Existence of Field of Quotients

Theorem
Let $\left({D, +, \circ}\right)$ be 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 $\displaystyle \frac {1_D} x$ 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:


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

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


 * Next, we see that:


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

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 $\displaystyle \frac a b, \frac c d \in K$.

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

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

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

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

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

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

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

Hence $\displaystyle \frac {-a} b$ is the inverse of $\displaystyle \frac a b$ 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 $\displaystyle \forall x \in D, y \in D^*: \frac x y \ne 0_D \implies x \ne 0_D$.

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

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

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

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