Existence of Field of Quotients

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Then there exists a field of quotients of $\struct {D, +, \circ}$.
== Proof ==
Let $\struct {D, +, \circ}$ 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 Exists, we can define the inverse completion $\struct {K, \circ}$ of $\struct {D, \circ}$.
Thus $\struct {K, \circ}$ is a commutative semigroup such that:
:$(1): \quad$ The identity of $\struct {K, \circ}$ is $1_D$
:$(2): \quad$ Every element $x$ of $\struct {D^*, \circ}$ has an inverse $\dfrac {1_D} x$ in $\struct {K, \circ}$
:$(3): \quad$ Every element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$ (which from the definition of division product, 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, $\struct {K^*, \circ}$ is closed.
Hence $\struct {K^*, \circ}$ 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 $\struct {K, +', \circ}$ is a field.
By Addition of Division Products, we define $+'$ as:
:$\forall a, c \in D, \forall b, d \in D^*: \dfrac a b +' \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$
where we have defined $\dfrac x y = x \circ y^{-1} = y^{-1} \circ x$ as $x$ divided by $y$.
Next, we see that:
:$\forall a, b \in D: a +' b = \dfrac {a \circ 1_D + b \circ 1_D} {1_D \circ 1_D} = a + b$
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 $\struct {K, +}$ is an abelian group
Taking the group axioms in turn:
==== ====
Let $\dfrac a b, \dfrac c d \in K$.
Then $a, c \in D$ and $b, d \in D^*$, and $\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$.
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 $\dfrac a b + \dfrac c d \in K$ and $+$ is closed.
==== ====
Hence $\dfrac a b + \dfrac c d \in K$ and $+$ is associative.
==== ====
The identity for $+$ is $\dfrac 0 k$ where $k \in D^*$:
Similarly for $\dfrac 0 k + \dfrac a b$.
==== ====
The inverse of $\dfrac a b$ for $+$ is $\dfrac {-a} b$:
From above, this is the identity for $+$.
Similarly, $\dfrac {-a} b + \dfrac a b = \dfrac 0 {b \circ b}$.
Hence $\dfrac {-a} b$ is the inverse of $\dfrac a b$ for $+$.
==== $\text C$: Commutativity ====
Therefore, $\struct {K, +, \circ}$ 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:
:$\forall x \in D, y \in D^*: \dfrac x y \ne 0_D \implies x \ne 0_D$
From Inverse of Division Product:
:$\forall x, y \in D^*: \paren {\dfrac x y}^{-1} = \dfrac y x$
Thus $\dfrac x y \in K$ has the ring product inverse $\dfrac y x \in K$.
=== Inverse Completion is a Field ===
We have that:
:the algebraic structure $\struct {K, +}$ is an abelian group
:the algebraic structure $\struct {K^*, \circ}$ is an abelian group
:the operation $\circ$ distributes over $+$.
Hence $\struct {K, +, \circ}$ is a field.
We also have that $\struct {K, +, \circ}$ contains $\struct {D, +, \circ}$ algebraically such that:
:$\forall x \in K: \exists z \in D, y \in D^*: z = \dfrac x y$
Thus $\struct {K, +, \circ}$ is a field of quotients of $\struct {D, +, \circ}$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.9$
* 
Category:Fields of Quotients
Category:Integral Domains