Abstract Algebra Theory and Applications - Computer Science ...

282 CHAPTER 16 INTEGRAL DOMAINSLet R be a commutative ring with identity, and let a and b be elementsin R. We say that a divides b, and write a | b, if there exists an elementc ∈ R such that b = ac. A unit in R is an element that has a multiplicativeinverse. Two elements a and b in R are said to be associates if there existsa unit u in R such that a = ub.Let D be an integral domain. A nonzero element p ∈ D that is not aunit is said to be irreducible provided that whenever p = ab, either a or bis a unit. Furthermore, p is prime if whenever p | ab either p | a or p | b.Example 2. It is important to notice that prime and irreducible elementsdo not always coincide. Let R be the subring of Q[x, y] generated by x 2 ,y 2 , and xy. Each of these elements is irreducible in R; however, xy is notprime, since xy divides x 2 y 2 but does not divide either x 2 or y 2 . The Fundamental Theorem of Arithmetic states that every positive integern > 1 can be factored into a product of prime numbers p 1 · · · p k , wherethe p i ’s are not necessarily distinct. We also know that such factorizationsare unique up to the order of the p i ’s. We can easily extend this resultto the integers. The question arises of whether or not such factorizationsare possible in other rings. Generalizing this definition, we say an integraldomain D is a unique factorization domain, or UFD, if D satisfies thefollowing criteria.1. Let a ∈ D such that a ≠ 0 and a is not a unit. Then a can be writtenas the product of irreducible elements in D.2. Let a = p 1 · · · p r = q 1 · · · q s , where the p i ’s and the q i ’s are irreducible.Then r = s and there is a π ∈ S k such that p i = q π(j)for j = 1, . . . , r = s.Example 3. The integers are a unique factorization domain by the FundamentalTheorem of Arithmetic.Example 4. Not every integral domain is a unique factorization domain.The subring Z[ √ 3 i] = {a + b √ 3 i} of the complex numbers is an integraldomain (Exercise 12, Chapter 14). Let z = a + b √ 3 i and defineν : Z[ √ 3 i] → N ∪ {0} by ν(z) = |z| 2 = a 2 + 3b 2 . It is clear that ν(z) ≥ 0with equality when z = 0. Also, from our knowledge of complex numberswe know that ν(zw) = ν(z)ν(w). It is easy to show that if ν(z) = 1, then zis a unit, and that the only units of Z[ √ 3 i] are 1 and −1.We claim that 4 has two distinct factorizations into irreducible elements:4 = 2 · 2 = (1 − √ 3 i)(1 + √ 3 i).