The construction
Definition (The proportionality relation on \(D \times (D \setminus \{0\})\))
Let \(D\) be an integral domain and let \(S = D \times (D \setminus \{0\})\). Define the relation \(\sim\) on \(S\) such that
\[(a, b) \sim (c, d) \iff a \cdot d = b \cdot c.\]We use the notation \(\frac{a}{b} := [(a, b)]\) for the equivalence class of \((a, b)\).
Lemma (The proportionality relation is an equivalence relation)
The relation \(\sim\) is an equivalence relation on \(S\).
Proof: Reflexivity and symmetry are immediate from the commutativity of \(D\): \(a \cdot b = b \cdot a\) gives \((a,b) \sim (a,b)\), and if \(a d = b c\), then \(c b = d a\), that is, \((c, d) \sim (a, b)\).
For transitivity we use that \(D\) is an integral domain. Suppose that \((a, b) \sim (c, d)\) and \((c, d) \sim (e, f)\), that is, \(a d = b c\) and \(c f = d e\). Multiplying the first equality by \(f\) and using the second,
\[a d f = b c f = b d e,\]and rearranging with commutativity, \(d (a f) = d (b e)\), that is, \(d (a f - b e) = 0\). Since \(d \neq 0\) and \(D\) has no zero divisors, we conclude that \(a f - b e = 0\), that is, \(a f = b e\), which is exactly \((a, b) \sim (e, f)\).
Definition (The field of fractions and its operations)
Let \(F := \left\{ \frac{a}{b} :\ (a, b) \in S \right\}\). Note that if \(x \in D \setminus \{0\}\), then \(\frac{a}{b} = \frac{a \cdot x}{b \cdot x}\). Moreover, we can define operations on \(F\) in the following way:
\[\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}, \qquad \frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}.\]Note that the denominators \(b d\) are non-zero because \(D\) is an integral domain, that \(\frac{0}{1}\) is the identity element for \(+\) and that \(\frac{1}{1}\) is the identity element for \(\cdot\). That these operations are well defined (their independence of the representatives) is verified in the proof of the following theorem.
Theorem (The field of fractions is a field)
\((F, +, \cdot, \frac{0}{1}, \frac{1}{1})\) is a commutative ring and, more interestingly, it is in fact a field.
Proof: The operations are well defined. Suppose that \(\frac{a}{b} = \frac{a'}{b'}\) and \(\frac{c}{d} = \frac{c'}{d'}\), that is, \(a b' = a' b\) and \(c d' = c' d\). For the sum we must show that \(\frac{a d + b c}{b d} = \frac{a' d' + b' c'}{b' d'}\), that is, that \((a d + b c) b' d' = (a' d' + b' c') b d\). Using commutativity and the two relations,
\[(a d + b c)\, b' d' = (a b')\, d d' + (c d')\, b b' = (a' b)\, d d' + (c' d)\, b b' = (a' d' + b' c')\, b d.\]For the product, \((a c)(b' d') = (a b')(c d') = (a' b)(c' d) = (a' c')(b d)\), that is, \(\frac{a c}{b d} = \frac{a' c'}{b' d'}\).
\(F\) is a commutative ring. The denominators \(b d\) are non-zero, since \(D\) is an integral domain. Associativity, commutativity of both operations and distributivity are verified with representatives and reduce to the corresponding properties of \(D\); the additive identity element is \(\frac{0}{1}\), the additive inverse of \(\frac{a}{b}\) is \(\frac{-a}{b}\), and the multiplicative identity element is \(\frac{1}{1}\).
\(F\) is a field. Note first that \(1 \neq 0\) in \(D\) (if \(1 = 0\), then \(D = \{0\}\) and there would be no denominators available), and hence \(\frac{1}{1} \neq \frac{0}{1}\) in \(F\), since \(1 \cdot 1 \neq 0 \cdot 1\). Now let \(\frac{a}{b} \neq \frac{0}{1}\); this means that \(a \cdot 1 \neq b \cdot 0\), that is, \(a \neq 0\). Then \(\frac{b}{a} \in F\) (the denominator \(a\) is non-zero) and
\[\frac{a}{b} \cdot \frac{b}{a} = \frac{a b}{b a} = \frac{1}{1},\]since \((a b) \cdot 1 = (b a) \cdot 1\). That is, every non-zero element of \(F\) is invertible, and \(F\) is a field.
Note (Examples of fields of fractions)
We write \(F = \operatorname{Frac}(D)\) for the field of fractions of \(D\).
- \(\operatorname{Frac}(\mathbb{Z}) \cong \mathbb{Q}\).
- If \(D\) is a field, then \(\operatorname{Frac}(D) \cong D\).
-
If \(D = \Bbbk[x_{1}, \dots, x_{n}]\) with \(\Bbbk\) a field, then
\[\operatorname{Frac}(D) = \Bbbk(x_{1}, \dots, x_{n}) = \left\{ \frac{p(x_{1}, \dots, x_{n})}{q(x_{1}, \dots, x_{n})} :\ q \neq 0 \right\},\]the field of rational functions in \(n\) variables.
Note (The domain embeds in its field of fractions)
We may assume that \(D \subseteq \operatorname{Frac}(D)\), since \(i : D \to \operatorname{Frac}(D)\), \(x \mapsto \frac{x}{1}\), is an injective homomorphism, so that \(D\) “embeds” in \(\operatorname{Frac}(D)\); that is,
\[D \cong \operatorname{Im}(i) = \left\{ \frac{x}{1} :\ x \in D \right\} \subseteq \operatorname{Frac}(D).\]The universal property
Theorem (Universal property of the field of fractions)
Let \(D\) be an integral domain, \(K\) a field and \(f : D \to K\) an injective homomorphism. Then there exists a unique injective homomorphism \(\tilde{f} : \operatorname{Frac}(D) \to K\) such that
\[\tilde{f}\left( \frac{x}{1} \right) = f(x) \quad \text{for all } x \in D.\]Explicitly,
\[\tilde{f}\left( \frac{a}{b} \right) = \frac{f(a)}{f(b)} := f(a) \cdot f(b)^{-1} \quad \text{for all } a, b \in D,\ b \neq 0.\]Proof: \(\tilde{f}\) is well defined. If \(\frac{a}{b} = \frac{c}{d}\), then \(a d = b c\), and applying \(f\) we obtain \(f(a) \cdot f(d) = f(b) \cdot f(c)\). Note that \(f(b) \neq 0\) and \(f(d) \neq 0\), since \(b, d \neq 0\) and \(f\) is injective (its kernel is trivial). Then, rearranging in the field \(K\), we have that
\[\frac{f(a)}{f(b)} = \frac{f(c)}{f(d)} \in K \implies \tilde{f}\left( \frac{a}{b} \right) = \tilde{f}\left( \frac{c}{d} \right).\]Proving that \(\tilde{f}\) is an injective homomorphism is left as an exercise.
Uniqueness. Suppose that \(h : \operatorname{Frac}(D) \to K\) is a homomorphism such that \(h\left( \frac{x}{1} \right) = f(x)\) for all \(x \in D\). Let \(b \neq 0\); since \(\frac{b}{1} \cdot \frac{1}{b} = \frac{b}{b} = \frac{1}{1}\) and homomorphisms preserve units, we have that \(h\left( \frac{1}{b} \right) = h\left( \frac{b}{1} \right)^{-1} = f(b)^{-1}\). Then
\[h\left( \frac{a}{b} \right) = h\left( \frac{a}{1} \cdot \frac{1}{b} \right) = h\left( \frac{a}{1} \right) \cdot h\left( \frac{1}{b} \right) = f(a) \cdot f(b)^{-1} = \frac{f(a)}{f(b)} = \tilde{f}\left( \frac{a}{b} \right),\]that is, \(h = \tilde{f}\).
Theorem (The prime field as the field of fractions of the prime ring)
Let \(\Bbbk\) be a field.
- If \(R\) is the prime subring of \(\Bbbk\) and \(F\) is the prime subfield of \(\Bbbk\), then \(F \cong \operatorname{Frac}(R)\).
- The prime subfield of \(\Bbbk\) is isomorphic to \(\mathbb{Q}\) or to \(\mathbb{Z}_{p}\) (also denoted \(\mathbb{F}_{p}\)), according as \(\operatorname{char}(\Bbbk) = 0\) or \(\operatorname{char}(\Bbbk) = p > 0\).
Proof: Part 1. Since every subfield of \(\Bbbk\) is in particular a subring, the prime subring \(R\) is contained in the prime subfield \(F\), and hence the inclusion \(i : R \to F\), \(x \mapsto x\), is an injective homomorphism from \(R\) into the field \(F\). Moreover, \(R\) is an integral domain, being a subring of a field. By the universal property of the field of fractions, there exists an injective homomorphism \(\tilde{\imath} : \operatorname{Frac}(R) \to F\) with \(\tilde{\imath}\left(\frac{a}{b}\right) = a b^{-1}\). Let
\[F_{0} = \operatorname{Im}(\tilde{\imath}) = \{ a b^{-1} :\ a, b \in R,\ b \neq 0 \} \subseteq F.\]Since \(\tilde{\imath}\) is an injective homomorphism from a field, \(F_{0} \cong \operatorname{Frac}(R)\) and \(F_{0}\) is a subfield of \(\Bbbk\): it is a subring (the image of a homomorphism), and the inverse of \(a b^{-1} \neq 0\) is \(b a^{-1} \in F_{0}\). By the minimality of the prime subfield, \(F \subseteq F_{0}\); and we have already seen that \(F_{0} \subseteq F\). We conclude that
\[F = F_{0} \cong \operatorname{Frac}(R).\]Part 2. By the prime subring theorem, \(R \cong \mathbb{Z}\) or \(R \cong \mathbb{Z}_{n}\); and since \(\Bbbk\) is a field, in particular an integral domain, its characteristic is \(0\) or a prime \(p\). If \(\operatorname{char}(\Bbbk) = 0\), then \(R \cong \mathbb{Z}\) and, by part 1 (an isomorphism of domains extends to the fields of fractions via the universal property),
\[F \cong \operatorname{Frac}(R) \cong \operatorname{Frac}(\mathbb{Z}) \cong \mathbb{Q}.\]If \(\operatorname{char}(\Bbbk) = p\) is prime, then \(R \cong \mathbb{Z}_{p}\), which is already a field (since \(p\) is prime), and the field of fractions of a field is itself:
\[F \cong \operatorname{Frac}(\mathbb{Z}_{p}) \cong \mathbb{Z}_{p} = \mathbb{F}_{p}.\]