solutions manual to accompany introduction to abstract algebra fourth edition pdf

www.TechnicalBooksPdf.com Solutions Manual to Accompany Introduction to Abstract Algebra www.TechnicalBooksPdf.com www.TechnicalBooksPdf.com Solutions Manual to Accompany Introduction to Abstract Algebra Fourth Edition W Keith Nicholson University of Calgary Calgary, Alberta, Canada www.TechnicalBooksPdf.com Copyright 2012 by John Wiley & Sons, Inc All rights reserved Published by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose No warranty may be created or extended by sales representatives or written sales materials The advice and strategies contained herein may not be suitable for your situation You should consult with a professional where appropriate Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002 Wiley also publishes its books in a variety of electronic formats Some content that appears in print may not be available in electronic formats For more information about Wiley products, visit our web site at www.wiley.com Library of Congress Cataloging-in-Publication Data: Nicholson, W Keith Introduction to abstract algebra / W Keith Nicholson – 4th ed p cm Includes bibliographical references and index ISBN 978-1-118-28815-3 (cloth) Algebra, Abstract I Title QA162.N53 2012 512’.02–dc23 2011031416 Printed in the United States of America 10 www.TechnicalBooksPdf.com Contents Preliminaries 0.1 0.2 0.3 0.4 Proofs / Sets / Mappings / Equivalences / Integers and Permutations 1.1 1.2 1.3 1.4 Induction / Divisors and Prime Factorization / Integers Modulo n / 11 Permutations / 13 Groups 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 17 Binary Operations / 17 Groups / 19 Subgroups / 21 Cyclic Groups and the Order of an Element / 24 Homomorphisms and Isomorphisms / 28 Cosets and Lagrange’s Theorem / 30 Groups of Motions and Symmetries / 32 Normal Subgroups / 34 Factor Groups / 36 The Isomorphism Theorem / 38 An Application to Binary Linear Codes / 43 v vi Contents Rings 3.1 3.2 3.3 3.4 3.5 108 Products and Factors / 108 Cauchy’s Theorem / 111 Group Actions / 114 The Sylow Theorems / 116 Semidirect Products / 118 An Application to Combinatorics / 119 Series of Subgroups 9.1 9.2 9.3 102 Modules / 102 Modules over a Principal Ideal Domain / 105 p-Groups and the Sylow Theorems 8.1 8.2 8.3 8.4 8.5 8.6 88 Vector Spaces / 88 Algebraic Extensions / 90 Splitting Fields / 94 Finite Fields / 96 Geometric Constructions / 98 An Application to Cyclic and BCH Codes / 99 Modules over Principal Ideal Domains 7.1 7.2 81 Irreducibles and Unique Factorization / 81 Principal Ideal Domains / 84 Fields 6.1 6.2 6.3 6.4 6.5 6.7 64 Polynomials / 64 Factorization of Polynomials over a Field / 67 Factor Rings of Polynomials over a Field / 70 Partial Fractions / 76 Symmetric Polynomials / 76 Factorization in Integral Domains 5.1 5.2 Examples and Basic Properties / 47 Integral Domains and Fields / 52 Ideals and Factor Rings / 55 Homomorphisms / 59 Ordered Integral Domains / 62 Polynomials 4.1 4.2 4.3 4.4 4.5 47 The Jordan-Hă older Theorem / 122 Solvable Groups / 124 Nilpotent Groups / 127 122 Contents 10 Galois Theory 10.1 10.2 10.3 10.4 vii 130 Galois Groups and Separability / 130 The Main Theorem of Galois Theory / 134 Insolvability of Polynomials / 138 Cyclotomic Polynomials and Wedderburn’s Theorem / 140 11 Finiteness Conditions for Rings and Modules 142 11.1 Wedderburn’s Theorem / 142 11.2 The Wedderburn-Artin Theorem / 143 Appendices Appendix A: Complex Numbers / 147 Appendix B: Matrix Arithmetic / 148 Appendix C: Zorn’s Lemma / 149 147 136 10 Galois Theory If G ∼ = Cp2 = σ , the lattices are: Here σ ◦ ⊇ F is Galois because σ G, and [ σ ◦ : F ] = If G ∼ = Cp × Cp = σ, τ the lattices are: Each of σ [ τ ◦ : F ] ◦ ⊇ F and τ ◦ ⊇ F are Galois because [ σ ◦ : F ] = /2 and By Example §10.1, G = gal(E : Zp ) = σ ∼ = Cn where σ : E → F is the Frobenius automorphism given by σ(u) = up By the Dedekind-Artin theorem [E : G◦ ] = |G| = n But [E : Zp ] = n so, since Zp ⊆ G◦ , G◦ = Zp Thus E ⊇ Zp is a Galois extension by Lemma Now the (inverted) lattice of subgroups of G = σ , when o(σ) = 12, is shown below at the right Write Ek = σ k ◦ = {u ∈ E | σ k (u) = u} for each divisor k of n Then the subfield lattice is shown below on the left Note that [Ek : Zp ] = [ σ k so |Ek | = pk , and Ek ∼ = GF (pk ) ◦ : G◦ ] = |G : σ k | = k (a) H = {σ1k1 σ2k2 · · · σnkn | n ≥ 1, σi ∈ X, ki ∈ Z} Thus u ∈ H is fixed by all τ ∈ H if and only if σ(u) = u for all σ ∈ X 10.2 The Main Theorem of Galois Theory (a) Let r = r(t) = f (t) g(t) 137 ∈ E Then r ∈ EG ⇔ σ(r) = r ⇔ r(t) = r(−t) ⇔ f (t)g(−t) = f (−t)g(t) If char F = 2, this always holds and K = EG = E Thus t ∈ K so m = x − t If char F = / 2, write h(t) = f (t)g(−t) Then h(t) = h(−t) so h(t) = k(t2 ) for some polynomial k (because char F = / 2) Thus f (t)g(−t) = k(t2 ) Similarly k(t2 ) ) and g(t)g(−t) = l(t ) for some polynomial l Thus r(t) = g(t)g(−t) = k(t l(t2 ) , from which K = F (t2 ) It follows that m = x2 − t2 Note that this shows [E : K] = = |G| in this case, as the Dedekind-Artin theorem guarantees K is a subgroup of the abelian group G, so it is normal Then the main theorem applies (a) If H → H ◦ is onto, and K is an intermediate field, then K = H ◦ for a subgroup H, so K ◦ = H ◦ ◦ = H = K Thus K is closed Conversely, if all intermediate fields K are closed then K = K ◦ is the image of K , and the map is onto 11 Write G = gal(E : F ) = {σ ∈ aut E | σ fixes F } Since K ⊇ F we have gal(E : K) = {σ ∈ aut E | σ fixes K} = {σ ∈ G | σ fixes K} = K Thus: E ⊇ K is Galois ⇔ ⇔ ⇔ ⇔ K K K K is the set of elements of E fixed by gal(E : K) = {u ∈ E | σ(u) = u for all σ ∈ K } =K◦ is closed in E ⊇ F 13 If [E : K] = then = [E : K] = |K : E | = |K : {1}| = |K |, so K is a subgroup of A4 of order There is no such subgroup (Exercise 34 §2.6) 15 (a) By its definition, K ∨ L is the smallest intermediate field containing both K and L Since the Galois connection K → K is order reversing, (K ∨ L) is the largest subgroup of G = gal(E : F ) contained in both K and L ; that is (K ∨ L) = K ∩ L 17 If K = σ(K1 ) let λ ∈ K1 If v ∈ K write v = σ(u), u ∈ K1 Then σλσ −1 (v) = σλ(u) = σ(u) = v so σK σ −1 ⊆ K On the other hand, if μ ∈ K then μ(v) = v for all v ∈ K, so μ[σ(u)] = σ(u) for all u ∈ K1 Hence σ −1 μσ ∈ K1 , so μ ∈ σK1 σ −1 This proves σK1 σ −1 = K Conversely, assume σ −1 K1 σ = K, that is K1 = σK σ −1 If u ∈ K1 then μ(u) = u for all μ ∈ K1 , so σ −1 λσ(u) = u for all λ ∈ K , that is λ[σ(u)] = σ(u) for all λ ∈ K , that is σ(u) ∈ K ◦ Since E ⊇ F is Galois, K ◦ = K, so this shows σ(u) ∈ K, u ∈ σ −1 (K) Hence K1 ⊆ σ −1 (K), so σ(K1 ) ⊆ K Similarly K ⊆ σ(K1 ) 19 (a) Let E = F (u1 , u2 , , um ) where X = {u1 , u2 , , um } is the set of distinct roots of f in E If σ ∈ G then σ(ui ) ∈ X for all i so σ : X → X is one-to-one (hence onto) Thus σ induces a permutation of these roots: σ(ui ) = uσ(i) , σ ¯ ∈ Sm 138 10 Galois Theory Now the map σ → σ ¯ is a group homomorphism G → Sm because uστ (i) = στ (ui ) = σ(uτ (i) ) = uσ¯ τ¯(i) for all σ, τ in G Moreover σ ¯ = ε means σ(ui ) = ui for all i, so σ = ε in G by Theorem §10.1 Thus σ → σ ¯ is an embedding 20 (a) If N (u) = v then, given τ ∈ G, τ (v) = τ σ(u) σ∈G = τ σ(u) σ∈G But τ σ traces all of G as σ does, so τ (v) = v This is true for all τ ∈ G, so v ∈ G◦ = F The proof that T (u) is in F is analogous (c) If K ⊇ F is Galois and H = gal(K : F ), then |H| = n by Theorem 3, Corollary Now p splits in K and has distinct roots u1 , u2 , , um (Theorem 3) Hence p = (x − u1 )(x − u2 ) · · · (x − un ) For each i there exists σi ∈ H such that σi (u) = ui , so p = (x − σ(u)) Thus the σ∈G coefficients of xn−1 and are, respectfully [−σ(u)] = −TK/F (u), an−1 = and σ∈H [−σ(u)] = (−1)n NK/F (u) a0 = σ∈H 21 If f = v0 + v1 x + · · · + vm xm define f τ = τ (v0 ) + τ (v1 ) x + · · · + τ (vn ) xm for τ ∈ G Thus f τ = σ∈G [x − τ σ(u)] = f because τ σ runs through G as σ does It follows that τ (vj ) = vj for all τ ∈ G, whence vj ∈ G◦ = F for each j Thus f ∈ F [x] Since f (u) = 0, p|f where p is the minimal polynomial of u over F Write f = pm g where p and g are relatively prime in F [x] If g = / 1, let q be an irreducible factor of g Since q|f , = q[σ(u)] = σ[q(u)] for some σ ∈ G, so q(u) = Since p(u) = this is a contradiction because gcd(p, q) = in F [x] In fact, = pr + qs gives = when we substitute x = u 10.3 INSOLVABILITY OF POLYNOMIALS √ √ √ √ √ √ (a) E = Q( 3, 5, 7) contains 3( − 7), and is radical over Q because √ √ √ √ √ 3 Q ⊆ Q( 3)( 5) ⊆ Q[ 3, 5]( 7) = E has the required properties (a) If f = x5 − 4x − then f = 5x4 − is zero at ±a, ±ai where a = 4 We find f = x(x4 − 4) − so f (a) = a( 45 − 4) − = −16a − < while f (−a) = −a( 45 − 4) − = 16a − > As in Example 1, this shows that f has three real roots and two (conjugate) nonreal roots Since f is irreducible by the Eisenstein criterion, its Galois group is S5 as in Example √ Take p = x7 − 14x + Then p = 7(x6 − 2) If a = 1.22, then p = if n = ±a, aw, aw2 , aw4 , aw5 where w = e2πi/6 Also p = x(x6 − 14) + so p(a) = a(2 − 12) + = − 10a < 10.3 Insolvability of Polynomials 139 and p(−a) = −a(2 − 12) + = + 10a > Thus p has three distinct real roots and the rest complex (conjugate pairs) If E ⊇ Q is the splitting field, view G = gal[E : Q] as a subgroup of SX where X ⊆ C is the set of roots If we identify G with as a subgroup of SX where X is the set of p distinct roots of f, then conjugation gives a transposition in C; if u is a real root, then [Q(u) : Q] = because p is the minimal polynomial of u over Q (it is irreducible by Eisenstein) Since [Q(u) : Q] = divides |G| = [E : Q], G has an element of order by Cauchy’s theorem (Theorem §8.2) Since the 7-cycles are the only elements in S7 of order 7, the proof in Example goes through Let X denote the set of roots of p in the splitting field E ⊇ F where p ∈ F [x] Then G = gal(E : F ) is isomorphic to a subgroup of SX Since |X| ≤ 4, G embeds in S4 Now S4 is solvable (S4 ⊇ A4 ⊇ K ⊇ {ε} has abelian factors) so every subgroup is solvable by Theorem §9.2 Since f = 3(x2 − 1), f (1) = −1 and f (−1) = 3, f has three real roots Here b = −3 and c = so, in the cubic formula, p3 and q are roots of x2 + x + which satisfy p3 + q = −1 and pq = The roots are w and w2 (w = e2πi/3 ) so take p3 = w, q = w2 Thus p=e q=e 2π i 4π i , , 2πi/9 We need pq = so take p = e e e 8π i 10π i , or e , or e 16πi/9 ,q=e u1 = p + q = p + p¯ = cos 16π i −2πi/9 =e 2π 10πi/9 u2 = wp + w2 q = e8πi/9 + e 14π i = cos u3 = w2 p + wq = e14πi/9 + e4πi/9 = cos = p¯ The roots are 8π 4π If E ⊆ C is the splitting field, then E ⊆ R and, since f is irreducible (the roots are not in Q) and separable |gal[E : Q]| = [E : Q] = deg f = Thus gal[E : Q] ∼ = C3 (a) σ(Δ2 ) = Δ2 for all σ ∈ G because σ permutes the roots ui Since E ⊇ F is Galois (f is separable because the ui are distinct) this means Δ2 ∈ G◦ = F (b) The transposition γ = (ui uj ) in G (regarding G ⊆ SX ) changes the sign of (ui − uj ) in Δ, interchanges (ui − uk ) and (uj − uk ), and fixes (uk − um ) Hence γ(Δ) = −Δ The even (odd) permutations are products of an even (odd) number of transpositions, so the result follows (c) If A = {σ ∈ G | σ is an even permutation of X} then (b) gives F (Δ) = gal(E : F (Δ)) = {σ ∈ G | σ(Δ) = Δ} ⊇ A (*) Write AX = {σ ∈ SX | σ even} so that A = G ∩ AX We have |SX : AX | = so (Exercise 15 §2.8) either G ⊆ AX or |G : G ∩ AX | = |G : A| = If 140 10 Galois Theory G ⊆ AX then A = G = F (Δ) If |G : A| = then (*) gives F (Δ) = G or F (Δ) = A But F (Δ) = G implies σ(Δ) = Δ for all σ ∈ G so G ⊆ AX a contradiction Thus F (A) = A in any case (d) Using (c): Δ ∈ F ⇐⇒ F (Δ) = F ⇐⇒ A = F (Δ) = G ⇐⇒ G ⊆ AX (e) If f = x2 + bx + c = (x − u1 )(x − u2 ) then u1 + u2 = −b, u1 u2 = c, so Δ2 = (u1 − u2 )2 = (u1 + u2 )2 − 4u1 u2 = b2 − 4c (f) If f = x3 + bx + c = (x − u)(x − v)(x − w) then u+v+w = uv + uw + vw = b uvw = −c (1) (2) (3) Now (1) and (2) give b = uv + (u + v)(−u − v) = uv − (u + v)2 ; so (u + v)2 = uv − b Thus (u − v)2 = (u + v)2 − 4uv = −b − 3uv Permuting u, v, w: (u − v)2 = −b − 3uv (u − w)2 = −b − 3uw (v − w)2 = −b − 3vw Hence: −Δ2 = (b + 3uv)(b + 3uw)(b + 3vw) = b3 + 3(uv + uw + vw)b2 + 9(u2 vw + uv w + uvw2 )b + 27(uvw)2 = b3 + 3(b)b2 + 9(uvw)(u + v + w)b + 27(−c)2 = 4b3 − 27c2 10.4 CYCLOTOMIC POLYNOMIALS AND WEDDERBURN’S THEOREM (a) Φ8 (x) = x8 − x8 − x8 − = = x4 + = Φ1 (x)Φ2 (x)Φ4 (x) (x − 1)(x + 1)(x + 1) (x − 1) (c) Φ12 (x) = x6 + x12 − x12 − = = x4 − x2 + = (Φ1 Φ2 Φ3 Φ6 )Φ4 (x − 1)(x2 + 1) x2 + x18 − x18 − x12 + x6 + = = (Φ1 Φ2 Φ3 Φ6 )Φ9 (x − 1)Φ9 Φ9 x9 − x9 −1 = x6 + x3 + = Now Φ9 (x) = Φ Φ x −1 x12 + x6 + = x6 − x3 + Finally Φ18 (x) = x + x3 + Since n is odd, d|2n if and only if either d|n or d = 2b, b|n Thus, by induction ⎛ ⎞⎛ ⎞ (e) Φ18 (x) = Φd (x) = ⎝ x2n − = d|2n Φd (x)⎠ ⎝ d|n Φ2d (x)⎠ = (xn − 1) d|n Φ2d (x) d|n 10.4 Cyclotomic Polynomials and Wedderburn’s Theorem 141 Now observe that Φ2 (x) = −Φ1 (−x), and (by induction) Φ2d (x) = Φd (−x) if d < n Hence ⎡ ⎤ x2n − = (xn − 1) ⎣− Φd (−x)⎦ · Φ2n (x) = d|n d= / n −(xn −1)Φ2n (x) Φn (−x) Φd (−x) d|n −(xn − 1)(−1)(xn + 1)Φ2n (x) −(xn − 1)Φ2n (x)((−x)n − 1) = Φn (−x) Φn (−x) Thus Φn (−x) = Φ2n (x), as required = If wk = e2πi/k , these fields are Q(wmn ) and Q(wm , wn ) respectively Now n m wmn = wm and wmn = wn , so Q(wm , wn ) ⊆ Q(wmn ) But gcd(m, n) = 1, so s s write = rm + sn; r, s ∈ Z Then mn = nr + m , so wmn = wnr · wm Thus Q(wmn ) ⊆ Q(wm , wn ) (a) If S ⊆ R is a finite subring of the division ring S, then char S = / 0, say char S = p Let Zp ⊆ S and let dimZp (S) = n Then if = / s ∈ S, 1, s, , sn are not linearly independent over Zp , so r0 + r1 s + · · · + rn sn = 0, ri ∈ Zp We may assume r0 = / (can cancel s) so s[r0−1 (−r1 − r2 s − · · · − rn sn−1 )] = Write σ(n) = d|n μ(d) If n = 1, σ(n) = μ(1) = If n = pn1 pn2 · · · pnr r then μ(d) = for any d|n with p2i |d If we write m = p1 p2 · · · pr , then σ(n) = σ(m) If d|m, then μ(d) = if and only if d is the product of an even number (possibly 0) of the pi , and μ(d) = −1 otherwise Since half the divisors d are in each category, σ(m) = (a) The sums are equal by replacing d by n/d throughout Use the hint: d|n μ(d)α n d = μ(d) = β(c) = μ(d)β(c) cd|n μ(d) = β(n), d| n c c|n But β(c) c| n d d|n μ(d) = if and only if n = c by the preceding exercise.We have d|n xn − = Φd (x) Fix x and take a formal logarithm: d|n log(xn − 1) = log(Φd (x)) d|n obius we get Let σ(n) = log(xn 1) and (n) = log(n (x)) By Mă n log(xd − 1) = log[Π(xd − 1)μ(n/d) ] log(Φn (x)) = μ d d|n The result follows If formal logarithms are distasteful, repeat Exercise 8(a) with Σ replaced by Π and all coefficients replaced by exponents Write n/m = k so n = km Thus d|m if and only if kd|n Now Exercise 8(b) gives m n Φm (xk ) = [(xk )d − 1]μ( d ) = [xkd − 1]μ( kd ) = (xb − 1)μ(n/b) d|m = Φn (x) kd|n b|n Chapter 11 Finiteness Conditions for Rings and Modules 11.1 WEDDERBURN’S THEOREM If R M is simple, then M = Rx for any = / x ∈ M Hence α : R → M given by α(r) = rx is an onto R-linear map, so M ∼ = R/L where L = ker(α), and L is maximal because M is simple Conversely, R/L is simple for every maximal left ideal L Conversely, R/L is simple for every maximal left ideal L again by Theorem §8.1 Define σ : Ra → Rb by σ(ra) = rb This is well defined because = implies rbR = r(aR) = 0, so rb = So σ is an onto homomorphism of left R-modules with σ(a) = b Finally if σ(ra) = then rb = so (ra)R = r(aR) = r(bR) = 0, so = Hence σ is one-to-one If L1 ⊆ L2 ⊆ · · · are left ideals of eRe, then RL1 ⊆ RL2 ⊆ · · · are left ideals of R so RLn = RLn+1 = · · · for some n by hypothesis If i ≥ n, the fact that Li ⊆ eRe for each i, gives Li = eLi ⊆ eRLi = eRLn = eReLn ⊆ Ln Hence Ln = Ln+1 = · · · , as required If R M is finitely generated, then M is an image of Rn for some n ≥ by Theorem §7.1 and its Corollaries Since Rn is noetherian as a left R-module (Corollary of Lemma 2), the same is true of M by Lemma (a) Observe that ann(X) is a left ideal for any X ⊆ M (note that ann(∅) = R) Hence use the artinian condition to choose ann(X) minimal in S = {ann(X) | X ⊆ M and X is finite} Then ann(M ) ⊆ ann(X) is clear; we prove equality If a ∈ ann(X), suppose that a ∈ / ann(M ), say am = / with m ∈ M Put Y = X ∪ {m} Then Student Solution Manual to Accompany Introduction to Abstract Algebra, Fourth Edition W Keith Nicholson © 2012 John Wiley & Sons, Inc Published 2012 by John Wiley & Sons, Inc 142 11.2 The Wedderburn-Artin Theorem 143 ann(Y ) ⊆ ann(X), so ann(Y ) = ann(X) by minimality But a ∈ ann(Y ) and so am = 0, a contradiction Hence a ∈ ann(M ) and we have proved (a) Let X = { pmn | m ∈ Z and p does not divide m} If Z ⊂ Y ⊂ X, where Y is / X, a subgroup of X, choose pmn ∈ Y with n maximal (n exists because Y = X = Z p1 ∪ Z p12 ∪ Z p13 ∪ · · · , and Z p1 ⊆ Z p12 ⊆ Z p13 ⊆ · · · ) Then p1n ∈ Y by Theorem §1.2 because m and pn are relatively prime Hence Z p1n ⊆ Y ; we claim that this is equality To see this, let y = m ∈ Y where p does not pk divide m Then k ≤ n by the maximality of n, so p1k = pn−k p1n ∈ Z p1n , whence y ∈ Z p1n 11 (a) If x ∈ M then x − π(x) ∈ ker π, and it follows that M = π(M )+ ker π If y ∈ π(M )∩ ker π write y = π(z), z ∈ M Since y ∈ ker π, we have = π(y) = π (z) = π(z) = y, so π(M )∩ ker π = This shows that M = π(M )⊕ ker π (c) We have ker π = (1 − π)(M ) because π(x) = if and only if (1 − π)(x) = x And π(M ) = ker(1 − π) because (1 − π)(x) = if and only if x = π(x) 13 (a) Suppose that K = K1 ⊕ K2 ⊕ · · · is an infinite direct sum of nonzero submodules of M Then K ⊃ K2 ⊕ K3 ⊕ · · · ⊃ K3 ⊕ K4 ⊃ · · · contradicts the DCC, and K1 ⊂ K1 ⊕ K2 ⊂ K1 ⊕ K2 ⊕ K3 ⊂ · · · contradicts the ACC 15 Here R = F F F is not simple as rings , and we consider e = F 0 0 Then e2 = e and Re = is a submodule But end(Re) ∼ = eRe = 0 F F F ∼ = F as 17 (a) We have ker(α) ⊆ ker(α2 ) ⊆ ker(α3 ) ⊆ · · · so, since M is noetherian, there exists n ≥ such that ker(αn ) = ker(αn+1 ) = · · · If x ∈ ker(α) then (since αn is onto) write x = αn (y) for some y ∈ M Then = α(x) = αn+1 (y), so y ∈ ker(αn+1 ) = ker(αn ) Hence x = αn (y) = 0, proving that ker(α) = 19 (a) The easy verification that θ is an F -linear homomorphism of additive groups is left to the reader To show that θ is one-to-one, suppose that θ(r) = 0; that is, ui r = for each i If = i ui , then r = · r = i ui r = 0, so ker θ = and θ is one-to-one Finally, let θ(s) = [sij ] so that ui s = nj=1 sij uj Then: ui rs = rik uk s = k Thus θ(rs) = [ rik k rik skj ] a −d (e) a + bu → b b a = rik skj uj j k = [r⎡ij ][sij ] = θ(r) · θ(s), ⎤ and the proof is complete ⎢ −b (c) a + bi + cj + dk → ⎣ −c a skj uj j k b c a d d −d a b c −b a −c ⎥ ⎦ 11.2 THE WEDDERBURN-ARTIN THEOREM The Z-modules are the (additive) abelian groups, so the simple ones are the simple abelian groups These are the prime cycles Zp (in additive notation) 144 11 Finiteness Conditions for Rings and Modules where p is a prime Hence the semisimple Z-modules are the the direct sums of copies of these Zp for various primes p The homogeneous semisimple Z-modules are the direct sums of copies of Zp for a fixed prime p (a) Since R R is complemented by Theorem 1, we have R = L ⊕ M for some left ideal M Write = e + f where e ∈ L and f ∈ M Hence Re ⊆ L, and we claim this is equality If x ∈ L then x − xe = xf ∈ L ∩ M = because x − xe ∈ L and xf ∈ M Hence x = xe ∈ Re, so L ⊆ Re, as required (a) Let α, β ∈ E Then (α + β) · x = (α + β)(x) = α(x) + β(x) = α · x + β · x so Axiom M2 holds (see Section 11.1) Similarly, (αβ) · x = (αβ)(x) = α[β(x)] = α · (β · x) proves Axiom M3 The other axioms are routinely verified Let N1 , N2 , , Nm be maximal submodules of M Define α:M → M N1 ⊕ M N2 ⊕ ··· ⊕ M Nm by α(x) = (x + N1 , x + N2 , , x + Nm ) for all x ∈ M Then α is R-linear and M ker(α) = {x ∈ M | x + Ni = for each i} = ∩i Ni Since each N is simple, i M/(∩i Ni ) ∼ = α(M ) is semisimple by Corollary of Theorem If M is a finitely generated semisimple module then (Lemma 4) M is a finite direct sum of simple submodules By Theorem M = H1 ⊕ · · · ⊕ Hn where the Hi are the homogeneous components of M By the preceding exercise, end(M ) ∼ = end(H1 ) × · · · × end(Hn ), so it suffices to show that end(H) is semisimple for any finitely generated, homogeneous, semisimple module H But then H is a finite direct sum of isomorphic simple modules (using Lemma 4), and so H ∼ = K n for some simple module K Now Lemma §11.1 gives end(H) ∼ = Mn (end K) Since end K is a division ring by Schur’s lemma, we are done Let K be a simple left ideal of R If R is a domain and then K = / so Brauer’s lemma shows that K = Re where e2 = e But e(1 − e) = so, again since R is a domain, e = or e = Since K = / we must have e = 1, whence R = Re is simple as a left R-module But then, given = / a ∈ R, we have Ra = R, say ba = Again, b = / so Rb = R, say cb = Now compute: a = 1a = (cb)a = c(ba) = c1 = c Hence ab = cb = and we have proved that a is a unit with inverse b Since a= / was arbitrary, this shows that R is a division ring 11 If A is an ideal of R, we must show that every left ideal L of the ring R/A is semisimple as an R/A-module But L is an R-submodule of the left R-module R/A because (r + A)(s + A) = rs + A = r(s + A) Hence L is semisimple as a left R-module by Corollary of Theorem (since R R is semisimple) Since the R-action and the R/A-action on L are the same, it follows that L is semisimple as a left R/A-submodule This is what we wanted 13 (2) ⇒ (1) Assume that eRe is a division ring, and that R is semiprime Choose = / x ∈ Re; we must show that Rx = Re, that is e ∈ Rx Since R is semiprime we have xRx = / 0, say xax = / 0, a ∈ R Since x ∈ Re, we have xe = x, so xe(ax) = / This means eax = / so eaxe = / (again since x = xe) As 11.2 The Wedderburn-Artin Theorem 145 eRe is a division ring, there exists t ∈ eRe such that t(eax) = e Thus e ∈ Rx, as required 15 Let R R be semisimple Then Lemma implies that R = K1 ⊕ K2 ⊕ · · · ⊕ Kn where the Ki are simple left ideals Hence R is left noetherian by the Corollary to Lemma §11.1 Since RR is also semisimple by Theorem 5, the same argument shows that R is right noetherian [Note that this also shows that R is left and right artinian.] 16 Let R be a semiprime ring (a) If LM = where L and M are left ideals, then (M L)2 = M LM L = M 0L = Since M L is a left ideal and R is semiprime, this gives M L = (c) If rA = then (Ar)2 = ArAr = A0r = 0, Hence Ar = because Ar is a left ideal and R is semiprime The converse is similar 17 Write R = R1 × R2 × · · · × Rn By Exercise 4, every ideal A of has the form A = A1 × A2 × · · · × An where each Ai an ideal of Ri Hence A2 = if and only if A2i = for each i The result follows 19 If R is semiprime, let A be an ideal of Mn (R) By Lemma §3.3, A has the form A = Mn (A) for some ideal A of R Hence if A2 = then A2 = 0, whence A = 0, so A = This shows that Mn (R) is semiprime for any n ≥ Conversely, assume that Mk (R) is semiprime for some k If A2 = where A is an ideal of R, then Mk (A) is an ideal of Mk (R) and Mk (A)2 = Hence Mk (A) = 0, whence A = This shows that R is semiprime 20 (a) If ab = in R then (Ra)(Rb) = because R is commutative Hence, if R is prime either Ra = or Rb = 0, that is a = or b = Thus R is a domain Conversely, if R is a domain and AB = then, if A = / 0= / B, choose = / a∈A and = / b ∈ B Then ab ∈ AB = 0, a contradiction So either A = or B = 0; that is R is a prime ring 21 Write P = P1 ⊕ P2 ⊕ · · · ⊕ Pn , and view P as an internal direct sum Then each Pi is isomorphic to a direct summand of a free module Fi ; by Lemma we may assume that Fi = Pi ⊕ Qi Then F = F1 ⊕ F2 ⊕ · · · ⊕ Fn is also free, and F = (P1 ⊕ P2 ⊕ · · · ⊕ Pn ) ⊕ (Q1 ⊕ Q2 ⊕ · · · ⊕ Qn ) Hence P1 ⊕ P2 ⊕ · · · ⊕ Pn is projective by Theorem 22 (a) If α : M → N is R-linear and K ⊆ M is simple then either α(K) = or γ(K) ∼ = K by Schur’s lemma Either way, α(K) ⊆ soc N Since soc M is a sum of simple submodules of M, it follows that α(soc M ) ⊆ soc N (c) If M = N1 ⊕ N1 it is clear that soc(N1 )⊕ soc(N2 ) ⊆ soc M because soc(Ni ) is semisimple for each i For the other inclusion, define the projection π1 : M → N1 by π1 (n1 + n2 ) = n1 , and define π2 : M → N2 analogously If K ⊆ M is any simple module then πi (K) ⊆ socNi for each i [πi (K) is either or isomorphic to K by Schur’s lemma] Hence if k ∈ K then k = π1 (k) + π2 (k) ∈ soc(Ni )⊕ soc(Ni ) It follows that soc(M ) ⊆ soc(N1 ) ⊕ soc(N2 ) Appendices APPENDIX A: COMPLEX NUMBERS (a) x = (c) x = 0, x = 4i (a) − 9i (c) − (a) z = − 3i 13 (c) z = + (e) − i 13 i ± √12 (1 − i) (g) − (e) z = + 3i Write z = a + bi and w = c + di (a) im (iz) = im (−b + ai) = a = re z (c) z + z¯ = 2a = re z (e) re (z + w) = a + c = re z + re w; re (tz) = ta = t re z (a) unit circle (c) line y = x (e) and the positive real axis |w + z| = (w + z)(w + z) = (w + z)(w ¯ + z¯) = ww ¯ + z z¯ + w¯ z + zw ¯ = |w|2 + |z|2 + w¯ z + z w ¯ (a) If z = a + bi and w = c + di, then z − w = (a − c) + (b − d)i, so |z − w|2 = (a − c)2 + (b − d)2 Take positive square roots √ 10 (a) e−πi/4 (c) 2e5πi/6 √ √ 11 (a) −3 (c) − + 2i √ (c) 16 12 (a) −2 − 3i 2iθ iθ (e) 7e−πi/2 (e) − √12 − √1 i (e) −64 = (cos θ + i sin θ) 13 (a) cos 2θ + i sin 2θ = e = e = (cos2 θ − sin2 θ) + i(2 cos θ sin θ) 14 (a) ± √12 (1 + i), ± √12 (1 − i) √ √ (c) 3i, 32 ( − i), 32 (− − i) Student Solution Manual to Accompany Introduction to Abstract Algebra, Fourth Edition W Keith Nicholson © 2012 John Wiley & Sons, Inc Published 2012 by John Wiley & Sons, Inc 147 148 Appendices 15 (a) z = r(cos θ + sin θ) so z¯ = r(cos θ − sin θ) But sin (−θ) = − sin θ and cos (−θ) = cos θ, so re−iθ = r[cos(−θ) + i sin(−θ)] = z¯ If z −1 = seiφ , then = zz −1 = rsei(θ+φ) Thus rs = 1, so s = 1r , and θ + φ = so (one choice for φ is) φ = −θ Hence z −1 = 1r e−iθ = 1r z¯ 16 (a) Let w = e2πi/n so that the k th root of unity is wk = wk by DeMoivre’s theorem Now (1 − wn ) = (1 − w)(1 + w + w2 + · · · + wn−1 ) by the Hint so, (1 − w)(1 + w + w2 + · · · + wn−1 ) = because wn = Since w = / this implies that + w + w2 + · · · + wn−1 = 0, as required 17 (a) Have zi = eiθi for angles θi The angles between them all equal β = 2π (because they are equally spaced) Let θ1 = α as in the diagram Then z1 = eiα , z2 = ei(α+β) , z3 = ei(α+2β) , z4 = ei(α+3β) , z5 = ei(α+4β) If we write z = eiβ , then z = Now use the hint: z1 + z2 + z3 + z4 + z5 = eiα (1 + z + z + z + z ) − z5 = = eiα 1−z 19 Let f (x) = ni=1 xi , ∈ R If z ∈ C is a root of f (x), then f (z) = Hence, 0=¯ = f (z) = ni=1 z i = ni=1 a ¯i z¯i = ni=1 z¯i = f (¯ z ) because a ¯i = for each i (being real) 21 Let z = reiθ (a) If t > 0, tz = treiθ has the same angle θ as z Hence tz is on the line through and z, on the same side of as z (b) If t = −s, s > 0, then tz = srei(θ+π) , and so is on the line through and z, on the other side of from z √ √ √ 23 (a) If b = a1 + b1 2, then a − a1 = (b1 − b) If b1 = / b, then √ a +a−a = b1 −b1 is rational, an impossibility Hence, b1 = b whence a = a1 √ (c) pq = (ac + 2bd) + (ad + bc) 2, so: √ √ √ pq = (ac + 2bd) − (ad + bc) = (a − b 2)(c − d 2) = p˜q˜ (e) Using (c) and (d) above: [pq] = (pq)(pq) = pq p˜q˜ = (p˜ p)(q q˜) = [p][q] APPENDIX B: MATRIX ARITHMETIC If A is invertible then B = IB = (A−1 A)B = A−1 = 0, contrary to assumption (a) By the definition of matrix multiplication column k of AB is exactly ABk Appendix C: Zorn’s Lemma AB = I2 but BA = A = 0 −9 −10 −6 −5 −30 −30 16 and B = B −1 = B), but A + B = −1 0 149 are both invertible (in fact A−1 = A and is not invertible by Theorem because det(A + B) = For a simpler example, take A = U and B = −U for any invertible matrix U (A + B)(A − B) = A2 + AB − BA − B , so (A − B)(A + B) = A2 − B if and only if AB − BA = It is routine that A3 = I, so AA2 = I = A2 A This implies that A−1 = A2 11 (a) AA−1 = I = A−1 A shows A is the inverse of A−1 , that is (A−1 )−1 = A Next, (AB)(B −1 A−1 ) = AIA−1 = AA−1 = I, and similarly (B −1 A−1 )(AB) = I Hence AB is invertible and (AB)−1 = B −1 A−1 13 (a) If AB = I then det A det B = det(AB) = det I = Hence det A is a unit so, by Theorem 7, A is invertible But then A−1 = A−1 I = A−1 (AB) = B, whence BA = A−1 A = I as required 15 (a) The only nonzero entry in Eik Emj must come from entry k of row i of Eik and column m of Emj The result follows (c) For each i and j, aij Eij has aij in the (i, j)-entry and zeros elsewhere So their sum is [aij ] APPENDIX C: ZORN’S LEMMA Let K ⊆ M be modules, M finitely generated, say M = Rx1 + Rx2 + · · · + Rxn If S = {X ⊆ M | K ⊆ X ⊂ M }, we must show that S contains maximal members Suppose {Xi | i ∈ I} is a chain from S, and put U = ∪i∈I Xi It is clear that U is a submodule and that K ⊆ U, and we claim that U = / M For if U = M then each xi ∈ U, and so each xi ∈ Xk for some k Since the Xi form a chain, this means that {x1 , x2 , , xn } ⊆ Xm for some m Since the xi generate M this means that M ⊆ Xm , contradicting the fact that Xm ∈ S This shows that U ∈ S, and so U is an upper bound for the Xi Hence S has maximal members by Zorn’s lemma, as required Let K ⊆ M be modules (a) Let S = {X ⊆ M | X is a submodule and K ∩ X = 0} Then S is nonempty because ∈ S, so let {Xi | i ∈ I} be a chain from S and put U = ∪i∈I Xi It is clear that U is a submodule, and K ∩ U = because K ∩ U ⊆ K ∩ Xi = for each i Hence U is an upper bound for the chain {Xi | i ∈ I}, so S contains maximal members by Zorn’s lemma Let S be the set of all prime ideals of R, and partially order S downward: Let P ≤ Q mean P ⊇ Q We must find a maximal element in S Let {Pi | i ∈ I} be a chain from S and put Q = ∩i∈I Pi We claim that Q is an upper bound 150 Appendices on {Pi | i ∈ I} Clearly Q ⊆ Pi for each i, so it remains to show that Q is a prime ideal If rs ∈ Q where r, s ∈ R; we must show that either r ∈ Q or s ∈ Q Suppose on the contrary that r ∈ / Q and s ∈ / Q Then r ∈ / Pi for some i, and s∈ / Pj for some j Since the Pi form a chain, one of Pi ⊆ Pj or Pj ⊆ Pi must hold; assume Pi ⊆ Pj Then r ∈ / Pj and s ∈ / Pj but rs ∈ Pj (because rs ∈ Q ⊆ Pj ) This contradicts the fact that Pj is a prime ideal Since we also obtain a contradiction if Pj ⊆ Pi , this proves that Q is a prime ideal ...www.TechnicalBooksPdf.com Solutions Manual to Accompany Introduction to Abstract Algebra www.TechnicalBooksPdf.com www.TechnicalBooksPdf.com Solutions Manual to Accompany Introduction to Abstract Algebra Fourth. .. = so x = or y = 0, contradicting our assumption Student Solution Manual to Accompany Introduction to Abstract Algebra, Fourth Edition W Keith Nicholson © 2012 John Wiley & Sons, Inc Published... 2k+3 = 9(7m − 2k+2 ) + 2k+3 = · 7m − 2k+2 (9 − 2) Student Solution Manual to Accompany Introduction to Abstract Algebra, Fourth Edition W Keith Nicholson © 2012 John Wiley & Sons, Inc Published