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 $$D^* = D - \left\{{0_D}\right\}$$ 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 Construction of Inverse Completion: This Equivalence Relation is a Congruence.

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 $$\left({\frac {D \times D^*} {\ominus}, \otimes_{\ominus}}\right)$$

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

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

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 $$\left({D', \circ'}\right)$$.

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