Algebra. Sample Solutions for Test 1


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1 EPFL  Section de Mathématiques Algebra Fall semester Sample Solutions for Test 1 Question 1 (english, 30 points) 1) Let n Find the number of units of the ring Z/nZ. 2) Consider the ring Z/423Z. Is [209] 423 a unit in Z/423Z? Why or why not? If it is a unit, what is its multiplicative inverse? 3) For each of the following congruences, decide if it has a solution. For the congurences having a solution, find a solution. (a) 77x 7 (mod 49) (b) 21y 6 (mod 48) (c) 81z 7 (mod 30) 4) Find x, y Z (if they exist) satisfying the following equation : 193 x + 63 y 2. 1) The number of units in the ring Z/nZ is given by ϕ(n). So the number of units in the ring Z/( )Z is ϕ( ) ϕ(11) ϕ(13) ϕ(17) (since 11, 13 and 17 are relatively prime) (since ϕ(p) p 1 for a prime p) ) [209] 423 is a unit in Z/423Z if and only if 209 and 423 are relatively prime. We apply the Euclidean algorithm to find their GCD : The last nonzero remainder is 1, so we have gcd(423, 209) 1. So [209] 423 is a unit in Z/423Z. To find its inverse, we substitute backwards : ( ) ( ) So we see that the multiplicative inverse of [209] 423 is [ 85] 423 [338] 423.
2 3) (a) Since gcd(77, 49) 7 7 we know that the congruence has a solution. We can cancel 7 and obtain : x 1 (mod gcd(49, 7) ), 11 x 1 (mod 7). Now 11 and 7 are relatively prime, so 11 is invertible modulo 7. Using the Euclidean algorithm we obtain Substituting backwards, we get (7 1 4) (11 1 7) So the inverse of 11 modulo 7 is 2. Multiplying the above congruence by 2 we obtain : So a solution is given for instance by x 2. x 2 (mod 7). (b) Since gcd(21, 48) 3 6, the congurence will have a solution. Using the Euclidean Algorithm, we get Substituting backwards, we obtain Multiplying this equation with 2 we get ( ) and hence (mod 48). So y 14 is a solution to the congruence. (c) We have gcd(81, 30) 3 which does not divide 7. So this congruence does not have a solution. 4) Applying the Euclidean Algorithm to the numbers 193 and 63 we obtain
3 By backwards substitution we get So ( ) ( ) 1 63 Multiplying this equation by 2 we obtain So a solution is given by x 32, y Question 2 (english, 30 points) 1) Let (G, ) be a finite group and let a, b G. Denote the order of an element g G by o(g). (a) Show that o(b a b 1 ) o(a). (b) Show that o(a b) o(b a). 2) Let a, n be positive integers. Consider the ring Z/(a n 1)Z. (a) Show that [a] a n 1 is a unit in Z/(a n 1)Z. (b) Find the smallest positive integer r such that [a] r a n 1 [1] a n 1. (c) Using Lagrange s theorem, conclude that n ϕ(a n 1). 3) Find the last two digits of ) (a) Since the group is finite, every element must have finite order. Suppose that o(a) n. Then we have (b a b 1 ) n (b a b 1 ) (b a b 1 ) (b a b 1 ) }{{} n b a (b 1 b) a (b 1 b) a b 1 b a n b 1 b e b 1 e. So (b a b 1 ) n e and hence o(b a b 1 ) n. If we would have o(b a b 1 ) m < n, then we have (with the same argument using the associativitiy of as above) e (b a b 1 ) m b a m b 1. Multiplying this from left by b 1 and from right by b we obtain b 1 e b b 1 b a m b 1 b and hence a m e. But this is a contradiction, since o(a) n and m < n. So we have o(b a b 1 ) o(a) n. (b) Suppose that o(a b) n. We have (b a) n b a b a b a b (a b) (a b) (a b) a. Multiplying this from the right by b we obtain (b a) n b b (a b) n b e b (since o(a b) n). Multiplying from the right by b 1 we get (b a) n e so o(b a) n o(a b). Similarly, interchanging the roles of a and b we get o(a b) o(b a). So we have o(b a) o(a b).
4 2) For this question to make sense, we must of course assume that a > 1. (a) To show that [a] a n 1 is a unit in Z/(a n 1)Z we have to show that a and a n 1 are relatively prime. But this follows since a n 1 a 1 (a n 1) 1. (b) a n 1 a n 1, so a n 1 (mod a n 1), i.e., [a] n a n 1 [a n ] a n 1 [1] a n 1. Since for 0 < r < n we have 1 < a r < a n 1, and hence [a] r a n 1 [a r ] a n 1 [1] a n 1, we conclude that n is the smalles such positive integer. (c) We consider the group (Z/(a n 1)Z). This group has ϕ(a n 1) elements. By part a) we know that [a] a n 1 is a unit in the ring Z/(a n 1)Z and hence an element of the multiplicative group ((Z/(a n 1)Z), ). In part b) we found that the element [a] a n 1 has order n in this group. By Lagrange s theorem, the order of every subgroup of a finite group divides the order of the group. In particular, the order of the cyclic subgroup generated by [a] a n 1 (the order of [a] a n 1) must divide the order of the group (Z/(a n 1)Z). So n ϕ(a n 1). 3) Since we are interested in the last two digits of , we have to consider it modulo 100. We have ϕ(100) ϕ( ) ϕ(2 2 ) ϕ(5 2 ) Since gcd(3, 100) 1, we can apply Euler s theorem, and obtain So we have So the last two digits of are ϕ(100) (mod 100) (3 40 ) (mod 100). Question 3 (english, 30 points) Let M 2 (Z/2Z) be the ring of 2 2 matrices with entries in Z/2Z. Consider the following subset of M 2 (Z/2Z) : R { a b } a, b Z/2Z. b a + b 1) Show that R is a subring of M 2 (Z/2Z) and hence is itself a ring under the usual operations of matrix addition and matrix multiplication. 2) Show that R is an abelian ring, i.e. for any two matrices A, B R we have A B B A. 3) Define the map f from R to R by f(a) A 2 for any A R. Show that f is a ring homomorphism from R to R. What is the kernel of f? Is f injective? 4) Write down all elements of R. 5) Write down the addition and multiplication tables for R. 6) Find all zero divisors and all units of R. 7) What is the characteristic of the ring R? 8) Is R a field? Is it isomorphic as a ring to Z/4Z? Is it isomorphic as a ring to Z/2Z Z/2Z? 1. To prove R is a subring of M 2 (Z/2Z) we must show that it is closed under addition and multiplication, that it contains all additive inverses and that 1 M2 (Z/2Z) R. a b a Let a, b, a, b Z/2Z and let A R and A b a + b b b a + b R.
5 We then have A + A a b a b + b a + b b a + b a + a b + b b + b (a + b) + (a + b ) a + a b + b b + b (a + a ) + (b + b R ) So R is closed under addition. We also have AA a b a b b a + b b a + b aa + bb ab + a b + bb a b + ab + bb bb + aa + ab + a b + bb aa + bb ab + a b + bb ab + a b + bb (aa + bb ) + (ab + a b + bb ) R So R is closed under multiplication. We note that a b b a + b + a b a b a b + b (a + b) b (a + b) b a + b R a b and that A R, so R contains additive inverses. b (a + b) Finally, 1 M2 (Z/2Z) R Therefore, R is a subring of M 2 (Z/2Z). ) ( aa 2. In the previous part we showed that AA + bb ab + a b + bb ab + a b + bb (aa + bb ) + (ab + a b + bb ) Using the fact that Z/2Z is commutative, we have A A a b a b b a + b b a + b aa + bb ab + a b + bb a b + ab + bb bb + aa + ab + a b + bb aa + bb ab + a b + bb ab + a b + bb (aa + bb ) + (ab + a b + bb ) AA Therefore R is commutative too. 3. To show that f is a ring homomorphism we need to show that f(aa ) f(a)f(a ), that f(a + A ) f(a) + f(a ) and that f(1 R ) 1 R. From part 2 we know A A AA, therefore, f(aa ) (AA ) 2 AA AA A 2 (A ) 2 f(a)f(a ). From the properties of M 2 (Z/2Z) we know that 2(aa 2AA + bb ) 2(ab + a b + bb ) 2(ab + a b + bb ) 2((aa + bb ) + (ab + a b + bb 0 )) R.
6 We therefore have f(a+a ) (A+A ) 2 A 2 +AA +A A+(A ) 2 A 2 +2AA +(A ) 2 A 2 +(A ) 2 f(a)+f(a ). We also have f(1 R ) 1 R.1 R 1 R. Therefore, f is a homomorphism {} of rings. 0 0 The kernel of f is which implies that f is injective { } R,,, Let 1 R,0 0 1 R,A and B R 1 R A B 0 R 0 R 1 R A B 1 R 1 R 0 R B A A A B 0 1 R B B A 1 R 0 R and. 0 R 1 R A B 0 R 0 R 0 R 0 R 0 R 1 R 0 R 1 R A B A 0 R A B 1 R B 0 R B 1 R A 6. From the multiplication table we see that for r, s R then rs 0 if and only if either r 0 R or s 0 R. Therefore, there are no zero divisors (note that, by definition, 0 R is not a zero divisors). Also from the multiplication table we see that AB BA 1 R so the units in R are {1 R, A, B}. 7. We have 2.1 R 0 R and n.1 R 0 R for n N and n < 2, so the characteristic of the ring is R is a ring, which is commutative and every element is a unit. Therefore it is a field. We normally denote this field as F 4. R is not isomorphic to either Z/4Z or Z/2Z Z/2Z. There are many ways to justify this. First, both these rings contain 4 elements, but neither are fields. Second, [2] 4 Z/4Z is a zero divisor and so is ([1] 2, [0] 2 ) Z/2Z Z/2Z. If an isomorphism existed it would have to map zero divisors onto zero divisors, since there areno zero divisors in R an isomorphism can not exist. Question 4 (english, 10 points) Let (R, +, ) be a ring. Show that there is no element of R, which is at the same time a unit and a zero divisor. For contradiction we assume that r R such that r is a zero divisor and a unit. As r is a zero divisor, we know that r 0 R and s R such that s 0 R and rs 0. or sr 0. As r is a unit, we know that r 1 R such that rr 1 1 R r 1 r. We then have rs 0 r 1 rs r 1.0 (r 1 r)s 0 s 0
7 or sr 0 srr 1 0.r 1 s(rr 1 ) 0 s 0 This is a contradiction, therefore, no such r R can exist.
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