[proofplan] We construct the fraction field explicitly as equivalence classes of pairs $(a,b)$ with $a\in R$ and $b\neq 0$, where $(a,b)$ represents the formal quotient $a/b$. The [integral domain](/page/Integral%20Domain) hypothesis is used twice: first to make the [equivalence relation](/page/Equivalence%20Relation) transitive and the arithmetic operations well-defined, and then to ensure that every nonzero class has a multiplicative inverse. Finally, the uniqueness statement follows because a field homomorphism compatible with the embedding of $R$ is forced to send the class of $(a,b)$ to $\iota'(a)\iota'(b)^{-1}$. [/proofplan]

[step:Construct formal fractions as equivalence classes] The symbols $0_R$ and $1_R$ denote the additive and multiplicative identities of $R$, respectively. Let \begin{align*} S := R\setminus\{0_R\}. \end{align*} Since $R$ is an integral domain, $1_R\in S$, and if $b,d\in S$, then $bd\in S$. Define a relation $\sim$ on $R\times S$ by declaring, for $(a,b),(c,d)\in R\times S$, \begin{align*} (a,b)\sim(c,d) \iff ad=bc. \end{align*} This relation is reflexive because $ab=ba$, and symmetric by symmetry of equality. To prove transitivity, suppose $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$, so \begin{align*} ad=bc \end{align*} and \begin{align*} cf=de. \end{align*} Multiplying the first equality by $f$ and the second by $b$ gives \begin{align*} adf=bcf \end{align*} and \begin{align*} bcf=bde. \end{align*} Hence $adf=bde$, so by commutativity, \begin{align*} d(af-be)=0_R. \end{align*} Because $d\in S$, we have $d\neq 0_R$, and because $R$ is an integral domain, cancellation against the nonzero factor $d$ gives \begin{align*} af=be. \end{align*} Thus $(a,b)\sim(e,f)$. Therefore $\sim$ is an equivalence relation. Let \begin{align*} F := (R\times S)/{\sim} \end{align*} be the set of equivalence classes. For $(a,b)\in R\times S$, denote its equivalence class by $[a,b]$. [/step]

[step:Define arithmetic on equivalence classes and verify it is well-defined]Define addition and multiplication on $F$ by \begin{align*} [a,b]+[c,d] := [ad+bc,bd] \end{align*} and \begin{align*} [a,b]\cdot[c,d] := [ac,bd]. \end{align*} These formulas are meaningful because $b,d\in S$ implies $bd\in S$. We verify well-definedness. Suppose $(a,b)\sim(a_1,b_1)$ and $(c,d)\sim(c_1,d_1)$, where all denominators lie in $S$. Thus \begin{align*} ab_1=a_1b \end{align*} and \begin{align*} cd_1=c_1d. \end{align*} For addition, we must prove \begin{align*} (ad+bc)b_1d_1=(a_1d_1+b_1c_1)bd. \end{align*} Expanding the left-hand side and using the two equivalence relations gives \begin{align*} (ad+bc)b_1d_1=adb_1d_1+bcb_1d_1. \end{align*} The equality $ab_1=a_1b$ gives \begin{align*} adb_1d_1=a_1bdd_1. \end{align*} The equality $cd_1=c_1d$ gives \begin{align*} bcb_1d_1=bb_1c_1d. \end{align*} Therefore \begin{align*} (ad+bc)b_1d_1=a_1bdd_1+bb_1c_1d=(a_1d_1+b_1c_1)bd. \end{align*} So addition is well-defined. For multiplication, we must prove \begin{align*} (ac)b_1d_1=(a_1c_1)bd. \end{align*} Using $ab_1=a_1b$ and $cd_1=c_1d$, we get \begin{align*} acb_1d_1=a_1bcd_1=a_1bc_1d=a_1c_1bd. \end{align*} So multiplication is well-defined.[/step]

[guided]The only possible ambiguity is that a formal fraction may have many representatives. We therefore check that changing representatives does not change the result of the addition and multiplication formulas. Assume \begin{align*} (a,b)\sim(a_1,b_1) \end{align*} and \begin{align*} (c,d)\sim(c_1,d_1). \end{align*} By definition of $\sim$, this means \begin{align*} ab_1=a_1b \end{align*} and \begin{align*} cd_1=c_1d. \end{align*} The sum of the first representatives is represented by $(ad+bc,bd)$, while the sum of the second representatives is represented by $(a_1d_1+b_1c_1,b_1d_1)$. To prove that these two pairs define the same class, we must prove exactly \begin{align*} (ad+bc)b_1d_1=(a_1d_1+b_1c_1)bd. \end{align*} Expanding the left-hand side gives \begin{align*} (ad+bc)b_1d_1=adb_1d_1+bcb_1d_1. \end{align*} The relation $ab_1=a_1b$ converts the first summand: \begin{align*} adb_1d_1=a_1bdd_1. \end{align*} The relation $cd_1=c_1d$ converts the second summand: \begin{align*} bcb_1d_1=bb_1c_1d. \end{align*} Combining these two identities gives \begin{align*} (ad+bc)b_1d_1=a_1bdd_1+bb_1c_1d=(a_1d_1+b_1c_1)bd. \end{align*} Thus the addition formula is independent of the chosen representatives. For multiplication, the product of the first representatives is $(ac,bd)$, and the product of the second representatives is $(a_1c_1,b_1d_1)$. The required equivalence is \begin{align*} (ac)b_1d_1=(a_1c_1)bd. \end{align*} Using $ab_1=a_1b$ first and $cd_1=c_1d$ second, we obtain \begin{align*} acb_1d_1=a_1bcd_1=a_1bc_1d=a_1c_1bd. \end{align*} Therefore multiplication is also independent of representatives. This is precisely where the cross-multiplication definition of $\sim$ is designed to match the usual arithmetic of fractions.[/guided]

[step:Show the formal fractions form a field] The zero and one elements of $F$ are \begin{align*} 0_F := [0_R,1_R] \end{align*} and \begin{align*} 1_F := [1_R,1_R]. \end{align*} The additive inverse of $[a,b]\in F$ is $[-a,b]$, since \begin{align*} [a,b]+[-a,b]=[ab-ba,b^2]=[0_R,b^2]=[0_R,1_R]. \end{align*} The associativity and commutativity of addition and multiplication, the distributive law, and the identity laws are checked on representatives by expanding the defining formulas above and applying the corresponding ring identities in $R$. Because addition and multiplication have already been proved well-defined on equivalence classes, these representative computations descend to identities in $F$. It remains to verify multiplicative inverses for nonzero elements. Let $[a,b]\in F$ and suppose \begin{align*} [a,b]\neq [0_R,1_R]. \end{align*} The equivalence $[a,b]=[0_R,1_R]$ is equivalent to $a1_R=0_Rb$, hence to $a=0_R$. Therefore $a\neq 0_R$, so $a\in S$, and $[b,a]\in F$ is defined. Then \begin{align*} [a,b]\cdot[b,a]=[ab,ba]=[1_R,1_R], \end{align*} because $(ab,ba)\sim(1_R,1_R)$ is the equality $ab=ba$. Thus every nonzero element has a multiplicative inverse, and $F$ is a field. [/step]

[step:Embed the integral domain into its constructed fraction field] Define the map \begin{align*} \iota: R &\to F \end{align*} by \begin{align*} \iota(r) := [r,1_R]. \end{align*} For $r,s\in R$, the definitions of addition and multiplication give \begin{align*} \iota(r+s)=[r+s,1_R]=[r,1_R]+[s,1_R] \end{align*} and \begin{align*} \iota(rs)=[rs,1_R]=[r,1_R]\cdot[s,1_R]. \end{align*} Also $\iota(1_R)=[1_R,1_R]=1_F$, so $\iota$ is a unital ring homomorphism. If $\iota(r)=\iota(s)$, then $[r,1_R]=[s,1_R]$, hence $r1_R=s1_R$, so $r=s$. Thus $\iota$ is injective. Finally, for any $[a,b]\in F$, with $b\in S$, we have \begin{align*} [a,b]=[a,1_R]\cdot[1_R,b]=\iota(a)\iota(b)^{-1}. \end{align*} Therefore every element of $F$ is a quotient of elements in the image of $R$ with nonzero denominator. [/step]

[step:Define the comparison map between two fraction fields] Let $(F,\iota)$ and $(F',\iota')$ be two fraction fields of $R$ in the sense of the statement. Define a map \begin{align*} \Phi: F &\to F' \end{align*} as follows. For $x\in F$, choose $a\in R$ and $b\in R\setminus\{0_R\}$ such that \begin{align*} x=\iota(a)\iota(b)^{-1}, \end{align*} and set \begin{align*} \Phi(x):=\iota'(a)\iota'(b)^{-1}. \end{align*} This formula is well-defined. Indeed, suppose also \begin{align*} x=\iota(c)\iota(d)^{-1} \end{align*} with $c\in R$ and $d\in R\setminus\{0_R\}$. Since $\iota(b)$ and $\iota(d)$ are nonzero in the field $F$, multiplying the equality \begin{align*} \iota(a)\iota(b)^{-1}=\iota(c)\iota(d)^{-1} \end{align*} by $\iota(b)\iota(d)$ gives \begin{align*} \iota(ad)=\iota(bc). \end{align*} Because $\iota$ is injective, $ad=bc$ in $R$. Applying the ring homomorphism $\iota'$ gives \begin{align*} \iota'(a)\iota'(d)=\iota'(b)\iota'(c). \end{align*} Since $\iota'(b)$ and $\iota'(d)$ are nonzero in $F'$, multiplying by their inverses gives \begin{align*} \iota'(a)\iota'(b)^{-1}=\iota'(c)\iota'(d)^{-1}. \end{align*} Thus $\Phi$ is well-defined. [/step]

[step:Verify the comparison map is the unique compatible field isomorphism] For $a,c\in R$ and $b,d\in R\setminus\{0_R\}$, the definition of $\Phi$ gives \begin{align*} \Phi(\iota(a)\iota(b)^{-1}+\iota(c)\iota(d)^{-1})=\Phi(\iota(ad+bc)\iota(bd)^{-1}). \end{align*} Therefore \begin{align*} \Phi(\iota(a)\iota(b)^{-1}+\iota(c)\iota(d)^{-1})=\iota'(ad+bc)\iota'(bd)^{-1}. \end{align*} Using that $\iota'$ is a ring homomorphism and that inverses distribute over products in the field $F'$, this equals \begin{align*} \iota'(a)\iota'(b)^{-1}+\iota'(c)\iota'(d)^{-1}. \end{align*} Hence $\Phi$ preserves addition. The same computation for products gives \begin{align*} \Phi(\iota(a)\iota(b)^{-1}\cdot\iota(c)\iota(d)^{-1})=\iota'(ac)\iota'(bd)^{-1} \end{align*} and hence \begin{align*} \Phi(\iota(a)\iota(b)^{-1}\cdot\iota(c)\iota(d)^{-1})=\iota'(a)\iota'(b)^{-1}\cdot\iota'(c)\iota'(d)^{-1}. \end{align*} Also $\Phi(1_F)=1_{F'}$. Thus $\Phi$ is a field homomorphism. The map is compatible with the embeddings because, for every $r\in R$, \begin{align*} \Phi(\iota(r))=\Phi(\iota(r)\iota(1_R)^{-1})=\iota'(r)\iota'(1_R)^{-1}=\iota'(r). \end{align*} Construct similarly a map \begin{align*} \Psi: F' &\to F \end{align*} by \begin{align*} \Psi(\iota'(a)\iota'(b)^{-1}) := \iota(a)\iota(b)^{-1}. \end{align*} To see that this definition is well-defined, suppose \begin{align*} \iota'(a)\iota'(b)^{-1}=\iota'(c)\iota'(d)^{-1}. \end{align*} Multiplying by \iota'(b)\iota'(d) in the field F' gives \begin{align*} \iota'(ad)=\iota'(bc). \end{align*} Since \iota' is injective, ad=bc in R, and therefore \begin{align*} \iota(a)\iota(b)^{-1}=\iota(c)\iota(d)^{-1}. \end{align*} The same quotient formulas as above then show that \Psi preserves addition and multiplication and sends 1_{F'} to 1_F. For every $a\in R$ and $b\in R\setminus\{0_R\}$, \begin{align*} (\Psi\circ\Phi)(\iota(a)\iota(b)^{-1})=\iota(a)\iota(b)^{-1} \end{align*} and \begin{align*} (\Phi\circ\Psi)(\iota'(a)\iota'(b)^{-1})=\iota'(a)\iota'(b)^{-1}. \end{align*} Since every element of $F$ and every element of $F'$ has one of these quotient forms, $\Psi\circ\Phi=\operatorname{id}_F$ and $\Phi\circ\Psi=\operatorname{id}_{F'}$. Thus $\Phi$ is a field isomorphism. Finally, let \begin{align*} \Theta: F &\to F' \end{align*} be any field homomorphism satisfying $\Theta\circ\iota=\iota'$. For $a\in R$ and $b\in R\setminus\{0_R\}$, \begin{align*} \Theta(\iota(a)\iota(b)^{-1})=\Theta(\iota(a))\Theta(\iota(b))^{-1}=\iota'(a)\iota'(b)^{-1}. \end{align*} Thus $\Theta$ agrees with $\Phi$ on every element of $F$, since every element of $F$ is such a quotient. Therefore $\Phi$ is the unique field isomorphism compatible with the embeddings of $R$. [/step]