Abstract Algebra Theory and Applications - Computer Science ...

278 CHAPTER 16 INTEGRAL DOMAINSelements in Q as ordered pairs in Z × Z. A quotient p/q can be writtenas (p, q). For instance, (3, 7) would represent the fraction 3/7. However,there are problems if we consider all possible pairs in Z × Z. There is nofraction 5/0 corresponding to the pair (5, 0). Also, the pairs (3, 6) and (2, 4)both represent the fraction 1/2. The first problem is easily solved if werequire the second coordinate to be nonzero. The second problem is solvedby considering two pairs (a, b) and (c, d) to be equivalent if ad = bc.If we use the approach of ordered pairs instead of fractions, then we canstudy integral domains in general. Let D be any integral domain and letS = {(a, b) : a, b ∈ D and b ≠ 0}.Define a relation on S by (a, b) ∼ (c, d) if ad = bc.Lemma 16.1 The relation ∼ between elements of S is an equivalence relation.Proof. Since D is commutative, ab = ba; hence, ∼ is reflexive on D.Now suppose that (a, b) ∼ (c, d). Then ad = bc or cb = da. Therefore,(c, d) ∼ (a, b) and the relation is symmetric. Finally, to show that therelation is transitive, let (a, b) ∼ (c, d) and (c, d) ∼ (e, f). In this casead = bc and cf = de. Multiplying both sides of ad = bc by f yieldsafd = adf = bcf = bde = bed.Since D is an integral domain, we can deduce that af = be or (a, b) ∼ (e, f).□We will denote the set of equivalence classes on S by F D . We now needto define the operations of addition and multiplication on F D . Recall howfractions are added and multiplied in Q:ab + c ad + bc= ;d bdab · cd = acbd .It seems reasonable to define the operations of addition and multiplicationon F D in a similar manner. If we denote the equivalence class of (a, b) ∈ S by[a, b], then we are led to define the operations of addition and multiplicationon F D by[a, b] + [c, d] = [ad + bc, bd]