1.4 Integers Modulo n

Transcription

1 1.4 J.A.Beachy Integers Modulo n from A Study Guide f Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 31. Find the multiplicative inverse of each nonzero element of Z 7. Solution: Since 6 1 (mod 7), the class [6] 7 is its own inverse. Furtherme, 2 4 = 8 1 (mod 7), and 3 5 = 15 1 (mod 7), so [2] 7 and [4] 7 are inverses of each other, and [3] 7 and [5] 7 are inverses of each other. 32. Find the multiplicative inverse of each nonzero element of Z 13. Comment: If ab 1 (mod n), then [a] n and [b] n are inverses, as are [ a] n and [ b] n. If ab 1 (mod n), then [a] n and [ b] n are inverses, as are [ a] n and [b] n. It is useful to list the integers with m with m ±1 (mod n), and look at the various ways to fact them. Solution: Note that 14, 27, and 40 are congruent to 1, while 12, 25, and 39 are congruent to 1. Facting 14, we see that [2] 13 and [7] 13 are inverses. Facting 12, we see that [3] 13 and [ 4] 13 are inverses, as are the pairs [4] 13 and [ 3] 13, and [6] 13 and [ 2] 13. Facting 40, we see that [5] 13 and [8] 13 are inverses. Here is the list of inverses: [2] 1 13 = [7] 13; [3] 1 13 = [9] 13; [4] 1 13 = [10] 13; [5] 1 13 = [8] 13; [6] 1 13 = [11] 13. Since [12] 1 13 = [ 1] 1 13 = [ 1] 13 = [12] 13, this answers the question f all of the nonzero elements of Z Find the multiplicative der of each element of Z 7. Solution: It helps to use Exercise in the text, which shows that if gcd(a,n) = 1, then the multiplicative der of [a] n is a divis of ϕ(n). In this problem, it follows that the possibilities f the multiplicative der of an element are the diviss of 6: 1, 2, 3 and 6. We have 2 2 = 4 (mod 7), and 2 3 = 8 1 (mod 7), so [2] 7 has multiplicative der 3. We have (mod 7), (mod 7), and 3 6 (3 3 ) 2 ( 1) 2 1 (mod 7), so [3] 7 has multiplicative der 6. Actually, after seeing that [3] 7 does not have multiplicative der 2 3, we can conclude that it has multiplicative der 6, since that is the only possibility left. We have (mod 7), (mod 7), so [4] 7 has multiplicative der 3. We have (mod 7), (mod 7), so [5] 7 has multiplicative der 6. We could have done the calculations with 2, since 5 2 (mod 7). This would have allowed us to use the calculations we had already made in the case of 2. Finally, it is clear that [6] 7 has multiplicative der 2, since [6] 7 = [ 1] Find the multiplicative der of each element of Z 9. Solution: We have ϕ(9) = 6, so [a] 6 9 = [1] 9 f all [a] 9 Z 9, and the possible multiplicative ders are 1, 2, 3 and 6. The elements are [1] 9,[2] 9,[4] 9,[5] 9,[7] 9,[8] 9.

2 1.4 J.A.Beachy 2 We have 2 2 = 4, 2 3 = 8, 2 4 = (mod 9), (mod 9), and (mod 9). This shows that [2] 9 has multiplicative der 6, and also shows that Z 9 is cyclic. We can use this infmation to shten our calculations f the other elements, by expressing them as the various powers of [2] 9. Since [4] 9 = [2] 2 9, it is easy to see that [4]3 9 = [1] 9, while [4] 2 9 [1] 9, so [4] 9 has multiplicative der 3. The element [8] 9 = [2] 3 9 must have der 2. This also obviously follows from the fact that [8] 9 = [ 1] 9. We have [7] 9 = [2] 4 9, so [7]3 9 = ([2]4 9 )3 = ([2] 6 9 )2 = [1] 9, and it then follows that [7] 9 has multiplicative der 3. The smallest positive power of [5] 9 = [2] 5 9 that will give us an exponent divisible by 6 is 30, so [5] 6 9 = ([2]5 9 )6 = ([2] 6 9 )5 = [1] 5 9 = [1] 9 and no smaller exponent gives [1] 9. We conclude that [5] 9 has multiplicative der Find [91] 1 501, if possible (in Z 501 ). Solution: We need to use the Euclidean algithm. [ ] [ ] [ ] [ Thus ( 11) = 1, so 91( 11) 1 (mod 501). Answer: [91] = [ 11] 501 = [490] Find [3379] , if possible (in Z 4061 ). Solution: The inverse does not exist. [ ] [ ] [ At the next step, , and so gcd(4061,3379) = ] [ ] In Z 20 : find all units (list the multiplicative inverse of each); find all idempotent elements; find all nilpotent elements. Comment: We know that Z n has ϕ(n) units. They occur in pairs, since gcd(a,n) = 1 if and only if gcd(n a,n) = 1. This helps in checking your list. Solution: TheunitsofZ 20 aretheequivalenceclasses representedby1, 3, 7, 9, 11, 13, 17, and 19. We have [3] 1 20 = [7] 20, [9] 1 20 = [9] 20, and [11] 1 20 = [11] 20 by trial and err, and then [13] 1 20 = [ 7] 1 20 = [ 3] 20 = [17] 20, and [19] 1 20 = [ 1] 1 20 = [ 1] 20 = [19] 20. The idempotent elements of Z 20 can be found by using trial and err. They are [0] 20, [1] 20, [5] 20, and [16] 20. If you want a me systematic approach, you can use the hint in Exercise of the text: if n = bc, with gcd(b,c) = 1, then any solution to the congruences x 1 (mod b) and x 0 (mod c) will be idempotent modulo n. The nilpotent elements of Z 20 can be found by using trial and err, by using the result stated in Problem They are [0] 20 and [10] 20. ]

3 1.4 J.A.Beachy Show that Z 17 is cyclic. Comment: Toshow thatz 17 iscyclic, weneedtofindanelement whosemultiplicative der is 16. The solution just uses trial and err. It is known than if p is prime, then Z p is cyclic, but there is no known algithm f actually finding the one element whose powers cover all of Z p. Solution: We begin by trying [2]. We have [2] 2 = [4], [2] 3 = [8], and [2] 4 = [16] = [ 1]. Exercise of the text shows that the multiplicative der of an element has to be a divis of ϕ(17) = 16, so the next possibility to check is 8. Since [2] 8 = [ 1] 2 = [1], it follows that [2] has multiplicative der 8. We next try [3]. We have [3] 2 = [9], [3] 4 = [81] = [ 4], and [3] 8 = [16] = [ 1]. The only divis of 16 that is left to try is 16 itself, so [3] does in fact have multiplicative der 16, and we are done. 39. Show that Z 35 is not cyclic but that each element has the fm [8]i 35 [ 4]j 35, f some positive integers i, j. Solution: We first compute the powers of [8]: [8] 2 = [ 6], [8] 3 = [8][ 6] = [ 13], and [8] 4 = [ 6] 2 = [1], so the multiplicative der of [8] is 4, and the powers we have listed represent the only possible values of [8] i. We next compute the powers of [ 4]: [ 4] 2 = [16], [ 4] 3 = [ 4][16] = [6], [ 4] 4 = [ 4][6] = [11], [ 4] 5 = [ 4][11] = [ 9], and [ 4] 6 = [ 4][ 9] = [1], so the multiplicative der of [ 4] is 6. There are 24 possible products of the fm [8] i [ 4] j, f 0 i < 4 and 0 j < 6. Are these all different? Suppose that [8] i [ 4] j = [8] m [ 4] n, f some 0 i < 4 and 0 j < 6 and 0 m < 4 and 0 n < 6. Then [8] i m = [ 4] n j, and since the only power of [8] that is equal to a power of [ 4] is [1] (as shown by our computations), this fces i = m and n = j. We conclude that since there are 24 different elements of the fm [8] i [ 4] j, every element in Z 35 must be of this fm. Finally, ([8] i [ 4] j ) 12 = ([8] 4 ) 3i ([ 4] 6 ) 2j = [1], so no element of Z 35 has multiplicative der 24, showing that Z 35 is not cyclic. 40. Solve the equation [x] [x] 11 [6] 11 = [0] 11. Solution: We can fact [x] 2 +[x] [6] = ([x]+[3])([x] [2]). Collary implies that either [x] + [3] = [0] [x] [2] = [0], and so the only possible solutions are [x] = [ 3] and [x] = [2]. Substituting these values back into the equation shows that they are indeed solutions, so the answer is [x] = [2] [x] = [ 3]. 41. Prove that [a] n is a nilpotent element of Z n if and only if each prime divis of n is a divis of a. Solution: First assume that each prime divis of n is a divis of a. If n = p α 1 1 pα 2 2 pαt t is the prime factization of n, then we must have a = p β 1 1 pβ 2 2 pβt t d, where 0 β j α j f all j. If k is the smallest positive integer such that kβ i α i f all i, then n a k, and so [a] k n = [0] k.

4 1.4 J.A.Beachy 4 Conversely, if [a] n is nilpotent, with [a] k n = [0], then n a k, so each prime divis of n is a divis of a k. But if a prime p is a divis of a k, then it must be a divis of a, and this completes the proof. 42. Show that if n > 1 is an odd integer, then ϕ(2n) = ϕ(n). Solution: Since n is odd, the prime 2 does not occur in its prime factization. The fmula in Proposition shows that to compute ϕ(2n) in terms of ϕ(n) we need to add 2 (1 1 2 ), and this does not change the computation. Alternate solution: Since n is odd, the integers n and 2 are relatively prime. Exercise of the text states that if m,n are relatively prime positive integers, then ϕ(mn) = ϕ(m)ϕ(n). It follows that ϕ(2n) = ϕ(2)ϕ(n) = ϕ(n). ANSWERS AND HINTS 43. Write out multiplication tables f the following sets. (a) Z 9 Answer: This set is cyclic, and each element can be expressed as a power of 2. Writing out a table in the der [1],[2],[4],[8],[7],[5] of consecutive powers of 2 means that each row is just a shift of the one above it. Other ders are interesting too (b) Z 10 Answer: The first table uses the fact that Z 10 is cyclic, and lists the elements as consecutive powers of 3. The second table shows a different pattern compare this to the pattern in part (c), which differs in a crucial way along the main diagonal. (c) Z 12 Answer: (d) Z 14 Hint: Problem 47 shows that each element of Z 14 is a power of 3, so a quick way to write out the table is to use the der [1],[3],[9],[13],[11],[5]. Then the table will have

5 1.4 J.A.Beachy 5 the same pattern as the one in part (a). 44. Find the multiplicative inverses of the given elements (if possible). (a) [12] in Z 15 Answer: There is no inverse, since 12 is not relatively prime to 15. (c) [7] in Z 15 Answer: [7] 1 = [13] 45. Find the multiplicative ders of the following elements. (b) [5] and [7] in Z 17 Answer: [5] and [7] both have multiplicative der Find the multiplicative der of each element of the following sets. (b) Z 10 Answer: [3] and [7] = [ 3] have der 4; [9] = [ 1] has der 2; [1] has der Is Z 14 cyclic? Answer: Yes; [3] has multiplicative der 6 = ϕ(14). 48. Is Z 16 cyclic? Answer: No Hint: Check that each element has multiplicative der less than Is Z 18 cyclic? Answer: Yes; [5] has multiplicative der 6 = ϕ(18). 50. Find all idempotent elements in the following sets. (a) Z 14 Hint: Exercise in the text states that if n = bc, with gcd(b,c) = 1, then any solution to the congruences x 1 (mod b) and x 0 (mod c) will be idempotent modulo n. First take b = 7, c = 2 and then b = 2, c = In Z 24 : find all units (list the multiplicative inverse of each); find all idempotent elements; find all nilpotent elements. Answer: The units of Z 24 are the equivalence classes represented by 1, 5, 7, 11, 13, 17, 19, and 23, and each element is its own inverse. The idempotent elements are [0] 24,[1] 24,[9] 24,[16] 24. The nilpotent elements are [0] 24,[6] 24,[12] 24,[18] Find {n Z + ϕ(n) = 2} and {n Z + ϕ(n) = 4}. Answer: {n Z + ϕ(n) = 2} = {3,4,6} and {n Z + ϕ(n) = 4} = {5,8,10,12}