverse Completion, this is done as follows:

Let $\ominus$ be the congruence relation defined on $D \times D^*$ by:


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

The fact that this is a congruence relation is proved in Equivalence Relation on Semigroup Product with Cancellable Elements.

Let $\left({D \times D^*, \otimes}\right)$ be the external direct product of $\left({D, \circ}\right)$ with $\left({D^*, \circ}\right)$, where $\otimes$ is the operation on $D \times D^*$ induced by $\circ$.

Let the quotient structure defined by $\ominus$ be $\left({\dfrac {D \times D^*} {\ominus}, \otimes_\ominus}\right)$.

where $\otimes_\ominus$ is the operation induced on $\dfrac {D \times D^*} \ominus$ by $\otimes$.

Let us use $D\,'$ to denote the quotient set $\dfrac {D \times D^*} {\ominus}$.

Let us use $\circ'$ to denote the operation $\otimes_\ominus$.

Thus $\left({D\,', \circ'}\right)$ is the inverse completion of $\left({D, \circ}\right)$.

An element of $D\,'$ is therefore an equivalence class of the congruence relation $\ominus$.

As the Inverse Completion is Unique up to isomorphism, it follows that we can define the structure $\left({K, \circ}\right)$ which is isomorphic to $\left({D\,', \circ'}\right)$.

An element of $D\,'$ is therefore an equivalence class of the congruence relation $\ominus$.

So an element of $K$ is the isomorphic image of an element $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\ominus$ of $\dfrac {D \times D^*} \ominus$.

Hence every element of $\left({K, \circ}\right)$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.

Alternatively, from the definition of divided by, of the form $\dfrac x y$.

Hence we can therefore interpret any element of $\left({K, \circ}\right)$ as equivalence classes of elements of the form $\dfrac x y$.