Integral Domains 1. Localization The intent of this section is to make formal the process of forming the rationals Q from the integers Z. This process, known as localization, is applicable to a large class of rings. We will define Q(x), the ring of rational functions in one variable (over Q), using this process. Another goal here is to extend notions of divisibility and factorization in Z to integral domains. One can associate any rational number p=q 2 Q with the ordered pair (p; q) 2 Z×Znf0g. This is not a bijection of sets, however, because 1=2 = 2=4 but (1; 2) 6= (2; 4). Thus, we place an equivalence relation on the latter set. Recall that a=b = c=d if and only if ad = bc. Hence, we say (a; b) ∼ (c; d)) if ad = bc. Lemma 1. Let D be any integral domain and define the set S = f(a; b): a; b 2 D and b 6= 0g: The relation ∼ on S given by (a; b) ∼ (c; d) if ad = bc is an equivalence relation. Proof. Let (a; b) 2 S. Then ab = ba because D is an integral domain (and hence commutative). Thus, (a; b) ∼ (a; b) and so ∼ is reflexive. Let (a; b); (c; d) 2 S such that (a; b) ∼ (c; d). Then ad = bc and so cb = da because D is an integral domain. Thus, (c; d) ∼ (a; b) and so ∼ is symmetric. Finally, suppose (a; b) ∼ (c; d) and (c; d) ∼ (e; f) in S. Then ad = bc and cf = de. Since D is an integral domain without zero divisors, then ade = bce so acf = bce. Thus, af = be so (a; b) ∼ (e; f) and ∼ is transitive. Let [a; b] denote the equivalence class of (a; b) 2 S under ∼. The set of such equivalence classes is denoted FD. We will prove that FD is in fact a field, justifying the name field of fractions of D. The operations on FD are defined to mimic the operations of adding and multiplying fractions. For [a; b]; [c; d] 2 FD, define [a; b] + [c; d] = [ad + bc; bd] and [a; b] · [c; d] = [ac; bd]: Lemma 2. The operations of addition and multiplication above are binary operations on FD. These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapter 18. Hungerford's Algebra was also consulted. Last Updated: December 6, 2019 1 Proof. That FD is closed under these operations is clear. We need only show that they are well- defined (independent of equivalence class). Let [a1; b1]; [a2; b2]; [c1; d1]; [c2; d2] 2 FD with [a1; b1] = [a2; b2] and [c1; d1] = [c2; d2]. Thus, a1b2 = b1a2 and c1d2 = d1c2. We claim [a1; b1] + [c1; d1] = [a2; b2] + [c2; d2]. Equivalently, [a1d1 + b1c1; b1d1] = [a2d2 + b2c2; b2d2] or (a1d1 + b1c1)(b2d2) = (b1d1)(a2d2 + b2c2): We have (a1d1 + b1c1)(b2d2) = (a1d1)(b2d2) + (b1c1)(b2d2) = (a1b2)(d1d2) + (b1b2)(c1d2) = (b1a2)(d1d2) + (b1b2)(d1c2) = (b1d1)(a2d2 + b2c2): Checking the operation of multiplication is left as an (easier) exercise. Lemma 3. The set FD with the above binary operations is a field. Proof. We claim the additive identity is [0; 1]. Let [a; b] 2 FD, then [a; b] + [0; 1] = [a · 1 + b · 0; b · 1] = [a; b]: Associativity is left as an exercise. We claim the additive inverse of [a; b] is [−a; b] (note −a 2 D because D is a ring). One checks that [a; b] + [−a; b] = [ab + b(−a); b2] = [0; b2] = [0; 1]. (Note that 2 this last equality follows because 0 · 1 = b · 0.) Thus, (FD; +) is an abelian group. Associativity and commutivity of multiplication are left as an exercise. We check the left distributive property. Let [a; b]; [c; d]; [e; f] 2 FD. Then [a; b][c; d] + [a; b][e; f] = [ac; bd] + [ae; bf] = [acbf + bdae; bdbf] = [acf + dae; bdf] = [a; b][cf + de; df] = [a; b]([c; d] + [e; f]): By commutivity, the right distributive property also holds. Hence, FD is a commutative ring. The multiplicative identity in FD is [1; 1]. This is an easy check: [a; b][1; 1] = [a; b]. Moreover, the multiplicative inverse of [a; b] is [b; a] as [a; b][b; a] = [ab; ba] = [1; 1]. Thus, FD a field. Definition 1. The field FD defined above is the field of fractions of D. We next aim to show that, in some sense, FD is the smallest field containing D. Lemma 4. Let D be an integral domain and FD its field of fractions. There is an injective homomorphism φ : D ! FD defined by φ(a) = [a; 1]. 2 Proof. Let a; b 2 D, then φ(a) + φ(b) = [a; 1] + [b; 1] = [a + b; 1] = φ(a + b) φ(a)φ(b) = [a; 1][b; 1] = [ab; 1] = φ(ab): Thus, φ is a ring homomorphism. Now suppose φ(a) = 0, then [a; 1] = [0; 1], so a = 0. Thus, ker φ = f0g and φ is injective. As a consequence of this lemma, D is isomorphic to a subring of FD. The next theorem is an example of a universal property. Theorem 5. Let D be an integral domain and FD its field of fractions. If E is any field containing D, then there exists a unique injective homomorphism : FD ! E such that (φ(a)) = a for all a 2 D. −1 Proof. Let E be any field containing D. Define a map : Fd ! E by [a; b] 7! ab . We must show that is well-defined. Suppose [a1; b1]; [a2; b2] 2 FD with [a1; b1] = [a2; b2]. Then a1b2 = b1a2. −1 −1 −1 −1 Thus, in E, a1b1 = a2b2 and so (a1b1 ) = (a2b2 ). Now we claim is an injective homomorphism. Let [a; b]; [c; d] 2 FD. Then ([a; b] + [c; d]) = ([ad + bc; bd]) = (ad + bc)(bd)−1 = ab−1 + cd−1 = ([a; b]) + ([c; d]) ([a; b][c; d]) = ([ac; bd]) = (ac)(bd)−1 = (ab−1)(cd−1) = ([a; b]) ([c; d]): For injectivity, suppose ([a; b]) = 0. Then ab−1 = 0. Multiplying both sides by b gives a = 0. Thus, [a; b] = [0; 1], the additive identity in FD, so is injective. Moreover, −1 (φ(a)) = ([a; 1]) = a1 = a: : The previous theorem can be visualized using the following commutative diagram where φ and are as in the theorem and ι is the inclusion map: ι D / E φ 9! FD Remark. It is possible to localize in noncommutative rings but there are several technical hurdles. Example. The field of fractions of Q[x], denoted Q(x), is the set of rational expressions p(x)=q(x) for polynomials p(x); q(x) 2 Q[x] with q(x) 6= 0. 3 2. Factorization Definition 2. Let R be a commutative ring with identity and let a; b 2 R. We say a divides b, denoted a j b, if there exists c 2 R such that b = ac. If there exists a unit u 2 R such that b = au, then a and b are said to be associates. Example. (1) The elements 2 and −2 are associates in Z. In fact, since the only units in Z are ±1, then a; b 2 Z are associates if and only if a = ±b. (2) The elements 2 and 1 are associates in Q since 1 = 2 · (1=2). In fact, if p; q 2 Q are nonzero, then p = q · (p=q) and so any two nonzero elements in Q are associates. Definition 3. Let D be an integral domain. A nonzero element p 2 D that is not a unit is irreducible provided that whenever p = ab, either a or b is a unit. Furthermore, p is prime whenever p j ab either p j a or p j b. 2 2 Example. Let R be the subring of Q[x; y] generated by x ; y , and xy. Then each element is irreducible in R, however xy is not prime because xy divides x2y2 but not x2 or y2. Definition 4. An integral domain D is a unique factorization domain (UFD) if any nonzero, nonunit element a 2 D can be written as a = p1 ··· pk for irreducible elements pi 2 D and if a = q1 ··· q` for irreducibles qj 2 D then k = ` and, up to reordering pi and qi are associates for all i. Example. (1) The integers are a UFD by the Fundamental Theorem of Arithmetic. p (2) The subring [i 3] of is an integral domain (exercise) and the only units are ±1 (also an Z C p p p exercise). The element 4 2 [i 3] has two factorizations: 4 = 2 · 2 = (1 − i 3)(1 + i 3). One can p Z p show that 2; 1 ± i 3 are irreducible. Hence Z[i 3] is not a UFD. Lemma 6. Let D be an integral domain and let a; b 2 D. Then (1) a j b if and only if hbi ⊂ hai. (2) a and b are associates if and only if hbi = hai. (3) a is a unit in D if and only if hai = D. Proof. (1) We have a j b if and only if there exists x 2 D such that ax = b if and only if b 2 hai if and only if hbi ⊂ hai. (2) Suppose a and b are associates. Then there exists a unit y 2 D such that b = ay, thus a j b and so hbi ⊂ hai.