Abstract Algebra Theory and Applications - Computer Science ...

280 CHAPTER 16 INTEGRAL DOMAINSIt is easy to show that [1, 1] is the multiplicative identity. Let [a, b] ∈ F Dsuch that a ≠ 0. Then [b, a] is also in F D and [a, b] · [b, a] = [1, 1]; hence,[b, a] is the multiplicative inverse for [a, b]. Similarly, [−a, b] is the additiveinverse of [a, b]. We leave as exercises the verification of the associative andcommutative properties of multiplication in F D . We also leave it to thereader to show that F D is an abelian group under addition.It remains to show that the distributive property holds in F D ; however,and the lemma is proved.[a, b][e, f] + [c, d][e, f] = [ae, bf] + [ce, df]= [aedf + bfce, bdf 2 ]= [aed + bce, bdf]= [ade + bce, bdf]= ([a, b] + [c, d])[e, f]The field F D in Lemma 16.3 is called the field of fractions or field ofquotients of the integral domain D.Theorem 16.4 Let D be an integral domain. Then D can be embedded ina field of fractions F D , where any element in F D can be expressed as thequotient of two elements in D. Furthermore, the field of fractions F D isunique in the sense that if E is any field containing D, then there existsa map ψ : F D → E giving an isomorphism with a subfield of E such thatψ(a) = a for all elements a ∈ D.Proof. We will first demonstrate that D can be embedded in the field F D .Define a map φ : D → F D by φ(a) = [a, 1]. Then for a and b in D,andφ(a + b) = [a + b, 1] = [a, 1] + [b, 1] = φ(a) + φ(b)φ(ab) = [ab, 1] = [a, 1][b, 1] = φ(a)φ(b);hence, φ is a homomorphism. To show that φ is one-to-one, suppose thatφ(a) = φ(b). Then [a, 1] = [b, 1], or a = a1 = 1b = b. Finally, any elementof F D can expressed as the quotient of two elements in D, sinceφ(a)[φ(b)] −1 = [a, 1][b, 1] −1 = [a, 1] · [1, b] = [a, b].Now let E be a field containing D and define a map ψ : F D → E byψ([a, b]) = ab −1 . To show that ψ is well-defined, let [a 1 , b 1 ] = [a 2 , b 2 ]. Thena 1 b 2 = b 1 a 2 . Therefore, a 1 b −11 = a 2 b −12 and ψ([a 1 , b 1 ]) = ψ([a 2 , b 2 ]).□