FIELD OF FRACTIONS 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 dene a binary relation ∼= on F¯, which we claim to be an equivalence relation by the following criteria: (p; q) ∼= (^p; q^), if and only if p · q^ =p ^ · q, for all (p; q) ; (^p; q^) 2 F¯. To show that ∼= is an equivalence relation, we must show that the reexive, symmetric, and transitive properties hold. To that end, let (p; q) ; (^p; q^) ; (~p; q~) 2 F¯. Reexive Since p · q = p · q, (p; q) ∼= (p; q). Symmetric Suppose that (p; q) ∼= (^p; q^). Then p · q^ =p ^· q, and so by the symmetric property of equality, p^ · q = p · q^. Thus, (^p; q^) ∼= (p; q). Transitive Now suppose that (p; q) ∼= (^p; q^) and (^p; q^) ∼= (~p; q~). Then p · q^ =p ^ · q and p^ · q~ =p ~ · q^. Multiplying these equations together, we obtain pq^p^q~ =pq ^ p~q^. Thus, pq^p^q~ − pq^ p~q^ = 0, and so p^q^(pq~ − qp~) = 0. Since q^ 6= 0, we must have p^(pq~ − qp~) = 0, as R has the zero-product rule. In the case that p^ 6= 0, we will have pq~−qp~ = 0, which implies pq~ =pq ~ and that (p; q) ∼= (~p; q~). Otherwise, if p^ = 0, then p · q^ = 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 dene 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 2 F p r pr , for all p r ; • q · s = qs q ; s 2 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-dened operations. To that end, let p , p^, r , and r^ be elements of , and suppose that p p^ and r r^. We q q^ s s^ F q = q^ s = s^ will now show that ps+qr p^s^+^qr^ and pr p^r^. qs = q^s^ qs = q^s^ Since p p^ and r r^, we have and . Then: q = q^ s = s^ pq^ =pq ^ rs^ =rs ^ (ps + qr)q ^s^ = psq^s^ + qrq^s^ = pqs^ s^ + qqr^ s^ =pqs ^ s^ + qq^rs^ = (^ps^ +q ^r^) qs and prq^s^ = pqr^ s^ =pq ^ rs^ =p ^rqs^ Therfore, ps+qr p^s^+^qr^ and pr p^r^. qs = q^s^ qs = q^s^ 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 2 F n 2 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 ^ r and the isomorphism is mapping ^, dened by r , R = 1 : r 2 R φ : R ! R φ (r) = 1 or all r 2 R. One-to-one Let and suppose that . Then r s , and so . Thus . r; s 2 R φ (r) = φ (s) 1 = 1 r · 1 = 1 · s r = s Onto Let r ^. Then r . 1 2 R φ (r) = 1 Preservation of structure Let . Then r+s r·1+1·s r s , and rs r·s r; s 2 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) .