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. {{qed|lemma}} === Addition on $K$ makes an Abelian Group === Now we verify that $\struct {K, +}$ is an abelian group Taking the group axioms in turn: ==== $\text G 0$: 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. {{qed|lemma}} ==== $\text G 1$: Associativity ==== {{begin-eqn}} {{eqn | l = \paren {\frac a b + \frac c d} + \frac e f       | r = \frac {a \circ d + b \circ c} {b \circ d} + \frac e f       | c =  }} {{eqn | r = \frac {\paren {a \circ d + b \circ c} \circ f + b \circ d \circ e} {b \circ d \circ f}       | c =  }} {{eqn | r = \frac {a \circ d \circ f + b \circ c \circ f + b \circ d \circ e} {b \circ d \circ f}       | c =  }} {{eqn | r = \frac {a \circ d \circ f + b \circ \paren {c \circ f + d \circ e} } {b \circ d \circ f}       | c =  }} {{eqn | r = \frac a b + \frac {c \circ f + d \circ e} {d \circ f}       | c =  }} {{eqn | r = \frac a b + \paren {\frac c d + \frac e f}       | c =  }} {{end-eqn}} Hence $\dfrac a b + \dfrac c d \in K$ and $+$ is associative. {{qed|lemma}} ==== $\text G 2$: Identity ==== The identity for $+$ is $\dfrac 0 k$ where $k \in D^*$: {{begin-eqn}} {{eqn | l = \frac a b + \frac 0 k       | r = \frac {a \circ k + b \circ 0} {b \circ k}       | c =  }} {{eqn | r = \frac {a \circ k} {b \circ k}       | c =  }} {{eqn | r = \frac a b       | c =  }} {{end-eqn}} Similarly for $\dfrac 0 k + \dfrac a b$. {{qed|lemma}} ==== $\text G 3$: Inverses ==== The inverse of $\dfrac a b$ for $+$ is $\dfrac {-a} b$: {{begin-eqn}} {{eqn | l = \frac a b + \frac {-a} b       | r = \frac {a \circ b + b \circ \paren {-a} } {b \circ b}       | c =  }} {{eqn | r = \frac {b \circ \paren {a + \paren {-a} } } {b \circ b}       | c =  }} {{eqn | r = \frac {b \circ 0} {b \circ b}       | c =  }} {{eqn | r = \frac 0 {b \circ b}       | c =  }} {{end-eqn}} 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 $+$. {{qed|lemma}} ==== $\text C$: Commutativity ==== {{begin-eqn}} {{eqn | l = \frac a b + \frac c d       | r = \frac {a \circ d + b \circ c} {b \circ d}       | c =  }} {{eqn | r = \frac {c \circ b + d \circ a} {d \circ b}       | c =  }} {{eqn | r = \frac c d + \frac a b       | c =  }} {{end-eqn}} Therefore, $\struct {K, +, \circ}$ is a commutative ring with unity. {{qed|lemma}} === Product Distributes over Addition === From Extension Theorem for Distributive Operations, it follows directly that $\circ$ distributes over $+$. {{qed|lemma}} === 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 $\dfrac y x \in K$. {{qed|lemma}} === 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 $\struct {D, +, \circ}$. {{Qed}}