《Discrete Mathematics II教学-华南理工》Section 16 Integral Domains.pdfVIP

IV.19 Integral Domains 1 Section IV.19. Integral Domains Note. In classical algebra, we solve polynomial equations by: setting equal to 0, factoring, and setting factors equal to 0. For example, x2 − 5x + 6 = 0 factors as (x −3)(x −2) = 0 which implies that x = 2 and x = 3 are (real) solutions. However, there are more solutions in different settings. Example 19.1 shows that in Z12 the solutions are 2, 3, 6, and 11. Notice that when x = 6 we have (x − 3)(x − 2) = (3)(4) = 12 ≡ 0 (mod 12). This example illustrates the unfamiliar result that the product of two “numbers” (elements of a group) can be “zero” (the additive identity) without one of the numbers being zero. This leads us to consider the following. Definition 19.2. If a and b are two nonzero elements of a ring R such that ab = 0 then a and b are divisors of 0. Example. In Z12, the divisors of 0 are 2, 3, 4, 6, 8, 9, and 10. Theorem 19.3. In the ring Zn , the divisors of 0 are precisely the nonzero elements that are not relatively prime to n. Corollary 19.4. If p is a prime, then Zp has no divisors of 0. IV.19 Integral Domains 2 Theorem 19.5. The left cancellation law states that “ab = ac with a = 0 implies b = c.” The right cancellation law states that “ba = ca with a = 0 implies b = c.” The cancellation laws hold in a ring R if and only if R has no divisors of 0. Note. If we are in a ring with no divisors of 0, then we can SOLVE THE ALGEBRAIC EQUATION (finally!!!) ax = b were a = 0 in at most one way. If R has unity 1 = 0 and a is a unit, then the solution is x = a−1b. Definition 19.6. An integral domain D is a commutative ring with unity 1 = 0 and containing no divisors of 0. Note.