Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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104 ADVANCED ABSTRACT ALGEBRA Definition. A non-zero element of integral domain D, which is not a unit and which has no proper divisors is called a prime or irreducible (indecomposible) element. Definition. An element a is said to be an associate of b if a is a divisor of b and b is a divisor of a. For example each of 3 and −3 is a divisor of the other in the domain I of integrals. Definition. A ring R is called a factorisation domain if every non-zero non-unit element of the same can be expressed as a product of irreducible elements. Thus if a is non-zero unit element of a F.D. then a = p 1 p 2 p 3 … p n , where p i ’s are irreducible elements. Definition. A F.D. is called a unique factorisation domain if whenever a = p 1 p 2 p 3 …p n = q 1 q 2 … q S then r = s and after rearrangement, if necessary, p 1 ~ q 1 , p 2 ~ q 2 …, p r ~ q s . Definition. An integral domain D is said to be principal ideal domain if every ideal A in D is principal ideal. Theorem. A principal ideal domain is a unique factorisation domain. Proof. Firstly we show that principal ideal domain is a factorisation domain. Let a be a non-zero non-unit element of a principal ideal domain D. If a is prime we are done. If a is not a prime, there exist two non-unit elements b and c such that a = bc a ∈ (b) (a) ⊂ (b) , (b) ≠ (a) . In case b, c are both irreducible, then again we have finished. If they are not prime, we continue as above. That is, there exists two non-unit elements c and d such that b = cd b ∈ (c) (b) ⊂ (c) (b) ≠ (c) Thus two cases arise : (i) After a finite number of steps, we arrive at an expression of a as a product of irreducible elements. (ii) Howsoever far we may continue, we always have a composite element occurring as a factor in the expression of a as product of elements of D. In case (i) we have finished. In case (ii), there exists an infinite system of elements a 1 , a 2 ,…, a n ,… such that (a 1 ) ⊂ (a 2 ) ⊂ (a 3 ) … ⊂ (a n ) ⊂ …. (I) no two of these principal ideals being the same. Consider the union A = U(a i ) We assert that A is an ideal of D. In fact 0 ∈ (a 1 ) 0 ∈ A A ≠ φ .
UNIT-IV 105 If x, y ∈ A , then there exist integers i and j such that x ∈ (a i ) , y ∈ (a j ) Without loss of generality suppose that i ≥ j . Then x, y ∈ (a i ) . This implies that x − y ∈ (a i ) and α ∈ (a i ) where α ∈ D. Hence x−y, αx ∈ A. Since D is a principal ideal domain, therefore ∃ an element β of D such that A = (β) . There exists, therefore, an ideal member (a m ) of the system such that β ∈ (a m ) and accordingly β ∈ (a n ) for all n ≥ m (β) ⊂ (a n ) for all n ≥ m Also since (β) is the union of the ideals, we have (β) ⊃ (a n ) for all n Thus from (II) and (III) (β) = (a n ) for all n ≥ m, (a m ) = (a m+1 ) = (a m+2 ) = …. which is a contradiction to (I). Hence case (ii) cannot arise. Thus we have proved that every non-zero non-unit element of a principal ideal domain is expressible as a product of prime element. Hence D is a F.d. To prove the uniqueness, let a = p 1 p 2 … p r = q 1 q 2 …q s (IV) where each p and q is prime. We shall prove the result by induction on r. The result is obvious if r = 1. Suppose now that the result is true for each natural number < r. Since D is a principal ideal domain, every prime element generates a prime ideal. Therefore p 1 p 2 … p n ∈ (p 1 ) which implies q 1 q 2 …. q s ∈ (p 1 ) Therefore one of the factors q 1 q 2 …. q s should belong to (p 1 ). Without loss of generality say, q 1 ∈ (p 1 ). Then p 1 |q 1 . As q 1 is prime, this implies q 1 /p 1 and therefore p 1 and q 1 are associates. Let q 1 = e 1 p 1 (V) where e 1 is a unit. From (IV) and (V) we have p 2 p 3 … p 2 = (e 1 q 2 ) q 3 … q s (VI) By the assumed hypothesis r−1 = s−1 and each factor, on the right of (VI) is an associate of some factor on the left and vice-versa. This proves the theorem. (II) (III) Theorem. In a principal ideal domain a prime element generates a maximal ideal. Proof. Let p be a prime element in a principal ideal domain R and let p = (p) be an ideal of R generated by p.
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1 Advanced Abstract Algebra M.A./M.
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Contents 3 Unit I 5 Unit II 49 Unit
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UNIT-I 5 Unit-I Definition Group A
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UNIT-I 7 2. In example 1, a 1 1 0 1
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UNIT-I 9 Solutions: 1. CGbg≠ a φ
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UNIT-I 11 ∴ Ha ⊆ H. To show let
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UNIT-I 13 Corollary 4: (Feremat's L
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UNIT-I 15 Definition: Group Homomor
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UNIT-I 17 Proof: Consider the diagr
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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 and 60: UNIT-IV 91 be the set of the ordere
Page 61 and 62: UNIT-IV 93 (iii) a + b − a b
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: UNIT-IV 103 deg f(x) = 0 and deg g(
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
Page 117 and 118: UNIT-V 131 As F = p dΘ F ≅ Z p i
Page 119 and 120: UNIT-V 133 In this case is called t
Page 121 and 122: + 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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