Math 541 Exam 2 Solutions

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1 Math 541 Exam 2 Solutions 1. Let G be a finite group and let K be a normal subgroup of G, Show that if ( K, [G : K]) = 1, then K is the only subgroup of G with order K. Proof: Suppose that H G with H = K. Let a H K. Then ak G/K. Suppose the order of a in G is n and that of ak in G/K is n. Then n n. By Lagrange, n divides H = K. Thus n divides K. Also by Lagrange, n divides G/K = [G : K]. Thus n also divides [G : K]. Since ( K, [G : K]) = 1, we must have n = 1. and so ak = K, contrary to the assumption that a H K. Thus H K = and so by H = K, we have H = K. 2. (10 %) Let H = 6 and K = 72 be two cyclic subgroups of Z, the additive group of the integers. (i) Show that K is a normal subgroup of H. (ii) Find an integer m such that H/K = Z m. (You must justify your answer. Solution with an answer only will receive zero credit.) Solution (i) Since H and K are subgroups of an abelian group, and since every subgroup in an abelian group is normal in the abelian group, it suffices to show that K H. For any n K, n = 72n for some integer n, and so n = 6n, where n = 12n. Thus n H. Thus K is a subgroup of H. Since H is abelian, K is normal in H. (ii) Note that H = {6n : n Z} and K = {72n : n Z}. Define a map f : H Z 12 by f(6n) = [n] Z 12, for any integer n Z. Then f((6n 1 ) + (6n 2 )) = f(6(n 1 + n 2 )) = [(n 1 + n 2 )] = [n 1 ] + [n 2 ] = f(6n 1 ) + f(6n 2 ), and so f : H Z 12 is a group homomorphism. For any [i] Z 12, 6i H and so f(6i) = [i]. Thus f is onto. Moreover, ker(f) = {6n : f(6n) = [n] = [0] Z 12 } = {6n : n 0 (mod 12)} = {72n : n Z} = K. By the first isomorphism theorem, H/K = Z 12.

2 3. (20 %) Let i denote the complex number such that i 2 = 1, and let Z and Q denote the ring of the integers and the field of the rational numbers, respectively. The binary operations involved in this problem are the addition and the multiplication of numbers. (i) Show that Z[i] = {a + bi : a, b Z} is a domain. (ii) Determine all the units in Z[i]. (iii) Show that Q[i] = {a + bi : a, b Q} is a field. (iv) Is F rac(z[i]) isomorphic to Q[i]? For a YES answer, display an isomorphism and justify it; for a NO answer, explain why not. Proof: (i) Since Z[i] is a subset of the complex number field C, it suffices to show that Z[i] is a subring, as verified below. (1) 1 = i Z[i]. (2) a + bi, a + b i Z[i], (a + bi) (a + b i) = (a a ) + (b b )i Z[i]. (3) a + bi, a + b i Z[i], (a + bi)(a + b i) = (aa bb ) + (ab + ab)i Z[i]. (ii) Let z = a + bi be a unit. Then for some z Z[i], zz = 1. Since z, z are complex numbers, we have zz = 1 and so z 2 z 2 = 1. Since z, z Z[i], both z 2 and z 2 must be integers. It follows that z 2 {1, 1} and so z {1, 1, i, i}. Conversely, each complex number in {1, 1, i, i} is a unit in Z[i]. Thus U(Z[i]) = {1, 1, i, i}. (iii) We will check that Q[i] is a suring of C in which every nonzero element has a multiplicative inverse, as shown below. (1) 1 = i Q[i]. (2) a + bi, a + b i Q[i], (a + bi) (a + b i) = (a a ) + (b b )i Q[i]. (3) a + bi, a + b i Q[i], (a + bi)(a + b i) = (aa bb ) + (ab + ab)i Q[i].

3 (4) a + bi, Q[i], (a + bi) 1 = a bi a 2 + b 2 Q[i]. (iv) The answer is YES. Consider the map f : F rac(z[i]) Q[i] by f([z, w]) = zw 1 Q[i], z Z[i] and w Z[i] {0}. (A) f is well-defined. In fact, if [z, w] = [z, w ], then zw = z w and so f([z, w]) = zw 1 = z (w ) 1 = f([z, w ]). (B) f is a ring homomorphism. In checking this, we let R denote Z[i] for simpler notation. [a 1, b 1 ], [a 2, b 2 ] F rac(r), f([a 1, b 1 ] + [a 2, b 2 ]) = f([a 1 b 2 + a 2 b 1, b 1 b 2 ]) = (a 1 b 2 + a 2 b 1 )(b 1 b 2 ) 1 = a 1 b a 2 b 1 2 = f([a 1, b 1 ]) + f([a 2, b 2 ]), f([a 1, b 1 ][a 2, b 2 ]) = f([a 1 a 2, b 1 b 2 ]) = (a 1 a 2 )(b 1 b 2 ) 1 = (a 1 b 1 1 )(a 2 b 1 2 ) = f([a 1, b 1 ])f([a 2, b 2 ]). (C) f is onto. z = a + bi Q[i], we can write z = (a 1 + b 1 i)/c where a 1, b 1 Z and c Z {0}. Thus a 1 + b 1 i, c Z[i], and so f([a 1 + b 1 i, c]) = z Q[i]. (D) f is injective. Let K be the kernel of f. Then K must be an ideal of Q[i]. But Q[i] is a field, and the only ideal of a field is either {0} or the field itself. Since f is not the zero homomorphism, the kernel cannot be the whole field and so K = {0}. 4. (10 %) Let R denote the set of all real valued continuous function on the interval [ 1, 1]. Define, for f(x), g(x) R, (f + g)(x) = f(x) + g(x) and (fg)(x) = f(x)g(x). (i) Show that R is a commutative ring. (ii) Is R a domain? You must have an explanation for your answer. Proof: (i) Let 0, 1 denote the constant valued functions. Then 0 1s the additive

4 identity and 1 1s the multiplicative identity of R. Since the additions and multiplications of real numbers are associative and commutative, and satisfy the distributive law, the additions and multiplications of R are also associative and commutative, and satisfy the distributive law. Therefore, R is a commutative ring. (ii) R is not a domain. Let 0 if x 0 f = x if x > 0, Then f, g R are nonzero elements but fg = 0. x if x 0 and g = 0 if x > (10 %) Let Z denote the ring of integers. Define R = Z Z and define addition and multiplication in R componentwise as follows: (a, b), (c, d) R, (a, b) + (c, d) = (a + c, b + d), (a, b)(c, d) = (ac, bd). (i) Show that R is a commutative ring. (ii) Is R a domain? You must have an explanation for your answer. Proof: (i) Since Z is a commutative ring and since the addition and multiplication of R are defined componentwise, R is also a commutative ring with (0, 0) and (1, 1) being the additive identity and the multiplicative identity, respectively. (ii) R is not a domain, as (0, 1)(1, 0) = (0, 0). 6. (10 %) Given the polynomials f(x) = 2x 3 + x and g(x) = x 3 3x + 2 in Z 5 [x], do each of the following. (i) Compute the greatest common divisor of f(x) and g(x) in Z 5 [x]. (ii) Find polynomials s(x) and t(x) in Z 5 [x] such that (f, g) = s(x)f(x) + t(x)g(x). Proof: (i) f(x) = 2x 3 + x = 2(x 3 3x + 2) + (x 2 + x + 2) g(x) = x 3 3x + 2 = (x 2 + x + 2)(x + 1) + ( x) x 2 + x + 2 = ( x)( x 1) + 2 ( x) = 2(2x) + 0

5 It follows that any common divisor of f and g must also divide 2. Since [2] is a unit in Z 5, (f(x), g(x)) = 1. (According to the definition of this textbook, it must be a monic polynomial). (ii) Reversing the process in (i) to get 2 = (x 2 + x + 2) + ( x)(x + 1) = (x 2 + x + 2) + (g(x) (x 2 + x + 2)(x + 1))(x + 1) = (x + 1)g(x) + (1 (x + 1) 2 )(x 2 + x + 2) = (x + 1)g(x) + x 2 (x 2 + x + 2) = (x + 1)g(x) + x 2 (f(x) 2g(x)) = x 2 f(x) + ( 2x 2 + x + 1)g(x). Since in Z 5, [2] [3] = [1], we have 1 = 3x 2 f(x) + ( x 2 + 3x + 3)g(x). 7. (10 %) Recall that in a commutative ring R, an element r R is irreducible if r U(R) {0}, and if whenever r = ab for some a, b R, either a U(R) or b U(R). (i) In the ring Z[i] (see Problem 3 above), determine if each of the elements {3, 1 + 3i, 2 + 3i} is irreducible or not. You must justify your answers. (ii) In the ring Z 7 [x], determine if each of the elements {x 2 +2x+4, x 3 +1} is irreducible or not. You must justify your answers. Proof: (i) As U(Z[i]) = {1, 1, i, i}, none of the elements in the problem is a unit. Suppose that 3 = (a + bi)(c + di) is a product of non unit elements. Then 9 = 3 2 = a + bi 2 c + di 2 = (a 2 + b 2 )(c 2 + d 2 ). Since a + bi, c + di are not units, we must have a 2 + b 2 = 3 and c 2 + d 2 = 3. Since a and b are integers, min{a 2, b 2 } 3. If a 2 = 0, then b 2 = 3; but no such integer b exists. If a 2 = 1, then b 2 = 3 1 = 2; again no such integer b exists. Thus by symmetry, for any case, no such integers a, b, c and d exist. Thus 3 is irreducible. Suppose that 1 + 3i = (a + bi)(c + di) is a product of non unit elements. Then 10 = 1 + 3i 2 = a + bi 2 c + di 2 = (a 2 + b 2 )(c 2 + d 2 ). Since a + bi, c + di are not units, we may assume a 2 + b 2 = 2 and c 2 + d 2 = 5. Thus a = ±1, b = ±1, and either c = ±2 and d = ±1 or c = ±1 and d = ±2. With these, one can find that 1+3i = (2+i)(1+i) and so not an irreducible element.

6 Suppose that 2 + 3i = (a + bi)(c + di) is a product of two other elements. Then 11 = 2 + 3i 2 = a + bi 2 c + di 2 = (a 2 + b 2 )(c 2 + d 2 ). Since 11 is a prime, either a 2 + b 2 = 1 or c 2 + d 2 = 1, and so one of the factors must be a unit. Thus 2 + 3i is irreducible. (ii) As Z 7 is a field, none of the elements in the problem is a unit. Let f(x) = x 2 +2x+4. Then f(1) = 7 0 (mod 7). Therefore, (x 1) f(x) and so f(x) = (x 1)(x + 3) is not irreducible in Z 7 [x]. As g(x) = x = (x + 1)(x 2 x + 1), g(x) is not irreducible. 8. (20 %) (i) If A and R are domains and φ : A R is a ring isomorphism, then [a, b] [φ(a), φ(b)], is a ring isomorphism F rac(a) F rac(r). (ii) Prove that if a field K contains Z (the integer ring) as a subring, then K must contain an isomorphic copy of Q (the field of the rational numbers). (iii) Let R be a domain and let φ : R K be a ring homomorphism, where K is a field. Prove that there exists a unique ring homomorphism Φ : F rac(r) K extending φ; that is, Φ R = φ. Proof: (i) Let ψ : F rac(a) F rac(r) be the map ψ([a, b]) = [φ(a), φ(b)]. The proof of this problem will be completed by the following steps. (A) ψ is well defined. (In validating (A), we only need that φ is injective). Suppose that [a 1, b 1 ] = [a 2, b 2 ] F rac(a). Then in A, b 1, b 2 A {0 A } and a 1 b 2 = a 2 b 1. Since φ : A R is injective, φ(b 1 ), φ(b 2 ) R {0 R }, and φ(a 1 )φ(b 2 ) = φ(a 1 b 2 ) = φ(a 2 b 1 ) = φ(a 2 )φ(b 1 ). Thus ψ([a 1, b 1 ]) = [φ(a 1 ), φ(b 1 )] = [φ(a 2 ), φ(b 2 )] = φ([a 2, b 2 ]), and so ψ is well defined. (B) ψ is a homomorphism. (In validating (B), we do not need the fact that φ is onto).

7 [a 1, b 1 ], [a 2, b 2 ] F rac(a), since φ : A R is a ring homomorphism, ψ([a 1, b 1 ] + [a 2, b 2 ]) = ψ([a 1 b 2 + a 2 b 1, b 1 b 2 ]) = [φ(a 1 b 2 + a 2 b 1 ), φ(b 1 b 2 )] = [φ(a 1 )φ(b 2 ) + φ(a 2 )φ(b 1 ), φ(b 1 )φ(b 2 )] = [φ(a 1 ), φ(b 1 )] + [φ(a 2 ), φ(b 2 )] = ψ([a 1, b 1 ]) + ψ([a 2, b 2 ]), ψ([a 1, b 1 ][a 2, b 2 ]) = ψ([a 1 a 2, b 1 b 2 ]) = [φ(a 1 a 2 ), φ(b 1 b 2 )] = [φ(a 1 )φ(a 2 ), φ(b 1 )φ(b 2 )] = [φ(a 1 ), φ(b 1 )][φ(a 2 ), φ(b 2 )] = ψ([a 1, b 1 ])ψ([a 2, b 2 ]). (C) ψ is onto. [a, b ] F rac(r), we have a, b R with b 0 R. Since φ : A R is an isomorphism (which is onto), a, b A with b 0 A such that a = φ(a) and b = φ(b). Since b 0 A, [a, b] F rac(a). It follows that ψ([a, b]) = [φ(a), φ(b)] = [a, b ]. (D) ψ is one-to-one. It suffices to check that ker(ψ) = {[0 A, 1 A ]}. Let [a, b] ker(ψ). Then ψ([a, b]) = [φ(a), φ(b)] = [0 R, 1 R ]. Thus in R, φ(a) = φ(a) 1 R = 0 R φ(b) = 0 R. Since φ : A R is an isomorphism, φ(a) = 0 R if and only if a = 0 A, and so [a, b] = [0 A, b] = [0 A, 1 A ]. This shows that ker(ψ) = {[0 A, 1 A ]}, and so ψ is one-to-one. Concluding Remark: We need φ to be injective to show that ψ is well defined in (A), and we need the fact that φ is a homomorphism to show that ψ is a homomorphism in (B). In checking (A) and (B), the fact that φ is surjective is not needed. (ii) Suppose that the field K contains Z as a subring. Then for any b Z {0}, b 1 K exists. It follows by a, b 1 K that ab 1 K, for any a, b Z with b 0. It follows that the set {ab 1 : a, b Z and b 0} K. But Q = {ab 1 : a, b Z and b 0}. (iii) φ must be an injective homomorphism. If φ is not injective, a counterexample is presented after the proof of (iii). Let R be a domain, and K be a field and φ : R K be an injective ring homomorphism. Then, for any a, b R with b 0 R, [a, b] F rac(r). The map ψ defined in (i) is a well defined ring homomorphism (as shown in (i)). By (i),

8 ψ : F rac(r) F rac(φ(r)) is also a ring homomorphism. Existence: Let R = φ(r) K. Define π : F rac(r ) K by π[x, y] = xy 1, x, y R with y 0 K. Then π is an injective ring homomorphism. Define Φ = π ψ([a, b]) : F rac(r) K. Then Φ([a, b]) = φ(a)(φ(b)) 1, [a, b] F rac(r). Then as composition of homomorphisms is also a homomorphism, Φ : F rac(r) K is a ring homomorphism. Moreover, a R, Φ(a) = Φ([a, 1 R ]) = φ(a). Uniqueness: Suppose that Ψ : F rac(r) K is also a ring homomorphism that extends φ. Then since Φ([a, 1]) = φ(a) = Ψ([a, 1]), a R, it follows that if b R {0 R }, then 1 = φ(1) = Φ([1, 1]) = Φ([b, b]) = Φ([b, 1][1, b]) = Φ([b, 1])Φ([1, b]) = φ(b)φ([1, b]), 1 = φ(1) = Ψ([1, 1]) = Ψ([b, b]) = Ψ([b, 1][1, b]) = Ψ([b, 1])Ψ([1, b]) = φ(b)ψ([1, b]). Thus in K, Φ([1, b]) = φ(b) 1 = Ψ([1, b]). Consequently, [a, b] F rac(r), Φ([a, b]) = Φ([a, 1][1, b]) = Φ([a, 1])Φ([1, b]) = φ(a)(φ(b)) 1 = Ψ([a, 1])Ψ([1, b]) = Ψ([a, 1][1, b]) = Ψ([a, b]), and so Φ = Ψ. Remark: In (iii), φ must be injective. Consider that map φ(n) = [n] Z 2. Then φ : Z Z 2 is a ring homomorphism from a domain to a field. Suppose that there is a ring homomorphism Φ : Q Z 2 extending φ. Then Φ(1) = Φ(1/2) + Φ(1/2) = [0] [1] = φ(1).