Every integral domain can be embedded in a eld (see proof below). That is, using concepts from set theory, given an arbitrary integral domain (such as the integers), one can construct a eld that contains a subset isomorphic to the integral domain. Such a eld is called the eld of fractions of the given integral domain.
1. Examples
• The rational numbers (Q) is the eld of fractions of the integers (Z); • The Gaussian rational numbers (Q [i]) is the eld of fractions of the integers (Z [i]) Theorem. Every integral domain can be embedded in a eld.
Proof. Let R be an integral domain. That is, a commutative ring with unity in which the zero-product rule holds. Now consider the set F¯ = R × R∗, the set of all ordered pairs of elements in R, excluding those in which the second element is 0. We will now de ne a binary relation ∼= on F¯, which we claim to be an equivalence relation by the following criteria: (p, q) ∼= (ˆp, qˆ), if and only if p · qˆ =p ˆ · q, for all (p, q) , (ˆp, qˆ) ∈ F¯. To show that ∼= is an equivalence relation, we must show that the re exive, symmetric, and transitive properties hold. To that end, let (p, q) , (ˆp, qˆ) , (˜p, q˜) ∈ F¯. Re exive Since p · q = p · q, (p, q) ∼= (p, q). Symmetric Suppose that (p, q) ∼= (ˆp, qˆ). Then p · qˆ =p ˆ· q, and so by the symmetric property of equality, pˆ · q = p · qˆ. Thus, (ˆp, qˆ) ∼= (p, q). Transitive Now suppose that (p, q) ∼= (ˆp, qˆ) and (ˆp, qˆ) ∼= (˜p, q˜). Then p · qˆ =p ˆ · q and pˆ · q˜ =p ˜ · qˆ. Multiplying these equations together, we obtain pqˆpˆq˜ =pq ˆ p˜qˆ. Thus, pqˆpˆq˜ − pqˆ p˜qˆ = 0, and so pˆqˆ(pq˜ − qp˜) = 0. Since qˆ 6= 0, we must have pˆ(pq˜ − qp˜) = 0, as R has the zero-product rule. In the case that pˆ 6= 0, we will have pq˜−qp˜ = 0, which implies pq˜ =pq ˜ and that (p, q) ∼= (˜p, q˜). Otherwise, if pˆ = 0, then p · qˆ = 0 · q = 0, which then implies that p = 0. Similarly, we will also nd that p˜ = 0. This is the special case that pq˜ = 0 =pq ˜ , and so (p, q) ∼= (˜p, q˜). Thus, ∼= is an equivalence relation, and so we will now de ne F = F/¯ ∼=, the set of all equivalence classes, and use the notation p to denote the element , the equivalence class q [p, q] containing (p, q). We now must determine addition and multiplication operations on F and show that F is a eld. We claim that addition and multiplication can be given by: 1 2 FIELD OF FRACTIONS
p r ps+qr , for all p r ; • q + s = qs q , s ∈ F p r pr , for all p r ; • q · s = qs q , s ∈ F Since p and r may not be unique representations of elements in , it is necessary to show q s F that the above rules for addition and multiplication provide for well-de ned operations. To that end, let p , pˆ, r , and rˆ be elements of , and suppose that p pˆ and r rˆ. We q qˆ s sˆ F q = qˆ s = sˆ will now show that ps+qr pˆsˆ+ˆqrˆ and pr pˆrˆ. qs = qˆsˆ qs = qˆsˆ Since p pˆ and r rˆ, we have and . Then: q = qˆ s = sˆ pqˆ =pq ˆ rsˆ =rs ˆ (ps + qr)q ˆsˆ = psqˆsˆ + qrqˆsˆ = pqsˆ sˆ + qqrˆ sˆ =pqs ˆ sˆ + qqˆrsˆ = (ˆpsˆ +q ˆrˆ) qs
and
prqˆsˆ = pqrˆ sˆ =pq ˆ rsˆ =p ˆrqsˆ
Therfore, ps+qr pˆsˆ+ˆqrˆ and pr pˆrˆ. qs = qˆsˆ qs = qˆsˆ Now we must show that is a eld. Take p and r as before, and 0 and 1 as the additive F q s and multiplicative identities in . Also take u . It should be noted that for any ∗, R v ∈ F n ∈ R since , pn p . (pn) q = p (qn) qn = q Additive identity We claim that 0 is the additive identity in . Observe that p 0 p·1+q·0 p . 1 F q + 1 = q·1 = q Additive inverses We claim that −p is the additive inverse of p . Observe that p −p pq+(−pq) 0 0q2 0 . q q q + q = qq = q2 = 1q2 = 1 Commutativity of addition Observe that p r ps+qr rq+sp r p . q + s = qs = sq = s + q Associativity of addition Observe that p r u p rv+su p(sv)+q(rv+su) psv+qrv+qsu (ps+qr)v+(qs)u q + s + v = q + sv = q(sv) = qsv = (qs)v = ps+qr u p r u . qs + v = q + s + v Multiplicative identity We claim that 1 is the multiplicative identity in . Observe that p 1 p·1 p . 1 F q · 1 = q·1 = q Multiplicative inverses FIELD OF FRACTIONS 3
On the supposition that p 0 , we nd that . We then claim that q is the multiplicative q 6= 1 p 6= 0 p inverse of p . Observe that p q pq 1pq 1 . q q · p = qp = 1pq = 1 Commutativity of multiplication Observe that p r pr rp r p . q · s = qs = sq = s · q Associativity of multiplication Observe that p r u p ru p(ru) (pr)u pr u p r u . q · s · v = q · sv = q(sv) = (qs)v = qs · v = q · s · v Distributive property Observe that p r u p rv+su prv+psu (prv+psu)q (pr)(qv)+(qs)(pu) pr pu q · s + v = q · sv = q(sv) = q(sv)q = (qs)(qv) = qs + qv = p r p u . q · s + q · v Now we are left to identify a subset of F that is ring-isomorphic to R. We claim that the subset is ˆ r and the isomorphism is mapping ˆ, de ned by r , R = 1 : r ∈ R φ : R → R φ (r) = 1 or all r ∈ R. One-to-one Let and suppose that . Then r s , and so . Thus . r, s ∈ R φ (r) = φ (s) 1 = 1 r · 1 = 1 · s r = s Onto Let r ˆ. Then r . 1 ∈ R φ (r) = 1 Preservation of structure Let . Then r+s r·1+1·s r s , and rs r·s r, s ∈ R φ (r + s) = 1 = 1·1 = 1 + 1 = φ (r) + φ (s) φ (rs) = 1 = 1·1 = r s . 1 · 1 = φ (r) φ (s)