Existence of Quotient Field

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:

$(1): \quad$ The identity of $\left({K, \circ}\right)$ is $1_D$

$(2): \quad$ Every element $x$ of $\left({D^*, \circ}\right)$ has an inverse $\displaystyle \frac {1_D} x$ in $\left({K, \circ}\right)$

$(3): \quad$ 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 $x$ divided by $y$.

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.

$\Box$

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.

$\Box$

G1: Associativity

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\frac a b + \frac c d}\right) + \frac e f\)	\(=\)	\(\displaystyle \frac {a \circ d + b \circ c} {b \circ d} + \frac e f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({a \circ d + b \circ c}\right) \circ f + b \circ d \circ e} {b \circ d \circ f}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {a \circ d \circ f + b \circ c \circ f + b \circ d \circ e} {b \circ d \circ f}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {a \circ d \circ f + b \circ \left({c \circ f + d \circ e}\right)} {b \circ d \circ f}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a b + \frac {c \circ f + d \circ e} {d \circ f}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a b + \left({\frac c d + \frac e f}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

$\Box$

G2: Identity

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac a b + \frac 0 k\)	\(=\)	\(\displaystyle \frac {a \circ k + b \circ 0} {b \circ k}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {a \circ k} {b \circ k}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac a b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

$\Box$

G3: Inverses

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac a b + \frac {-a} b\)	\(=\)	\(\displaystyle \frac {a \circ b + b \circ \left({-a}\right)} {b \circ b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {b \circ \left({a + \left({-a}\right)}\right)} {b \circ b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {b \circ 0} {b \circ b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 0 {b \circ b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $+$.

$\Box$

C: Commutativity

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac a b + \frac c d\)	\(=\)	\(\displaystyle \frac {a \circ d + b \circ c} {b \circ d}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {c \circ b + d \circ a} {d \circ b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac c d + \frac a b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

$\Box$

Product Distributes over Addition

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

$\Box$

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 $\displaystyle \frac y x \in K$.

$\Box$

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

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 23$: Theorem $23.9$