Existence of Quotient Field

From ProofWiki
Jump to: navigation, search

Theorem

Let $\struct {D, +, \circ}$ be an integral domain.


Then there exists a quotient field 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.

$\Box$


Addition on K makes an Abelian Group

Now we verify that $\struct {K, +}$ is an abelian group

Taking the group axioms in turn:


G0: Closure

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.

$\Box$


G1: Associativity

\(\displaystyle \paren {\frac a b + \frac c d} + \frac e f\) \(=\) \(\displaystyle \frac {a \circ d + b \circ c} {b \circ d} + \frac e f\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {a \circ d + b \circ c} \circ f + b \circ d \circ e} {b \circ d \circ f}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {a \circ d \circ f + b \circ c \circ f + b \circ d \circ e} {b \circ d \circ f}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {a \circ d \circ f + b \circ \paren {c \circ f + d \circ e} } {b \circ d \circ f}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac a b + \frac {c \circ f + d \circ e} {d \circ f}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac a b + \paren {\frac c d + \frac e f}\) $\quad$ $\quad$

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

$\Box$


G2: Identity

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

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

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

$\Box$


G3: Inverses

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

\(\displaystyle \frac a b + \frac {-a} b\) \(=\) \(\displaystyle \frac {a \circ b + b \circ \paren {-a} } {b \circ b}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {b \circ \paren {a + \paren {-a} } } {b \circ b}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {b \circ 0} {b \circ b}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 0 {b \circ b}\) $\quad$ $\quad$

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 $+$.

$\Box$


C: Commutativity

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


Therefore, $\struct {K, +, \circ}$ 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:

$=\forall x \in D, y \in D^*: \dfrac x y \ne 0_D \implies x \ne 0_D$

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 $\dfrac y x \in K$.

$\Box$


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 quotient field of $\struct {D, +, \circ}$.

$\blacksquare$


Sources