MATH Algebra for High School Teachers Integral Domains

Transcription

1 MATH Algebra for High School Teachers Integral Domains Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 8

2 Cancellation We showed previously that if ab ac (mod m) and (a, m) = 1), then b c (mod m). In Z m, this translates to if [a][b] = [a][c] and (a, m) = 1, then [b] = [c]. Of course, in Z m, (a, m) = 1 implies [a] is a unit; e.g., if [2]x = [6] in Z 15, then [2] 1 = [8], and so [8][2]x = [8][6], hence x = [48] = [3]. But we also solved the equation 2x = 6 over the ring of integers, Z, without resorting to multiplicative inverses: If 2x = 6, then 2x 6 = 0, and so 2(x 3) = 0. For a, b Z, ab = 0 implies a = 0 or b = 0. Therefore, 2(x 3) = 0 implies x 3 = 0, and so x = 3. The key over Z is that 2 is not 0 and is not a zero divisor. The same method works over Z 15, because (2, 15) = 1 implies [2] is not [0] and is not a zero divisor. D.L. White (Kent State University) 2 / 8

3 Cancellation In a general ring, the key to cancelling a common factor is that the common factor must not be 0 or a zero divisor. It need not be a unit, but if it is a unit, then it is not 0 or a zero divisor. We do, however, need to be careful to allow for the possibility of non-commutativity in a general ring. We have the following theorem. Theorem Let R be a ring and let a, b, and c be elements of R, such that a 0 R and a is not a zero divisor. If a b = a c, then b = c. If b a = c a, then b = c. NOTE: If R is non-commutative and a b = c a, then even if a is a unit we cannot conclude b = c. For example, in M 2 (Z), [ ] [ ] [ ] [ ] [ ] =, but is a unit D.L. White (Kent State University) 3 / 8

4 Cancellation Proof of Theorem: We prove the first statement and leave the second as an exercise. Let a b = a c and assume a 0 R and a is not a zero divisor. Adding (a c) to both sides of the equation, we have a b a c = 0. By the left distributive law, we obtain a (b c) = 0. But a 0 R and a is not a zero divisor. Therefore, b c = 0; that is, adding c to both sides, b = c. D.L. White (Kent State University) 4 / 8

5 Definition We now generalize the very nice property of having no zero divisors enjoyed by the ring of integers Z (as well as fields). Definition An integral domain is a commutative ring with 1, with no zero divisors. Thus in an integral domain, if ab = 0 then a = 0 or b = 0. Equivalently, if a 0 and b 0, then ab 0. The next result is a corollary of the cancellation theorem and the definition. Corollary Let R be an integral domain and let a, b, c R such that a 0 R. If a b = a c, then b = c. Note that since an integral domain is commutative, the order of multiplication on each side does not matter. D.L. White (Kent State University) 5 / 8

6 Examples 1 The ring of integers, Z, is an integral domain. In fact, Z is the prototype for integral domains. All integral domains behave much like Z in essential ways. Even the name reflects the fact that they are integer-like rings. 2 Every field is an integral domain; e.g., Q, R, Z p with p prime. 3 Z m is an integral domain if and only if it is a field; i.e., if and only if m is prime. D.L. White (Kent State University) 6 / 8

7 Examples 4 Let R be a commutative ring with 1. The polynomial ring R[x] is an integral domain if and only if R is an integral domain. Proof: Since R is a commutative ring with 1, so is R[x]. Hence we need to prove that R[x] has no zero divisors if and only if R has no zero divisors. Suppose R[x] has no zero divisors. Since R is contained in R[x], R cannot have any zero divisors. Suppose R has no zero divisors. Let f (x) = a m x m + + a 1 x + a 0 and g(x) = b n x n + + b 1 x + b 0 be non-zero polynomials in R[x], with a m 0 and b n 0. The leading term of f (x) g(x) is a m b n x m+n. Since R has no zero divisors and a m 0, b n 0, we have a m b n 0, and so f (x) g(x) 0. Therefore, R[x] has no zero divisors. D.L. White (Kent State University) 7 / 8

8 Finite Integral Domains We have seen that if Z m is an integral domain, then Z m is actually a field. In general, we have: Theorem If R is a finite integral domain, then R is a field. Proof: If R is an integral domain, then R is a commutative ring with 1. We need to show that every non-zero element is a unit. Let 0 R u R. Since R is finite, we can write R = {r 1, r 2,..., r n }, where the r i are distinct elements of R, and r k = 1 R for some k. Consider the subset R = {u r 1, u r 2,..., u r n } of R. Suppose u r i = u r j for some i, j. Since R is an integral domain and u 0 R, this implies r i = r j, so i = j. Hence all of the u r i are distinct and R contains n distinct elements of R. Thus R = R, and so 1 R is in R ; that is, 1 R = u r i for some i. Therefore u is a unit. D.L. White (Kent State University) 8 / 8