A Brief Introduction to Classical and Adelic Algebraic ... - William Stein

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58 CHAPTER 9. CHINESE REMAINDER THEOREM Theorem 9.1.3 (Chinese Remainder Theorem). Suppose I1, . . .,Ir are ideals of OK such that Im + In = OK for any m = n. Then the natural homomorphism OK → r n=1 (OK/In) induces an isomorphism r r OK/ → (OK/In). n=1 In Thus given any an ∈ In then there exists a ∈ OK such that a ≡ an (mod In) for n = 1, . . .,r. Proof. First assume that we know the theorem in the case when the In are powers of prime ideals. Then we can deduce the general case by noting that each OK/In is isomorphic to a product OK/pem m , where In = pem m , and OK/( n In) is isomorphic to the product of the OK/pe , where the p and e run through the same prime powers as appear on the right hand side. It thus suffices to prove that if p1, . . .,pr are distinct prime ideals of OK and e1, . . .,er are positive integers, then r ψ : OK/ p en r n → (OK/p en n ) n=1 is an isomorphism. Let ϕ : OK → ⊕r en n=1 (OK/pn ) be the natural map induced by reduction mod pen n . Then kernel of ϕ is ∩r n=1pen n , which by Lemma 9.1.1 is equal to r n=1 pen n , so ψ is injective. Note that the projection OK → OK/pen n of ϕ onto each factor is obviously surjective, so it suffices to show that the element (1, 0, . . .,0) is in the image of ϕ (and the similar elements for the other factors). Since J = r n=2 pen n is not divisible by p1, hence not contained in p1, there is an element a ∈ J with a ∈ p1. Since p1 is maximal, OK/p1 is a field, so there exists b ∈ OK such that ab = 1 − c, for some c ∈ p1. Then 1 − c n1 = (1 − c)(1 + c + c 2 + · · · + c n1−1 ) = ab(1 + c + c 2 + · · · + c n1−1 ) n=1 n=1 is congruent to 0 mod pen n for each n ≥ 2 since it is in r to 1 modulo p n1 1 . n=2 pen n , and it is congruent Remark 9.1.4. In fact, the surjectivity part of the above proof is easy to prove for any commutative ring; indeed, the above proof illustrates how trying to prove something in a special case can result in a more complicated proof!! Suppose R is a ring and I, J are ideals in R such that I + J = R. Choose x ∈ I and y ∈ J such that x + y = 1. Then x = 1 − y maps to (0, 1) in R/I ⊕ R/J and y = 1 − x maps to (1, 0) in R/I ⊕ R/J. Thus the map R/(I ∩ J) → R/I ⊕ R/J is surjective. Also, as mentioned above, R/(I ∩ J) = R/(IJ). Example 9.1.5. The Magma command ChineseRemainderTheorem implements the algorithm suggested by the above theorem. In the following example, we compute a prime over (3) and a prime over (5) of the ring of integers of Q( 3√ 2), and find an element of OK that is congruent to 3√ 2 modulo one prime and 1 modulo the other.
9.1. THE CHINESE REMAINDER THEOREM 59 > R := PolynomialRing(RationalField()); > K := NumberField(x^3-2); > OK := MaximalOrder(K); > I := Factorization(3*OK)[1][1]; > J := Factorization(5*OK)[1][1]; > I; Prime Ideal of OK Two element generators: [3, 0, 0] [4, 1, 0] > J; Prime Ideal of OK Two element generators: [5, 0, 0] [7, 1, 0] > b := ChineseRemainderTheorem(I, J, OK!a, OK!1); > b - a in I; true > b - 1 in J; true > K!b; -4 The element found by the Chinese Remainder Theorem algorithm in this case is −4. The following lemma is a nice application of the Chinese Remainder Theorem. We will use it to prove that every ideal of OK can be generated by two elements. Suppose I is a nonzero integral ideals of OK. If a ∈ I, then (a) ⊂ I, so I divides (a) and the quotient (a)/I is an integral ideal. The following lemma asserts that (a) can be chosen so the quotient (a)/I is coprime to any given ideal. Lemma 9.1.6. If I, J are nonzero integral ideals in OK, then there exists an a ∈ I such that (a)/I is coprime to J. Proof. Let p1, . . .,pr be the prime divisors of J. For each n, let vn be the largest power of pn that divides I. Choose an element an ∈ pvn n that is not in pvn+1 n (there is such an element since pvn n = pvn+1 n , by unique factorization). By Theorem 9.1.3, there exists a ∈ OK such that for all n = 1, . . .,r and also a ≡ an (mod p vn+1 n ) a ≡ 0 (mod I/ p vn n ). (We are applying the theorem with the coprime integral ideals p vn+1 n , for n = 1, . . .,r and the integral ideal I/ p vn n .)
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A Brief Introduction to Classical a
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4 CONTENTS 8 Factoring Primes 49 8.
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6 CONTENTS
Page 8 and 9: 8 CHAPTER 1. PREFACE Acknowledgemen
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Page 15 and 16: Chapter 3 Finitely generated abelia
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Page 57: Chapter 9 Chinese Remainder Theorem
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Page 63 and 64: Chapter 10 Discrimannts, Norms, and
Page 65 and 66: 10.2. DISCRIMINANTS 65 Lemma 10.2.1
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Page 73 and 74: Chapter 11 Computing Class Groups I
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Page 79 and 80: 12.1. THE GROUP OF UNITS 79 Lemma 1
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108 CHAPTER 15. VALUATIONS Note tha
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110 CHAPTER 15. VALUATIONS To say t
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112 CHAPTER 15. VALUATIONS Proof. F
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114 CHAPTER 15. VALUATIONS 1 so |u|
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116 CHAPTER 15. VALUATIONS of a val
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118 CHAPTER 16. TOPOLOGY AND COMPLE
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130 CHAPTER 17. ADIC NUMBERS: THE F
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138 CHAPTER 18. NORMED SPACES AND T
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146CHAPTER 19. EXTENSIONS AND NORMA
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154 CHAPTER 20. GLOBAL FIELDS AND A
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168 CHAPTER 21. IDELES AND IDEALS E
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170 CHAPTER 21. IDELES AND IDEALS T
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172 CHAPTER 21. IDELES AND IDEALS P
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174 CHAPTER 22. EXERCISES 6. Find t
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176 CHAPTER 22. EXERCISES 22. (*) S
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178 CHAPTER 22. EXERCISES (a) The p
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180 CHAPTER 22. EXERCISES 60. Find
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182 CHAPTER 22. EXERCISES
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184 BIBLIOGRAPHY [Fre94] G. Frey (e
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186 INDEX complex n-dimensional rep
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188 INDEX lies over, 73 local compa
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190 INDEX valuations such that |a|
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A Brief Introduction to Classical and Adelic Algebraic ... - William Stein

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