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 Inverse Completion of Integral Domain, we can define the 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 $$x$$ of $$\left({D^*, \circ}\right)$$ has an inverse $$\frac {1_D} x$$ 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 $$x \in D, y \in D^*$$.

It can also be noted that from Inverse Completion Less Zero of Integral Domain is Closed, $$\left({K^*, \circ}\right)$$ is closed.

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

Additive Operation on K
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 an operation $$+'$$ on $$K$$, so that $$\left({K, +', \circ}\right)$$ is a field.

By Addition of Division Products, we define $$+'$$ as:


 * $$\forall a, c \in D, \forall b, d \in D^*: \frac a b +' \frac c d = \frac {a \circ d + b \circ c} {b \circ d}$$

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


 * Next, we see that:


 * $$\forall a, b \in D: a +' b = \frac {a \circ 1_D + b \circ 1_D} {1_D \circ 1_D} = a + b$$

So $$+$$ induces the given operation $+$ on its substructure $$D$$, and we are justified in using $$+$$ for both operations.

Addition on K makes an Abelian Group
Now we verify that $$\left({K, +}\right)$$ is an abelian group

Taking the group axioms in turn:

G0: Closure
Let $$\frac a b, \frac c d \in K$$.

Then $$a, c \in D$$ and $$b, d \in D^*$$, and $$\frac a b + \frac c d = \frac {a \circ d + b \circ c} {b \circ d}$$.

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

By the fact of closure of $$+$$ and $$\circ$$ in $$D$$, $$a \circ d + b \circ c \in D$$.

Hence $$\frac a b + \frac c d \in K$$ and $$+$$ is closed.

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

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

$$ $$ $$

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

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.

Product Distributes over Addition
From Extension Theorem for Distributive Operations, it follows directly that $$\circ$$ distributes over $$+$$.

Product Inverses in K
From Ring Product with Zero, we note that $$\forall x \in D, y \in D^*: \frac x y \ne 0_D \Longrightarrow x \ne 0_D$$.

From Inverse of Division Product, $$\forall x, y \in D^*: \left({\frac x y}\right)^{-1} = \frac y x$$.

Thus $$\frac x y \in K$$ has the product inverse $$\frac y x \in K$$.

Inverse Completion is a Field
We have that:
 * the algebraic structure $$\left({K, +}\right)$$ is an abelian group;
 * the algebraic structure $$\left({K^*, \circ}\right)$$ is an abelian group;
 * the operation $$\circ$$ distributes over $$+$$.

Hence $$\left({K, +, \circ}\right)$$ is a field.

We also have that $$\left({K, +, \circ}\right)$$ contains $\left({D, +, \circ}\right)$ algebraically such that:


 * $$\forall x \in K: \exists z \in D, y \in D^*: z = \frac x y$$

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