MTH Abstract Algebra I and Number Theory S17. Homework 6/Solutions
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1 MTH Abstract Algebra I and Number Theory S17 Homework 6/Solutions Exercise 1. Let a, b and e be integers. Suppose a and b are not both zero and that e is a positive common divisor of a and b. Put d = gcd (a, b). (a) Prove that gcd ( a e, b e ) = d e. (b) Prove that gcd ( a d, b d ) = 1. (a) By Corollary 2.9.8(II), since e is a common divisor of a and b we have e d. Also e divides a and b, there exists integers r, s, t with ( ) a = er, b = es, and d = et Then a e = r, b e = s and d = t. So need to show that gcd (r, s) = t. I will give two proofs for this. e Proof 1: By definition of gcd, d divides a and b. Thus a = dk and b = dl for some k, l Z. Now (1) implies that er = etl and es = etl. Since e is positive, we have e 0. As Z is an integral domain, the multiplicative cancelation law holds in Z (see Theorem 2.8.7). We conclude that Hence t divides r and s, that is r = tl and s = tl ( ) t is a common divisor of r and s. Now let c be a common divisor of r and s. Then ( ) r = cu and s = cv for some u, v Z. Hence a ( ) ( ) = er = e(cu) = (ec)u, and b ( ) ( ) = es = e(cv) = (ec)v. Thus ec divides a and b. Since d = gcd (a, b), this implies ec d. As d = et this gives ec et. By hypothesis e is positive and we conclude that c t. We proved that (+) If c is common divisor of r and s, then c t By ( ) and (+) we see that t = gcd (r, s). 1
2 Proof 2: Since d = gcd (a, b), the Euclidean Algorithm shows that there exist u, v Z with d = au + bv. Thus (1) implies et = eru + esv = e(ru + sv). Since e is positive, e 0. As Z is an integral domain, the multiplicative cancelation law holds in Z (see Theorem 2.8.7). We conclude that (++) t = ru + sv. Since d = et and both d and e are positive also t is positive. Put f = gcd (r, s). By Theorem f is the smallest positive integer of the form rũ + sṽ with ũ, ṽ Z and so (++) shows that (+++) f t. Since f = gcd (r, s), the Euclidean Algorithm implies there exist x, y Z with f = rx + sy. Hence (#) ef = e(rx + sy) = (er)x + (es)y ( ) = ax + by. Since f and e are positive, also ef is positive. By Theorem d is the smallest positive integer of the form a x + bỹ with x, ỹ Z and so (#) shows that d ef. As d = et this gives et ef and since e is positive we conclude that (##) t f. By (+++) and (##) we have f t and t f. Hence f = t, that is gcd (r, s) = t. (b) Since d = gcd (a, b), d is a positive common divisor of a and b. Thus by (a) applied to e = d we have gcd ( a d, b d ) = d d = 1. Exercise 2. Let p be an integer with p / {0, ±1}. Suppose that p has the following property: ( ) Whenever a and b are integers with p ab, then p a or p b. Show that p is a prime. We will first show that ( ) Let n, m be integers with n m and m n. Then n = ±m If n = 0, then 0 m and Theorem shows that m = 0, so n = m. Hence we may assume n 0 and by symmetry also m 0. Since n m and m 0, Theorem gives n m. Similarly from m n and n 0 we get m n. Hence n = m, so n = ±m. Suppose now that ( ) holds. We need to show that p is a prime. By assumption p {0, ±1} and so by the definition of a prime we need to show: 2
3 (X) If d is integer with d p then d = ±1 or d = ±p. So let d be an integer with ( ) d p. Then (+) p = dk for some integer k. Since p = 1 p we know that p p. As p = dk (by (+)) this gives p dk. Hence we can apply ( ) with b = d and c = k and conclude that (++) p d or p k. We will treat these two cases separately. Case 1: p d. Since d p by ( ) we have p d and d p and ( ) gives d = ±p. So (X) holds in Case 1. Case 2: p k. By (+) p = dk and so k p. Thus p k and k p and ( ) implies k = ±p. By (+) we have p = dk and thus d = p k = p = ±1. So (X) also holds in Case 2. ±p We proved that (X) holds and so p is a prime. Exercise 3. Let L = {[ a 0 b c ] a, b, c Z}. Given that L is a subring of M 2(Z). Show that the function f L Z given by f ([ a 0 b c ]) = a is surjective homomorphism but is not an isomorphism. Let a, b, c, d, e, g Z. Then Thus f is a homomorphism. f ([ a b 0 c ] + [ d e g 0 ]) = f ([ a+b b+e c+g 0 ]) = a + b = f ([ a 0 d 0 b c ]) + f ([ e g ]) f ([ a 0 b c ] [ d 0 e g ]) = f ([ ab 0 bd+cg cf Let m Z. Then f ([ m ]) = m and so f is surjective. ]) = ab = f ([ a 0 d 0 b c ]) f ([ e g ]) We have f ([ ]) = 0 = f ([ 0 0 ) but [ 0 1 [ ]. So f is not injective and hence f is not an isomorphism. Exercise 4. Let f R S be a ring homomorphism. Let B be a subring of S and define Show that A is a subring of R. A = {r R f(r) B}. 3
4 We will verify the four conditions of the Subring Theorem. Let r R. Note first that by definition of A: ( ) r A f(r) B. Let a, b A. Then by (*) ( ) f(a) B and f(b) B. (I) By Theorem (a), f(0 R ) = 0 S. By the Subring Theorem, 0 S B and so by (*) 0 R A. (II) By (**) f(a) B and f(b) B. Since B is a subring of S, the Subring theorem implies f(a) + f(b) B. Since f is a homomorphism, f(a + b) = f(a) + f(b) and so f(a + b) B. Thus by (*), a + b A. (III) By (**) f(a) B and f(b) B. Since B is a subring of S, the Subring Theorem implies f(a)f(b) B. Since f is a homomorphism, f(ab) = f(a)f(b) and so f(ab) B. Thus by (*), ab A. (IV) By (**), f(a) B. Since B is a subring of S, the Subring Theorem implies f(a) B. By Theorem b, f( a) = f(a) and so f( a) B. Thus by (*) a A. We verified the four conditions of the Subring Theorem and so A is a subring of R. Exercise 5. Let S = {[ a 0 a+b b ] a, b Z 2}. Given that S is a subring of M 2 (Z 2 ). Show that S is isomorphic to the ring R in Exercise 5 of Homework 2. By Theorem 2.5.6(b) Z 2 = {0, 1} and so S has the following four elements: a = 0, b = 0 [ 0 0, a = 1, b = 0 [ ], a = 0, b = 1 [ ], a = 1, b = 1 [ 1 1. Through straightforward calculation we obtain the following addition and multiplication table: + [ 0 0 [ 1 0 [ 0 0 [ 1 0 [ ] [ 1 0 [ 1 1 [ ] [ 0 1 [ ] [ 1 1 [ 1 1 [ ] [ 1 0 [ 0 0 [ 1 0 [ 0 0 [ 1 0 [ 1 0 [ 0 1 [ 0 1 [ 1 1 [ 1 1 [ 1 1 Put 0 = [ 0 0, e = [ ], b = [ ] and c = [ 1 1. Then the above tables become 4
5 + 0 e b c 0 0 e b c e e 0 c b b b c 0 e c c b e 0 0 e b c e 0 e b c b 0 b b 0 c 0 c 0 c Recall that the ring R in Exercise 5 of Homework 2 has tables: + 0 e b c 0 0 e b c e e 0 c b b b c 0 e c c b e 0 0 e b c e 0 e b c b 0 b b 0 c 0 c 0 c Remark now shows that the function f S R defined by f(0 ) = 0, f(e ) = e, f(b ) = b and f(c ) = c is an isomorphism. Exercise 6. Let f R S and g S T be homomorphism of rings. (a) Show that g f R T is a homomorphism of rings. (b) If f and g are isomorphisms, show that g f is an isomorphism. (c) Suppose f is an isomorphism. For s S let s be the unique element of R with f(s ) = s. Show that the function h S R, s s is an isomorphism of rings. (a): Let a, b R. Then Thus g f respects addition. Also (g f)(a + b) = g(f(a + b)) definition of g f = g(f(a) + f(b)) f is a homomorphism = g(f(a)) + g(f(b)) g is a homomorphism = (g f)(a) + (g f)(b) definition of g f (g f)(a b) = g(f(a b)) definition of g f = g(f(a) f(b)) f is a homomorphism = g(f(a)) g(f(b)) g is a homomorphism = (g f)(a) (g f)(b) definition of g f Hence g f also respects multiplication and so g f is a homomorphism. 5
6 (b) Since g and f are isomorphism, g and f are bijective and so (by Theorem a) f and g are injective and surjective. Since g and f are homomorphism, we conclude from (a) that g f is a homomorphisms. To show that g f is injective let a, b R with (g f)(a) = (g f)(b). The definition of g f implies that g(f(a)) = g(f(b)). Since g is injective we conclude from b that f(a) = f(b). As f is injective another application of b implies that a = b. We proved that (g f)(a) = (g f)(b) implies a = b and so a third application of b shows that g f is injective. To show that g f is surjective, let t T. Since g is surjective, there exists s S with g(s) = t. Since f is surjective, there exists r R with f(r) = s. Thus (g f)(r) = g(f(r)) = g(s) = t. Hence g f is surjective. We proved that g f is a injective and surjective homomorphism and so g f is an isomorphism. (c) Let r R and s S. Since s is the unique element of R with f(s ) = s we have r = s if and only f(r) = s. By definition of h, s = h(s). So the Principal of Substitution shows that ( ) r = h(s) if and only f(r) = s. Since h(s) = s, we have f(h(s)) = f(s ) = s. Hence ( ) f(h(s)) = s. Let a, b S. To show that h respects addition, put r = h(a) + h(b) and s = a + b. Then f(r) = f(h(a) + h(b)) f is hom = f(h(a)) + f(h(b)) (**) = a + b = s and so ( ) implies that r = h(s), that is h(a) + h(b) = h(a + b). So h respect addition. To show that h respect multiplication, put r = h(a) h(b) and s = a b. Then f(r) = f(h(a) h(b)) f is hom = f(h(a)) f(h(b)) ( ) = a b = s and so ( ) implies that r = h(s), that is h(a) h(b) = h(a b). So h respect multiplication. Thus h is a homomorphism. To show that h is injective, suppose that h(a) = h(b). Then f(h(a)) = f(h(b)) and ( ) implies that a = b. Hence b shows that h is injective. To show that h is surjective let r R and put s = f(r). Then ( ) shows that r = h(s) and so h is surjective. We proved that h is a injective and surjective homomorphism and so h is an isomorphism. Exercise 7. Let f R S be an isomorphism of rings. If R is an integral domain, show that S is an integral domain. 6
7 Suppose that R is an integral domain. As R is an integral domain, R has an identity. Since f is an surjective homomorphism we conclude from a that S has an identity and f(1 R ) = 1 S. Since f is a homomorphism a shows that f(0 R ) = 0 S. As R is an integral domain, 0 R 1 R. Since f is injective this gives f(0 R ) f(1 R ) and so 0 S 1 S. Now let a, b S. Since f is surjective there exists c, d R with f(c) = a and f(d) = b. Since R is an integral domain, R is commutative. Hence cd = dc and so f(cd) = f(dc). Since f is a homomorphism this implies that f(c)f(d) = f(d)f(c) and so ab = ba. We proved that ab = ba for all a, b S and so S is commutative. To show that Axiom 11 holds in S, suppose that ab = 0 S. Then f(c)f(d) = 0 S. Since f is a homomorphism we know that f(cd) = f(c)f(d) and f(0 R ) = 0 S. Thus f(cd) = f(0 R ). Since f is injective, this implies cd = 0 R. As R is an integral domain, Axiom 11 holds in R and we conclude that c = 0 R or d = 0 R. Thus also f(c) = f(0 R ) or f(d) = f(0 R ) and so a = 0 S or b = 0 S. Thus Axiom 11 holds in S. We proved that S is a commutative ring with identity, that 1 S 0 S and that Axiom 11 holds in S. Thus S is an integral domain. Exercise 8. Show that the two rings are not isomorphic. (a) 2Z and Z. (b) R R R R and M 2 (R). (c) Z 2 Z 14 and Z 16. (d) Q and R. (e) Z Z 2 and Z. (f) Z 4 Z 4 and Z 16. In each case let R be the first ring in question and S the second and suppose for a contradiction that f R S is an isomorphism. (a) Since f is surjective, there exists a 2Z with f(a) = 1. By a f(0) = 0 and so a 0. Note that f(a 2 ) = f(aa) = f(a)f(a) = 1 1 = 1 = f(a). Since f is injective this gives a 2 = a and so a a = a 1. Since a 0 this implies a = 1, a contradiction since a 2Z and 1 2Z. (b) By Example 2.1.4(g) S = M 2 (R) is not commutative and so there exist x, y S with xy yx. Since f is surjective there exists a, b R with f(a) = x and f(b) = y. Since R is commutative, Theorem 2.1.7d shows that R R is commutative. Applying 2.1.7d one more times now shows that R = R R R R is commutative. Thus ab = ba. Hence also f(ab) = f(ba). Since f is a homomorphism this gives f(a)f(b) = f(b)f(a) and so xy = yx, a contradiction. (c) By Theorem 2.5.6(c) we have Z n = n. So R = Z 2 Z 14 = Z 2 Z 14 = 2 14 = 28 and S = Z 16 = 16. Hence R S, a contradiction since f is bijective and so R = S. (d) I will provide three proofs that Q is not isomorphic to R. Proof 1: Let x = 2 R = S. Then x 2 = 2. Since f is surjective, f(a) = x for some a R = Q. As f is surjective Theorem a gives f(1) = 1. So f(2) = f(1 + 1) = f(1) + f(1) = = 2. Hence f(a 2 ) = f(a)f(a) = xx = x 2 = 2 = f(2). Since f is injective this gives a 2 = 2, a contradiction to a Q. Proof 2: Since f is a bijection we get Q = R. But Q is countable and R is uncountable and so Q R, a contradiction. 7
8 Proof 3: We claim that f(n) = n for all n N. By Theorem a f(0) = 0 Since f is surjective homomorphism, Theorem a shows that f(1) = 1. Suppose f(k) = k. Then f(k + 1) = f(k) + f(1) = k + 1. So the Principal of Induction shows that f(n) = n for all n N. By Theorem b, f( n) = f(n) = n and so f(m) = m for all m Z. Let q Q. Then q = a for some b a, b Z with b 0. By Theorem b f(b 1 ) = f(b) 1 = b 1 and so f(q) = f ( a b ) = f(ab 1 ) = f(a)f(b 1 ) = ab 1 = a b = q. We proved that f(q) = q for all q Q. Since f is surjective, this shows that Q = R, a contradiction. (e) Put a = (0, [1] 2 ). Then a Z Z 2 = R, a 0 R and a + a = (0, [2] 2 ) = (0, [0] 2 ) = 0 R. By Theorem (a) f(0 R ) = 0 S = 0. Hence f(a) + f(a) = f(a + a) = f(0 R ) = 0. Since f(a) S = Z this gives f(a) = 0. As also f(0 R ) = 0 and f is injective this implies that a = 0 R, a contradiction. (f) Since f is surjective, a shows that f(1 R ) = 1 S, that is f(([1] 4, [1] 4 )) = [1] 16. Thus and so [2] 16 = [1] 16 + [1] 16 = f(([1] 4, [1] 4 )) + f(([1] 4, [1] 4 )) = f(([1] 4, [1] 4 ) + ([1] 4, [1] 4 )) = f(([2] 4, [2] 4 )) a contradiction. [4] 16 = [2] 16 + [2] 16 = f(([2] 4, [2] 4 )) + f(([2] 4, [2] 4 )) = f(([2] 4, [2] 4 ) + ([2] 4, [2] 4 )) = f(([4] 4, [4] 4 )) = f(([0] 4, [0] 4 )) = f(0 R ) = 0 S = [0] 16 8
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