Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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92 ADVANCED ABSTRACT ALGEBRA Now (ad + bc) (b 1 d 1 ) = adb 1 d 1 + bcb 1 d 1 = a(db 1 )d 1 + b(cb 1 )d 1 = ab 1 dd 1 + bb 1 cd 1 (by commutativity of D) = ba 1 dd 1 + bb 1 c 1 d (using (i)) = bd (a 1 d 1 + b 1 c 1 ) Therefore addition is well defined. If then that is Now a a = 1 c c , = 1 b b1 d d1 ac a = 1c1 bd b1d1 acb 1 d 1 = bda 1 c 1 acb 1 d 1 = ab 1 cd 1 = ba 1 dc 1 = bda 1 c 1 ∴ multiplication is also well defined. We now prove that F is a field under these operations of addition and multiplications. Let (i) and Therefore (ii) Similarly a c e , , ∈ F . Then b d f a c e + + = b d f a + b c d ad + bc + bd e + f e f = = a cf + de = + b df = a c e + + = b d f 0 a 0 .b + ba + = b b 2 b a 0 a + = b b b ( ad + bc)f + bde bdf adf + bcf + bde bdf a + b c d adf + bcf + bde bdf e + f ab a = 2 = . b b Therefore 0 is additive identity. b
UNIT-IV 93 (iii) a + b − a b ab − ab = 2 b Thus every element of F is invertible. a c ad + (iv) + = b d bdbc and c a cb + + = d b dbda = 0 0 a a = 2 = = − + b b b b bc + ad bd (by commutativity of D) (v) (vi) If a c e , , ∈ F , then b d f a b a c e . . = b d f c d ace bdf = a c e . b d f e a cf + de + = . f b df = acf ade + bdf bdf ac ae = + . bd bf Similarly it can be shown that (vii) and Hence (viii) a c e + . = b d f a a b a a a . a b = ae + bf ce df aa a = = ba b a . b a = 1 is multiplicative identity. a a b b a b a = a b ab = = 1 ba ba ba = = 1 ab ab Thus every element of F is invertible. (ix) a c = b d ac bd c a = . d b Hence F is a field. This field F is called Quotient field or field of fractions. We define a function f : D → F
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1 Advanced Abstract Algebra M.A./M.
Page 3 and 4:
Contents 3 Unit I 5 Unit II 49 Unit
Page 5 and 6:
UNIT-I 5 Unit-I Definition Group A
Page 7 and 8:
UNIT-I 7 2. In example 1, a 1 1 0 1
Page 9 and 10: UNIT-I 9 Solutions: 1. CGbg≠ a φ
Page 11 and 12: UNIT-I 11 ∴ Ha ⊆ H. To show let
Page 13 and 14: UNIT-I 13 Corollary 4: (Feremat's L
Page 15 and 16: UNIT-I 15 Definition: Group Homomor
Page 17 and 18: UNIT-I 17 Proof: Consider the diagr
Page 19 and 20: UNIT-I 19 ( G ) N G ( H ) ≅ H N D
Page 21 and 22: UNIT-1 21 Theorem 8. Let H be a sub
Page 23 and 24: UNIT-1 23 From (i) and (ii), we hav
Page 25 and 26: UNIT-I 25 1 Note that αβ = 2 2 1
Page 27 and 28: UNIT-1 27 In this case the length o
Page 29 and 30: UNIT-1 29 | n Theorem 18. The set o
Page 31 and 32: UNIT-I b = φ ( φ φ ) ( ) g h t x
Page 33 and 34: UNIT-I 33 This is impossible. So no
Page 35 and 36: UNIT-I 35 integers x and y. Note th
Page 37 and 38: UNIT-I 37 O(7) = 4, as 7 1 = 7, 7 2
Page 39 and 40: UNIT-I 39 Put a x mt , = b = x Then
Page 41 and 42: UNIT-I n iii. N = ∈Gs, show that
Page 43 and 44: UNIT-I 43 abab------ab = a n b n
Page 45 and 46: UNIT-I Q.20. Q.21. Q.22. Q.23. Q.24
Page 47 and 48: UNIT-I 47 Remark : If G is abelian,
Page 49 and 50: UNIT-IV 81 Definition. Let S be a s
Page 51 and 52: UNIT-IV 83 (r+s) − (r 1 +s 1 ) =
Page 53 and 54: UNIT-IV 85 and = ψ(r+K) + ψ(s+K)
Page 55 and 56: UNIT-IV 87 M ⊂ M′ ⊂ R M′ =
Page 57 and 58: UNIT-IV 89 We define addition and m
Page 59: UNIT-IV 91 be the set of the ordere
Page 63 and 64: UNIT-IV 95 and Z/A ~ Im(f) = f (z)
Page 65 and 66: UNIT-IV 97 If a n = 0 for all n ≥
Page 67 and 68: UNIT-IV 99 x 3 = (0, 0, 0, 1,…) .
Page 69 and 70: UNIT-IV 101 Then either (i) or n i
Page 71 and 72: UNIT-IV 103 deg f(x) = 0 and deg g(
Page 73 and 74: UNIT-IV 105 If x, y ∈ A , then th
Page 75 and 76: UNIT-IV 107 The possibility φ(r) <
Page 77 and 78: UNIT-IV 109 Lemma 4. If f(x) is an
Page 79 and 80: UNIT-IV 111 Proof. Let, if possible
Page 81 and 82: UNIT-II 49 Unit-II Definition: Comp
Page 83 and 84: UNIT-II 51 If r = 1, then G is simp
Page 85 and 86: UNIT-II 53 group of the Commutator
Page 87 and 88: UNIT-II di + 1 idi + 1 i di + 1 idi
Page 89 and 90: UNIT-II 57 we write the refinement
Page 91 and 92: UNIT-II 59 of length e in which eac
Page 93 and 94: UNIT-II 61 of nilpotency of G) , so
Page 95 and 96: UNIT-III 63 n.v = R i.e. abelian gr
Page 97 and 98: UNIT-III 65 Theorem 1 Let M, N be R
Page 99 and 100: UNIT-III 67 F ( BA ad 1'= = , 1a d(
Page 101 and 102: UNIT-III 69 Vector space over F. Le
Page 103 and 104: UNIT-III 71 1. J = only one linearl
Page 105 and 106: UNIT-III Hence J = −L NM 4 0 0 0
Page 107 and 108: UNIT-III 75 Example 5 The direct su
Page 109 and 110: UNIT-IV 77 A. B ≠ B. A. If IR, th
Page 111 and 112:
UNIT-IV 79 Example: (D, +, .) is a
Page 113 and 114:
UNIT-IV 81 Examples 1. The polynomi
Page 115 and 116:
UNIT-V 129 (Characteristic of a rin
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UNIT-V 131 As F = p dΘ F ≅ Z p i
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UNIT-V 133 In this case is called t
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+ c c+ 0 0 1 c c+ 1 1 1 0 1+ c c c
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UNIT-V 137 element of defines a per
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Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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