Existence of Field of Quotients

Theorem
If $$\left({D, +, \circ}\right)$$ is an integral domain, then there exists a quotient field of $$\left({D, +, \circ}\right)$$.

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

Inverse Completion is an Abelian Group
By the Inverse Completion Theorem, there exists an inverse completion $$\left({K, \circ}\right)$$ of $$\left({D, \circ}\right)$$.

Thus $$\left({K, \circ}\right)$$ is a commutative semigroup such that:


 * 1) The identity of $$\left({K, \circ}\right)$$ is $$1_D$$;
 * 2) Every element of $$\left({D^*, \circ}\right)$$ has an inverse in $$\left({K, \circ}\right)$$;
 * 3) Every element of $$\left({K, \circ}\right)$$ is of the form $$x \circ y^{-1}$$ (which from the definition of divided by, we can also denote $$x / y$$), where $$x \in D, y \in D^*$$.

Hence $$\left({K^*, \circ}\right)$$ is an abelian group.

Inverse Completion is a Field
Having established that $$\left({K, \circ}\right)$$ is an abelian group, we now want to show that

In what follows, we take for granted the rules of associativity, commutativity and distributivity of $$+$$ and $$\circ$$ in $$D$$.


 * We require to extend the operation $$+$$ on $$D$$ to a composition $$+'$$ on $$K$$, so that $$\left({K, +', \circ}\right)$$ is a field.

By Addition of Division Products, it makes sense to define $$+'$$ as:

$$\forall x, y \in D, \forall z, w \in D^*: \frac x z +' \frac y w = \frac {x \circ w + y \circ z} {z \circ w}$$

where we have defined $$\frac a b = a \circ b^{-1} = b^{-1} \circ a$$ as here.


 * We need to ensure that $$+'$$ is well-defined.

Let $$x, y, x', y' \in D, z, w, z', w' \in D^*$$ such that:

$$\frac x z = \frac {x'} {z'}, \frac y w = \frac {y'} {w'}$$

Then:

$$ $$ $$

Similarly, $$y \circ w' = y' \circ w$$.

(Compare the result in Construction of Inverse Completion: Equivalence Relation on Semigroup Product with Cancellable Elements.)

Hence:

$$ $$ $$

Thus:

$$ $$ $$

showing that $$+'$$ is indeed well-defined.


 * Next, we see that:

$$\forall x, y \in D: x +' y = \frac {x \circ 1_D + y \circ 1_D} {1_D \circ 1_D} = x + y$$

So $$+'$$ induces the given operation $+$ on its substructure $$D$$.


 * It is easy to verify that $$\left({K, +'}\right)$$ is an abelian group, and that $$\circ$$ distributes over $$+'$$:

Taking the group axioms in turn:

G0: Closure
Let $$\frac x z, \frac y w \in K$$.

Then $$x, y \in D$$ and $$z, w \in D^*$$, and $$\frac x z +' \frac y w = \frac {x \circ w + y \circ z} {z \circ w}$$.

As $$z, w \in D^*$$ it follows that $$z \circ w \in D^*$$ because $$D$$ is an integral domain.

By the fact of closure of $$+$$ and $$\circ$$ in $$D$$, $$x \circ w + y \circ z \in D$$.

Hence $$\frac x z +' \frac y w \in K$$ and $$+'$$ is closed.

G1: Associativity
$$ $$ $$ $$ $$ $$

G2: Identity
The identity for $$+$$ is $$\frac 0 r$$ where $$r \in D^*$$:

$$ $$ $$

Similarly for $$\frac 0 r + \frac a b$$.

G3: Inverses
The inverse of $$\frac a b$$ for $$+$$ is $$\frac {-a} b$$:

$$ $$ $$ $$

From above, this is the identity for $$+$$.

Similarly, $$\frac {-a} b + \frac a b = \frac 0 {b \circ b}$$.

Hence $$\frac {-a} b$$ is the inverse of $$\frac a b$$ for $$+$$.

C: Commutativity
$$ $$ $$

Therefore, $$\left({K, +', \circ}\right)$$ is a commutative ring with unity.


 * As a consequence, from Ring Product with Zero:

$$\forall x \in D, y \in D^*: x / y \ne 0_D \Longrightarrow x \ne 0_D$$

Thus $$x / y$$ has the product inverse $$y / x$$ in $$K$$.


 * Thus $$\left({K, +', \circ}\right)$$ is a quotient field of $$\left({D, +, \circ}\right)$$.