# Section 19 Integral domains

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1 Section 19 Integral domains Instructor: Yifan Yang Spring 2007

2 Observation and motivation There are rings in which ab = 0 implies a = 0 or b = 0 For examples, Z, Q, R, C, and Z[x] are all such rings There are also ring in which there exist some a, b such that a, b 0, but ab = 0 For( example, ) ( in Z) 6 we( have 2 ) 3 = Also, in M 2 (R) we have = In the first cases, an equation of the form (x a)(x b) = 0 has exactly two solutions a and b since (x a)(x b) = 0 implies x a = 0 or x b = 0 In the second cases, an equation (x a)(x b) = 0 may have more than two solutions For example, 2, 3, 6, 11 are all solutions of (x 2)(x 3) = 0 in Z 12 This shows that the these two classes of rings are fundamentally different

3 Divisors of zero Definition If a and b are two nonzero elements of a ring R such that ab = 0, then a and b are divisors of zero (or zero divisors) Example 1 2, 3, 4, are all zero divisors in Z 6 ( ) ( ) The matrices and are zero divisors in M 2 (R)

4 Zero divisors of Z n Theorem (193) The zero divisors of Z n are precisely the nonzero elements that are not relatively prime to n Corollary (194) If p is a prime, then Z p has no zero divisors

5 Proof of Theorem 193 Case gcd(m, n) = d > 1 We have and m is a zero divisor ( n ) ( m ) m = n = 0, d d Case gcd(m, n) = 1 Assume that mk = 0 in Z n, ie, n mk Since m is relatively prime to n, we have n k, ie, k = 0 in Z n We see that m is not a zero divisor

6 Cancellation law Theorem (195) The cancellation law (ie, ab = ac, a 0 b = c) holds for a ring R if and only if R has no zero divisors Proof Assume that the cancellation law holds, but ab = 0 for some a, b 0 Then we have ab = 0 = a0, but b 0, which is a contradiction Assume that R has no zero divisors If ab = ac and a 0, then ab ac = 0, which, by the distributive law, gives a(b c) = 0 Since R has no zero divisors and a is assumed to be nonzero, we have b c = 0 and thus b = c

7 Remarks Let R be a ring with zero divisors Even if ab = ac and a 0 do not imply b = c for general a, b, c R, the cancellation law still holds for the cases when a has a multiplicative inverse For example, in Z 15, 2a = 2b still implies a = b since 2 is relatively prime to 15 Also, if A is an invertible matrix in M 2 (R), then AB = AC still implies B = C If R is a ring with no zero divisors, then the equation ax = b has at most one solution in R

8 Notation b/a Suppose that R is a commutative ring with no zero divisors, and that a is a unit in R Then the equation ax = b has exactly one solution a 1 b For convenience, we let b/a denote this element a 1 b The notation 1/a will denote a 1 in this case However, when R is not commutative, we do not use this notation because we do not know whether b/a means a 1 b or ba 1

9 Integral domains Definition A commutative ring R with unity 1 0 that has no zero divisors is an integral domain Example 1 The ring of integers Z is an integral domain In fact, this is why we call such rings integral domains 2 If p is a prime, then Z p is an integral domain On the other hand, if n is composite, then Z n is not an integral domain 3 The direct product R S of two nonzero rings R and S is never an integral domain since (r, 0)(0, s) = (0, 0) for all r R and s S

10 Fields are integral domains Theorem (199) Every field F is an integral domain Proof Suppose that a, b F is such that ab = 0 We need to show that if a 0, then b = 0 By the associativity of multiplication, we have 0 = a 1 (ab) = (a 1 a)b = 1b = b This proves the theorem

11 Finite integral domains are fields Theorem (1911) Every finite integral domain D is a field Corollary (1912) If p is a prime, then Z p is a field

12 Proof of Theorem 1911 We need to show that every nonzero element a of D has a multiplicative inverse Let 0, 1, a 1,, a n be all the elements of the finite integral domain D Consider the products 0 = a0, a = a1, aa 1,, aa n These products are all distinct since ab = ac implies b = c by Theorem 195 Thus, 0, a, aa 1,, aa n must be all the elements of D One of these must be 1, ie, aa i = 1 for some a i This proves the theorem

13 Relations between various rings rings commutative rings 2Z rings with unity M n (R) Z n integral domains Z Q[x] fields Q, Z p

14 In-class exercises 1 Find all solutions of x 3 2x 2 3x = 0 in Z 12 2 Characterize all the zero divisors of M 2 (R) 3 Let F be the set of all continuous functions f : R R with addition and multiplication given by f + g : x f (x) + g(x), f g : x f (x)g(x) Find the proper place for F in the diagram on the last page

15 The characteristic of a ring Definition Let R be a ring Suppose that there is a positive integer n such that n a = 0 for all a R The least such positive integer is the characteristic of the ring R If no such positive integer exists, then R is of characteristic 0 Example 1 The rings Z n are of characteristic n 2 The rings Z, Q, R, and C are all of characteristic 0

16 The characteristic of a ring Theorem (1915) Let R be a ring with unity If n 1 0 for all n Z +, then R has characteristic 0 If n 1 = 0 for some n Z +, then the smallest such integer n is the characteristic of R Proof If n 1 0 for all n Z +, then R is clearly of characteristic 0, by definition If n 1 = 0 for some n Z +, then for all a R, we have n a = (a + + a) = a( ) = a(n 1) = a0 = 0 Then the theorem follows

17 Homework Problems 2, 10, 12, 23, 27, 28, 29, 30 of Section 19

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### 8 Appendix: Polynomial Rings

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### Prime Rational Functions and Integral Polynomials. Jesse Larone, Bachelor of Science. Mathematics and Statistics

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