Advanced Number Theory with Applications (Solutions, Instructor Solutions Manual) [1 ed.] 9781439824825, 9781420083286, 1420083287

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SOLUTIONS MANUAL FOR Advanced Number Theory with Applications

by Richard A. Mollin

SOLUTIONS MANUAL FOR Advanced Number Theory with Applications

by Richard A. Mollin Univeristy of Calgary

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2010 by Taylor and Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number: 978-1-4398-2482-5 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

v

Solutions to Even-Numbered Exercises

Introduction This manual for the main text, Advanced Number Theory, is intended to be a teaching aid for an instructor who adopts the main text for a second course in number theory. With nearly 170 unworked even-numbered exercises therein, this will come as a welcome additional tool. The solutions that are provided have been worked by this author in detail. In any case, suggestions or comments are welcome, and if errors or typos are found, then a communication via the address below is welcome.

Richard A. Mollin, Calgary, June 30, 2009 http://www.math.ucalgary.ca/˜ ramollin/ [email protected]

1

Solutions to Even-Numbered Exercises

Solutions to Even-Numbered Exercises Section 1.1 1.2 Since 0 ∈ Rj for all j ∈ I, then 0 ∈ ∩j∈I Rj , so ∩j∈I Rj #= ∅. For any a, b ∈ ∩j∈I Rj , a, b ∈ Rj for all j ∈ I, so a + b, ab ∈ Rj for all such j. Hence, a + b, ab ∈ ∩j∈I Rj . Thus, the latter is a ring in R. To see that ∪j∈I Rj is a subring, we first note that it is nonempty since 0 ∈ ∪j∈I Rj , since 0 is in every Rj . If a, b ∈ ∪j∈I Rj , then there are k, ! ∈ J such that a ∈ Rk , b ∈ R! . If k ≤ !, then Rk ⊆ R! , so a + b ∈ R! , and we have additive closure since a + b ∈ ∪j∈I Rj . Lastly, since each Rj is closed under multiplication, then so is ∪j∈I Rj and we have the result.

1.4 It suffices to prove the result for n = 2 since we may extrapolate by induction from this case. Q α Consider mα,Q (x) = dj=1 (x − αj ), where the αj are all of the conjugates of Qdβ α1 = α over Q, and let mβ,Q (x) = j=1 (x − βj ), where the βj are all of the conjugates of β1 = β over Q . Also, αj #= αk for any j #= k, and βi #= β! for any i #= !, by Corollary 1.3 on page 11. Select a q ∈ Q such that q #= (α−αk )/(βj −β) for any k = 1, 2, . . . , dα and any j = 1, 2, . . . , dβ , and let γ = α + qβ,

(ES1)

with Since and

f (x) = mα,Q (γ − qx) ∈ Q(γ)[x]. f (β) = mα,Q (γ − qβ) = mα,Q (α) = 0, mβ,Q (β) = 0,

then β is a common root of f (x) and mβ,Q (x). We now show that this is the only common root. If there exists a σ ∈ C, with σ #= β, such that f (σ) = 0 = mβ,Q (σ) = 0, then σ = βj for some j > 1. Since 0 = mα,Q (α) = mα,Q (γ − qβj ) = f (βj ), then there is a k ∈ {1, 2, . . . , dβ } such that γ − qβj = αk . Thus, by (ES1), αk + qβj = γ = α + qβ, so

α − αk , βj − β contradicting the choice of q. We have shown that β is the only common ˛ root of f (x) and m˛β,Q (x). Therefore, by Theorem 1.6 on page 10, mβ,Q(γ) (x) ˛ f (x) and mβ,Q(γ) (x) ˛ mβ,Q (x). However, since f (x) and mβ,Q (x) have only one root in common, then deg(mβ,Q(γ) (x)) = 1. Thus, mβ,Q(γ) (x) = x+δ for some δ ∈ Q(γ). Since mβ,Q(γ) (β) = 0 = β + δ, then β = −δ ∈ Q(γ), so α = γ − qβ ∈ Q(γ). This shows that Q(α, β) ⊆ Q(γ). However, since γ = α + qβ ∈ Q(α, β), then Q(γ) ⊆ Q(α, β). We have completed the proof that Q(α, β) = Q(γ), as required. q=

2

Advanced Number Theory

1.6 Clearly, Q(

√

2 (1 2

+ i)) ⊆ Q(i,

√

2). To see that equality holds, we observe that

„√

«2 2 (1 + i) = i = ζ4 , 2

so

„√

« 2 (1 + i) 2

is a primitive eighth root of unity, and so is any odd power thereof. Since ˛ „√ ˛ « √ ˛ ˛ 2 |Q(i, 2) : Q| = 4 = ˛˛Q (1 + i) : Q˛˛ , 2 then we must have

Q(i,

√

2) = Q

„√

« 2 (1 + i) . 2

Section 1.2 √ √ √ 1.8 (a) If α ∈ F = Q( D), then α = e + f D ∈ Q( D). Thus, √ √ N (α) = (e + D)(e − D) = e2 − f 2 D ∈ Q. √ √ (b) If α = e + f D, β = g + h D ∈ F , then √ N (αβ) = N ((eg + f hD) + (gf + eh) D) = √ √ ((eg + f hD) + (gf + eh) D)((eg + f hD) − (gf + eh) D) =

(eg + f hD)2 − (gf + eh)2 D = e2 g 2 + f 2 h2 D2 − g 2 f 2 D − e2 h2 D = √ √ √ √ (e −f 2 D)(g 2 −h2 D) = (e+f D)(e−f D)(g +h D)(g −h D) = N (α)N (β), 2

as required. (c) If N (α) = 0, then N (α) = e2 − f 2 D = 0. By part (a), e = a/b and f = c/d for a, b, c, d ∈ Z, then we may assume that gcd(a, b) = 1 = gcd(c, d). Thus, a2 d2 = c2 b2 D, which implies |a| = |c| and |b| = |d| since D may be assumed to be squarefree by Theorem 1.2 on page 5. Thus, if e #= 0 or√f #= 0, then√D = 1, a contradiction, so e = f = 0. Conversely, if α = 0 = e + f D = 0 + 0 D, then clearly, N (α) = 02 + 02 D = 0. (d) If √ « „ √ σ−1+ D α = (a + b D)/σ ∈ OF = Z + Z , σ in the notation of the proof of Theorem 1.3 on page 6, then N (α) =

a2 − b2 D ∈ Z, σ2

since, again by the proof of Theorem 1.3, a, b must have the same parity if σ = 2. ˛ (e) If α ˛ β, then there is˛ a δ ∈ OF such that β = αδ, so by part (b), NF (β) = NF (α)NF (δ), so NF (α) ˛ NF (β) in Z by part (d).

1.10 Since mα,F (αj ) = 0 for each j = 1, 2, . . . , d, then by Definition 1.1, αj ∈ OF for each such j.

Solutions to Even-Numbered Exercises

3

1.12 α = βσ + δ where: (a) σ = i, δ = −3i. (b) σ = 1, δ = −5i. (c) σ = 3, δ = −1 − i. (d) σ = −i, δ = 4. 1.14 α = 2x − y + (x + 2y)i for any x, y ∈ Z, since α = 2x − y + (x + 2y)i = (2 + i)(x + yi) = βσ. 1.16 If γ˛1 and γ2 are associates of one another, ˛ then γ1 = uγ2 , for ˛a unit u. Thus, ˛ γ2 ˛ γ1 , and since u−1 γ1 = γ2 , then γ1 ˛ γ2 . Conversely, if γ1 ˛ γ2 and γ2 ˛ γ1 , then there are δj ∈ OF , for j = 1, 2 such that γ1 = δ1 γ2 and γ2 = δ2 γ1 . Therefore, γ1 = δ1 γ2 = δ1 δ2 γ1 . Hence, dividing through by γ1 , we get that 1 = δ1 δ2 , so δj for j = 1, 2 are units. Thus, γ1 and γ2 are associates of one another. 1.18 Since 1 clearly divides both α = 1 · α and u = 1 · u, then by Definition 1.14 on page˛ 21, we need ˛ only show that any other common divisor of α and u divides 1. If δ ˛ α and δ ˛ u, then there exist σ1 , σ2 ∈ OF such that α = σ1 δ and u = σ2 δ, so˛ δ ∈ U F . However, since 1 is an associate of any unit, then by Exercise 1.16, δ ˛ 1. ˛ ˛ 1.20 If δ ˛ α and δ ˛ β in Z[i], then there exist σ1 , σ2 ∈ Z[i] such that N (α) = N (δ)N (σ1 ) and N (β) = N (δ)(N (σ2 ). ˛ If N (δ) > 1, then N (δ) ˛ gcd(N (α), N (β)) = 1, so δ ∈ UZ[i] by Exercise 1.19. In other words, δ ∼ 1, which secures the result.

1.22 N (1 − 2i) = N (1 + 2i) = 5. However, if 2 + i = u(2 − i) for some u ∈ {±1, ±i}, then a check shows that all of these values of u lead to a contradiction. Hence, this is a counterexample to the equality of norms implying that the preimages are associates. 1.24 (a) 1 + 5i where 7 + 9i = (2 − i)(1 + 5i) (b) 1 + i where 71 + 9i = (40 − 31i)(1 + i) and 111 + 7i = (59 − 52i)(1 + i). 1.26 (a) 3 + i where 111 + 7i = (34 − 9i)(3 + i) and 7 + 9i = (3 + 2i)(3 + i) (b) 1 + 2i where 1 + 7i = (3 + i)(1 + 2i) and 7 + 4i = (3 − 2i)(1 + 2i).

1.28 It suffices to prove the result for n = 2 since the general case then follows by ˛ induction. If ρ ˛ α1 α2 , then there exists a δ ∈ Z[i] such that α1 α2 = ρδ. Assume that ρ # |α1 . Then since ρ is prime, α1 and ρ are relatively prime. Therefore, by Theorem 1.10 on page 21, there are τ, σ ∈ Z[i] such that 1 = ρσ + τ α1 , so multiplying through by α2 we get α2 = ρσα2 + τ α1 α2 = ρσα2 + τ ρδ = ρ(σα2 + τ δ). ˛ Hence, ρ ˛ α2 , and the proof is secured.

1.30 Since UZ[i] = {±1, ±i} and α = a + bi ∈ Z[i] is primary only when a is odd and b is even with a + b ≡ 1 (mod 4), then only α = 1 satisfies all three conditions. 1.32 By Exercise 1.31, a + bi is primary if and only if a + bi ≡ 1 (mod 2 + 2i) in Z[i], so if a + bi ≡ 0 (mod 1 + i), then a + ˛ bi = 1 + (2 + 2i)α and a + bi = (1 + ˛ i)β for some α, β ∈ Z[i]. Hence, (1 + i) ˛ a + bi − 1 = (1 + i)β − 1, so (1 + i) ˛ 1, a contradiction since 1 + i is not a unit.

4

Advanced Number Theory

1.34 If α is a primary prime, then by Exercises 1.30–1.32, and Definition 1.12 on page 19, α = α · 1 is the only factorization of α into a product of primary primes, which must necessarily be odd. This is the induction step. Assume now the induction hypothesis, namely that any primary Gaussian integer, δ, with 2 < N (δ) < N (α), may be uniquely factored into a product of primary primes. Since this means that if α is not a prime, then α = σ1 σ2 for σj ∈ Z[i], and 2 < N (σj ) < N (α) for j = 1, 2. By the induction hypothesis, σj may be uniquely factored into a product of primary primes, so after possibly renumbering the factorizations to respect the non-decreasing nature of the norms, we have the result. 1.36 (a) (2+i)(3+2i)3 (b) (5+2i)2 (1−2i)3 (c) (2+i)(2+5i)(3−2i)3 (d) (2+5i)(3−2i)3 . 1.38 Suppose that N (α) = a2 + b2 = p is prime. If α = βγ for β, γ ∈ OF , then by part (d) of Exercise 1.8 on page 28, NF (β), NF (γ) ∈ Z so they both divide p. Hence, one of them, say NF (β) = 1, so by Exercise 1.19 on page 29, this means that β is a unit in Z[i]. Thus, any divisor of α is a unit or associate. Therefore, α is a Gaussian prime. Since N (α) = N (a + bi) = N (a − bi) = (a + bi)(a − bi) = p, then p is not a Gaussian prime since it is divisible by both a + bi and a − bi neither of which is a unit or an associate of p. Indeed, both a + bi and a − bi are Gaussian primes by Example 1.11 on page 19. Also, by Example 1.15 on page 28, p ≡ 1 (mod 4) or p = 2.

Section 1.3 1.40 The ring R = 2Z has 2 ∈ R but 2 has no divisors in R. Hence, 2, 4 ∈ R have no greatest common divisor. ˛ 1.42 Since 1R ˛ α for all nonzero α ∈ R, then by Exercise 1.41, f (1R ) ≤ f (α). 1.44 (a) Since α ∼ β, there exists a u ∈ UR such that α = uβ so β = u−1 α. Therefore, by condition (a) of Definition 1.17, f (α) = f (uβ) ≥ f (β) = f (u−1 α) ≥ f (α). Hence, f (α) = f (β). (b) By condition (b) of Definition 1.17, ˛ there exist ˛ σ, δ ∈ R such that α = βσ + δ and f (δ) < f (β) = f (α). Since α ˛ β, then α ˛ δ. If δ #= 0, then by Exercise 1.41, ˛ f (δ) ≥ f (α) > f (δ), a contradiction, so δ = 0. Therefore, α = βσ, so β ˛ α. Hence, by Exercise 1.16 on page 29, α ∼ β. (c) This is Exercise 1.43. (d) By condition (b) of Definition 1.17, there exist σ, δ ∈ R such that 0 = ασ + δ and f (δ) < f (α). If δ #= 0, then by condition (a), f (δ) = f (−ασ) ≥ f (α) > f (δ), a contradiction. Hence, δ = 0 and f (δ) = f (0) < f (α). 1.46 We require the following elementary result. Claim ES.1 For a number field F and any ρ ∈ F , there exists z ∈ Z such that zρ ∈ OF .

Solutions to Even-Numbered Exercises

5

Since ρ ∈ F , then by Definition 1.4 on page 2, ρ is a root of an irreducible, monic polynomial, f (x) = q0 + q1 x + · · · + qd−1 xd−1 + xd ∈ Q[x]. If z is the gcd of the denominators of the qj for j = 0, 1, . . . , d − 1, then z d f (x) ∈ Z[x] and if we set y = zρ, then h(y) = z d q0 + z d−1 q1 y + · · · + zqd−1 y d−1 + y d ∈ Z[x], so since y = zρ is a root of h(y), then zρ ∈ OF by Definition 1.1 on page 1, which is the claim. Using Claim ES.1, and applying the condition in Definition 1.18 with α = zρ ∈ OF , and β = z ∈ OF where z #= 0, there exist σ, δ ∈ OF such that zρ = zσ + δ with |N (δ)| < |N (z)|. Thus, |N (δ/z)| = |N (ρ − σ)| < 1. Conversely assume condition (c) holds. Then for α, β ∈ OF , select ρ = α/β ∈ F . Thus, there exists a σ ∈ OF such that ˛ „ «˛ ˛ α − σβ ˛˛ |N (ρ − σ)| = ˛˛N ˛ < 1. β

By selecting δ = α − σβ, we get,via multiplication through by |N (β)|, α = σβ + δ where |N (δ)| < |N (β)|, which is the entire result. 1.48 If N (4 +

p √ √ 10) = (a + b 10)(c + d 10),

for a, b, c, d ∈ Z, then √ √ √ √ √ √ 2 · 3 = (4 + 10)(4 − 10) = (a + b 10)(c + d 10)(a − b 10)(c − d 10) = (a2 − 10b2 )(c2 − 10d2 ). (ES2) √ √ Thus, if a ± b 10 and c ± d 10 are not units in Z[ 10], then it follows that one of the factors in (ES2) is ±2 and the other is ±3. Without loss of generality, we may assume a2 − 10b2 = ±2 and c2 − 10d2 = ±3. √

If a2 − 10b2 = 2, then b is odd, and a is even. so 2(a/2)2 − 5b2 = 1, which implies that 2(a/2)2 ≡ 6 (mod 8), so (a/2)2 ≡ 3 (mod 4), a contradiction. Thus, a2 − 10b2 = −2. Therefore, by (ES2), c2 − 10d2 = −3, so c is odd. Therefore, −3 = c2 − 10d2 ≡ 1 − 10d2 (mod 8), which implies that 10d2 ≡ 4 (mod 8), or d2 ≡ 5d2 ≡ 2 (mod 4), again a contradiction. √ √ √ Hence, one of a + b 10 or c + d 10 is a unit, √ which proves that 4 + 10 is irreducible. The same argument shows that 4 − 10 is irreducible. It follows that 2 and 3 are irreducible since otherwise one would factor into a √ product of two non-units contradicting that 4 ± 10 are irreducible. Moreover, √ since N (2) = 4, N (3) = 9, N (4 ± √ 10) = 6, then by Exercise 1.21 on page 29, 2, 3 cannot be associates of 4 ± 10.

6

Advanced Number Theory

1.50 If α = βγ for β, γ ∈ OF , then NF (α) = p = NF (β)NF (γ), so one of NF (β) or NF (γ) is equal to ±1. Thus, by Exercise 1.19 on page 29 one of them is a unit. In other words, α is divisible only by units and associates so α is irreducible. 1.52 By Theorem 1.16 on page 38, α is prime if and only if α is irreducible, so by Exercise 1.50, α is a prime in OF . P j 1.54 Let x = p−1 j=0 ζp . Then xζp = x. If x #= 0, then ζp = 1, a contradiction. Thus, x = 0 as required.

Section 1.4 1.56 If p ∈ Z is prime, then by Exercise 1.19 on page 29, p is not a unit in OF since N (p) = p2 , so p = αβ where α is a prime in OF and β ∈ OF . Since p2 = NF (p) = NF (α)NF (β), and |NF (α)| = # 1, then |NF (β)| = 1 or |NF (β)| = p. In the former case, β is a unit so α is an associate of p which must also be prime in OF . In the latter case, β is a prime in OF by Exercise 1.52 on page 46, so p is a product of two primes in OF , as required. 1.58 (a) If ( D ) = 1, then there exists a rational integer x such that D ≡ x2 (mod p), ˛ p2 ˛ √ √ so p ˛ (x − D) = (x − D)(x + D). If p is also a prime in OF , then either p ˛ ˛ √ √ √ (x − D) or p ˛ (x + D), which implies one of (x ± D)/p ∈ OF , contradicting Theorem 1.3 on page 6, since p is odd. Thus, by Exercise 1.56, p = αβ where α, β are primes in OF . Conversely, if p " D is an odd rational prime, and is not a prime in OF , then by the solution of Exercise 1.56 above, p = αβ where |N (α)| = p = |N (β)|. If √ α = a + b D, then by˛ Theorem 1.3, 2a,˛2b ∈ Z. If gcd(p, b) = g, then since 4a2 ≡ 4b˛2 D (mod p), g ˛ a if and only if g ˛ b since gcd(p, 4D) = 1. This implies that g 2 ˛ p so g = 1. Therefore, there exists a multiplicative inverse (4b)−1 ) = 1. modulo p, so D ≡ (ab−1 )2 (mod p), namely ( D p (b) Suppose that p = αβ for some rational prime p, with √ NF (α) = NF (a + b D) = a2 − b2 D = ±p. Then, √ √ √ p(a − b D) p(a − b D) p p √ = = = = ±(a − b D) = ±α$ . 2 2 α a −b D ±p a+b D (ES3) Therefore, β = ±α$ so β ∼ α$ , so by taking conjugates via the properties outlined in Example 1.5 on the next page, α ∼ β $ . Lastly if α ∼ β, then by (ES3), √ √ √ ±(a − b D) ±(a − b D)2 ±(a − b D)2 β ±α$ √ ±1 = = = = = = α α (a2 − b2 D ±p a+ D √ (a2 + b2 D − 2ab D) a2 + b2 D 2ab √ = − D, p p p but the latter is not even an integer in OF much less a unit, so α #∼ β. β=

7

Solutions to Even-Numbered Exercises 1.60 Since D ≡ 3 (mod 4), then D2 − D ≡ 2 (mod 4), so √ √ ˛ 2 ˛ (D2 − D) = (D + D)(D − D),

but 2 cannot divide either factor by Theorem 1.3, so 2 cannot be prime in OF . Therefore,√by Exercise 1.56, 2 = αα$ where α and α$ are primes in OF and if √ α = a + b D ∈ Z[ D], then ±2 = a2 − b2 D where a and b must be odd. Thus, ±α$ α−1 = ±(α$ )2 (a2 − b2 D)−1 =

(α$ )2 = 2

√ √ a2 + b2 D − 2ab D a2 + b2 D = − ab D ∈ OF , 2 2 and by taking conjugates, ±α(α$ )−1 ∈ OF . However, α(α$ )−1 = (α$ α−1 )−1 so α$ α−1 ∈ UF . Hence, α$ ∼ α so α$ = uα for some u ∈ UF . Thus, 2 = αα$ = α2 u, which implies that 2 ∼ α2 , as required. √ √ 1.62 If D ≡ 1 (mod 8), then (1−D)/4 is even and (1−D)/4 = (1− D)/2)(1+ D)/2, where 2 cannot divide either factor by Theorem 1.3, so 2 cannot be prime in OF . Thus, √ by Exercise 1.56,√2 = αα$ where α and α$ are primes in OF and α = (a + b D)/2 ∈ Z[(1 + D)/2] with a, b having the same parity. Since ±8 = a2 − b2 D, then a and b cannot both be even since in that case, ±2 = (a/2)2 − (b/2)2 D, where a/2 and b/2 are both odd, so ±2 ≡ 1 − D ≡ 0 (mod 4), a contradiction. Thus, both a and b are odd. If α ∼ α$ , then α(α$ )−1 ∈ UF . however. √ √ √ a+b D a2 + b2 D + 2ab D a2 + b2 D ab D √ = = ± ± #∈ OF , α(α$ )−1 = a2 − b2 D 8 4 a−b D

so in particular, α(α$ )−1 #∈ UF . We have shown that 2 is a product of two nonassociated primes as was the task.

1.64 We calculate that √ √ 3 = α2 · u = (9 + 2 21)2 (55 − 12 21), √ √ where u = 55 − 12 21 ∈ UF and α = 9 + 2 21 ∈ OF .

Section 2.1 2.2 We prove only the case where σ = 1, since the other case is similar. √ Given that I ⊆ Z[ D], then only two basis elements for I are required. Let I = Zα1 + Zα2 = [α1 , α2 ]. Then any element γ ∈ I is represented in the form √ √ √ γ = x(a1 + b1 D) + y(a2 + b2 D) = (xa1 + ya2 ) + (xb1 + yb2 ) D, for some x, y ∈ Z. Therefore, in order for a rational integer to be in I, we must have xb1 + yb2 = 0. Hence, the smallest positive rational integer a ∈ I is a = x1 a1 + y1 a2 , where x1 and y1 are the smallest integers, in absolute value not both zero, such that x1 b1 + y1 b2 = 0. Also, let √ β =b+c D

8

Advanced Number Theory be the element of I such that c ∈ N is smallest possible. Therefore, bj = ctj + rj ,

0 ≤ rj < c,

where rj , tj ∈ Z for j = 1, 2. However, √ √ √ αj − tj β = aj + bj D − (btj + tj c D) = aj − btj + rj D ∈ I, so rj = 0. In other words, c|bj for j = 1, 2. However, gcd(b1 , b2 )|c since c can be expressed as a linear combination√ of b1 and b2 , so c = gcd(b1 , b2 ), by the minimality of c. Hence, I = [a, b + c D], √ where c ∈ N and 0 ≤ b < a since we may reduce b modulo a, given that b + c D − na ∈ I for any n ∈ Z.

2.4 We only prove this for σ = 1 since √ the other case is similar. First, we show that in the Z-module I = [a, b + c D], the value c is the minimum value such that √ √ b + c D ∈ I. If c√ is not minimal, there exists a b + c D ∈ I with 0 < c1 < c. 1 1 √ Therefore, b1 + c1 √ D = ax + (b + c D)y for some x, y ∈ Z. Thus, by comparing the coefficients of D, we get that c1 = cy, so c1 ≥ c, a contradiction. This establishes the minimality of c in I. Similarly, a is the least positive rational integer in I. √ If I is an ideal, then a D ∈ I, so c|a by the minimality of c. Also, √ √ √ D(b + c D) = b D + cD ∈ I, so c|b. Moreover, since (b/c −

√

√ D)(b + c D) = (b2 − c2 D)/c ∈ I,

then a|(b2 − c2 D)/c by the minimality of a. Conversely, assume that√I satisfies the conditions. √ √ To verify that I is an ideal, we need to show that a D ∈ I, and (b + c D) D ∈ I. However, √ √ a D = −(b/c)a + (a/c)(b + c D) ∈ I, and

√ √ b D + cD = (−b2 + c2 D)/c + b(b + c D)/c ∈ I,

so we are done.

2.6 In Exercise 2.5, let I = (a, α), J = (a, na ± α) for any n ∈ Z, and „ « 1 0 X= . n ±1 Then a X α

!

=

! a , na ± β

so I = J. 2.8 Clearly, g = gcd(a1 , a2 , (b1 + b2 )/2) is the rational integer factor of I1 I2 . Thus, √ « √ √ „ a1 a2 a1 b2 + a1 ∆ a2 b1 + a2 ∆ b1 b2 + ∆ + (b1 + b2 ) ∆ , , , . I1 I2 = (g) g2 g g 2g

9

Solutions to Even-Numbered Exercises

√ Therefore, a3 = a1 a2 /g 2 . Since the coefficients of ∆ are a1 /g, a2 /g, and (b1 + b2 )/(2g), we must ensure a linear combination of them that sums to 1. By the Euclidean algorithm, there exist integers δ, µ, and ν, such that δa2 + µa1 +

ν (b1 + b2 ) = g. 2

Hence, having made this choice of integers, it follows that the constant terms must satisfy, b3 ≡

1 ν (δa2 b1 + µa1 b2 + (b1 b2 + ∆)) (mod 2a3 ). g 2

Section 2.2 2.10 Since J + H is an ideal, then given α ∈ I, β ∈ J, γ ∈ H, we have α(β + γ) = αβ + αγ ∈ IJ + IH and α(β + γ) ∈ I(J + H).

2.12 Let S be the set of all R-ideals that are not finite intersections of irreducible ideals. Assume that S #= ∅. By Corollary 2.3 on page 72, S has a maximal element M . Thus, M cannot be irreducible. Since M = ∩I⊇M I where the intersection ranges over all sets (including M ) that contain M , there are nonzero ideals I, J #= R, M such that M = I ∩ J. Since M is properly contained in both I and J, then I and J are both finite intersections of irreducible ideals. Hence, so is M , a contradiction, so S is empty. −1 2.14 Since an invertible fractional R-ideal Pn I satisfies II = R, then there P exist ai ∈ −1 I and bi ∈ I such that 1R = i=1 ai bi . Thus, if α ∈ I, then α = n i=1 (αai )bi . Also, αai ∈ I for i = 1, 2, . . . , n and since ai ∈ I −1 = {β ∈ F : βI ⊆ R}, then I is finitely generated as an R-module by the bi for i = 1, 2, . . . , n.

2.16 Suppose that I = Rb1 + Rb2 + · · · + RbnQwhere bj = cj /aj ∈ F , for aj , cj ∈ R, with aj #= 0 for j = 1, 2, . . . , n. Set α = n j=1 aj for which α #= 0 and αI = Rc1

n Y

j=2

aj + · · · + Rcn

n−1 Y j=1

aj ⊆ R

that makes I a fractional R-ideal. 2.18 That (i) implies (ii) is Theorem 2.12 on page 77 and Theorem 2.11 on page 76. That (ii) implies (iii) is clear. That (iii) implies (iv) follows from Remark 2.8 on page 75 and Theorem 2.11 on page 76. That (iv) is equivalent to (v) is Exercise 2.17. Now we show that (iv) implies (vi). By Exercise 2.14, every fractional R-ideal is finitely generated. Thus, by Lemma 2.2 on page 71, R is Noetherian. If F is the quotient field of R and α ∈ F is integral over R, then by Exercise 2.15, R[α] is a finitely-generated R-module. Therefore, by Exercise 2.16, R[α] is a fractional R-ideal. Hence, by (iv), R[α] is invertible. Accordingly, since R[α]R[α] = R[α], R[α] = RR[α] = (R[α])−1 (R[α])R[α] = (R[α])−1 (R[α]R[α]) = (R[α])−1 )R[α] = R,

10

Advanced Number Theory so α ∈ R. We have shown that R is integrally closed. It remains to show that every nonzero prime R-ideal P is maximal. By Corollary 2.3 on page 72, there is a maximal R-ideal M such that P ⊆ M , so by (iv), M is invertible. Thus, M −1 P = I is a fractional R-ideal and M −1 P ⊆ M −1 M = R, so M −1 P is an integral R-ideal. Moreover, since M (M −1 P) = RP = P, and P is prime, then by Theorem 2.2 on page 57, either M ⊆ P or M −1 P ⊆ P. If M −1 P ⊆ P, then R ⊆ M −1 = M −1 R = M −1 PP−1 ⊆ PP−1 ⊆ R, which shows that M −1 = R. However, R = M M −1 = M R = M , contradicting that M is maximal. Hence, M ⊆ P, which means that M = P, which is maximal. From the above equivalence of (iv) and (v), we have therefore shown that (v) implies (vi). To complete the logical circle, we must show that (vi) implies (i), but this is immediate from Remark 2.7 on page 73.

2.20 If I = (0) or I = R = (1), then one element suffices so ˛assume that (0) ⊂ I ⊂ R and let α ∈ I such that α #= 0, 1. Then (α) ⊆ I so I ˛ (α) by Corollary 2.5 on page 76. Thus, there exists an R-ideal J such that (α) = IJ. Let S be the set of distinct prime R-ideals Pj for j = 1, 2, . . . , n such that either ordP (I) #= 0 or ordP (IJ) #= 0, or possibly both. Since I #= R, then S #= ∅. By part (c) of Exercise 2.19, there exists a β ∈ F , the quotient field of R such that ordP ((β)) = ˛ordP (I) for all prime R-ideals P dividing I. Therefore, for all prime R-ideals P ˛ I,

ordP (I) = min(ordP (I), ordP ((α))) = min(ordP ((β)), ordP ((α))) = ordP ((α)+(β)), by part (b) of Exercise 2.19. Hence, I = (α) + (β) = (α, β), as required.

Section 2.3 2.22 Maintaining the notation from Theorem 2.19 on page 88, for j = 1, 2, . . . , n, θj (F ) = {θj (q0 + q1 α + · · · + qn−1 αn−1 ) : q0 , q1 , . . . , qn−1 ∈ Q} = {q0 + q1 αj + · · · + qn−1 αjn−1 : q0 , q1 , . . . , qn−1 ∈ Q} = Q(αj ).

This shows that

θj : Q(α) → Q(αj )

is an isomorphism, so all fields Q(αj ) are isomorphic.

11

Solutions to Even-Numbered Exercises 2.24 Since Q(β) is a subfield of Q(α), then by Exercise 2.23, n = |Q(α) : Q| = |Q(α) : Q(β)| · |Q(β) : Q| = |Q(α) : Q(β)| · d.

Thus, n/d ∈ N, and Q(α) = Q(β) if and only if d = n. Let θj for j = 1, 2, . . . , n be the n monomorphisms of F into C. By the definition of a monomorphism of F into C, the restriction of θj to Q(β) is a monomorphism of Q(β) into C. Therefore, we may arrange the θj so that the first d of them are the distinct monomorphisms of Q(β) into C, when considered as restrictions to it. Hence, d d Y Y mβ,Q (x) = (x − θj (β)) = (x − βj ). (ES4) j=1

j=1

˛ We observe that by Theorem 1.6 on the preceding page, mβ,Q (x) ˛ f (x), since f (β) = f (θ1 (β)) = 0, where θ1 (α) = α. Thus, for some N ∈ N and some monic polynomial g(x) ∈ Q(α)[x], we have f (x) = mN β,Q (x)g(x), where gcd(mβ,Q (x), g(x)) = 1. If g(x) is not constant, there is an algebraic integer γ such that g(γ) = 0. Hence, f (γ) = 0, so γ = θj (β) for ˛ some j = 1, 2, . . . , d. However, by Equation (ES4), θj (β) = βj , so mβ,Q (x) ˛ g(x), a contradiction. N Thus, f (x) = mN β,Q (x) ∈ Q[x], so n = deg(f ) = deg(mβ,Q ) = dN . Therefore, N = n/d.

2.26 Since every nonzero ideal in a Dedekind domain R is a product of prime R-ideal, then we need only show that a prime R-ideal P always has a prime element in it. Since all ideals are finitely generated,then P = (α1 , . . . , αn ). If n = 1, then P = (α1 ) so by Theorem 2.3 on page 58, α1 is a prime element. The result now follows by induction. 2.28 If α1 , . . . , αn are the F -conjugates of α = α1 and β1 , . . . , βn are the F -conjugates of β = β1 , then NF (α)NF (β) = α1 · · · αn · β1 · · · βn = α1 β1 · · · αn βn = NF (αβ), and TF (α)+TF (β) = α1 +· · · αn +β1 +· · ·+βn = α1 +β1 +· · ·+αn +βn = TF (α+β). 2.30 F6 = 274177 · 67280421310721. √ 149 + 1) √ = 7150 + 7 = x3 + 7, where√x = 750 . Observe that for 2.32 In Z[ 3 −7], √ 7(7 F = Q( 3 −7), NF (a+b 3 −7) = a3 −7b3 , so NF (x− 3 −7) = x3 +7, and we may apply the gcd method. An initial run shows that gcd(a3 − 7b3 , 7150 + 7) = 10133 for a = 47, b = 1804, so 10133 is a factor of 7149 + 1.

Section 3.1 3.2 Let f (x, y) = ax2 + bxy + cy 2 . If f (p, q) = n for relatively prime p, q ∈ Z, then by the Euclidean algorithm, there exist r, s ∈ Z such that ps − qr = 1, so f (px + ry, qx + sy) = (px + ry)2 a + b(px + ry)(qx + sy) + c(qx + sy)2 , and after expanding and simplifying, this equals x2 (p2 a + bpq + q 2 c) + xy(2pra + bps + brq + 2qsc) + y 2 (r2 a + brs + s2 c) =

12

Advanced Number Theory x2 f (p, q) + xyB + y 2 C, where B = 2pra + bps + brq + 2qsc and C = r2 a + brs + s2 c = f (r, s).

3.4 If f (x, y) = g(px + qy, rx + sy) with ps − qr = −1,

(ES5)

and the reflexive property holds, then f (x, y) = g(x, y), so p = 1 = s and q = r = 0, which contradicts (ES5). 3.6 If f (x, y) = n and gcd(x, y) = g, then n = g 2 n1 for some integer n1 . Also, „ « x y f , = n1 g g with gcd(x/g, y/g) = 1 so f properly represents n1 . 3.8 Take f (x, y) = x2 + 6y 2 and g(x, y) = 2x2 + 3y 2 , where the discriminant is D = −24 for both. To see that they are not equivalent we invoke Exercise 3.1. Since f (1, 0) = 1, but g(x, y) #= 1 for any integers x, y, then f and g cannot be equivalent. 3.10 This is a consequence of Exercise 3.9 since −4n is a quadratic residue modulo p if and only if the Legendre symbol equality (−4n/p) = (−n/p) = 1. 3.12 This follows immediately from Exercise 3.10 and Theorem 3.1.

Section 3.2 3.14 If (−1, 0, 3) ∼ (1, 0, 3), then −x2 + 3y 2 = (px + qy)2 − 3(rx + sy)2 ,

(ES6)

for some p, q, r, s ∈ Z with ps − qr = 1. However, by comparing coefficients of x2 in (ES6), we get that p2 − 3r2 = −1, which is a contradiction since −1 is not a quadratic residue modulo 3. This shows that C∆F #= {1}. On the other hand, Remark 1.19, and Theorem 1.21 on page 50, OF = O12 is a Euclidean domain, so by Theorem 1.17 on page 39, it is a UFD. Thus, by Theorem 2.17 on page 83, it is a PID, from which we get that CO F = {1}.

3.16 Since we have that

C+ OF = then

I∆F ∼ I∆F P∆F · + , = + P∆F P∆ P∆ F F C+ O F = CO F

(ES7)

+ if and only if P∆F = P∆ . When F is real, by Exercise 3.15, this occurs if and F only if OF has a unit of norm −1. In the case where F is complex, (ES7) holds since all elements of OF have positive norm. In the case where F a real field + with no such unit, then |P∆F : P∆ | = 2 by Exercise 3.15. The result follows. F

3.18 Since hO F = |CO F |, then IhOF = OF = 1 for all I ∈ CO F . Also, since IhOF = 1 for all integral OF -ideals, then IhOF ∼ 1 for all integral OF -ideals I.

Solutions to Even-Numbered Exercises

13

Section 3.3 3.20 We have

I 2 ∼ (1) if and only if I ∼ I −1 if and only if I ∼ I $ ,

where the last equivalence comes from Exercise 3.19. √ 3.22 Since (15 + 221)/2 has norm 1 and there can be no smaller positive unit in OF , then the result follows.

Section 3.4 3.24 By the solution to Exercise 3.13, any reduced form ax2 + bxy + cy 2 must satisfy ac ≤ −D/3 = 5. For this inequality together with the inequalities in Definition 3.4 on page 100, the only solutions are for (a, b, c) = (1, 1, 4) and (a, b, c) = (2, 1, 2). Also, if n ∈ Z satisfies n = x2 + xy + 4y 2 , then 4n ≡ (2x + y)2 (mod 15)

so the only values modulo 15 which satisfy this quadratic residuacity are n ≡ 1, 4, 6, 9, 10 (mod 15) provided that n is not divisible by 15. 3.26 Given forms f, g ∈ P that represent n and n1 respectively, then f ◦ g represents nn1 . Thus, since the principal form represents n and n1 , given that f, g ∈ P , then the principal form represents nn1 , so f ◦ g ∈ P . This is the first statement. Also, if G is a genus of forms of discriminant ∆F , then given g ∈ G, we have that g ◦ g −1 ∈ P so G is a coset of P in C∆F . Since each coset has the same cardinality, then there are an equal number of classes in each genus. 3.28 Since any p representable in the form p = 3x2 ± 2xy + 5y 2 satisfies 3p = (3x + y)2 + 14y 2 ,

then we invoke Corollary 3.8 on page 138 to conclude, via Exercise 3.25, that

−1

Since 3 we get

3p ≡ 1, 9, 15, 23, 25, 39 (mod 56). ≡ 19 (mod 56) then multiplying each of these values by 19 modulo 56

as required. 3.30 Let

p ≡ 3, 5, 13, 19, 27, 45 (mod 56), f (xj , yj ) = ax2j + bxj yj + cyj2

for j = 1, 2, be two forms of discriminant D = −4n. Since D = b2 − 4ac = −4n, then b = 2b1 for some b1 ∈ Z. Thus, f (x1 , y1 )f (x2 , y2 ) = (ax21 + 2b1 x1 y1 + cy12 )(ax22 + 2b1 x2 y2 + cy22 ) = (ax1 x2 + b1 (x1 y2 + y1 x2 ) + cy1 y2 )2 + n(x1 y2 − y1 x2 )2 = X 2 + nY 2 ,

where X = ax1 x2 + b1 (x1 y2 + y1 x2 ) + cy1 y2 and Y = x1 y2 − y1 x2 .

14

Advanced Number Theory

3.32 By Theorem 3.8 on page 120, f has order at most 2 if and only if f ∼ f −1 . Now if |b| < a < c, then f −1 is also reduced, so by Claim 3.1 on page 101, f ∼ f −1 if and only if b = 0. If a = b or a = c, then the restriction that b ≥ 0 in Definition 3.4 on page 100 yields the result. 3.34 They are f = (a, b, c) for the values (1, 1, 18), (2, 1, 9), (2, −1, 9), (3, 1, 6), (3, −1, 6), (4, 3, 5), and (4, −3, 5).

3.36 Three inequivalent reduced forms are f = (1, 1, −57) for which f (−803, 1) = 643949 = 1 + 2812 · 229 ∈ 1 is prime, g = (−9, 7, 5) for which g(37, 56) = 17863 = 1 + 78 · 229 ∈ 1 is prime, and h = (−9, 11, 3) for which h(1, 89) = 24733 = 1 + 108 · 229 ∈ 1 is prime.

Section 3.5 3.38 We have that (−3/p) = (−1/p)(3/p) = 1 if and only if (−1/p) = (3/p) = −1 or (−1/p) = (3/p) = 1. Thus, from the hint, (−3/p) = 1 if and only if either p ≡ −1 (mod 4) and p ≡ ±5 (mod 12), or else p ≡ 1 (mod 4) and p ≡ ±1 (mod 12). In other words, (−3/p) = 1 if and only if either p ≡ −5 (mod 12) or p ≡ 1 (mod 12), and this holds if and only if p ≡ 1 (mod 3). 3.40 Since (−11/p) = (−1/p)(11/p) = 1 if and only if (−1/p) = (11/p) = −1 or (−1/p) = (11/p) = 1, then (−11/p) = 1 if and only if either p ≡ −1 (mod 4) and p ≡ 1, 3, 4, 5, 9 (mod 11), or else p ≡ 1 (mod 4) and p ≡ 1, 3, 4, 5, 9 (mod 11). In other words, (−11/p) = 1 if and only if either p ≡ 3, 15, 23, 27, 31 (mod 44) or p ≡ 1, 5, 9, 21, 25 (mod 44), and this holds if and only if p ≡ 1, 3, 5, 9, 15, 21, 23, 25, 27, 31 (mod 44). By Example 3.10, Theorem 1.3 on page 6, and (3.6), we have that h−11 = hZ[(1+√−11)/2] = 1. Thus, by Theorem 3.15, if (∆F /p) = (−11/p) = 1, then p = a2 + ab + 3b2 for some integers a, b. Also 11 = (−1)2 − 1 · 2 + 3 · 22 . Conversely, by Exercise 3.9 on page 104, if p #= 11 and p = a2 + ab + 3b2 , then (−11/p) = 1. 3.42 Given that (−43/p) = (−1/p)(43/p) = 1 if and only if (−1/p) = (43/p) = −1 or (−1/p) = (43/p) = 1, then (−43/p) = 1 if and only if either p ≡ −1 (mod 4) and p ≡ 1, 4, 6, 9, 10, 11, 13, 14, 15, 16, 17, 21, 23, 24, 25, 31, 35, 36, 38, 40, 41 (mod 43),

(ES8)

or else p ≡ 1 (mod 4) and (ES8) holds. This implies that (−43/p) = 1 if and only if either p ≡ 11, 15, 23, 31, 35, 47, 59, 67, 79, 83, 87, 95, 99, 103, 107, 111, 127, 135, 139, 143, 167 (mod 172),

(ES9)

or p ≡ 1, 9, 13, 17, 21, 25, 41, 49, 53, 57, 81, 97, 101, 109, 117, 121, 133, 145, 153, 165, 169 (mod 172),

(ES10)

Lastly, (ES9)–(ES10) hold if and only if p ≡ 1, 9, 11, 13, 15, 17, 21, 23, 25, 31, 35, 41, 47, 49, 53, 57, 59, 67, 79, 81,

15

Solutions to Even-Numbered Exercises 83, 87, 95, 97, 99, 101, 103, 107, 109, 111, 117, 121, 127, 133, 135, 139, 143, 145, 153, 165, 167, 169 (mod 172).

Now, as in the solution of Exercise 3.40, we have that h−43 = hZ[(1+√−43)/2] = 1. Thus, by Theorem 3.15, if (∆F /p) = (−43/p) = 1, then p = a2 + ab + 11b2 for some integers a, b. Also 43 = (−1)2 − 1 · 2 + 11 · 22 . Conversely, by Exercise 3.9 on page 104, if p #= 43 and p = a2 + ab + 11b2 , then (−43/p) = 1.

3.44 The following are all the prime values for class number one negative discriminants via Rabinowitsch. −∆F 3 7 11 19 43 67

x2 + x + (1 − ∆F )/4 x2 + x + 1 x2 + x + 2 x2 + x + 3 x2 + x + 5 x2 + x + 11 x2 + x + 17

163

x2 + x + 41

for x = 0, 1, . . . , /|∆F |/4 − 10 − 2. 3, 5. 5, 7, 11, 17. 11, 13, 17, 23, 31, 41, 53, 67, 83, 101. 17, 19, 23, 29, 37, 47, 59, 73, 89, 107, 127, 149, 173, 199, 227, 257. 41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 173, 197, 223, 251, 281, 313, 347, 383, 421, 461, 503, 547, 593, 641, 691, 743, 797, 853, 911, 971, 1033, 1097, 1163, 1231, 1301, 1373, 1447, 1523, 1601.

3.46 First we note that, using the notation in the proof of Theorem 3.5, √ τ : J = (2, 1 + −5) 1→ (2, 2, 3), and τ : I = (1,

√

−5) 1→ (1, 0, 5)

where J #∼ 1 in CO F . The latter holds since (1, 0, 5) and (2, 2, 3) are reduced forms so if they were properly equivalent, then they would be identical by Claim 3.1 on page 101. Also, we note that J2 = (2). For part (a), If p ≡ 1, 9 (mod 20), then (−5/p) = 1 so by Theorem √ 2.4 on page 60 and Remark√2.2 on page 63 (p) = PP$ , where P = (p, (b + −20)/2) and P$ = (p, (−b + −20)/2). We denote the norm of P by N √(P) = p and similarly for J, N (J) = 2. Now, if P is principal, then P = (a + b −5) for some integers a, b. Thus, √ √ (p) = PP$ = (a + b −5)(a − b −5) = (a2 + 5b2 ), so since N (P) = p, then p = a2 + 5b2 , as required. If P is not principal,√then P ∼ J, so PJ ∼ J2 ∼ 1. Hence, there are integers x, y so that PJ = (x+y −5), so N (PJ) = N (P)N (J) = 2p = x2 + 5y 2 . Thus, both x and y are odd, so 2p ≡ 6 (mod 8), whence, p ≡ 3 (mod 4), a contradiction. We have established one direction for part (a). Conversely, if p = a2 + 5b2 , then (p/5) = (a2 /5) = 1, so p ≡ 1, 4 (mod 5). Also, since one of a, b must be even, then p ≡ 1 (mod 4). Hence, p ≡ 1, 9 (mod 20), as required. For part (b), first assume that p ≡ 3, 7 (mod 20). Then (−5/p) = 1 since (−1/p) = −1 = (5/p). As above (p) = P P $ . If P is principal, then as above

16

Advanced Number Theory p = a2 + 5b2 , which means that p ≡ 1 (mod 4), a contradiction. Thus, as above 2p = x2 + 5y 2 for some integers x, y. Thus, x and y must have the same parity, so we may select an integer z such that x = y + 2z. Therefore, 2p = (y + 2z)2 + 5y 2 = 4z 2 + 4yz + 6y 2 , and dividing through by 2, we get p = 2z 2 + 2yz + 3z 2 . We have established one direction for part (b). Conversely, assume that there are integers a, b with p = 2a2 + 2ab + 3b2 . Then 2p = x2 + 5y 2 , where x = 2a + b and y = b. Thus, as above, p ≡ 3 (mod 4). Also,(−5/p) = 1 by Exercise 3.9 on page 104. Therefore, 1 = (−5/p) = (−1/p)(5/p) = −(5/p), so (5/p) = (5/p) = −1. Thus, p ≡ 2, 3 (mod 5), whence p ≡ 3, 7 (mod 20), which secures part (b) . For part (c), we know from Remark 3.8 on page 112 that h∆F = hO F = 2, so there are two classes of forms and by (a)–(b), two genera of forms, each having one class, the principal one with the unique reduced form (1, 0, 5) and the unique class of order 2 given by the reduced form (2, 2, 3).

Section 3.6 3.48 Since 2 " ∆F , then 2 " b. If 2 " (ac), then ax2 + bxy + cy 2 ≡ x2 + xy + y 2 (mod 2), so (a, b, c) ∼ (1, 1, 1). ˛ ˛ ˛ ˛ ˛ ˛ ˛ If 2 (ac), then 2 a or 2 c. If 2 ˛ a, then

ax2 + bxy + cy 2 ≡ xy + cy 2 ≡ y(x + cy) ≡ XY (mod 2),

where X = x + cy and Y = y, so (a, b, c) ∼ (0, 1, 0) (mod 2). ˛ The argument for 2 ˛ c is similar.

3.50 This is an immediate consequence of Exercise 3.48.

Section 4.1 4.2 If the subset is of finite cardinality then by Definition 4.1 on page 163, it is countable. Assume that S = {s1 , s2 , . . .} is an infinite subset of a countable set. Then define f (sj ) = j for all sj ∈ S, which is clearly countable since f (S) = N.

17

Solutions to Even-Numbered Exercises 4.4 We may assume, without loss of generality, that |x| ≤| y|.

If y = 0, then there is at most one solution, so we assume that y > 0. Suppose that αj for j = 1, 2, . . . , n are all solutions of f (x, 1) = 0. Also, set M=

max

0≤k+!≤n−3

Since f (x, y) = g(x, y), then it follows that ˛ ˛ n ˛ Y ˛ ˛ ˛ (x − αj y)˛ ≤ |f (x, y)| ≤ ˛a0 ˛ ˛ j=0

≤M

X

0≤k+!≤n−3

{|bk! |}.

X

0≤k+!≤n−3

|bk! ||xk y ! |

|y k+! | ≤ n2 M y n−3 .

(ES11)

GIven that n and M are fixed, we cannot have that ˛ n ˛ ˛Y ˛ ˛ ˛ n−3 K ˛ (x − αj y)˛ ≥ y ˛ ˛ j=0

for any constant K ∈ R+ , then there must exist a nonnegative integer m such that |x − αm y| < k0 y 1−n/3 for some k0 ∈ R+ . It t #= m and y is sufficiently large, say y ≥ k1 ∈ R+ , then |x − αt y| = |(αm − αt )y + (x − αm y)| > |(αm − αt )|y − k0 y 1−n/3 > k3 y, where k3 ∈ R+ . Hence,

Y

j'=m +

|x − αj y| > k4 y n−1

where k4 ∈ R . Therefore, by (ES11), |x − αm y| < where

k5 n2 M y n−3 = 2, k4 y n−1 y

k5 = (n2 M )/k4 ,

so |αm −

x k5 | < 3. y y

By Roth’s result (4.2) there can be only finitely many solutions. It remains to show that the above holds when y < k1 . However, since |x| ≤| y| < k1 there can only be finitely many solutions. This completes the proof of the result. 4.6 This is an immediate consequence of the application of Roth’s result provided in Exercise 4.4.

18

Advanced Number Theory

Section 4.2 4.8 Order the monomials according to the hint with αxa1 1 · · · xann < βxb11 · · · xbnn if and only if aj < bj where j ∈ N is the least such that aj #= bj . Let axa1 1 xa2 2 · · · xann be the largest monomial in f . Select any permutation σ of 1 2 n {1, 2, . . . , n}. Given that f is symmetric, then axaσ(1) xaσ(2) · · · xaσ(n) is a summand of f . Therefore, aj ≥ aj+1 for j = 1, 2, . . . , n − 1. Set g1 = asa1 1 −a2 sa2 2 −a3 · · · sann which is symmetric in x1 , x2 , . . . , xn . By writing g1 as a sum of monomials in the xi , the largest of these is axa1 1 −a2 (x1 x2 )a2 −a3 · · · (x1 · · · xn )an which is the largest monomial in f . Then f1 = f − g1 is symmetric, and the largest monomial in f1 is smaller than that in f . Now repeat the above process to get an f2 in a similar manner, with largest monomial smaller than that P of f1 . Clearly the process terminates with say M steps involved where M ≤ n j=1 aj . We ultimately get f − g1 − g2 − · · · − gM = 0, so

f=

M X

gj

j=1

where each gj is similar in its type to g1 so f ∈ Q[s1 , . . . , sn ].

Section 4.3 4.10 If {h1 , . . . , hn } forms a basis for G, then gi =

n X

bi,j hj

j=1

(bi,j ∈ Z; i = 1, 2, . . . , n).

Therefore, AB = (ai,j )(bi,j ) = In , so det(A) det(B) = 1. This implies that det(A) = ±1, so A ∈ GLn (Z). Conversely, if A ∈ GLn (Z), then det(A) #= 0, so the hj are linearly independent. Since there exists A−1 = (bi,j ) ∈ GLn (Z), and A−1 =

Aa det(A)

which is a basic property of matrices–see [68, Theorem A.5, p. 292], for instance. then n X gi = bi,j hj (i = 1, . . . , n). j=1

Thus, the gi generate G.

19

Solutions to Even-Numbered Exercises

Section 5.1 5.2 Compare the coefficients of xn in ∞ X

xj x = (e − 1) = Bj j! j=0 x

∞ X xi i! i=1

!

∞ X

xj Bj j! j=0

!

.

5.4 From the hint,

! ( n X n 1 + B1 = 1/2 Bn (1) = Bj = Bn j j=0

if n = 1, if n > 1.

5.6 Taking the hint, we get $ $ Bn+1 (x + 1) − Bn+1 (x) = (n + 1)nxn−1 = (n + 1)(Bn (x + 1) − Bn (x)).

Therefore, $ $ Bn+1 (x) − (n + 1)Bn (x) = Bn+1 (x + 1) − (n + 1)Bn (x + 1),

which shows that

$ f (x) = Bn+1 (x) − (n + 1)Bn (x)

is periodic with period length 1. Since f (x) is a polynomial, then it is constant. Hence, $ f (x) = f (0) = Bn+1 (0) − (n + 1)Bn (0) = (n + 1)Bn − (n + 1)Bn = 0,

so for any x,

$ Bn+1 (x) = (n + 1)Bn (x),

which is what we sought to prove.

Section 5.2 5.8 For j = 1, 2, . . . /x0, set

˛ Sj (x) = {n ∈ N : j 2 ˛ n but k2 " n for any k ∈ N with j < k < x}.

Then S1 (x) consists of all squarefree n ≤ x and „ « x |Sj (x)| = SF . j2 Thus, /x0 =

√ ) x*

Now for x = y 2 /y 2 0 =

X

SF

„

x j2

«

.

SF

„

y2 j2

«

.

j=1

)y* X j=1

20

Advanced Number Theory Using the M¨ obius inversion formula we get, — 2! X X µ(j) y SF (y 2 ) = µ(j) 2 = y 2 + O(y), j j2 1≤j≤y

1≤j≤y

and by (5.24) on page 214, and (5.5) on page 198, this equals ! X 1 6y 2 6y 2 2 + y O + O(y) = + O(y), π2 j2 π2 j>y so SF (x) =

√ 6x + O( x). 2 π

5.10 By the hint, ∞ jxk X g(d) 1 X 1X f (n) = g(d) = +O x x d d n≤x

d=1

d≤x

P

X |g(d)| d

d>x

and since limx→∞ d>x |g(d)| = 0, given that d hint the last term also goes to zero. Hence, lim

x→∞

P∞

n=1

!

0

1 X 1 +O @ |g(d)|A , x d≤x

|g(n)|n < ∞, and by the

∞ X g(n) 1 X f (n) = . x n n=1 n≤x

Section 5.3 5.12 Let N ∈ N and let pj for j = 1, 2, . . . , m be all the primes no bigger than N with pj < pj+1 for j = 1, 2, . . . , m − 1. Also, via the hint, set SN = {n ∈ N : n Therefore,

∞ Y X

p≤N j=0

=

j=1

paj where aj ≥ 0}.

X

f (pj ) = X

n Y

a1 ,a2 ,...,am ≥0

a1 ,a2 ,...,am ≥0

f (pa1 1 ) · · · f (pamm )

f (pa1 1 · · · pamm ) =

X

f (n),

n∈S N

by the multiplicativity of f . Thus, by the absolute convergence of S ˛ ˛ ˛ ˛ ˛X ˛ ˛X ˛ ∞ Y X X ˛∞ ˛ ˛ ˛ |S − P | = ˛˛ f (n) − f (pj )˛˛ = ˛˛ f (n)˛˛ ≤ |f (n)|, ˛n=1 ˛ ˛n'∈S N ˛ n>N p≤N j=0 and

lim

n→∞

X

n>N

|f (n)| = 0,

since n #∈ SN implies n > N . Therefore, we have shown that S = P , from which we get the absolute convergence of P .

21

Solutions to Even-Numbered Exercises

5.14 Since f (n)n−s is a multiplicative function, then the result follows from Exercise 5.12. 5.16 This is an immediate consequence of the answer provided in the solution of Exercise 5.15 on page 421. 5.18 This is an immediate consequence of Theorem 5.10 on page 219. 5.20 By the Euler product (5.26) on page 218, using the fact that if j is not prime then π(j) = π(j − 1), we get that loge ζ(s) = −

X

p=prime

loge (1 − p−1 ) = − lim

n→∞

n X (π(j) − π(j − 1)) loge (1 − j −s ). j=2

Thus, by rearranging terms, " n X loge ζ(s) = − lim π(j)[loge (1 − j −s ) − loge (1 − (j + 1)−s )] n→∞

j=2

−s

−π(1) loge (1 − 2 = lim

n→∞

"

n X j=2

ˆ

π(j) loge (1 − (j + 1)

) + π(n) loge (1 − (n + 1)

−s

) − loge (1 − j

−s

−s

#

)

# ˜ ) − lim π(n) loge (1−(n+1)−s ). n→∞

However, for sufficiently large n, since Re(s) > 1, then we get, ˛ „ «˛ ˛ ˛ 1 1 2n ˛< π(n) loge (1 − (n + 1)−s ) < ˛˛n + + · · · , ˛ (n + 1)s 2(n + 1)2s |(n + 1)s |

and this latter goes to 0 as n goes to infinity. Hence, loge ζ(s) =

∞ X

π(j)

j=2

=s

Z

j+1 j

∞ Z X j=2

as required.

∞

X d(loge (1 − x−s )) dx = π(j) dx j=2

j+1

j

π(x) dx = s x(xs − 1)

Z

∞ 2

Z

j+1

j

s dx x(xs − 1)

π(x) dx, x(xs − 1) !

5.22 We have, from Definition 5.6, that Γ(s) =

Z

∞

−t s−1

e

t

0

(s − 1) as required.

`

dt = −t Z

∞ 0

˛∞ ˛ s−1 −t ´ ˛ e

˛ ˛

0

+

Z

∞ 0

(s − 1)ts−2 e−t dt =

ts−2 e−t dt = (s − 1)Γ(s − 1),

` ´ 5.24 By Theorem 5.15 on page 225, ζ(s) = 2s π s−1 Γ(1 − s)ζ(1 − s) · sin πs , so 2 “s” “s” “ ” πs π −s/2 Γ ζ(s) = π −s/2 Γ 2s π s−1 Γ(1 − s)ζ(1 − s) · sin . (ES12) 2 2 2

22

Advanced Number Theory Now using (5.35) with z = (1 − s)/2, we get „ « “ 1−s s” Γ(1 − s) = (2π)−1/2 21/2−s Γ Γ 1− , 2 2 so (ES12) equals π (s−3)/2 Γ

„ « “ “ πs ” 1−s s” ζ(1 − s) · sin Γ Γ 1− , 2 2 2 2

“s”

(ES13)

and by using the formula in the hint with z = s/2, Γ(s/2)Γ(1 − s/2) = π/ sin(πs/2). Thus, (ES13) equals π (s−1)/2 ζ(1 − s)Γ

„

1−s 2

«

= ξ(1 − s),

as required.

Section 6.1 6.2 x ≡ 13 (mod 53 ).

6.4 x ≡ 2525 (mod 173 ).

6.6 7 + 5 · 11 + 5 · 112 + 5 · 113 + 5 · 114 + · · ·

6.8 8+7·13+11·132 +12·133 +7·134 +7·135 +11·136 +12·137 +11·138 +4·139 +· · ·

Section 6.2 6.10 Since υ(1) = υ(1 · 1) = υ(1)υ(1), and υ(1) #= 0, then υ(1) = 1. Also, υ(−1)2 = υ((−1)2 ) = υ(1) = 1 and υ(−1) > 0, so υ(−1) = 1. If x ∈ Q, then υ(−x) = υ(−1)υ(x) = υ(x), by the above. If n ∈ N, then υ(n) = υ(1 + 1 + · · · + 1) ≤ nυ(1) = n. | {z } n terms

6.12 For each rational E > 0, there exists an integer N = N (E) such that |qj − qk |p < E for all j, k > N, and there exists an integer N $ = N $ (E) such that |qj$ − qk$ |p < E for all j, k > N $ . Therefore, for any ε = 2E > 0, there exists n = max{N, N $ }, such that |qj + qj$ − (qk + qk$ )|p ≤ |qj − qk |p + |qj$ − qk$ |p < E + E = ε, for any j, k > n

Solutions to Even-Numbered Exercises

23

so {qj } + {qj$ } = {qj + qj$ } is a Cauchy sequence. Similarly, |qj qj$ − qk qk$ |p = |qj qj$ − qk qj$ + qk qj$ + qk qj$ − qk qk$ |p ≤ |qj$ |p |qj − qk |p + |qk |p |qj$ − qk$ |p < (|qj$ |p + |qk |p )E for all j, k > n.

However, by Exercise 6.11, there is an M ∈ R+ such that |qj$ |p , |qk |p < M , so for ε = 2M E, |qj qj$ − qk qk$ |p < ε for j, k > n. Hence, {qj } · {qj$ } = {qj } · {qj$ } is a Cauchy sequence.

6.14 By the hint, there is a convergent subsequence with limit M , say. Now given, ε > 0, in the subsequence, we have, for sufficiently large N that |qN − M | < ε. Also, by the definition of a Cauchy sequence, we may assume that N is chosen large enough so that for any qk , |qk − qj | < ε. Thus, |qk − M | ≤ |qk − qj | + |qj − M | < 2ε, Hence, since k was arbitrarily chosen, M is the limit of the Cauchy sequence as well. 6.16 Suppose that x = apνp (x) /b and y = a$ pνp (y) /b$ , where aba$ b$ #≡ 0 (mod p). We may assume, without loss of generality, that νp (x) > νp (y) . Then ˛ ˛ ˛ νp (x) „ «˛ ˛ νp (y) ab$ pνp (x)−νp (y) + a$ b ˛ ˛ ap a$ pνp (y) ˛˛ ˛p ˛ |x + y|p = ˛˛ + = ˛ ˛ ˛ b b$ bb$ p p = p−νp (y) = max{p−νp (x) , p−νp (y) } = max{|x|p , |y|p }.

6.18 If there exists an n ≥ 0 such that z ≡ w (mod pn ), then ˛ ˛ ˛ ˛ ˛ z − w˛ ˛z − w˛ |z − w|p = ˛˛pn n ˛˛ = p−n ˛˛ n ˛˛ ≤ p−n . p p p p

˛˛ Conversely, assume that |z−w|p ≤ p−n for some n ≥ 0 and that pν ˛˛ (z−w)—see page 102. Then −ν ≤ −n, so n ≤ ν and z ≡ w (mod pn ).

6.20 Without loss of generality assume that |x| > |y|. Since | · | is non-Archimedean, then |x + y| ≤ max{|x|, |y|} = |x|. Also,

|x| = |x + y − y| ≤ max{|x + y|, |y|} = |x + y|

where the last inequality follows from the fact that |x| > |y|. Hence, from the two inequalities, |x| ≥ |x + y| ≥| x|, so

|x + y| = |x|.

Section 6.3 6.22 This is immediate from Exercise 6.21.

24

Advanced Number Theory

Section 6.4 6.24 If α, β ∈ P, then from Exercises 6.16- 6.17 on page 238, |α + β|p ≤ max{|α|p , |β|p } < 1, and if γ ∈ Op , then

|γα|p = |γ|p |α|p < |γ|p ≤ 1,

since γ ∈ Op if and only if νp (γ) ≥ 0 if and only if |γ|p ≤ 1 by Theorem 6.1. Hence, P is an ideal in Op . 6.26 Let I be a non-zero ideal of Op and let p−n = maxx∈I {|x|p }. Suppose that α ∈ I with |α|p = p−n . We claim that I = Pn = {x : |x| ≤ p−n }. Let β ∈ Pn , so |β|p = p−i where i ≥ n and so |βα−1 |p = p−i+n ≤ 1, which means that βα−1 ∈ Op . Therefore, β ∈ αOp ⊆ I, so Pn ⊆ I. If the inclusion is proper, then there exists a γ ∈ I with γ #∈ Pn and |γ|p > p−n contradicting the choice of α. Hence, I = Pn . Next we show by induction, that Pn = Pn . If n = 0, then P0 = P0 = Op by Theorem 6.1 on page 236. Suppose n > 0 and Pk = Pk for all k < n. Since (m ) X Pn = Pn−1 P = αj βj : |αj |p ≤ p−n+1 , |βj |p ≤ p−1 , j=1

then α ∈ Pn , so |α|p ≤ p−n , which means that Pn ⊆ Pn . Conversely, if α ∈ Pn , then |α|p ≤ p−n , so |xp−1 |p = p−n+1 , which shows that αp−1 ∈ Pn−1 = Pn−1 . Thus, α ∈ pPn−1 ⊆ PPn−1 = Pn . Hence, we have established that I = Pn . It remains to establish that Pn = pn Op . Since p ∈ P, then pn Op ⊆ Pn . Conversely, since |pn |p = p−n , then pn ∈ Pn = Pn , so Pn ⊆ pn Op , which is the entire result.

Section 7.1 7.2 This follows from Theorem 5.10 since |χ(n)n−s | ≤ n−s , from which the terms of L(s, χ) are dominated in absolute value by the corresponding terms of ζ(s).

Section 7.2 7.4 By Corollary 7.3, lim (s − 1)L(s, χ0 ) = lim (s − 1)ζ(s)

s→1+

s→1+

=

Y

p|D

Y

p|D

(1 − p−s )

(1 − p−1 ) = φ(D)/D,

by Exercise 5.18 and the fact about φ(D) given in the hint. Clearly then, lims→1+ L(s, χ0 ) = ∞.

25

Solutions to Even-Numbered Exercises

7.6 Form the sum of F (s) over all Dirichlet characters modulo D, using Corollary 7.2, we get, X X 1 −ns G(s, χ) = φ(D) p ≥ 0, n χ∈G χ (mod D)

where the right-hand sum is over all prime p and all integers n such that pn ≡ 1 (mod D). Thus, 0 1 „ X X 1 −ns « exp @ G(s, χ)A = exp φ(D) p ≥ 1. n χ∈G χ (mod D)

`P∞ 1 n ´ 1 = 1−z Using the fact in the hint, exp , and substituting z = n=1 n z −s χ(p)p , we get Y exp (G(s, χ)) = (1 − χ(p)p−s )−1 . p

Consequently, exp(G(s, χ)) = L(s, χ) for s > 1. Hence, F (s) ≥ 1

as required. (Note that further reasons for G(s, χ) to be continuous for 3(s) > 1 comes from the Since |(1/n)χ(pn )p−ns | ≤ p−ns and since G(s, χ) = P following P∞ 1 observations. n −ns , then the convergence of ζ(s) for 3(s) > 1 implies that p n=1 n χ(p )p the same is true for G(s, χ). Hence, the continuity assertion follows.) 7.8 By Example 7.2 on page 248 χ(n) is a root of unity for gcd(n, D) = 1, and since χ(n) is real for all n ∈ N, then χ(n) = ±1. If gcd(n, D) > 1, then χ2 (n) = χ0 (n) = 0. For gcd(n, D) = 1, then χ2 (n) = (±1)2 = 1 = χ0 (n).

Section 7.3 7.10 By Exercise 5.18 on page 227, (s − 1)ζ(s) = (s − 1)

∞ X

n−s = 1 + o(1).

n=1

so by taking logs, we see that as s → 1+ ,

loge ζ(s) + loge (s − 1) = loge (s − 1) + and since

X

∞ X

X

∞ X

p−ns = O(1),

p=prime n=1

p−ns = O(1),

p=prime n=2

then

X

p=prime

Hence,

p−s = − loge (s − 1) + O(1).

D(N) = lim

P

p=prime

p−s

= 1. | loge (s − 1)−1 | Hence, it follows immediately that any set S having all but finitely many primes must have the same Dirichlet density. s→1+

26

Advanced Number Theory

7.12 We have by Definition 7.3 on page 263 that P P −s −s p∈S p p∈W p D(S) + D(W) = lim + lim −1 −1 s→1+ loge (s − 1) s→1+ loge (s − 1) and the fact that S ∩ W = ∅ implies this equals P −s p∈(S∪W) p lim = D(S ∪ W). −1 s→1+ loge (s − 1) 7.14 The summation over the principal character is clear since |Fp | = p. Thus, assume ∗ χ #= χ 0 . Thus, there exists a value b ∈ Fp such that χ(b) #= 1. Now, if we set P X = a∈Fp χ(a), then χ(b)X =

X

χ(b)χ(a) =

a∈Fp

X

χ(ab) = X,

a∈Fp

where the last equality holds since ab runs over the values of Fp as does a. Hence, X(χ(b) − 1) = 0, and dividing by χ(b) − 1 #= 0, we get X = 0 as we sought to prove. P 7.16 Let X = χ∈G χ(a). Since a #= 1p , then with reference to the solution of Exercise 7.15 provided on page 427, a = g j for some j with (p − 1) " j. Then j α(a) = α(g)j = ζp−1 #= 1. Hence, we have a character γ = αj for which γ(a) #= 1. Thus, X X γ(a)X = γ(a)χ(a) = γχ(a) = X, χ∈G

χ∈G

so

X(γ(a) − 1) = 0,

and since γ(a) #= 1, we may divide by (γ(a) − 1) to get the result.

7.18 The sum is over all characters on Fp whose order divides m. We show there are exactly m such characters. Since the values of the characters are m-th roots of unity, there are at most m of them. However, by Exercise 7.15, there is a character of order m so there must be m distinct characters χ0 , χj for jP= 1, 2, . . . , m − 1. Now we establish the summation formula. If a = 0, then m =a χm =χ0 χ(0) = 1 since χ0 (0) = 1 and χ(0) = 0 for χ #= χ0 . If a #= 0, and x ∗ ∗ m m is solvable in Fp , then there exists a z ∈ Fp suchP that z = a. If χ = χ0 , then χ(a) = χ(z m ) = χ(z)m = χ0 (z) = 1. Hence, χm =χ0 χ(a) = m = N (m, a). Lastly, assume that a #= 0 and xm = a is not solvable. Then by Exercise 7.17, there is a character α on Fp such that αm = χ0 and α(a) #= 1. Therefore, X X X X α(a) χ(a) = α(a)χ(a) = αχ(a) = χ(a), χm =χ0

χm =χ0

χm =χ0

since αχ ranges over all values as does χ. Hence, X (α(a) − 1) χ(a) = 0, χm =χ0

and we get the result by dividing through by α(a) − 1 #= 0.

χm =χ0

27

Solutions to Even-Numbered Exercises 7.20 We have χ(a)Ga (χ) = χ(a)

X

χ(j)ζpaj =

χ(aj)ζpaj =

j∈Fp

j∈Fp

so

X

X

χ(k)ζpk = G1 (χ),

k∈Fp

Ga (χ) = χ(a−1 )G1 (χ).

7.22 Now since a #= 0, then Ga (χ0 ) =

X

χ0 (j)ζpaj =

j∈Fp

X

ζpaj = 0,

j∈Fp

where the last equality is Exercise 7.21. Now if a = 0, X X G0 (χ) = χ(j)ζp0j = χ(j), j∈Fp

j∈Fp

and the latter sum is p if χ = χ0 , and if χ #= χ0 it equals 0 by Exercise 7.14.

7.24 If j #= 0, then by Exercise 7.20,

Gj (χ) = χ(j −1 )G1 (χ) = χ(j)G1 (χ), so

Gj (χ) = χ(j −1 )G1 (χ).

Hence,

Gj (χ)Gj (χ) = χ(j −1 )χ(j)G1 (χ)G1 (χ) = |G1 (χ)|2 .

(ES14)

By Exercise 7.21, G0 (χ) = 0, so X Gj (χ)Gj (χ) = (p − 1)|G1 (χ)|2 . j∈Fp

Also, Gj (χ)Gj (χ) =

X X

χ(a)χ(b)ζpja−jb .

a∈Fp b∈Fp

Thus, by Exercise 7.23, if we set

d(a, b) = 1 if a = b and d(a, b) = 0 if a #= b,

then X

j∈Fp

Gj (χ)Gj (χ) =

X X

a∈Fp b∈Fp

χ(a)χ(b)d(a, b)p = (p − 1)p.

Hence, from (ES14)–(ES15), we have (p − 1)|G1 (χ)|2 = (p − 1)p, so |G1 (χ)| =

√

p.

(ES15)

28

Advanced Number Theory

Section 8.1 8.2 By the (8.2) on page 271, p √ R + R − 4Q = (α + β) + (α − β) = 2α, and

√

R−

8.4 We have Un V m − V n Um =

p R − 4Q = (α + β) − (α − β) = 2β.

(αn − β n )(αm + β m ) (αn + β n )(αm − β m ) − = (α − β) (α − β)

αn+m + αn β m − αm β n − β m+n − αn+m + αn β m − αm β n + β m+n = α−β 2

(αn β m − αm β n ) (αn−m − β n−m ) = 2(αβ)m = 2Qm Un−m . (α − β) (α − β)

8.6 We use induction on n. In n = 1 it is clear since U1 = 1. Assume that gcd(Uj , Q) = 1 for all positive j < n. Let p be a prime dividing gcd(Un , Q). Then by part (d) of Theorem 8.1, p must divide Vn . However, by part (b) of Theorem 8.1, √ Vn = RVn−1 − QVn−2 , ˛ ˛ ˛ √ so p ˛ Vn−1 and using part (b) again, p ˛ Vn−2 and so on until we get p ˛ V1 = R forcing p = 1, a contradiction. Similarly, gcd(Vn , Q) = 1. If p| gcd(Un , Vn ), then p|2Q, by part (d) of Theorem 8.1. Since gcd(Q, Un ) = 1, then p|2, and 4 " gcd(Un , Vn ). 8.8 In any case, if Un is even, then Q is odd by Exercise 8.7. If Un is odd and Q is even, this contradicts Exercise 8.4 with m = 1. The balance is as follows. √ (a) In this case, R ≡ 0 (mod 2). By part (b) of Theorem 8.1,

Since V0 = 2, V1 = (b) Define:

√

Vn+2 ≡ Vn (mod 2). R, then 2|Vn for all n. √ $ $ V2n+1 = V2n+1 / R, and V2n = V2n .

By part (b) of Theorem 8.1, $ $ $ (mod 2), V2n+1 ≡ V2n + V2n−1

with V0$ = 2, V1$ = 1. Then Vn$ ≡ 0 (mod 2) if and only if n ≡ 0 (mod 2). (c) Since,

$ $ Vn+2 ≡ Vn+1 + Vn$ (mod 2),

with V1$ = 1, and V2$ = R − 2Q ≡ 1 (mod 2), then

Vn$ ≡ 0 (mod 2) if and only if n ≡ 0 (mod 3).

29

Solutions to Even-Numbered Exercises

8.10 By the same technique as that used in the solution of Exercise 8.5, we get that, for s odd, and m ∈ N ! !! (s−1)/2 X s−j s−j−1 s s−2j j Vms = Vm + Vm (−1) + Qjm . j j−1 j=1 Hence, if s = n/m, is odd then Vm |Vn .

8.12 (a) If Un = 0, then Vn2 = 4Qn by part (d) of Theorem 8.1. However, by Exercise 8.6, gcd(Vn , Q) = 1. Thus, |Q| = 1, contradicting the hypothesis. √ √ (b) If R √= 1 = Q, then α = (1 + −3)/2 and β = (1 − −3)/2. Thus, α − β = −3, and α6 = 1 = β 6 , so U6 = (α6 − β 6 )/(α − β) = 0. If Q = −1, and R = 0, then α = 1, β = −1, and α − β = 2. Thus, U2 = (α2 − β 2 )/(α − β) = 0. (c) If n > m, then by Exercise 8.4, 2Qm Un−m = Un Vm −Vn Um = Un Vn −Vn Un = 0, so Un−m = 0, since Q #= 0, by hypothesis. This contradicts part (a).

Section 8.2 8.14 The equation has no solution since 22 + 49 = 53, so a = 2, but D #= −3a2 ± 1.

8.16 The equation has the solution x = 52, and n = 3, namely 522 + 2209 = 173 . There is no other solution by Theorem 8.3.

Section 8.3 8.18 By Theorem 2.12,

K = Pa1 1 Pa2 2 . . . Par r ,

where the Pj are the distinct prime R-ideals containing K, and aj ∈ N for j = 1, 2, . . . , r. Thus, IJ = K n = Pa1 1 n Pa2 2 n . . . Par r n . However, I, J are relatively prime so any prime dividing I cannot divide J and vice-versa, so we may relabel the primes as follows, a

n

t+1 I = Pa1 1 n . . . Pat t n and J = Pt+1 . . . Par r n ,

with 0 ≤ t ≤ r. Now se let a

t+1 I = Pa1 1 Pa2 2 . . . Pat t and J = Pt+1 . . . Par r .

Hence, as required.

I = In , J = Jn , and K = IJ,

30

Advanced Number Theory

8.20 If α = βu for a unit u in R, then for any γ ∈ (α), there exists a δ ∈ R such that γ = αδ. Therefore, γ = βδu. This shows that (α) ⊆ (β). Since β = αu−1 , then we may reverse the roles of α and β in the above argument to show that (β) ⊆ (α), so we have equality. Conversely, if (α) = (β), then α = βγ for some γ ∈ R. and β = αδ for some δ ∈ R. Hence, β = βγδ, but we know that β #= 0, so we may cancel to get, 1 = γδ. Therefore, both γ and δ are units in R. Hence, α = uβ for a unit u in R. 8.22 In Theorem 8.4, let k = −193 = −1 − 3u2 with u = 8, for which x = 4 · 82 + 1 = 257 and

y = ±8(3 + 8 · 82 ) = ±4120.

Thus,

(x, y) = (257, ±4120).

8.24 By Theorem 8.4 there can be no solutions since k = −57 #= −3u2 ± 1 for any integer u.

Section 8.4 8.26 By Exercise 1.3 on page 16, the minimal polynomial of ζp is mζp ,Q (x) = xp−1 + xp−2 + · · · + x + 1 =

p−1 Y xp − 1 = (x − ζpj ). x−1 j=1

By Letting x = 1, we get p−1

NF (1 − ζp ) =

Y

j=1

(1 − ζpj ) = p,

which is the first part of the exercise. Now we use this fact to get that for i = 1, 2, . . . , p − 1, « „ NF (1 − ζpi ) 1 − ζpi p NF = = = 1, 1 − ζp NF (1 − ζp ) p so by Exercise 1.5 on page 17, 1 − ζpi ∈ UF , 1 − ζp which means, by Definition 1.12 on page 19, that 1 − ζp and 1 − ζpi are associates for i = 1, 2, . . . , p − 1.

8.28 By Exercise 2.11, with there exists an R-ideal H, relatively prime to I, such that HJ = (α) for some α ∈ R. Define the map ψ : R 1→

J , by r 1→ rα + IJ, IJ

Solutions to Even-Numbered Exercises

31

which is a surjective group homomorphism since (α) + IJ = HJ + IJ = (H + I)J = RJ = J, by Exercise 2.10 since gcd(H, I) = H + I = R.

(ES16)

By the theorem in the hint it remains to show that ker(ψ) = I. If β ∈ ker(ψ), then ψ(β) = αβ + IJ = IJ, so αβ ∈ IJ. Therefore, (α)(β)H ⊆ IJH ⊆ I(α), so (β)H ⊆ I, by Corollary 2.7 on page 77. Given (ES16), there exist h ∈ H and γ ∈ I such that 1R = h + γ, so β = βh + βγ ∈ I. We have shown that ker(ψ) ⊆ I. Conversely, if β ∈ I, then ψ(β) = αβ + IJ = IJ, since α ∈ HJ ⊆ J implies that αβ ∈ IJ. Therefore, I ⊆ ker(ψ), so ker(ψ) = I as required.

8.30 This is an immediate consequence of Exercise 8.29 and Theorem 2.18 on page 84.

8.32 This is immediate from Exercise 8.31 since the different cosets of I in OF form different residue classes modulo I. 8.34 This will follow from the fact, elucidated in Exercise 8.32 that N (IJ) = N (I)N (J) once we establish this exercise for any prime power. The integers in OF that are not relatively prime to Pa are those divisible by P. There are N (Pa−1 ) = (N (P))a−1 of these that are incongruent modulo Pa . Thus, „ « 1 Φ(Pa ) = N (P)a − N (P)a−1 = N (Pa ) 1 − . N (P) 8.36 Since f is the product of d linear factors in its algebraic closure, then it has exactly d roots there. 8.38 This follows from Exercises 8.35 and 8.37 for the following reason. Since the residue classes modulo I, relatively prime to I, form a group of order Φ(I), then αΦ(I) ≡ 1 (mod I), for any α ∈ OF relatively prime to I. In particular, if I = P, a prime OF -ideal, then Φ(P) = N (P) − 1, so αN (P)−1 ≡ 1 (mod P).

Section 8.5 8.40 If xn + y n = z n for relatively prime integers x, y, z, then set a = xn , b = y n and c = z n in the ABC-conjecture. Without loss of generality x, y, z ∈ N may be assumed. Therefore, S(xn y n z n ) = S(xyz) ≤ xyz.

Thus, the ABC-conjecture implies that for any κ > 1, we have with finitely many exceptions, c < S(xn y n z n )κ = S(xyz)κ ≤ (xyz)κ Therefore,

(xyz)n ≤ c3 ≤ (xyz)3κ ,

32

Advanced Number Theory so

(xyz)n−3κ ≤ 1,

and by taking logarithms,

(n − 3κ) loge (xyz) ≤ 0. Thus, if we take κ = 1.01, then n ≤ 3, so with finitely many exceptions, FLT holds for n > 2, since we know if holds for n = 3 by Theorem 1.18 on page 41. We have shown that there is a bound N ∈ N such that FLT holds for all n > N . 8.42 If (n − 1, n, n + 1) is a triple of powerful numbers, then clearly n2 − 1 is powerful. If n is odd, then n ± 1 are both even, but then one of n − 1 or n + 1 must be congruent to 2 modulo 4, a contradiction. Hence, n is even. Conversely, if n is even powerful and n2 −1 is powerful, then n #≡˛2 (mod 4), so gcd(n−1, n+1) = 1. If p||(n−1) say, then since n2 −1 is powerful, p ˛ (n+1), a contradiction. A similar argument works for a prime properly dividing n + 1. Hence, (n − 1, n, n + 1) is a powerful triple. 8.44 Let x, y ∈ Z such that 1 + x2 y 3 has largest prime factor p and let Q be the product of all primes no bigger than p. If a = 1, b = x2 y 3 , then c = a + b, implies S(abc) ≤ xyQ. Thus, by the ABC-conjecture, for any κ > 1, with finitely many exceptions, we have that 1 + x2 y 3 < S(abc)κ ≤ (xyQ)κ , which implies x2−κ y 3−κ < Qκ . 2 In particular for κ = 2, y < Q , which bounds y and therefore x, and so x + |y| as well. This is the result we sought since as p → ∞, then x + |y| → ∞.

Section 9.1 9.2 Suppose that Y = A/B, X = C/D with nonzero A, B, C, D ∈ Z such that Y 2 = X 3 − 4X holds. Set x = AD2 , y = 2BCD, and z = 2C 3 B 2 D − A2 D4 .

Then

z 2 = 4B 4 C 6 D2 − 4A2 B 2 C 3 D5 + A4 D8 ,

so

z 2 = 4B 2 C 3 D2 (B 2 C 3 − A2 D3 ) + A4 D8 . However, since X − Y 2 = 4X, then

(ES17)

3

B 2 C 3 − A2 D3 = 4B 2 CD2 .

Therefore, (ES17) becomes z 2 = (4B 2 C 3 D2 )(4B 2 CD2 ) + A4 D8 = (2BCD)4 + (AD2 )4 = y 4 + x4 . Conversely, if x4 + y 4 = z 2 with nonzero x, y, z ∈ Z, then by setting X= we get Y 2 = X 3 − 4X.

4x(x2 + z) 2(x2 + z) , and Y = , y2 y3

33

Solutions to Even-Numbered Exercises

Section 9.2 9.4 By Theorem 1.19, P1 = (3, 5) and P2 = (3, −5) are the only rational integer solutions of y 2 = x3 − 2. However, 2P1 = (129/100, −383/1000) and 2P2 = (129/100, 383/1000). Hence, by the Nagell-Lutz Theorem P1 and P2 are points of infinite order. 9.6 By Mazur’s theorem and the discussion surrounding (9.2) on page 303, we get that E(Q)t ∼ = Z/3Z. 9.8 Since the points (1, 0), (0, 0), (−1, 0) are all of order 2, and the sum of any two of them is the third one, Moreover, there is no other point (x, y) with x, y ∈ Z and y 2 dividing dividing ∆(E(Q)) = 4. Hence, E(Q)t ∼ = Z/2Z × Z/2Z.

Section 9.3 9.10 We provide the calculations for only the (E, P ) pair (y 2 = x3 + 2x + 1, (0, 1)), since the others are similar. We check that indeed the discriminant of E is relatively prime to n and proceed. s 1 2 22 23 24 25 26 27 26 3 27 3 26 32 27 32 26 33 27 33 26 34 27 34

m −− 1 2662 2294 1687 3029 1104 2342 3486 3087 2508 3050 1232 3535 1375 751

sP (0, 1) (1, 3549) (1997, 2497) (2962, 3168) (2796, 3447) (567, 1294) (3240, 2146) (2842, 3159) (1694, 343) (2399, 85) (701, 850) (1029, 132) (3368, 1632) (622, 595) (1054, 1973) (835, 2701)

This exhausts the divisors of M , so we perform the algorithm on a = 3, 4, 5, 6, 7, 8 in the same fashion, each time exhausting the same divisors. A succesful run is provided for a = 9 in Example 9.4.

34

Advanced Number Theory

9.12 (a) 31 · 71 (b) 97 · 167 (c) 43 · 211 (d) 97 · 331

Section 9.4 9.14 Assume that n = ab/2 for a Pythagorean triple (a, b, c), set x = c2 /4. Thus,

(a − b)2 /4 = x − n

and

(a + b)2 /4 = x + n,

so x and x ± n are squares of rational numbers. Conversely, if x and x ± n are all squares of rational numbers, we set √ √ a = x + n + x − n, √ √ b = x + n − x − n, and

√ c = 2 x.

Then a, b, c ∈ Q and

a2 + b2 = c2 .

9.16 Suppose that (2) of Exercise 9.14 holds. Then (x − n)x(x + n) = x3 − n2 x = y 2 for some y ∈ Q. However, x $= ±n, 0 since n is squarefree. This value of x yields the desired point. Conversely, if E has a point P = (x1 , y1 ) with x1 , y1 ∈ Q and P $= (±n, 0), (0, 0)

or o, then y1 $= 0 and

2P = (x2 , y2 ) $= o.

By Exercise 9.15, x2 ± n and x are all squares of rational numbers, as required. 9.18 We first compute ac0 = 5(12, 8) = (10, 11) = (y1 , y2 ). Then we achieve the plaintext via: (c1 y1−1 , c2 y2−1 ) = (2 · 10−1 , 8 · 11−1 ) = (2 · 4, 8 · 6) = (8, 9) = m.

35

Solutions to Even-Numbered Exercises

Section 10.1 10.2 By (10.1), %(αz) =

%(z) . |cz + d|2

Since the numbers of pairs of integers (c, d) such that |cz + d| is less than any given N ∈ N is finite, then there exists an α ∈ Γ such that %(αz) is a maximum. We may select an n ∈ Z such that −1/2 ≤ '(T n αz) ≤ 1/2. If β = T n αz and |β| < 1, then %(−β −1 ) > %(β).

(S18)

However, if β = x + yi, then −1/β = (yi − x)/|β|2 , so (S18) implies that y > y, |β|2

a contradiction. Hence, |β| ≥ 1, which means that β ∈ D. In other words, T n α = γ is the required element. 10.4 This follows from the solution to Exercise 10.3 on page 438

Section 10.2 10.6 We have that |Γ :Γ 0 (n)| = p + 1. If γp = for 0 ≤ j ≤ p − 1, then we have that

!

1 0

0 1

"

and γj =

!

j −1

1 0

"

∪pj=0 γj Γ0 (p) ⊆ Γ, so we merely have to show that these γj represent distinct cosets, namely that γp−1 γj = γj $∈ Γ0 (p) for j < p, and that γi−1 γj $∈ Γ0 (p) unless i = j. The first assertion is clear since γp is the identity and γj $∈ Γ0 (p) for j < p. For the last assertion, ! "! " ! " 0 −1 j 1 1 0 γi−1 γj = = , 1 i −1 0 j−i 1 which is the required result. 10.8 From the solution to Exercise 10.6 above, we have that ! " ! " ! 0 1 1 1 1 γ0 = , γ1 = , and γ2 = −1 0 −1 0 0

0 1

"

.

36

Advanced Number Theory

10.10 If both γi−1 (∞) and γj−1 (∞) represent the same cusp, then for some α ∈ Γ0 (n), we have that γj−1 (∞) = γi−1 α(∞). Thus, γj γi−1 α fixes ∞, so γj γi−1

!

=±

1 0

b 1

"

(S19)

for some b ∈ Z, which is the sufficient condition. Applied to the case where p = n is prime all pairs of distinct γj for j = 1, 2, . . . , p satisfy (S19) since, ! " 1 i−j −1 γj γi = . 0 1 10.12 We have that f (x + 1) = Γ(x + 1)Γ(−x) sin(π(x + 1)),

(S20)

so via the hint, (5.34) tells us that Γ(x + 1) = xΓ(x),

(S21)

Γ(1 − x) , −x

(S22)

Γ(−x) = and since

sin(π(x + 1)) = sin(πx) cos(π) + cos(πx) sin(π) = − sin(πx),

(S23)

then, via (S21)–(S23), (S20) equals ! " Γ(1 − x) (xΓ(x)) (− sin(πx)) = Γ(x)Γ(1 − x) sin(πx) = f (x). −x 10.14 By the hint, z cot z = 1 − and sin z = z

∞ #

Bk

k=1

∞ ! $

n=1

22k z 2k , (2k)!

z2 1− 2 2 n π

"

.

(S24)

(S25)

By taking the logarithmic derivative of (S25), we achieve, z cot(z) = 1 + 2

∞ #

n=1

To proceed, we need the following.

z2

z2 . − n2 π 2

(S26)

37

Solutions to Even-Numbered Exercises Claim 10.2 For z ∈ C, ∞

# z 2k z2 =− . 2 2 2 z −n π n2k π 2k k=1

We have −

∞ #

k=1

(k ∞ N %& # # z 2k nπ '−2 −2k . = 1 − (nπ/z) = 1 − lim N →∞ n2k π 2k z k=0

k=0

However, by the standard geometric formula—see [68, Theorem 1.2, p.2], this equals  &, - 'N +1  nπ −2 − 1 z 1 z2  1 − lim  = 2 ,  = 1 + , nπ -−2 , nπ -−2 N →∞ z − n2 π 2 −1 −1 z z which is Claim 10.2.

Now by plugging the result of Claim 10.2 into (S26), and equating the result with (S24), we get 1− so Bk

∞ #

k=1

Bk

∞ # ∞ # 22k z 2k z 2k =1−2 , (2k)! n2k π 2k n=1 k=1

∞ ∞ # 22k−1 z 2k z 2k z 2k # 1 z 2k = = = 2k ζ(2k), 2k 2k 2k 2k (2k)! n π π n=1 n π n=1

which implies that

Bk

22k−1 π 2k = ζ(2k), (2k)!

as required. 10.16 By Theorem 10.2 on page 337, G4 (z) = and G6 (z) = Since

π4 (2π)4 , + q + 9q 2 + 28q 3 + 73q 4 + · · · , 45 3

2π 6 (2π)6 , − q + 33q 2 + 244q 3 + 1057q 4 + · · · . 945 60 ∆ = g23 − 27g32 ,

38

Advanced Number Theory then

∆(z) = 603 G4 (z)3 − 27 · 1402 G6 (z)2 ,

so the constant term is

603 π 12 27 · 1402 (2π 6 )2 64π 12 64π 12 − = − = 0, 453 9452 27 27 and the coefficient of q which is also a factor of all other powers of q is 3 · 603 (2π)4 · π 8 2 · 27 · 1402 · 2π 6 (2π)6 5120π 12 7168π 12 + = + = (2π)12 . 3 · 452 60 · 945 3 3

Hence,

∆(z) = (2π)12 (q − 24q 2 + 252q 3 − · · · ),

which means ∆ is a modular form of weight 12, as required to show. 10.18 We have η(z + 1) = exp(πi(z + 1)/12)

∞ $

n=1

(1 − exp(2πi(z + 1)n)),

but exp(2πi(z + 1)) = exp(2πiz), so η(z + 1) = exp(πi(z + 1)/12)

∞ $

n=1

(1 − exp(2πi(z)n))

= exp(πi(z + 1)/12) exp(−πiz/12) exp(πiz/12)

∞ $

n=1

(1 − exp(2πi(z)n))

= exp(πi(z + 1)/12) exp(−πiz/12)η(z) = exp(πi/12)η(z), which is the first part of (10.14). For the second part, we have, ∆(z) = (2π)12 η(z)24 , and by Exercise 10.5,

∆(−z −1 ) = z 12 ∆(z).

So by taking 24-th roots, we get √ η(−z −1 ) = u −izη(z), for some root of unity u. Since both sides are chosen to take positive real values on the imaginary axis, then u = 1 is forced which gives us our result, √ η(−z −1 ) = −izη(z).

39

Solutions to Even-Numbered Exercises

Section 10.3 10.20 From Exercise 2.25 on page 96, we conclude that 3 #

ej = 0

(S27)

j=1

since this is the coefficient of x2 in the expansion of f (x) = 4x3 − g2 x − g3 ; $ (ei − ej ) = −g2 /4 1≤ei |c4 |p , then since |u|4p |c%4 |p = |c4 |p , we have by (S30) that |c4 |p = |u|4p |c%4 |p ≤ p−4 , so if |c4 |p > p−4 , we are done, so this is (2).

If 1 ≥ |c%6 |p > |c6 |p , then since u6 c%6 = c6 , as above, |c6 |p = |u|6p |c%6 |p ≤ p−6 , so if |c6 |p > p−6 , we are done, and this is (3).

Now assume that p > 3, then |1728|p = |26 33 |p = 1. If |∆|p ≤ p−12 and |c4 |p ≤ p−4 , then given that 1728∆ = c34 − c26 , it follows that |c6 |2p = |c34 − 1728∆|p ≤ max{|c4 |3p , |∆|p } = p−12 , so |c6 |p ≤ p−6 . However, we showed how to use an admissible change of variables to get (10.27) on page 366, namely y 2 = x3 − 27c4 x + 54c6 with discriminant ∆% = 26 39 ∆ via Remark 10.9. Now if we make the admissible change of variables x ,→ p2 X and y ,→ p3 Y , then (10.27) becomes Y 2 = X 3 − 27(c4 p−4 )X − 54c6 p−6 .

(S31)

42

Advanced Number Theory Now since |c4 |p ≤ p−4 , then |c4 p−4 |p ≤ 1 and since |c6 |p ≤ p−6 , then |c6 p−6 |p ≤ 1, namely the coefficients of (S31) are p-adic integers. Lastly, the discriminant of (S31) is given by ∆%% = p−12 ∆% , then |∆%% |p = p12 |∆% |p = p12 |∆|p , so if |∆%% |p = p−j , then |∆|p = p−j−12 , so we have reduced the power of p in the discriminant from the original equation, which therefore could not have been minimal for p.

10.32 If E is singular at P = (x0 , y0 ), then the admissible change of variables in Definition 10.14 on page 355 with u = 1 and s = 0 leaves c4 and ∆ invariant—see Remark 10.10 on page 355. Hence, we may assume that x0 = y0 = 0. Now via (10.37) we have that a6 = f (0, 0) = 0, a4 = ∂f /∂x(0, 0) = 0, and a3 = ∂f /∂y(0, 0) = 0. Hence, (10.37) becomes f (x, y) = y 2 + a1 xy − a2 x2 − x3 = 0. Since ∆ = (c34 − c26 )/1728, c4 = b22 , and c6 = −b32 , then ∆ = 0. By Remark 10.10, E has a node at (0, 0) if f (0, 0) = 0 = (y − αx)(y − βx) − x3 = y 2 + a1 xy − a2 x2 − x3 , with α $= β, However, this means that the discriminant a21 + 4a2 $= 0 for y 2 + a1 xy − a2 x2 . Since c4 = b22 = (a21 + 4a2 )2 , then we have shown that if E has a node, then c4 $= 0. By Exercise 10.31 ∆ = 0. Conversely, if ∆ = 0, then by Exercise 10.31, E has a singular point P . If c4 $= 0, then by the above argument α $= β so E has a node. If E has a cusp at (0, 0), then by Remark 10.10 again,

f (0, 0) = 0 = (y − αx)2 − x3 = y 2 + a1 xy − a2 x2 − x3 , so the discriminant of y 2 + a1 xy − a2 x2 is a21 + 4a2 = 0, so c4 = 0 and by Exercise 10.31, ∆ = 0. Conversely, if ∆ = 0 = c4 , then by Exercise 10.31, E has a singular point and by the above it is a cusp. 10.34 Via sending y to (y − 1)/2, y 2 + y = x3 − x2 − 10x − 20 becomes y 2 = 4x3 − 4x2 − 40x − 79. Then using the techniques in the solution of Exercise 10.31, we get that reduction modulo 11 yields y 2 = 4x3 − 4x2 − 40x − 79 = (x − 5)2 (4x − 8), so at (x, y) = (5, 0), there is a node, illustrated by the graph of the righthand side:

43

Solutions to Even-Numbered Exercises

10.36 Since a1 = a3 = a4 = 0, a2 = 1 and a6 = p, then b2 = 4, b4 = 0, and b6 = 4p = b8 , then ∆ = b22 b8 − 8b34 − 27b26 + 9b2 b4 b6 = −16p(4 + 27p), so E has bad reduction at p. Moreover, modulo p, y 2 = x2 (x + 1), so E has a node at (0, 0) over Fp . 10.38 If (x, y) satisfies then set

E1 : y 2 + y = x3 − x2 , X =x+

and

1 2 1 + + , x2 x − 1 (x − 1)2

Y = y − (2y + 1) which satisfies

!

1 1 1 + + 3 3 x (x − 1) (x − 1)2

E2 : Y 2 + Y = X 3 − X 2 − 10X − 20. Thus, the map

h : E1 ,→ E2 ,

via (x, y) ,→ (X, Y ) is the isogeny.

"

,

44

Advanced Number Theory

10.40 First we show that the curve E % y 2 = x3 − 4kx2 − 160k 2 x − 1264k 3 = x3 + a%2 x2 + a%4 x + a%6

(S32)

arises from a global minimal Weierstrass equation for E by an admissible change of variables with u = 2. We have b%2 = (a%1 )2 + 4a%2 = 02 + 4(−4k) = −16k, b%4 = 2a%4 + a%1 a%3 = 2(−160k 2 ) + 0 = −26 · 5k 2 ,

so

c%4 = (b%2 )2 − 24b%4 = (−16k)2 − 24(−26 · 5k 2 ) = 24 (24 · 31k 2 ) = 24 c4 . Also, and

b%6 = (a%3 )3 + 4a%6 = 0 + 4(−1264k 3 ) = −26 · 79k 3 c%6 = −(b%2 )3 + 36b%2 b%4 − 216b%6

= −(−16k)3 + 36(−16k)(−26 · 5k 2 ) − 216(−26 · 79)k 3 = 26 (23 · 2501k 3 ) = 26 c6 ,

and ∆% =

(c%4 )3 − (c%6 )2 = 212 (−115 k 6 ) = 212 ∆. 1728

Hence, (∆, c4 , c6 ) goes to (∆% , c%4 , c%6 ) = (212 ∆, 24 c4 , 26 c6 ), which shows, via Remark 10.10 on page 355, that E is taken to E % under an admissible change of variables under u = 2 with r = s = t = 0. Now we3 show that k has no square odd factor. Assume, to the contrary, that p2 3 K for an odd prime k. Then we may make an admissible change of variables on (S32) with u = 1/p, which shows E not to be minimal at p. Hence, k has no square odd prime factor. 3 Now we provide the congruences modulo 16 for k. If 16 3 k, then as above, we may use u = 1/4 to determine that E is not minimal at 2. Hence, k $≡ 0 (mod 16). If k ≡ 1 (mod 4), then (S32) has an admissible change of variables given by x = 4X and y = 8Y + 4, namely for u = 2, t = 4 and r = s = 0, to take it to the curve y 2 + y = x3 − kx2 − 10k 2 x −

79k 3 + 1 , 4

(S33)

with (∆, c4 , c6 ) given by (10.40). Therefore, by Exercise 10.30, (S33) is global minimal. If k ≡ 2 (mod 4), then an admissible change of variables with u = 2, and r = s = t = 0, namely x = 4X, y = 8Y , takes (S32) to y 2 = x3 − kx2 − 10k 2 x − 79k 3 /4

45

Solutions to Even-Numbered Exercises

which again has (∆, c4 , c6 ) given by (10.40), so is also global minimal. Now assume that k ≡ 3 (mod 4). By Exercise 10.30, we know that (S33) is minimal at all odd primes. Now we show that it is minimal at 2, since Exercise 10.30 does not apply for p = 2 here. Assume that we have made an admissible change of variables to get the situation given in Definition 10.14 on page 355. Then in the notation of that definition, each of the a%j must be 3divisible by 2j since this is a reduction at p = 2. From % 3 a%3 we get 3 that 4 t since a32 = 0 = a1 in (S33). Thus,% from a4 we get6 that 4 3 r since a4 = −160k and a2 = −4k. Hence, a6 tells us that 2 2 3 3 % must 3 divide t + 1264k , since a6 = −1264k . By setting t = 4t , we get 4 3 ((t% )2 − 79k 3 ). However, since k ≡ 3 (mod 4), then 79k 3 ≡ 1 (mod 4). Therefore, (t% )2 ≡ −1 (mod 4), a contradiction. 3 Hence, (10.40) is global minimal when k ≡ 3 (mod 4). The case where 4 3 k remains. Set k = 4k % . Then (10.40) becomes 2

y 2 = x3 − 4k % x2 − 160k % x − 1264k %

3

after an admissible change of variables. This is global minimal since (∆, c4 , c6 ) is given by (10.40). Since 16 ! k, then 4 ! k % . Hence, by the above argument k % ≡ 3 (mod 4). In conclusion, we have shown that (10.40) yields a global minimal Weierstrass equation exactly when k has no odd square factor, and k ≡ r (mod 16) where r ∈ {1, 2, 5, 6, 9, 10, 12, 13, 14}.

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