Galois Theory and Applications: Solved Exercises and Problems 9813238305, 9789813238305

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GALOIS

THEORY AND APPLICATIONS Solved Exercises and Problems

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GALOIS

THEORY AND APPLICATIONS Solved Exercises and Problems

Mohamed Ayad Université du Littoral, Calais, France

World Scientific NEW JERSEY

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LONDON

10939_9789813238305_TP.indd 2

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SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TAIPEI

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CHENNAI

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TOKYO

27/3/18 9:59 AM

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Names: Ayad, Mohamed (Mathematician), author. Title: Galois theory and applications : solved exercises and problems / by Mohamed Ayad (Université du Littoral, Calais, France). Description: New Jersey : World Scientific, 2018. | Includes bibliographical references. Identifiers: LCCN 2018020885 | ISBN 9789813238305 (hardcover : alk. paper) Subjects: LCSH: Galois theory--Problems, exercises, etc. | Algebra, Abstract--Problems, exercises, etc. Classification: LCC QA162 .A93 2018 | DDC 512/.32--dc23 LC record available at https://lccn.loc.gov/2018020885

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. Cover image: Factorizations of 90 in [ 14 ]. Copyright © 2018 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher. For any available supplementary material, please visit http://www.worldscientific.com/worldscibooks/10.1142/10939#t=suppl

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Preface

This book arose from lectures and tutorials given by the author at Algiers University (U.S.T.H.B.) and at Universit´e du Littoral (Calais). To the best of my knowledge, there is no book of exercises and problems in Galois Theory except the author’s books, published in French. The material contained in the present book is completely different from that which makes up the French one. This book is intended for several types of audiences: First, students who want to learn the material around the arithmetic of polynomials and field extensions. It can be used by beginners (3rd year students), but also more advanced students (e.g. Master or Ph. D. students working in the area). Second, teachers who want to work on problems with their students. Third, researchers can use it as a reference: the book contains some useful results generally known by specialists, but for which a detailed proof is not always in the literature. Besides Galois Theory, the book contains chapters devoted to fields which make use of this theory: Finite fields, Permutation polynomials and an introduction to Algebraic Number theory. This last field occupies the longest chapter of the book. In the solutions of the exercises or problems, when it is useful, I refer to Lang’s book (Algebra) if it concerns results in Galois theory, and to Marcus’s book (Number Rings) for results in Number theory. Occasionally references are made to other books cited in the bibliography. I am indebted to several colleagues who each read some chapters and suggested corrections on a preliminary draft of the book. In alphabetical order: B. Bensebaa, D. Bitouz´e, R. Bouchenna, A. Bouhamidi, H. Chapdelaine, P. D`ebes, P. Feischman, O. Kihel, F. Recher and C. Woodcock. v

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I express my thanks to the following students of Brock University for reading parts of the book or for their help in the use of Latex: C. Asselin, Z. Schedler, J. Larone, B. Earp-Lynch and S. Earp-Lynch. The responsibility for any errors is solely mine and I thank in advance vigilant readers who will point out errors or will indicate alternative solutions. I learned a lot during the preparation of this work, which was going on for seventeen years. I hope that the readers will profit from its use. M. Ayad [email protected]

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Contents

Preface

v

1.

Polynomials, Fields, Generalities

1

2.

Algebraic extension, Algebraic closure

27

3.

Separability, Inseparability

71

4.

Normal extensions

83

5.

Galois extensions, Galois groups

91

6.

Finite fields

159

7.

Permutation polynomials

177

8.

Transcendental extensions, Linearly disjoint extensions, Luroth’s theorem

197

Multivariate polynomials

243

9.

10. Integral elements, Algebraic number theory

279

11. Derivations

411

Notations

439

Bibliography

443

vii

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Chapter 1

Polynomials, Fields, Generalities

Exercise 1.1. Let p be a prime number and
s be a positive integer. Show that for any s i ∈ {0, 1, . . . , ps − 1}, p i−1 ≡ (−1)i (mod p). Solution 1.1. Let x be an indeterminate over Q, then s pX −1  s

i=0

 s s s X p −1 i x = (1 + x)p −1 = (1 + x)p /(1 + x) = (1 + x)p (−1)i xi i i≥0 s X ≡ (1 + xp ) (−1)i xi (mod p) i≥0

X X s ≡ (−1)i xi + (−1)i xp +i

≡ ≡

i≥0

i≥0

i≥0

j≥ps

X X s (−1)i xi + (−1)j−p xj s pX −1

(−1)i xi

(mod p) (mod p)

(mod p),

i=0

hence the result. Exercise 1.2. Let n be a positive integer.
Pn (1) Show that k=0 (−1)k nk = 0.

(2) Show that for any k ∈ {1, . . . , n}, k nk = n n−1 k−1 . (3) Show that for any commutative ring R and
any polynomial f (x) ∈ R[x] Pn of degree d < n, we have k=0 (−1)k nk f (k) = 0. 1

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(4) Show that the condition on d cannot be omitted. Solution 1.2. (1) The assertion follows immediately from the identity   n X n k n (x − 1) = (−1) xn−k k k=0

after substitution of 1 for x. (2) We have     n−1 n kn(n − 1)! =n . k = k(k − 1)!(n − 1 − (k − 1))! k−1 k

(3) We fix the ring R, the integer n and we prove the statement by induction on d. For d = 0, the result follows from (1). Suppose that the statement is true for all polynomials with coefficients in R of degree less than d. To prove it for polynomials of degree d it is sufficient to prove it for f (x) = xd . Using (2) and the inductive hypothesis, we obtain     n n X n d X n d (−1)k k = (−1)k k k k k=0 k=1   n X n − 1 d−1 k = (−1) n k k−1 k=1   n−1 X n−1 k+1 = (−1) n (k + 1)d−1 k k=0   n−1 X n−1 = −n (−1)k (k + 1)d−1 k k=0

= 0.

(4) For n = 1, the identity in (3) reads f (0) − f (1) = 0. Therefore any polynomial f (x) ∈ R[x] such that f (0) 6= f (1) does not satisfy this identity. For instance f (x) = xd , with d ≥ 1 is such a polynomial. Exercise 1.3. Let 0 ≤ k ≤ n be integers, n 6= 0 and let p be a prime number. Let Pr Pr n = i=0 ni pi and k = i=0 ki pi with 0 ≤ ki , ni < p, nr 6= 0 and kr ≤ nr .
Qr

(1) Show that nk ≡ i=0 nkii (mod p), where (by definition) nkii = 0 if
ki > ni . Compute the residue of 23 6 modulo 35.

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3


(2) Show that nk ≡ 0 (mod p) if and only if there exists i ∈ {1, . . . , r} such that ki > ni .

(3) Show that the number of nk , 0 ≤ k ≤ n such that nk 6≡ 0 (mod p) is Qr equal to i=0 (ni +1). Deduce that, reducing modulo 2, the coefficients in Pascal’s triangle, then on each row the number of one’s is a power of 2. (4) Let p be a fixed prime number. Represent any non negative integer n in the form n = nk pk + nk−1 pk−1 + · · · + n0 , where ni ∈ {0, . . . , p − 1} for i = 0, . . . , k and nk 6= 0 if n 6= 0. Denote this representation by n = [nk nk−1 · · · n0 ]. Define in N the following law of composition: [0] ? [nk nk−1 · · · n0 ] = [nk nk−1 · · · n0 ],

[nk nk−1 · · · n0 ] ? [0] = [nk nk−1 · · · n0 0]

and

[nk nk−1 · · · n0 ] ? [mh nh−1 · · · m0 ] = [nk nk−1 · · · n0 mh · · · m0 ]. We admit that (N, ?) is a semigroup. Let G be the multiplicative semigroup, G = Z[x]/(xp−1 − 1). Fix a generator ξ of (Z/pZ)? and let σ : N → G be the map such that σ(0) = 1 and σ(n) =

p−2 X

ri (n)xi

(Eq 1)

i=0

if n 6= 0,
where ri (n) denotes the number of integers l, 0 ≤ l ≤ n such that nl ≡ ξ i (mod p). Show that σ is a morphism of semigroups. (5) For any positive integer m let Rm (x) =

p−2 X

ri (m)xi .

(Eq 2)

i=0

Show that Rn (x) ≡

p−1 Y j=1

Rj (x)tj (n)

(mod xp−1 − 1),

(Eq 3)

where tj (n) denotes the number of times the digit j appears in the p-ary expansion of n. (6) Let m and n be positive integers. Suppose that any non zero digit j has the same number of apparitions in the p-ary expansions of m and of n. Show that reducing modulo p the m-th and the n-th line of Pascal’s triangle, any integer l ∈ {1, . . . , p − 1} has the same number of occurrences in both lines.

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(7) Fix c and d in {1, . . . , p − 1}. For any (i, j) ∈ {1, . . . , p − 1}2 , i < j let nij = cpi + dpj . Show that reducing modulo p the elements of Pascal’s triangle, the number of apparitions of l ∈ {1, . . . , p − 1} in the line nij does not depend on (i, j). Qk (8) Let n = [nk · · · n0 ] with nk 6= 0. Show that Rn (1) = i=0 (ni + 1). Solution 1.3.

Q i (1) We have (1 + x)n ≡ (1 + xp )ni (mod p). Identifying, in both sides of this identity, the coefficient of xk , we obtain     X n0 n1  n nr ≡ ··· (mod p). k l0 l1 lr P i li p =k

Since k has a unique representation in base p this sum (of products) reduces to one product and the proof of the first part of (1) is complete. We have 23 = 4.5 + 3 and 6 = 1.5 + 1, hence      23 4 3 ≡ (mod 5) ≡ 2 (mod 5). 6 1 1 Similarly, we have 23 = 3.7 + 2 and 6 = 0.7 + 6, hence      23 3 2 ≡ (mod 7) ≡ 0 (mod 7). 6 0 6
Now Chinese remainder theorem shows that 23 6 ≡ 7 (mod 35).
(2) By (1),
nk ≡ 0 (mod p) if and only if there exists i ∈ {0, . . . , r}
such that nkii ≡ 0 (mod p). But it is clear that if ki ≤ ni , then nkii 6≡ 0 (mod p), hence
the result. (3) We have
nk 6≡ 0 (mod p) if and only if for any i ∈ {0, . . . , r}, ki ≤ ni , hence nk 6≡ 0 (mod p) if and only if k0 = 0, . . . , n0 , k
1 = 0, . . . , n1 , . . . and kr
= 0, . . . , nr . It follows that the number of nk , 0 ≤ k ≤ n such Qr that nk 6≡ 0 (mod p) is equal to i=0 (ni + 1). We deduce that if p = 2, then the number of coefficients congruent to 1 Q modulo 2 in the n-th row of the Pascal’s triangle is equal to (ni + 1). Q Since ni = 0 or ni = 1, then (ni + 1) = 2e with e ≥ 1. (4) For fixed n, we extend the definition of ri (n) for any i ≥ 0 as follows. We write i in the form i = (p − 1)qi + ui with ui < p − 1. We set     n ri (n) = l, 0 ≤ l ≤ n such that ≡ ξ i (mod p − 1) l     n = l, 0 ≤ l ≤ n such that ≡ ξ ui (mod p − 1) l = rui (n).

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Let m1 = [ak · · · a0 ], m2 = [bh · · · b0 ] and n = m1 ? m2 . We may write any integer g ≤ n in one and only one way in the form 0 0 0 g = g1 ? g2 with g1 = [ak · · · a0 ] and g2 = [bh · · · b0 0 ]. Here some first 0 digits aj of g1 may be equal to 0. The same remark applies for g2 . We have σ(m1 ? m2 ) =

p−2 X

ri (m1 ? m2 )xi

i=0

 p−2   X m1 ? m2 ≡ ξi = g; g i=0

p−2  X

 (mod p) xi

    i m2 i g; m1 x ≡ ξ (mod p) g1 g2 i=0        p−2 X p−2  X m1 m2 j  = g1 ; g1 ≡ ξ (mod p) . g2 ; g2

=

i=0

j=0

i=0

j=0

  ≡ ξ i−j (mod p) xi   p−2 X p−2 X  = rj (m1 )ri−j (m2 ) xi ≡

p−2 p−2 X X i=0 j=0



rj (m1 )xj . rp−1+i−j (m2 )xp−1+i−j (mod xp−1 − 1)

≡ σ(m1 )σ(m2 )

(mod xp−1 − 1).

(5) Set n = [nk · · · n0 ]. Since R0 (x) = 1, then by (4) we have Rn (x) = σ(n) ≡ σ([nk ])σ([nk−1 ]) · · · σ([n0 ])

(mod xp−1 − 1)

≡ Rnk (x)Rnk−1 (x) · · · Rn0 (x)

(mod xp−1 − 1)

≡

p−1 Y j=1

Rj (x)tj (n)

(mod xp−1 − 1).

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(6) Our assumption on the apparition of the digit j means that tm (j) = tn (j). Since this equality is true for any j ∈ {1, . . . , p − 1}, then Rn (x) ≡

p−1 Y

≡

p−1 Y

Rj (x)tj (n)

(mod xp−1 − 1)

Rj (x)tj (m)

(mod xp−1 − 1)

j=1

j=1

≡ Rm (x)

(mod xp−1 − 1).

Since the polynomials Rn (x) and Rm (x) have their degrees less than p − 1, then Rn (x) = Rm (x). It follows that for any i = 0, . . . , p − 2, ri (n) = ri (m), that is     l ∈ {0, . . . , n}; n ≡ ξ i (mod p) l     m i = l ∈ {0, . . . , m}; ≡ ξ (mod p) . l Since for any digit j 6= 0, there exists i ∈ {0, . . . , p − 2} such that j ≡ ai (mod p), then     l ∈ {0, . . . , n}; n ≡ j (mod p) l     m ≡ j (mod p) . = l ∈ {0, . . . , m}; l

(7) Let ni1 j1 = cpi1 + dpj1 and ni2 j2 = cpi2 + dpj2 , then tc (ni1 j1 ) = tc (ni2 j2 ) = 1

and

td (ni1 j1 ) = td (ni2 j2 ) = 1. By (6), the number of apparitions of l ∈ {1, . . . , p−1} in the line ni1 j1 is the same as in the line ni2 j2 . (8) The definition of Rn (x) (see (Eq 2)), implies that    n Rn (1) = ri (n) = l ∈ {0, . . . , p − 1} such that 6≡ 0 l i=0 p−2 X

By (3), we get Rn (1) =

Qk

i=0 (ni

+ 1).

 (mod p) .

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Exercise 1.4. (1) For
any non
negative integers n and k define the binomial coefficient
(−n)(−n−1)···(−n−(k−1)) −n −n −n for n > 0. Show by 0 = 1 and k = k! k that n(n + 1) · · · (n + k − 1) k! n(n + 1) · · · (n + k − 2) n + + ··· + + 1 (k − 1)! 1! (n + 1) · · · (n + k) . = k!
k P Deduce that (1 + x)−n = k≥0 −n k x . (2) Let p be a prime number, r, s, a, b be rational
integers
such that r ≥ 1, b a ≡ m (mod ps+1 ) for s ≥ 0 and a ≡ b (mod pr+s ). Show that m r any integer m such that 1 ≤ m < p . Solution 1.4. n (1) We proceed by induction on k. If k = 1, then 1! + 1 = n + 1 = n+1 1! , hence the proof in this case. Suppose that the result holds for k − 1, that is n(n + 1) · · · (n + k − 2) n (n + 1) · · · (n + k − 1) + ··· + + 1 = , (k − 1)! 1! (k − 1)! then n(n + 1) · · · (n + k − 1) k! n n(n + 1) · · · (n + k − 2) + ··· + + 1 + (k − 1)! 1! n(n + 1) · · · (n + k − 1) = k! (n + 1) · · · (n + k − 1) + (k − 1)! (n + 1) · · · (n + k − 1) (1 + n/k) = (k − 1)! (n + 1) · · · (n + k) = . k!
We prove the second part of (1). Notice that if n = 1, then −1 = k (−1)k . We have X X −1 (1 + x)−1 = 1/(1 + X) = (−1)k xk = xk , k k≥0

k≥0

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so that the formula holds for n = 1. Suppose that it holds for the integer n, then X −n X (1 + x)−n−1 = xk (−1)k xk k k≥0 k≥0  ! X X t −n = (−1) xk . l
k≥0 l+t=k P Let σn,k = l+t=k (−1)t −n l , then by the first part of (1),   n n · · · (n + k − 1) n · · · (n + k − 2) + + ··· + + 1 σn,k = (−1)k k! (k − 1)! 1! (n + 1) · · · (n + k) = (−1)k k!   −n − 1 = , k hence the result. (2) Let c ∈ Z such that a = b + c. Since (1 + x)a = (1 + x)b (1 + x)c , then from (1), we have      X a b c = , m t l l≥0,t≥0,l+t=m

for
any non
negative integer m, in particular for 1 ≤ m < pr . We have c c c−1 b ≡ 0 (mod pr+s ) and 1 ≤ m < pr , then l = l l−1 . Since c = a −

a b c s+1 , thus m ≡ m (mod ps+1 ) l ≡ 0 (mod p)

Exercise 1.5. Let d1 , . . . , dn be rational integers and d = gcd(d1 , . . . , dn ). Recall that Pn there always exist integers k1 , . . . , kn ∈ Z such that d = i=1 ki di . Show that gcd(k1 , . . . , kn ) = 1. Solution 1.5. Let x = gcd(k1 , . . . , kn ), then xd | ki di for i = 1, . . . , n, hence xd | d and therefore x = 1. Exercise 1.6. Let f (x) and g(x) ∈ Z[x] be coprime. Let u(x), v(x) ∈ Z[x] and m ∈ N such that u(x)f (x) + v(x)g(x) = m. Let d be the greatest positive divisor of m such that there exists an integer x0 such that gcd(f (x0 ), g(x0 )) = d. Let a be an integer such that 0 ≤ a < d and x0 ≡ a (mod d). Set 1 1 f1 (x) = f (dx + a), g1 (x) = g(dx + a), d d u1 (x) = u(dx + a) and v1 (x) = v(dx + a).

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Show that u1 (x)f1 (x) + v1 (x)g1 (x) = m/d, f1 (x), g1 (x) ∈ Z[x] and that for any integer x, gcd(f1 (x), g1 (x)) = 1. Solution 1.6. Replacing x by dx + a in the Bezout’s Identity relating f (x) and g(x) and dividing by d, we obtain u1 (x)f1 (x) + v1 (x)g1 (x) = m/d. We show that f1 (x) ∈ Z[x] (the same argument applies for the polynomial g1 (x)). We have (n)

0

f (a) + dx f 1!(a) + · · · + (dx)n f n!(a) f1 (x) = d f (a) f 0 (a) f (n) (a) = +x + · · · + dn−1 xn d 1! n! and since d | f (a), then f1 (x) ∈ Z[x]. Suppose by contradiction that there exist an integer x1 and an integer δ > 1 such that gcd(f1 (x1 ), g1 (x1 )) = δ. Then δ | m/d, hence dδ | m, which contradicts the definition of d. Exercise 1.7. Let K be a field, f (x, y) and g(x, y) be relatively prime polynomials with coefficients in K, m = degx f and n = degx g. Let x1 , . . . , xk be new variables, A(x1 , . . . , xk ) and B(x1 , ..., xk ) be relatively prime polynomials with coefficients in K such that A/B 6∈ K. Show that B m f (A/B, y) and B n g(A/B, y) are coprime in K[x1 , . . . , xk , y]. Solution 1.7. By Bezout’s identity applied in k(x)[y], there exist u0 , v0 ∈ k(x)[y] such that u0 f + v0 g = 1. Multiplying this identity by the common denominator of u0 and v0 , we obtain the following new identity uf + vg = w, where u, v ∈ K[x, y] and w ∈ K[x]. Substituting A/B for x and multiplying by a power of B, say B e , we obtain: u ˆ(x, y)B m f (A/B, y) + vˆ(x, y)B n g(A/B, y) = B e w(A/B), where u ˆ and vˆ ∈ K[x, y]. Any common divisor D(x1 , . . . , xk , y) of B m f (A/B, y) and B n g(A/B, y) must divide B e w(A/B), hence it belongs to K[x1 , . . . , xk ], that is free from y. It follows that expressing B m f (A/B, y) and B n g(A/b, y) as polynomials in y, then all their coefficients are divisible by D. Set f (x, y) =

t X j=0

fj (x)y j and g(x, y) =

l X i=0

gi (x)y i ,

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then B m f (A/B, y) =

t X

B m fj (A/B)y j and B n g(A/B, y) =

j=0

l X

B n gi (A/B)y i .

i=0

Since f (x, y) and g(x, y) are relatively prime, then so are the polynomials ft (x), . . . , f0 (x), gl (x), . . . , g0 (x). Writing a Bezout’s identity relating these polynomials, substituting A/B for x and multiplying by some power of B, we obtain a new identity, which shows that D divides a power of B. Since m = degx f , there exists j ∈ {0, . . . , t} such that deg fj = m. Set fj (x) = am xm + · · · + a1 x + a0 , then B m fj (A/B) = am Am + · · · + a1 AB m−1 + a0 B m .

Let p be a prime factor of D. Then p divides B and the left hand side, hence also A. This contradiction shows that D = 1. Exercise 1.8. Let K be a field of characteristic 0, f (X) be a polynomial of degree n and m > n be an integer. Show that    n X m m−j−1 f (m) = (−1)n−j f (j). j n−j j=0 Solution 1.8. Using Lagrange’s interpolation formula, [Bourbaki (1950), Application: Formule d’interpollation de Lagrange, Chap. 4.2] or [Ayad (1997), Exercice 1.14] we obtain: Q n X i6=j (x − i) . f (j) Q f (X) = i6=j (j − i) j=0

We deduce that Q n X i6=j (m − i) f (j) Q f (m) = i6=j (j − i) j=0 =

n X

f (j)

j=0

=

n X

m(m − 1) · · · (m − (j − 1))(m − (j + 1)) · · · (m − n) j(j − 1) . . . (j − (j − 1))(j − (j + 1)) · · · (j − n)

m(m − 1) · · · (m − (j − 1)) (m − (j + 1)) · · · (m − n) j! (−1)n−j (n − j)(n − (j − 1)) · · · 1 j=0    n X m m−j−1 = (−1)n−j f (j) j n−j j=0 f (j)

and the proof is complete.

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Exercise 1.9. Let K be a field, a0 , . . . , an be distinct elements of K, b0 , . . . , bn be elements of K, distinct or not. Let f (x) ∈ K[x] such that deg f ≤ n and f (ai ) = bi for i = 0, . . . , n. Let g(x) and h(x) ∈ K[x], both of degree at most n − 1 and satisfying the conditions g(ai ) = bi and h(aj ) = bj for i = 0, . . . , n − 1 and j = 1, . . . , n. Show that f (x) =

(x − a0 )h(x) − (x − an )g(x) . an − a0

Solution 1.9. n )g(x) . One verifies easily that F (ai ) = f (ai ) for Let F (x) = (x−x0 )h(x)−(x−x xn −x0 i = 0, . . . , n. We also have deg f and deg F ≤ n, hence F (x) = f (x) by Lagrange’s interpolation theorem [Bourbaki (1950), Application: Formule d’interpollation de Lagrange, Chap. 4.2] or [Ayad (1997), Exercice 1.14]. Exercise 1.10. Let A be a factorial ring, K be its fraction field. Let g(x), h(x) ∈ K[x] and f (x) = g(x)h(x). Let a and b be coefficients of g and h respectively. Suppose that f (x) ∈ A[x]. Prove that ab ∈ A. Solution 1.10. We have a = a0 cont(g) and b = b0 cont(h) where a0 and b0 are elements of A. Then ab = cont(g) cont(h)a0 b0 = cont(f )a0 b0 ∈ A. Exercise 1.11. Let K be a field, n, m be positive integers, a0 , . . . , an , b0 , . . . , bm be algebraically independent variables over K. Let f (x) = an xn + · · · + a0 and g(x) = bm xm + · · · + b0 be polynomials with coefficients in K(a0 , . . . , bm ) and R(a0 , . . . , an , b0 , . . . , bm ) = Resx (f (x), g(x)). Show that R is homogeneous of weight nm under either of the following assignments of weights, w(ai ) and w(bj ) for the ai and the bj . (1) w(ai ) = i for i = 0, . . . , n and w(bj ) = j for j = 0, . . . , m. (2) w(ai ) = n − i for i = 0, . . . , n and w(bj ) = m − j for j = 0, . . . , m. Solution 1.11. (1) Let t a new variable, we must show that R(a0 , ta1 , . . . , tn an , b0 , tb1 , . . . , tm bm ) = tnm R(a0 , . . . , an , b0 , . . . , bm ).

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Let α1 , . . . , αn (resp. β1 , . . . , βm ) be the roots of f (x) (rep. g(x)) in an algebraic closure of K(t, a0 , . . . , an , b0 , . . . , bm ), then we have R(a0 , ta1 , . . . , tn an , b0 , tb1 , . . . , tm bm ) = Resx (f (tx), g(tx)) Y = (tn an )m (tm bm )n (αi /t − βj /t) (i,j)

=

n tnm am n bm

Y

(αi − βj )

(i,j)

= tnm Resx (f (x), g(x))) = tnm R(a0 , . . . , an , b0 , . . . , bm ). (2) Here we must show that R(tn a0 , tn−1 a1 , . . . , an , tm b0 , tm−1 b1 , . . . , bm ) = tnm R(a0 , . . . , an , b0 , . . . , bm ). We have R(tn a0 , tn−1 a1 , . . . , an , tm b0 , tm−1 b1 , . . . , bm ) = Resx (tn f (x/t), tm g(x/t)) Y = (an )m (bm )n (tαi − tβj ) (i,j)

=

n tnm am n bm

Y

(αi − βj )

(i,j)

=t

nm

Resx (f (x), g(x)))

=t

nm

R(a0 , . . . , an , b0 , . . . , bm ),

hence the result. Exercise 1.12. Let K be a field, f (x) = a0 xn + · · · + an and g(x) = b0 xm + · · · + bm be polynomials with coefficients in K of degree n and m respectively. Recall that the resultant of f (x) and g(x) is the determinant of the (m+n)×(m+n) matrix:   a0 a1 ... an 0 ... 0 0 a0 ... an−1 an ... 0    ... ... ... ... ... ... ...      ... 0 a0 ... ... an  0 S= ,  b0 b1 ... bm 0 ... 0   0 b0 ... bn−1 bm ... 0    ... ... ... ... ... ... ...  0 ... 0 b0 ... ... bm

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where the first m rows are devoted to the coefficients of f (x) and the remaining ones to the coefficients of g(x). Let D(x) = gcd(f (x), g(x)) and d = deg D. Let Km−1 [x] (resp. Kn−1 [x], Km+n−1 [x]) be the K-vector space constituted by the polynomials with coefficients in K of degree at most m−1 (resp. n−1, m+n−1). Let φ : Km−1 [x]×Kn−1 [x] → Km+n−1 [x] be the map such that φ(A(x), B(x)) = A(x)f (x) + B(x)g(x). (1) Show that φ is linear. Let f1 (x) = f (x)/D(x) and g1 (x) = g(x)/D(x). Show that n     o B= g1 (x), −f1 (x) , xg1 (x), −xf1 (x) , . . . , xd−1 g1 (x), −xd−1 f1 (x) is a basis of Ker φ over K. (2) Deduce that rank(S) = m + n − d. Solution 1.12. (1) Obviously, φ is linear over K. Let (A(x), B(x)) ∈ Km−1 [x] × Kn−1 [x]. Suppose that (A(x), B(x)) ∈ Ker φ, then A(x)f (x) + B(x)g(x) = 0 hence A(x)f1 (x) = −B(x)g1 (x). It follows that g1 (x)|A(x). Set A(x) = g1 (x)h(x) where deg h = deg A − deg g1 ≤ m − 1 − (m − d) = d − 1; then g1 (x)h(x)f1 (x) = −B(x)g1 (x), hence B(x) = −f1 (x)h(x). Set h(x) = h0 + h1 x + · · · + hd−1 xd−1 , where hi ∈ K for i = 0, . . . , d. Then (A(x), B(x)) = ((h0 + h1 x + ... + hd−1 xd−1 )g1 (x) − (h0 + h1 x + ... + hd−1 xd−1 ))f1 (x)

= h0 (g1 (x), −f1 (x)) + h1 (xg1 (x), −xf (x)) + · · ·   + hd−1 xd−1 gd−1 (x) − xd−1 fd−1 (x)

hence Ker φ is contained in the vector space generated by B. Obviously the elements of B belong to Ker φ. We conclude that Ker φ is generated by B. Obviously the elements of B are linearly independent over K. Therefore B is a basis of Ker φ over K. (2) Let A = {(1, 0), (x, 0), . . . , (xm−1 , 0), (0, 1), (0, x), . . . , (0, xn−1 )}. It it easy to see that A is a basis of Km−1 [x] × Kn−1 [x]. Moreover one verifies that the matrix of φ relatively to A and the canonical basis of Km+n−1 [x] over K is equal to the transpose of S. Therefore rank(S) = rank(S T ) = m + n − DimK Ker φ = m + n − d.

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Exercise 1.13. Let E and F be two fields of characteristic 6= 2. Suppose that there exists a map σ : E → F satisfying the following conditions. (1) For any x, y ∈ E, σ(x + y) = σ(x) + σ(y). (2) For any x ∈ E, σ(x2 ) = (σ(x))2 . Show that σ = 0 or σ is a field homomorphism. Solution 1.13. Let x and y ∈ E, then σ((x + y)2 ) = (σ(x + y))2 , hence σ(x2 ) + σ(y 2 ) + 2σ(xy) = (σ(x)))2 + (σ(y))2 + 2σ(x)σ(y). Since the characteristic is different from 2, then the result follows. Exercise 1.14. Let K be a field and p be a prime number. Show that the following propositions are equivalent. (i) p is at least equal to the characteristic of K. (ii) For any f (x) ∈ K[x], f (p) (x) = 0, where f (p) (x) denotes the p-th derivative of f (x). Solution 1.14. (1) (i) ⇒ (ii). It is sufficient to show that for any integer k ≥ 0, (xk )(p) = 0. If k < p, then (xk )(k) = k!, hence the result in this case. If k ≥ p, then (xk )(p) = k(k − 1) · · · (k − (p − 1))xk−p . Clearly one of the integers k, k − 1, . . . , k − (p − 1) is divisible by the characteristic of K, hence the result. (2) (ii) ⇒ (i). Let l be the characteristic of K. Suppose that p < l and let f (x) = xp , then f (x)(p) = p! 6= 0. Exercise 1.15. Let f (x) = an xn + an−1 xn−1 + · · · + a0 be a polynomial with integral coefficients. Suppose that there exists a prime number p|a0 such that p > Pn a0 i−1 . i=1 |ai || p |

(1) Show that any root α of f (x) satisfies the condition |α| > | ap0 |. (2) Deduce that if f (x) is primitive then it is irreducible in Z[x]. (3) Show that the preceding result applies for the polynomial f (x) = x2 + 1 x + 32 , where 1 and 2 ∈ {−1, 1}.

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Solution 1.15. (1) Suppose that |α| ≤ | ap0 |, then |a0 | = |an αn + an−1 αn−1 + · · · + a1 α| i n X a0 ≤ |ai | p i=1 X i−1 n a0 a0 = |ai | p i=1 p < a0 ,

hence a contradiction. We conclude that |α| > | ap0 |. (2) Suppose that f (x) is reducible in Z[x] and let f (x) = (bm xm + · · · + b0 )(ck xk + · · · + c0 ) be a factorization in this ring with m and k ≥ 1. Since b0 c0 = a0 , then p|b0 or p|c0 . We may suppose that p|b0 . Let α1 , . . . , αk be the roots of Qk the second factor, then i=1 αi = (−1)k cck0 . Hence by (1), k Y k a0 a0 b0 = |c0 | ≥ c0 = |αi | > , ck p p p i=1

which is a contradiction, thus f (x) is irreducible over Z. (3) Here p = 3 and |a1 | + |a2 || ap0 | = 2.

Exercise 1.16. Let n ≥ 2 be an integer, p ≥ 5 be a prime number and f (x) = xn −xn−1 −2p. (1) Suppose that f (x) = g(x)h(x) in Z[x], where g(x) and h(x) are monic of degree at least 1. Show that deg g = 1 or deg h = 1. (2) Show that f (x) is irreducible over Z. (3) Show that the condition p ≥ 5 is necessary. Solution 1.16. (1) We have f (x) ≡ xn−1 (x − 1)

(mod p) ≡ g(x)h(x)

(mod p).

(mod p) and h(x) ≡ xs

(mod p),

We may suppose that g(x) ≡ xr (x − 1)

where r and s are nonnegative integers, s ≥ 1 and r + s = n − 1. Suppose that r ≥ 1, then g(0) ≡ 0 (mod p) and h(0) ≡ 0 (mod p),

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hence f (0) ≡ 0 (mod p2 ), which contradicts the relation f (o) = −2p. We deduce that r = 0, g(x) ≡ x−1 (mod p) and h(x) ≡ xn−1 (mod p). Therefore, there exists a ∈ Z, a ≡ 1 (mod p) such that g(x) = x − a. (2) Suppose that f (x) is reducible over Z and let f (x) = g(x)h(x) be a nontrivial factorization, then g and h are monic and we may suppose that g(x) = x − a, with a ≡ 1 (mod p) and h(x) ≡ xn−1 (mod p). It follows that f (a) = 0 = an − an−1 − 2p, hence an−1 (a − 1) = 2p. We deduce that an−1 a−1 p = 2. This implies that a − 1 = p with  = ±1 and |a| = 2. These conditions are incompatible with the assumption p ≥ 5. Therefore f (x) is irreducible over Z. (3) Let p = 3, n = 2 and f (x) = x2 − x − 6, then f (x) is reducible over Z, since f (x) = (x + 2)(x − 3). Exercise 1.17. Determine the set of the couples (a, b) ∈ Z2 such that the polynomial f (x) = x3 − ax − b is reducible over Q. Solution 1.17. Let (a, b) be a couple of integers such that the polynomial f (x) is reducible, then it has a root d ∈ Z. Moreover d = 0 or d | b. In the first case we conclude that b = 0. Conversely if b = 0, then f (x) is reducible. In the second case, set b = de, then d3 − ad − de = 0, hence d2 − a − e = 0. It follows that a = d2 − e and b = de for some d, e ∈ Z. Conversely it is easy to verify that for these values of a and b, f (x) is reducible over Q. Exercise 1.18. Let p be a prime number, f (x) ∈ Z[x] be monic and non constant. Denote by u(x), the reduction modulo p of the polynomial u(x). Suppose that f (x)
e may be written in the form f (x) = g(x) + ph(x), where g(x) and h(x) ∈ Z[x], g(x) is monic, irreducible over Fp , deg h < deg f and g(x) - h(x). Show that f (x) is irreducible over Q. Show that the condition deg h < deg f could not be omitted. Solution 1.18. Suppose that f (x) = f1 (x)f2 (x) in Z[x], where f1 and f2 are monic and e e1 non constant, then f (x) = f1 (x)f2 (x) = g(x) , hence f1 (x) = g(x) and e2 f2 (x) = g(x) . It follows that f1 (x) = g(x)e1 + ph1 (x) f2 (x) = g(x)

e2

+ ph2 (x),

and

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where h1 (x) and h2 (x) ∈ Z[x] and e1 , e2 are positive integers such that e1 + e2 = e. Therefore
f (x) = g(x)e + p h2 (x)g(x)e1 + h1 (x)g(x)e2 + p2 h1 (x)h2 (x) , and

h(x) = h2 (x)g(x)e2 + h1 (x)g(x)e1 + ph1 (x)h2 (x), which shows that g(x) | h(x), hence a contradiction. Through the following example it is seen that the condition on the degree of h(x) is necessary. Let p = 2, f (x) = (x2 +2)(2x+1) = x2 +2(x3 +2x+1), g(x) = x, and h(x) = x3 + 2x + 1, then f (x) is reducible in Z[x] and g(x) - h(x). Exercise 1.19. Let f (x) = x4 + 3x + 7x + 4. Factorize this polynomial into its irreducible factors in Z/2Z[x] and in Z/11Z[x]. Show that f (x) is irreducible in Z[x]. Solution 1.19. We have f (x) ≡ x(x3 + x + 1) 2

(mod 2) 2

and

f (x) ≡ (x + 5x − 1)(x − 5x − 4)

(mod 11).

In these factorizations, the factors over F2 (resp. F11 ) are irreducible. Suppose that f (x) = f1 (x)f2 (x), where f1 and f2 are monic polynomials with integral coefficients and 1 ≤ deg f1 ≤ deg f2 ≤ 3, then f1 (x) ≡ x (mod 2) and f2 (x) ≡ x3 + x + 1 (mod 2), so that deg f1 = 1 and deg f2 = 3. We also have f1 (x) ≡ x2 + 5x − 1 (mod 11) and f2 (x) ≡ x2 − 5x − 4 (mod 11) or conversely. In any case deg f1 = 2. Therefore we realized a contradiction. It follows that f (x) is irreducible over Z. Exercise 1.20. Let p be a prime number, f (X, Y ) ∈ Z[X, Y ] be a homogeneous polynomial of degree n ≥ 1. Suppose that f is primitive and that for any (a, b) ∈ Z2 , p | f (a, b). Show that p < n. Solution 1.20. (1) First proof. We have f = Y n f˜(X/Y ) with f˜(X) = f (X, 1) and where f˜ induces the zero function on Fp . It follows that X p −X | f˜(X), hence Y p (X p /Y − X/Y ) | f (X, Y ), thus X p −XY p−1 | f (X, Y ). By symmetry the same holds with X, Y swapped, hence XY (X p−1 − Y p−1 ) divides f (X, Y ), so n ≥ p + 1.

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(2) Second proof. We first show that if g(X, Y ) ∈ Fp [X, Y ] induces the zero function over Fp , then XY (X p−1 − Y p−1 ) | g(X, Y ) in Fp [X, Y ]. Let c ∈ Fp . Application of the Euclidean algorithm yields: g(X, Y ) = (X − cY )Q(X, Y ) + d, where Q(X, Y ) ∈ Fp [X, Y ] and d ∈ Fp . Putting Y = 1, X = c in this identity leads to d = 0, hence X − cY | g(X, Y ). Using the same argument, we obtain Y | g(X, Y ). Therefore XY (X p−1 − y p−1 ) | g(X, Y ). Denote by f¯(X, Y ) the polynomial obtained from f by reducing its coefficients modulo p. Since f is primitive, then f¯ 6= 0. According to what has been proved, we conclude that XY (X p−1 − Y p−1 ) divides f¯(X, Y ), hence p + 1 ≤ deg f¯ ≤ deg f = n and the proof is complete. Exercise 1.21. Let Nq (n) be the number of monic irreducible polynomials over Fq . Recall P that Nq (n) = n1 d|n µ(n/d)q d , where µ is the Mobius function. If q1 < q2 show that Nq1 (n) < Nq2 (n). Solution 1.21. We fix n and we consider Nq (n) as a polynomial function of q. We show that this function is strictly increasing for q ≥ 2. Let m = bn/2c, then we have 1X µ(n/d)dq d−1 (Nq (n))0 = n d|n

= q n−1 +

1 n

X

µ(n/d)dq d−1

d|n,d6=n m

≥ q n−1 −

1 X d−1 dq n d=1 m X

m q d−1 n d=1 m m n−1 =q − (q − 1)/(q − 1) n m m n−1 ≥q − q n 1 n−1 n−1 ≥q − q 2 1 n−1 = q > 0. 2 ≥ q n−1 −

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Exercise 1.22. Let s and n be positive integers, K be a field, P0 , P1 , . . . , Ps be s + 1 distinct points in K n . Show that there exists a polynomial of degree s, f (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] having the form f (x1 , ..., xn ) =

n Y

e

(ai xi + bi ) i ,

i=1

where ai , bi ∈ K, ai 6= 0, 0 ≤ ei ≤ s for i = 1, ..., n and e1 + e2 + · · · + en = s such that f (Pi ) = 0 for i = 1, . . . , s and f (P0 ) = 1. Solution 1.22. For i = 0, 1, . . . , s, set Pi = (ci1 , . . . , cin ), where cij ∈ K for any (i, j). Fix h ∈ {1, . . . , s}. Since P0 6= Ph , then there exists j ∈ {1, . . . , n} such that chj 6= c0j . Therefore we may determine ah and bh ∈ K such that ah 6= 0, ah c0j + bh = 1 and ah chj + bh = 0, that is the polynomial fh (x1 , . . . , xn ) = ah xj + bh , satisfies the condition fh (P0 ) = 1 and fh (Ph ) = 0. That ah and bh could be determined may be easily verified since the determinant of the system of equations is equal to c0j − chj , hence nonzero. Now the polynomial f (x1 , . . . , xn ) =

s Y

fh (x1 , ..., xn )

h=1

satisfies the stated assertions. Exercise 1.23. Let K be a field, E be an extension of K, I be an ideal of K[x1 , . . . , xn ] and ˆ Iˆ be the ideal of E[x1 , . . . , xn ] generated by I. Show that I∩K[x 1 , . . . , xn ] = I. Solution 1.23. Let (ej )j∈J be a basis of E over K. We may suppose that ej0 = 1. We have M M E[x1 , . . . , xn ] = K[x1 , ..., xn ]·ej = K[x1 , . . . , xn ] K[x1 , ..., xn ]·ej , j∈J

j∈J\{ej 0 }

hence Iˆ = I · E(x1 , . . . , xn ) = I Thus Iˆ ∩ K[x1 , . . . , xn ] = I.

M

M

j∈J\{ej 0

Iej .

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Exercise 1.24. Let K be a field, f (x) ∈ K[x] be non constant and let α ∈ K. (1) Let Aα = {u(x)/v(x) ∈ K(x), v(α) 6= 0 and gcd(u, v) = 1}. Show that Aα is a local ring. Describe explicitly its maximal ideal M. Show that Aα /M ' K. (2) Let mα be the unique nonnegative integer such that f (x) = (x − α)mα g(x), where g(x) ∈ K[x] and g(α) 6= 0. Show that Aα /f Aα is a vector space over K of dimension mα . Solution 1.24. (1) Let S = {v(x) ∈ k[x], v(α) 6= 0}, then S is a multiplicative subset of K[x] containing 0 and 1 and clearly Aα = S −1 K[x]. We first determine the units of Aα . Let u(x)/v(x) ∈ A∗α , then there exists a(x)/b(x) ∈ Aα a(x) v(x) such that u(x)a(x) u(x)b(x) = 1. We deduce that b(x) = u(x) , hence u(α) 6= 0. ∗ Conversely if u(α) 6= 0, then u(x)/v(x) ∈ Aα . We conclude that A∗α = {u(x)/v(x) ∈ k(x), gcd(u, v) = 1, v(α) 6= 0

and u(α) 6= 0}.

To show that Aα is local it is sufficient to prove that the complementary set M of A∗α in Aα is an ideal. We have M = {u(x)/v(x) ∈ K(x), gcd(u, v) = 1, v(α) 6= 0

and u(α) = 0}

and we omit the proof that this set is an ideal of Aα . Let φ : Aα → K be the map defined for any u(x)/v(x) ∈ Aα by φ(u/v) = u(α)/v(α). Clearly φ is a morphism of rings. For any c ∈ K, we have c ∈ Aα and φ(c) = c, hence φ is surjective. Let u(x)/v(x) ∈ Aα , then φ(u/v) = 0 if and only if u(α) = 0, that is if and only if u(x)/v(x) ∈ M . Therefore Ker φ = M and then, the first isomorphism theorem implies Aα /M ' K. (2) Substituting y for x − α if necessary, we may suppose that α = 0, so that we have f (x) = xm g(x) where g(x) ∈ K[x], g(0) 6= 0 and m ≥ 0. Since g(x) is a unit in A0 , then f (x)A0 = xm A0 and A0 /f A0 = A0 /xm A0 . Obviously A0 is a vector space over K and xm A0 is a subspace, hence A0 /xm A0 is a vector space over K. We show that {¯1, x ¯, . . . , xm−1 } is m a basis of A0 /x A0 over k. Let u(x)/v(x) ∈ A0 , then v(0) 6= 0. We divide u(x) by v(x) according to the ascending powers of x at the order

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m. We obtain: u(x) = v(x)q(x) + xm r(x), where q(x) and r(x) ∈ K[x], hence u(x) r(x) = q(x) + xm ≡ q(x) v(x) v(x)

(mod xm A0 ),

so that {¯ 1, x ¯, ..., xm−1 } generates xm A0 over K. Suppose that a0 ¯ 1 + a1 x ¯ + · · · + am−1 xm−1 = 0, with a0 , a1 , . . . , am−1 ∈ K. Then a0 + a1 x + · · · + am−1 xm−1 ∈ xm A0 , hence a0 + a1 x + · · · + am−1 xm−1 = xm

u(x) , v(x)

where u(x), v(x) ∈ K[x], gcd(u, v) = 1 and v(0) 6= 0. We deduce that xm divides a0 + a1 x + · · · + am−1 xm−1 in K[x]. Therefore a0 = a1 = . . . = am−1 = 0, which implies that {¯1, x ¯, . . . , xm−1 } is a basis of m A0 /x A0 over K. Exercise 1.25. Let n be a positive integer, K be a field, E be an extension of K and A ∈ Mn (K). Show that the characteristic polynomials (resp. minimal polynomials) of A over K and E are equal. Solution 1.25. The proof of the assertion on the characteristic polynomials is obvious and will be omitted. Let f (x) and g(x) be the minimal polynomials of A over K and E respectively. It is clear that g(x) | f (x) in E[x]. Let B be a basis of E over K. We may suppose that 1 ∈ B. Set g(x) = e0 + e1 x + · · · + en−1 xm−1 + xm , where m ≤ deg f and ei ∈ E for i = 0, . . . , m−1. There is a finite set C ⊂ B say C = {1, w1 , . . . , wk } such that each ei , for i = 0, . . . , m − 1, is a linear Pk combination of the elements of C. Set w0 = 1 and ei = j=0 λij wj for Pk i = 0, . . . , m − 1. The identity g(A) = 0 takes the form: j=0 Bj wj , where Pm−1 Bj is the n × n matrix with coefficients in K given by Bj = i=0 λij Ai for Pm−1 j = 1, . . . , k and B0 = i=0 λi0 Ai + Am . We deduce that Bj = 0 for j = Pm−1 Pm−1 0, . . . , k. This implies that f (x) | i=0 λi0 xi + xm and f (x) | i=0 λij xi for j = 1, . . . , k. Since m ≤ deg f and f (x) is monic, then m = deg f , Pm−1 f (x) = i=0 λi0 xi + xm and λij = 0 for i = 0, . . . , m − 1 and j = 1, . . . , k. We conclude that g(x) = f (x).

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Exercise 1.26. Let E be a finite dimensional vector space over K, σ : E → E be a linear map. Let mσ (x) and χσ (x) be the minimal polynomial and characteristic polynomial of σ respectively. (1) Let α ∈ E and let I = {f (x) ∈ K[x], f (σ)(α) = 0}. Show that I is a principal ideal of K[x] generated by some monic polynomial. Denote this polynomial by m(σ,α) (x). Show that m(σ,α) (x) divide mσ (x) and χσ (x). (2) Let {α1 , . . . , αn } be a basis of E of K. Show that n

mσ (x) = lcm m(σ,α) (x) = lcm m(σ,αi ) (x). α∈E

i=1

Solution 1.26. (1) Let φ : K[x] → E be the map such that for any f (x) ∈ K[x], φ(f (x)) = f (σ)(α). Then obviously φ is a K-linear map. Since I 6= Ker(φ), then I is an additive subgroup of K[x]. Let f (x) ∈ I and g(x) ∈ K[x], then φ(g(x)f (x)) = (g(σ)f (σ))(α) = g(σ)(f (σ)(α)) = g(σ)(0) = 0, hence g(x)f (x) ∈ I. Therefore I is an ideal of K[x]. Thus, I is principal. Since mσ (x) ∈ I, then I 6= (0) and I is generated by a monic polynomial. Clearly m(σ,α) (x) | mσ (x) | ξσ (x). (2) Since for any α ∈ E, m(σ,α) (x) | mσ (x), then lcm m(σ,α) (x) | mσ (x). On the other hand, let g(x) = lcm m(σ,α) (x), then for any β ∈ E, we have α∈E

g(σ)(β) = 0, hence g(σ) = 0. Therefore mσ (x) | g(x). We conclude that mσ (x) = g(x) = lcm m(σ,α) (x). In the same way, we may prove n

α∈E

that mσ (x) = lcm m(σ,αi ) (x). i=1

Exercise 1.27. (1) Show that the number of monomials xi y j with 0 ≤ i, j ≤ m and i + j ≤ m is equal to ( m+2 2 ). (2) Let K be a field, f (t), g(t) be polynomials with coefficients in K of degree r and s respectively. Show that for m large enough, the family of polynomials f (t)i g(t)j with i + j ≤ m is dependent over K. (3) Deduce that there exists F (x, y) ∈ K[x, y]\{0} irreducible such that F (f (t), g(t)) = 0.

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Solution 1.27. (1) We make a partition of the set of couples (i, j) with i, j ≥ 0 and i + j ≤ m by putting in the same class all the couples having the same first component. The number of couples whose first component equals i is equal to m − i + 1. Therefore the total number of couples is equal to (m + 1) + m + ... + 1 = (m + 1)(m + 2)/2 = ( m+2 2 ).

(2) If f (t)i g(t)j = f (t)h g(t)k , with i + j ≤ m and h + k ≤ m, then the result is true. Therefore, we may suppose that the set {f (t)i g(t)j

( m+2 2 )

with

i + j ≤ m}

polynomials with coefficients in K. We have: contains exactly i j deg f (t) g(t) = ri+sj ≤ (r+s)m. Therefore all the elements of the set belong to the K-vector space Km [t] whose elements are the polynomials h(t) ∈ K[t] such that deg h ≤ (r + s)m. The dimension over K of this space is equal to (r + s)m + 1. Therefore if ( m+1 2 ) > (r + s)m + 1, that is m > 2(r + s) − 3, then the family is linearly dependent over K. (3) If m > 2(r + s) − 3 the family f (t)i g(t)j is linearly dependent over K, hence there exist a family (aij )(i,j) with aij ∈ K not all 0 such P P that aij f (t)i g(t)j = 0. Let G(x, y) = aij xi y j , then clearly G 6= 0 and G(f (t), g(t)) = 0. Let G(x, y) = G1 (x, y) · · · Gk (x, y) be the factorization into irreducible factors in K[x, y] of G(x, y). Then there exist i ∈ {1, . . . , k} such that Gi (f (t), g(t)) = 0. Exercise 1.28. Let K be a field, E be an extension of K, x be an indeterminate over E and α1 , . . . αn be elements of E, linearly independent over K. Show that these elements are linearly independent over K(x). Solution 1.28. Pn Suppose that there exist λ1 (x), . . . , λn (x) ∈ K(x) such that i=1 λi (x)αi = 0. We may suppose that λi (x) ∈ K[x]. Substituting 0 for x leads to λ1 (0) = · · · = λn (0) = 0. This shows that x | λi (x) for i = 1, . . . , n. Pn Suppose that xk | λi (x), then i=1 (λi (x)/xk )αi = 0. Substituting again 0 for x shows that xk+1 | λi (x). It follows that all the polynomials λi (x) are 0. Exercise 1.29. Let f (x) and g(x) be polynomials with integral coefficients. Show that g(x) | f (x) in Z[x] if and only if cont(g) | cont(f ) and there exist infinitely many n ∈ Z such that g(n) 6= 0 and g(n) | f (n).

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Solution 1.29. The necessity of the conditions is obvious. We prove the sufficiency of the conditions. First proof. Let (un ) be a sequence of integers such that lim |un | = ∞, g(un ) 6= 0

n→∞

and g(un ) | f (un ).

Using the Euclidean algorithm, we obtain f (x) = g(x)q(x) + r(x), where q(x), r(x) ∈ Q[x] and deg r < deg g. Suppose that r(x) 6= 0 and let m be the lcm of all the denominators of the coefficients of g(x) and r(x). Then mf (x)/g(x) − mq(x) = mr(x)/g(x).

There exists a rank n0 such that for n > n0 , r(un ) 6= 0. For any of these n, we have mr(un )/g(un ) ∈ Z and |mr(un )/g(un )| ≥ 1. On the other hand since deg r < deg g, then limn→∞ |mr(un )/g(un )| = 0 and we have reached a contradiction. Therefore r(x) = 0 and f (x) = g(x)q(x), hence cont(q) = cont(f )/ cont(g) ∈ Z which shows that q(x) ∈ Z[x] and g(x) | f (x) in Z[x]. Second proof. Let D(x) = gcd(f (x), g(x)) and let f1 (x), g1 (x) ∈ Z[x] such that f (x) = D(x)f1 (x) and g(x) = D(x)g1 (x). Let r = Resx (f1 (x), g1 (x)), then r ∈ Z and there exist u(x), v(x) ∈ Z[x] such that u(x)f1 (x) + v(x)g1 (x) = r. By hypothesis the set A = {m ∈ Z, g1 (m) 6= 0 and g1 (m) | f1 (m)}

is infinite. Define in A the relation R by nRm if g1 (n) = g1 (m). Clearly this relation is an equivalence relation. Let C be an equivalence class and m ∈ C, then g1 (m) = d, for some nonzero integer d dividing r, hence the number of classes is finite. If deg g1 ≥ 1, then |C| ≤ deg g1 , which implies that A is finite, contradicting our assumption. It follows that there exists a divisor d of r such that g1 (x) = d. We have cont(f ) = cont(D) cont(f1 ) and cont(g) = d cont(D) and by hypothesis the later of these integers divides the former, hence d | cont(f1 ). It follows that f (x) = f1 (x)D(x)


= f1 (x)/ cont(f1 ) cont(f1 )D(x)

= f1 (x)/cont(f1 ) cont(f1 )/d dD(x)

= f1 (x)/ cont(f1 ) cont(f1 )/d g(x),

hence g(x) | f (x) in Z[x].

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Exercise 1.30. Let q1 , . . . , qn be positive integers such that gcd(qi , qj ) = 1 for i 6= j. Let g1 (x), . . . , gn (x) be monic polynomials with integral coefficients all of degree m. Show that there exists a monic polynomial g(x) of degree m with integral coefficients such that g(x) ≡ gi (x) (mod qi Z[x]), for i = 1, . . . , n. Solution 1.30. First proof. For j fixed in {0, . . . , m − 1}, let aj ∈ Z such that aj ≡ c(xj , gi (x)) (mod qi ) for i = 1, . . . , n, where c(xj , gi (x)) denotes the coefficient of xj in the polynomial gi (x). Such elements aj are determined Pm−1 by the Chinese remainder theorem. Let g(x) = xm + j=0 aj xj . Then clearly g(x) ≡ gi (x) (mod qi Z[x]), for i = 1, . . . , n. Second proof. We use again the Chinese remainder theorem but in a different way. For i = 1, . . . , n, let yi ∈ Z such that yi ≡ 1 (mod qi ) and Q Pn yi ≡ 0 (mod j6=i qj ). Let g0 (x) = i=1 yi gi (x), then deg g0 (x) = m. Its Pn leading coefficient c is given by c = i=1 yi and we have c ≡ 1 (mod qi ) for i = 1, . . . , n. Now the polynomial g(x) = g0 (x)/c satisfies the required conditions. Exercise 1.31. Let f (x) = an xn + an−1 xn−1 + · · · + a0 be a polynomial with integral coefficients, α1 , . . . , αn be the roots of f (x) and let H(f ) = max(|an |, . . . , |a0 |). maxn i=1 (αi ) . Show that H(f ) ≥ 2 Solution 1.31. We show that H(f ) ≥ αi /2 for any i ∈ {1, . . . , n}. Let α be any of these αi . If α ≤ 2, then the result is true since H(f ) ≥ 1. Suppose that α > 2, then |αn | ≤ |an αn |

= |a0 + · · · + an−1 αn−1 |

≤ |a0 | + |a1 ||α| + · · · + |an−1 ||α|n−1 ≤ H(f )(1 + |α| + · · · + |α|n−1 ),

hence

 1 1 + + 1 |α|n−1 |α|   1 1 ≤ H(f ) + · · · + + 1 2n−1 2

α ≤ H(f )



≤ 2H(f ).

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Exercise 2.1. Let f (x) = x4 − 2x2 + 2x + 2 and α be a root of f (x) in C. Show that f (x) is irreducible over Q and express β = 1/(α2 + α + 1) as a polynomial in α with rational coefficients. Solution 2.1. The polynomial f (x) is irreducible over Q by application of Eisenstein’s irreducibility criterion [Lang (1965), Eisenstein Criterion, Chap. 5.7]. It follows that {1, α, α2 , α3 } is a basis of Q(α) = Q[α] over Q. Therefore β may be expressed as a polynomial in α with rational coefficients. Let g(x) = x2 + x + 1, then gcd(f (x), g(x)) = 1, hence there exist u(x) and v(x) in Q[x] such that u(x)f (x) + v(x)g(x) = 1. Substituting α for x in this identity, we obtain v(α)g(α) = 1. Thus β = 1/g(α) = v(α). We compute v(x) as follows. The Euclidean division of f (x) by g(x) leads to f (x) = g(x)(x2 − x − 2) + 5x + 4. Similarly, we have g(x) = (5x + 4)(x/5 + 1/25) + 21/25. It follows that 21/25 = g(x) − (5x + 4)(x/5 + 1/25)   = g(x) − (x/5 + 1/25) f (x) − g(x)(x2 − x − 2)   = g(x) 1 + (x2 − x − 2)(x/5 + 1/25) − (x/5 + 1/25)f (x).

Substituting α for x, we obtain   21 = g(α) 25 + (α2 − α − 2)(5α + 1) . 27

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We conclude that β = 1/g(α) 25 + (α2 − α − 2)(5α + 1) 21 23 11 4 5 = − α − α2 + α3 . 21 21 21 21 =

Exercise 2.2. Show that the following identities hold in the complex field. p p √ √ (1) 11 + √ 6 2 + 11 − 6 √2 = 6. √ (2) (10 + 6 3)1/3 + (10 − 6 3)1/3 = 2 3. Here the square roots are the positive ones and the cubic roots are the real ones. Solution 2.2. (1) We have q

√ 11 + 6 2 +

q q q √ √ √ 11 − 6 2 = (3 + 2)2 + (3 − 2)2 √ √ = 3 + 2 + 3 − 2 = 6.

(2) Let √ α = (10 + 6 3)1/3 ,

√ β = (10 − 6 3)1/3

and γ = α + β.

Then √ γ 3 = α3 + β 3 + 3αβγ = 12 3 + 6γ. √ Hence γ 3 − 6γ = 12 3. Squaring both sides of this equality shows that γ is√a root of f (x) = x6 − 12x4 + 36x2 − 3.122 . One may check that f (2 3) = 0. From the above calculations it is clear that {γ, −γ, jα + j 2 β, −jα − j 2 β, j 2 α + jβ, −j 2 α − jβ} is the set of the roots of f (x). One and only √ one element of this set is real and positive, namely γ. Therefore γ = 2 3. Exercise 2.3. Let f (x) and g(x) ∈ Z[x] such that f (x) is monic and irreducible, α be a root of f (x) in C. Let p be a prime number and let f (x) and g(x) denote the reduced polynomials modulo p of f (x) and g(x) respectively. Show that g(α) ≡ 0 (mod p) in Z[α] if and only if f (x) | g(x) in Fp [x].

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Solution 2.3. We have g(α) ≡ 0

(mod p) in Z[α] ⇔ g(α) = ph(α)

with

⇔ f (x) | g(x) − ph(x)

h(x) ∈ Z[x]

in Z[x]

⇔ g(x) − ph(x) = f (x)q(x), with q(x) ∈ Z[x] ⇔ f (x) | g(x)

in Fp [x].

Exercise 2.4. Let K be a field of characteristic not equal to 2, a, b, d ∈ K such that d 6∈ K 2 . Show that the following statements are equivalent (i) There exist a1 , . . . , an ∈ K such that q √ √ √ √  a+b d∈K d, a1 , . . . , an . q √ √ (ii) There exists s ∈ K \ {0} such that s(a + b d) ∈ K( d). √ (iii) a2 − b2 d ∈ K. Solution 2.4. (i) ⇒ (ii). The proof is by induction on n, n being the number of ai . If n = 0, then (ii) holds with s = 1. Suppose that the implication is true for √ n − 1 elements ai . Without loss of generality, we assume that an 6∈ E, p √ √ √ √ where E = K( d, a1 , . . . , an−1 ). We may write a + b d in the form p √ √ a + b d = A + B an , where A, B ∈ E. Then √ √ a + b d = A2 + B 2 an + 2AB an . √ Since a + b d, A2 + B 2 an ∈ E, then AB = 0. Case 1. Ap= 0. q √ √ √ We have a + b d = B an , hence an (a + b d) = Ban ∈ E. Therefore inductive hypothesis, there exists s1 ∈ K \ {0} such that q by the √ √ s1 an (a + b d) ∈ K( d). We conclude that (ii) holds for s = s1 an . Case 2. Bp= 0. √ We have a + b d = A ∈ by the inductive hypothesis, there qE, hence √ √ exists s ∈ K \ {0} such that s(a + b d) ∈ K( d). √ √ (ii) ⇒ (iii). Let u, v ∈ K such that s(a + b d)√= (u + v d)2 , then sa = u2 + dv 2 and sb = 2uv. If v = 0, then b = 0, a2 − b2 d = ±a ∈ K 2 2 b and (ii) holds. If v 6= 0 then sa = s4v2 + dv 2 . Therefore we obtain the 2 2 following equation satisfied by s: s b − 4av 2 s + 4dv 4 = 0. Since s ∈ K,

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it √ follows that the discriminant of this equation is a square in K, that is a2 − b2 d ∈ K. √ 2 2 let a1 = 2(a + e). If (iii) ⇒ (i). Suppose that √ a − b d := e ∈2 K and 2 2 2 a1 = 0, then a = −e = − a − b d, hence a = a − db2 , whence b = 0. p √ √ √ √ Therefore a + b d = a ∈ K( d, a). Suppose that a1 6= 0. We have √ √ !2 a21 + 4db2 + 4a1 b d a1 + 2b d = √ 2 a1 4a1 √ 4(a + e)2 + 4(a2 − e2 ) + 8(a + e) db = 8(a + e) √ = a + b d, hence

q

√ √ √ √ a1 + 2b d a+b d= ∈ K( d, a1 ). √ 2 a1

Exercise 2.5. Let K be a field, E be an algebraic extension of K of degree n and B = {ω1 , . . . , ωn } be a basis of E over K. Let α ∈ E and β = u(α), where u(x) ∈ K[x]. Let M be the matrix relative to the base B of the K-endomorphism mα of E defined by mα (γ) = αγ. Show that the matrix in the same basis of the K-endomorphism multiplication by β is equal to u(M ). Solution 2.5. Let mβ be the endomorphism of E representing the multiplication by β. To get the result it is equivalent to prove that mu(α) = u(mα ). It is sufficient to prove this identity, when u(x) is a monomial, say u(x) = axk . In this case for any γ ∈ E, we have a(mα )k (γ) = a(mα )k−1 (αγ) = aαk γ = maαk (γ) = mu(α) (γ),

hence the result. Exercise 2.6. P∞ n Let p be a prime number. Show that the series f (x) = n=0 xp is algebraic over Fp (x). Solution 2.6. Since f (x) = x + f (xp ) = x + f (x)p , then f (x) is a root of the polynomial g(y) = y p − y + x which is nonzero and has its coefficients in Fp (x), hence f (x) is algebraic over Fp (x).

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Exercise 2.7. Let f (x) = x3 − x2 − 2x − 8, θ be a root of f (x) and K = Q(θ). (1) Show that f (x) is irreducible over Q. (2) Let α = (θ2 − θ)/2. Show that α is a primitive element of K over Q. (3) Set θ = b0 + b1 α + b2 α2 , where b0 , b1 , b2 ∈ Q. By solving a linear system of equations, compute b0 , b1 , b2 . (4) Compute the minimal polynomial g(x) of α and the dual basis {ω0 , ω1 , ω2 } of {1, α, α2 }. Using this basis, compute again b0 , b1 , b2 . Solution 2.7. (1) The polynomial f (x) is irreducible in F3 [x], hence irreducible over Q. (2) Since Q(α) ⊂ Q(θ), then [Q(α) : Q] | 3, hence [Q(α) : Q] = 1 or [Q(α) : Q] = 3. In the first case, we conclude that α ∈ Q, which implies θ is a root of a monic polynomial of degree 2 with rational coefficients. This statement is false, hence [Q(α) : Q] = 3 and then α generates K over Q. (3) Replacing α by its expression in terms of θ, leads to     1 1 1 3 b2 − b1 θ + b2 + b1 θ2 , θ = b0 + b1 α + b2 α2 = (b0 − 2b2 ) + 2 2 2 2 hence b0 − 2b2 = 0, 3b2 − b1 = 2 and b2 + b1 = 0. Therefore b2 = 1/2, b1 = −1/2, b0 = 1 and θ = 1 − α/2 + α2 /2. (4) We have α=

1 2 1 θ − θ, 2 2

αθ = θ + 4

and

αθ2 = θ2 + 4θ.

We conclude that (θ2 , θ, 1) is a nonzero solution of the following linear system of equations 1/2x − 1/2y − αz = 0, (1 − α)y + 4z = 0, (1 − α)x + 4y = 0. It follows that the determinant of this system is 0. We thus obtain 1/2 −1/2 −α 0 (1 − α) 4 = 0. (1 − α) 4 0

Therefore the minimal polynomial of α over Q is given by g(x) = x3 − 2x2 + 3x − 10. Performing the Euclidean division of g(x) by x − α, we obtain g(x) = (x − α))(x2 + (α − 2)x + α2 − 2α + 3).

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Therefore the dual basis {ω0 , ω1 , ω2 } of {1, α, α2 } is given by ω0 = (α2 − 2α + 3)/(3α2 − 4α + 3), ω1 = (α − 2)/(3α2 − 4α + 3) and ω2 = 1/(3α2 − 4α + 3),

see [Lang (1965), Prop. 1, Chap. 8.6]. It follows that bi = TrK/Q (θωi ) for i = 0, 1, 2. We have 1/(3α2 − 4α + 3) = (5α2 + 32α − 18)/(2.503), hence b2 = TrK/Q (θω2 )   θ(5α2 + 32α − 18) = TrK/Q 2.503 = TrK/Q =


! θ 5(θ2 − θ)2 /4 + 32(θ2 − θ)/2 − 18 2.503

1 TrK/Q (40θ2 + 36θ + 592). 8.503

Since TrK/Q (θ) = 1 and TrK/Q (θ2 ) = 5, then 9 74.3 5 TrK/Q (θ2 ) + TrK/Q (θ) + = 1/2. 503 2.503 503 Similarly, we have b2 =

b0 = TrK/Q (θω0 )   α2 − 2α + 3 = TrK/Q θ 2 3α − 4α + 3   θ(α2 − 2α + 3)(5α2 + 32α − 18) = TrK/Q 2.503
1 = TrK/Q θ(−18α2 + 86α + 266) 2.503
1 = TrK/Q θ(−18(θ2 − θ)2 )/4 + 86(θ2 − θ)/2 + 266 2.503 1 TrK/Q (−36θ2 + 370θ + 272) = 2.503 −36 370 3.272 = TrK/Q (θ2 ) + TrK/Q (θ) + 2.503 2.503 2.503 −5.36 370 816 = + + = 1. 2.503 2.503 2.503

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By the same method, we compute b1 . We have b1 = TrK/Q (θω1 )   α−2 = TrK/Q θ 2 3α − 4α + 3   θ(α − 2)(5α2 + 32α − 18) = TrK/Q 2.503
1 = TrK/Q θ(32α2 − 97α + 86) 2.503
1 TrK/Q θ(32(θ2 − θ)2 )/4 − 97(θ2 − θ)/2 + 86 = 2.503   128θ2 − 86θ − 520 1 TrK/Q = 2.503 2 128 86 3.520 = TrK/Q (θ2 ) − TrK/Q (θ) − 4.503 4.503 4.503 5.64 43 3.260 = − − = −1/2. 2.503 2.503 2.503 Exercise 2.8. Let K be a field, α be algebraic of degree n over K. Let r and s be non negative integers such that r + s = n − 1. Show that any β ∈ K(α) may be written in the form β = u(α)/v(α), where u(x) and v(x) are polynomials with coefficients in K satisfying the conditions deg u ≤ r and deg v ≤ s. Moreover if we suppose that the polynomials u and v are coprime, then this representation is unique (up to a multiplication of u and v by a same nonzero constant). Solution 2.8. If β = 0, the result is clear. Suppose next that β 6= 0. Let A and B be the linear spaces generated by {1, α, . . . , αr } and {β, βα, . . . , βαs } respectively. The dimensions of the spaces A and B over K are equal to r + 1 and s + 1 respectively. Since the family {1, α, . . . , αr , β, βα, . . . , βαs } contains n + 1 elements, then this family is not free. It follows that A ∩ B 6= {0}. Let γ be a non zero element of this intersection, then γ = a0 + · · · + ar αr = b0 β + · · · + βαs and the result follows. Suppose that some β has two representations β = u1 (α)/v1 (α) = u2 (α)/v2 (α),

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where gcd(u1 (x), v1 (x)) = gcd(u2 (x), v2 (x)) = 1 and with the conditions on the degrees. Then u1 (α)v2 (α) − u2 (α)v1 (α) = 0, hence u1 (x)v2 (x) − u2 (x)v1 (x) = 0. We deduce that u1 (x) | v1 (x) and v1 (x) | u1 (x). It follows that v1 (x) = cu1 (x) and v2 (x) = cu2 (x) with c ∈ K ? . Exercise 2.9. (1) Let K be a field of characteristic p ≥ 0, u(x) be a non constant polynomial with coefficients in K such that u(0) = 0, f (x) ∈ K[[x]] satisfying the identity f (u(x)) = f (x) − v(x) for some non zero polynomial v(x) with coefficients in K with deg v < deg u. Show that f (x) is transcendental over K(x) or p > 0, f (x) is algebraic over K(x) and p | [K(x, f (x)) : K(x)]. (2) Suppose that K is of characteristic 0 and let d ≥ 2 be an integer. Show P k that f (x) = k≥0 xd is transcendental over K(x). Solution 2.9.

(1) Suppose that f (x) is algebraic over K(x) and let φ(x, y) = y n + an−1 (x)y n−1 + · · · + a0 (x) ∈ K(x)[y] be its minimal polynomial over K(x). From the following identity (f (x))n + an−1 (x)(f (x))n−1 + · · · + a0 (x) = 0 and the property of f (x), we obtain (f (x) − v(x))n + an−1 (u(x))(f (x) − v(x))n−1 + · · · + a0 (u(x)) = 0. We deduce that an−1 (x) = an−1 (u(x)) − nv(x).

(Eq 1)

Set an−1 (x) = a(x)/b(x), where a(x) and b(x) ∈ K[x]. We may suppose that b(x) is monic and gcd(a(x), b(x)) = 1. We have a(x)b(u(x)) = a(u(x))b(x) − nv(x)b(x)b(u(x)), hence b(u(x)) | b(x). It follows that b(x) = 1 and a(x) = a(u(x)) − nv(x). The assumption on the degrees of u and v implies that a(x) ∈ K and then by (Eq 1), we conclude that n = 0 if p = 0 and p | n if p > 0. (2) Here, we have f (xd ) = f (x) − x, so that (1) applies with u(x) = xd and v(x) = x.

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Exercise 2.10. Find the gap in the following reasoning. Let K be a field. Let α be a root of the polynomial f (x, y) = y 3 − x2 in an algebraic closure of K(x). f (x, y) is irreducible over K(x), hence [K(x, α) : K(x)] = 3. The element x is algebraic over K(x2 ) since it is a root of the polynomial g(x, y) = y 2 − x2 irreducible over K(x2 ), hence [K(x) : K(x2 )] = 2. Considering the chain of fields K(x2 ) ⊂ K(x) ⊂ K(x, α), we can conclude that the degree of the minimal polynomial of α over K(x2 ) is equal to 6. But α is a root of f (x, y), irreducible over K(x2 ). Therefore we have reached a contradiction. Solution 2.10. The gap in the proof lies in the fact that α is not a primitive element of K(x, α) over K(x2 ), that is K(x, α) 6= K(x2 , α). To better understand the situation look at the following diagram of fields. K(x, α)

K(x2 , α)

K(x)

K(x2 )

Exercise 2.11. Let f (x) be a monic irreducible polynomial of degree n with coefficients in Z and let p be a prime number. Let α1 , . . . , αn be the roots of f and g(x) = (x − α1p ) · · · (x − αnp ). Show that pn | Resx (g(x), f (x)). Solution 2.11. Let h(x) = f (xp ) − (f (x))p , then h(x) = pu(x)) with u(x) ∈ Z[x]. We have Resx (g(x), f (x)) =

n Y

f (αip ) =

i=1

= pn

n Y

i=1

hence the result.

n Y

i=1

n
Y f (αip ) − f (αi )p = h(αi )

u(αi ) = pn Resx (f (x), u(x)),

i=1

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Exercise 2.12. Let m be a positive integer. Show that the polynomial with integral coefficients fm (x) = (x2 + 2)m − 2m x is irreducible over Q. Solution 2.12. Let α be a root of fm (x) and let β = (α2 + 2)/2. Since α = β m , then Q(α) = Q(β). Easy computations show that β 2m − 2β + 2 = 0, hence β is a root of the polynomial g(x) = x2m − 2x + 2 which is irreducible over Q by Eisenstein’s criterion. We deduce that [Q(β) : Q] = 2m and then [Q(α) : Q] = 2m. We conclude that fm (x) is irreducible over Q. Exercise 2.13. Let f (x) and g(x) be monic polynomials with integral coefficients such that gcd(f (x), g(x)) = 1. (1) Show that there exist c ∈ Z \ {0}, u(x) and v(x) ∈ Z[x] with deg u < deg g, deg v < deg f such that u(x)f (x) + v(x)g(x) = c. (2) Let I = {c ∈ Z, there exist u(x) and v(x) ∈ Z[x]; u(x)f (x)+v(x)g(x) = c}.

Show that I is an ideal of Z and that Resx (f (x), g(x)) ∈ I. (3) Let c0 be a positive integer such that I = c0 Z and let p be a prime factor of Resx (f (x), g(x)). Show that p | c0 . (4) Show that we may have c0 < | Resx (f (x), g(x))|. Pn Pm (5) Recall that if we set f (x) = i=0 ai xn−i and g(x) = j=0 bj xm−j , then Resx (f x), g(x) is the following (m + n) × (m + n) determinant, where the first m-th rows are devoted to the coefficients of f (x) and the remaining one’s to the coefficients of g(x), a0 a1 . . . an 0 ... 0 0 a0 . . . an−1 an . . . 0 . . . . . . . . . ... . . . . . . . . . 0 ... 0 a0 . . . a n 0 R= b0 b1 . . . bm 0 ... 0 0 b0 . . . bm−1 bm . . . 0 . . . . . . . . . ... . . . . . . . . . 0 0 ... 0 b0 . . . bm and that

Resx (f (x), g(x)) = (A0 + A1 x + · · · + Am−1 xm−1 )f (x)

+ (B0 + B1 x + · · · + Bn−1 xn−1 )g(x),

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where Ai (resp. Bj ) is the cofactor of the coefficient appearing in the first column at the i + 1 (resp. m + j + 1)-th row. Show that c0 = | Resx (f (x), g(x))|/d, where d = gcd(A0 , . . . , Am−1 , B0 , . . . , Bn−1 ). Solution 2.13. (1) Since f (x) and g(x) are relatively prime in Q[x], then by Bezout’s Identity, there exist u1 (x) and v1 (x) ∈ Q[x] with deg u1 < deg g, deg v1 < deg f such that u1 (x)f (x) + v1 (x)g(x) = 1. If we multiply the two sides of this identity by the common denominator, say c, of all the coefficients of u1 (x) and v1 (x), then we obtain the required identity. (2) Easy. (3) For any polynomial h(x), with integral coefficients, denote by h(x) the reduction modulo p of h(x). Since p | Resx (f (x), g(x)), then f (x) and g(x) have a common root in an algebraic closure of Fp . Let u0 (x) and v0 (x) ∈ Z[x] such that u0 (x)f (x) + v0 (x)g(x) = c0 . Reducing modulo p the two sides of this identity shows that c0 = 0, hence p | c0 . (4) Let f (x) = x2 + 1 and g(x) = x2 + 3, then √ 2
2 Resx (f (x), g(x)) = −1 + 3 = 4. On the other hand, we have 2 = (x2 + 3) − (x2 + 1) so that c0 | 2. Since the polynomials f (x) and g(x) have a common root, then c0 6= 1, hence c0 = 2. (5) It is clear that Resx (f (x), g(x))/d ∈ I, thus c0 | Resx (f (x), g(x))/d. Let λ ∈ Z such that Resx (f (x), g(x))/d = λc0 . Write c0 in the form

0 c0 = (A00 +A01 x+· · ·+A0m−1 xm−1 )f (x)+(B00 +B10 x+· · ·+Bn−1 xn−1 )g(x),

where the coefficients A0i , Bj0 are integers. Multiply this identity by dλ and subtract it from the following identity Resx (f (x), g(x)) = (A0 + A1 x + · · · + Am−1 xm−1 )f (x)

+ (B0 + B1 x + · · · + Bn−1 xn−1 )g(x).

We obtain 0=

m−1 X i=0

(Ai − dλA0i )xi

!



f (x) + 

j=0

Since f (x) and g(x) are coprime, then f (x) |

n−1 X j=0



n−1 X

(Bj − dλBj0 )xj and g(x) |

(Bj − dλBj0 )xj  g(x). m−1 X i=0

(Ai − dλA0i )xi .

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It follows that Ai = dλA0i for i = 1, . . . , m − 1 and Bj = dλBj0 for j = 1, . . . , n − 1. According to the definition of d, we conclude that λ = ±1, so that | Resx (f (x), g(x))|/d = c0 . Exercise 2.14. Let K be a field of characteristic 0, f (x) = xn + · · · + a1 x + a0 and g(x) = xm + · · · + b1 x + b0 ∈ K[x]. Let y a new variable and R(y) = Resx (xn + · · · + a1 x + y, g(x)). Let e be the exact number of roots β of g(x), each of them counted with its multiplicity, such that f (β) = 0. Show that e is the multiplicity of a0 as a root of R(y). Solution 2.14. Let β1 , . . . , βm be the roots (distinct or not) of g(x) in an algebraic closure of K and let F (x, y) = f (x) − a0 + y = xn + · · · + a1 x + y. Qm Let P (x, y) = i=1 (x − F (βi , y)), then P (x, y) has the form:

P (x, y) = xm − Am−1 (y)xm−1 + · · · + (−1)m−1 A1 (y)x + (−1)m A0 (y), P Q where Ai (y) = |J|=m−i j∈J F (βj , y), for i = 0, . . . , m − 1. We have A0 (y) =

m Y

F (βj , y) = Resx (g(x), F (x, y)) = R(y).

j=1

Since

d dy F (βj , y)

= 1, then R0 (y) =

m Y X

F (βj , y) = A1 (y).

i=1 j6=i

By induction we may prove that R(k) (y) = Ak (y) for k = 0, . . . , m − 1. This preparation being done, let e0 be the multiplicity of a0 as a root of 0 0 R(y). We have R(a0 ) = R0 (a0 ) = · · · = R(e −1) (a0 ) = 0 and R(e ) (a0 ) 6= 0, hence A0 (a0 ) = A1 (a0 ) = · · · = Ae0 −1 (a0 ) = 0 and Ae0 (a0 ) 6= 0. It follows that 0 is a root of P (x, a0 ) of order e0 . But the roots of P (x, a0 ) are F (β1 , a0 ), . . . , F (βm , a0 ). Since F (βi , a0 ) = f (βi ) for i = 1, . . . , m, then there are exactly e0 elements among β1 , . . . , βm such that f (βi ) = 0, hence e = e0 . Exercise 2.15. Let K be a field, x1 , x2 , x3 be the roots of P (X) = X 3 − s1 X 2 + s2 X − s3 ∈ K[X] in an algebraic closure of K. Let D(s1 , s2 , s3 ) be the discriminant of P (X). Recall that D(s1 , s2 , s3 ) = −4s31 s3 + s21 s22 + 18s1 s2 s3 − 27s23 − 4s32 .

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Let A = x2 x21 + x3 x22 + x1 x23 and B = x1 x22 + x2 x23 + x3 x21 . (1) Show that A + B = s1 s2 − 3s3 and AB = 9s23 + s32 − 6s1 s2 s3 + s31 s3 . (2) Show that A = B if and only if D(s1 , s2 , s3 ) = 0. Solution 2.15. (1) We first compute σj = s21 − 2s2 ,

P3

i=1

σ3 =

xji for j = 2 and j = 3. We have σ2 =

3 X i=1

(s1 x2i − s2 xi + s3 )

= s1 σ2 − s1 s2 + 3s3 = s31 − 3s1 s2 + 3s3 .

We have A + B = s1 σ2 − σ3 = s1 s2 − 3s3 and X X AB = x4i xj xk + x3i x3j + 3s3 . i,j,k distinct

It is clear that X

i,j,k distinct

We have

X i 1, thus ω∈Z |2 − ω|2 > 1 which is a contradiction. Remark. We may exclude the possibility φd (2) = ±1, by using Bang’s Theorem [Roitman (1997), Th. 3] which asserts that for any integer d > 1, d 6= 6, there is a prime l (called primitive prime divisor) dividing 2d − 1 but not 2k − 1 for any k < d. Since in our situation d is odd then this theorem applies and we conclude that the prime l divides φd (2). It follows that φd (2) 6= ±1. Exercise 2.20. Let K be a field, a(x) and f (x, y) be polynomials with coefficients in K. Suppose that f (x, y) ∈ K(a(x))[y]. Show that f (x, y) ∈ K[a(x), y]. Solution 2.20. We may write f (x, y) in the form f (x, y) =

n X i=0

ci (a(x))y

i

!

/d(a(x)),

where d(x) and ci (x)
∈ K[x] for i = 0, . . . , n and with gcd c0 (x), . . . , cn (x), d(x) = 1. We may suppose also that d(x) is monic. First proof. We make use of the contents of polynomials. We have cont(f ) ∈ K[x], hence d(a(x)) | ci (a(x)) for i = 0, . . . , n. Since
gcd c0 (x), . . . , cn (x), d(x) = 1, then there exist u0 (x), . . . , un (x), v(x) ∈ Pn K[x] such that i=0 ui (x)ci (x) + v(x)d(x) = 1. It follows that n X ui (a(x))ci (a(x)) + v(a(x))d(a(x)) = 1. i=0

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Therefore d(a(x)) = 1 and then f (x, y) =

n X i=0

ci (a(x))y i ∈ K[a(x), y].

Second proof. Suppose that d(x) is not constant. We have d(a(x))f (x, y) =

n X

ci (a(x))y i .

i=0

¯ such that a(α) = β and d(β) = 0, then Let α and β ∈ K n X i=0

i

ci (a(α))y =

n X

ci (β)y i = 0.

i=0

It follows that c0 (β) =, . . .
, = cn (β) = d(β) = 0 which is a contradiction to gcd c0 (x), . . . , cn (x), d(x) = 1.

Exercise 2.21. Let K be a field of characteristic p ≥ 0, P (x) be a polynomial with coefficients in K of degree n ≥ 2. If p > 0, suppose that p - n. For any m ≥ 1, define Pm (x) by P1 (x) = P (x) and for m ≥ 2, Pm (x) = Pm−1 (P (x)). Let Q(x) ∈ K[x]. For any positive integer m such that deg Q < nm and any P z ∈ K, we set fp (Q) = n−m Pm (ξ)=z Q(ξ), where the sum runs over all the roots (distinct or not) of the polynomial pm (x) − z in an algebraic closure of K. (1) Show that fP (Q) does not depend on z nor on m (satisfying nm > deg Q). (2) Show that the map fP : K[x] → K which maps Q(x) onto fp (Q) is K-linear and surjective. L (3) Show that K[x] = K Ker fP . Deduce that, given Q(x) ∈ K[x], there exists one and only one λ ∈ K such that fP (Q − λ) = 0. Solution 2.21. (1) Set Q(x) = ak xk + ak−1 xk−1 + · · · + a0 where a0 , a1 , . . . , ak ∈ K, ak 6= 0 and k < nm . Let ξ1 , . . . , ξnm be the roots of Pm (x) − z in an algebraic

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closure of K, then m

fP (Q) = n

−m

n X

Q(ξj )

j=1 m

=n

−m

n X k X

ai ξji

j=1 i=0

= n−m

k X i=0

= n−m

k X



ai 

m

n X j=1



ξji 

ai Si ,

i=0

Pnm where Si = j=1 ξji . For any h ≥ 1, let σh be the elementary symmetric function of the ξj . Using the formulas of Newton [Small (1991), Prop. 2.3, Chap. 2]: Sd − σ1 Sd−1 + σ2 Sd−2 − · · · + (−1)

d−1

s1 − σ1 = 0,

σd−1 S1 + (−1)d dσd = 0,

for 2 ≤ d ≤ n, we see that Sd is a homogeneous polynomial in σ1 , . . . , σd of degree d if we attribute the weight h for σh . Since k = deg Q < nm , then fP (Q) is independent of σnm . Therefore fP (Q) is independent of z. We next show that fP (Q) is independent of m. For any l ∈ N, we have: X X 1 1 1 Q(ξ) = l m Q(ξ) l+m n n n Pl ◦Pm (ξ)=z

Pl+m (ξ)=z

=

1 nl

Pl

X

(ξ 0 )=z

1 nm

X

Pm

Q(ξ) =

(ξ)=ξ 0

1 nl

Pl

X

fP (Q)

(ξ 0 )=z

= fP (Q). (2) Let Q1 (x), Q2 (x) ∈ K[x], m be a positive integer such that nm > M ax(deg Q1 , deg Q2 ) and let ξ1 , ...,x inm be the roots of Pm (x), then m

fP (Q1 + Q2 ) = n

−m

n X j=1 m

= n−m

n X j=1

(Q1 (ξj ) + Q2 (ξj )) 

Q1 (ξj ) + n−m

= fp (Q1 ) + fp (Q2 ).

m

n X j=1



Q2 (ξj )

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We also have for any λ ∈ K m

fP (λQ1 ) = n

−m

n X

m

λQ1 (ξj ) = λn

−m

j=1

n X

Q1 (ξj ) = λfP (Q).

j=1

We conclude that fP is a linear map. Obviously fP (λ) = λ for any λ ∈ K, hence fP is surjective. (3) Let Q(x) ∈ K[x], then fP (Q − fP (0)) = 0, hence Q1 (x) := Q(x) − fP (0) ∈ Ker fP . Therefore Q(x) = fP (Q) + Q1 (x) ∈ K + Ker fP . Since fP (λ) = λ for any λ ∈ K, then K ∩ Ker fP = {0}. It follows that K[x] is the direct sum of K and Ker φ. The last statement is therefore trivial. Exercise 2.22. ¯ be an algebraic closure of K, f (x) be a non constant Let K be a field, K polynomial with coefficients in K. Suppose that there exist g(x) ∈ K[x], ¯ y]. 1 ≤ deg g < deg f such that g(x) − cg(y) divides f (x) − f (y) in K[x, (1) Show that c is a root of unity and there exists h(x) ∈ K[x] such that f (x) = h(g(x)). ¯ ? . Show that h(x) ∈ K[xd ]. (2) Let d be the order of c in K Solution 2.22. Pk n c. Let f (x) = i=0 ai (x)g(x)i be (1) Let n = deg f , m = deg g and k = b m the g-adic expansion of f (x) in K[x] [Lang (1965), Th. 9, Chap. 5.5]. We have 0 ≡ f (x) − f (y) ≡ ≡ hence

k X i=0

k X i=0

(mod g(x) − cg(y))

ai (x)g(x)i −

k X

ai (y)g(y)i

i=0

g(y)i (ci ai (x) − ai (y))

(mod g(x) − cg(y))

(mod g(x) − cg(y)),

k k X X i (cg(y)) ai (x) − g(y)i ai (y) = (g(x) − cg(y))F (x, y), i=0

i=0

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where F (x, y) ∈ K[x, y]. Since the degree in x of the polynomial appearing in the left side of this identity is at most equal to m − 1, then F (x, y) = 0. We deduce that k k X X (cg(y))i ai (x) = g(y)i ai (y). i=0

i=0

Equating the degrees in y of the two sides of this equality, we obtain mk = mk + deg ak , hence deg ak = 0, thus ak ∈ K. It follows that ck = 1. Let l be an integer such that 0 ≤ l < k. We prove by induction that al = 0 or al ∈ K ? and cl = 1. This claim is proved for l = k. Fix an integer l and suppose that the claim holds for all t such that l < t ≤ k. Then the above equation reads: (cg(y))l al (x) + · · · + cg(y)a1 (x) = g(y)l al (y) + · · · + cg(y)a1 (y). It follows that al (y) = 0 or lm = lm + deg al (y). In this last case, we have al (y) ∈ K ∗ and cl = 1. We conclude that f (x) = h(g(x)), where Pk h(x) = i=0 ai xi . (2) We have seen in the preceding proof that if aj 6= 0, then cj = 1, hence d | j. It follows that h(x) ∈ K[xd ]. Exercise 2.23. Let K be a field not algebraically closed, n be a positive integer. For any subset S of K[x1 , . . . , xn ] denote by V (S), the set V (S) = {(a1 , . . . , an ) ∈ K n , g(a1 , . . . , an ) = 0 for any g ∈ S}. (1) Show that there exists f ∈ K[x1 , . . . , xn ] such that V (f ) = {(0, ..., 0)}. Deduce that there exists f ∈ K[x1 , . . . , xn ] irreducible such that V (f ) = {(0, . . . , 0)}. (2) Let m be a positive integer and let f1 (x1 , . . . , xn ), . . . , fm (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ]. Show that there exists f (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] such that V (f1 , . . . , fm ) = V (f ). Solution 2.23. (1) We proceed by induction on n. If n = 1, the result is clear by taking f (x1 ) = x1 . Suppose the result is true for n − 1 indeterminates. Let φ(t) ∈ K[t], deg φ ≥ 1 such that φ has no root in K. Set φ(t) = ak tk + · · · + a1 t + a0 with a0 , . . . , ak ∈ K and ak 6= 0. Let   x k ψ(x, y) = y φ y = ak xk + ak−1 xk−1 y + · · · + a1 xy k−1 + a0 y k .

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It is clear that V (ψ) = {(0, 0)} (in K 2 ). Let g(x1 , . . . , xn−1 ) ∈ K[x1 , . . . , xn−1 ] such that V (g) = {0} (in K n−1 ). We show that the polynomial f (x1 , . . . , xn ) = ψ(g(x1 , . . . , xn−1 ), xn ) satisfies the required property. For any (a1 , . . . , an ) ∈ K n , we have f (a1 , . . . , an ) = 0 ⇔ g(a1 , . . . , an−1 ) = 0

and

an = 0

⇔ a1 = · · · = an−1 = an = 0,

hence V (f ) = {(0, . . . , 0)} (in K n ). We prove the existence of F ∈ k[x1 , . . . , xn ] irreducible such that V (f ) = {(0, . . . , 0)}. Let f ∈ K[x1 , . . . , xn ] such that V (f ) = {0} and let f (x1 , . . . , xn ) = F1 (x1 , . . . , xn ) · · · Fk (x1 , . . . , xn ) be the factorization of f into a product of irreducible factors over K. For any i = 1, . . . , k, V (Fi ) = ∅ or V (Fi ) = {0, . . . , 0} and at least for one index i0 , V (Fi ) = {0, . . . , 0}. The conclusion follows by taking F = Fi0 . (2) By (1), let g(x1 , . . . , xm ) ∈ K[x1 , . . . , xm ] such that V (g) = (0, . . . , 0) (in K m ). Let f (x1 , . . . , xn ) = g(f1 (x1 , . . . , xn ), . . . , fm (x1 , . . . , xn )), then for any (a1 , . . . , an ) ∈ K n we have f (a1 , . . . , xn ) = 0 ⇔ f1 (x1 , . . . , xn ) = · · · = fm (x1 , . . . , xn ) = 0 ⇔ (x1 , . . . , xn ) ∈ V (f1 , . . . , fm ).

Question. Is the polynomial f in (2) unique? Exercise 2.24. Let K be a field, A be a subring of K[x1 , . . . , xn ] and F be its fraction field. We say that A satisfies the property (P) if for any f (x1 , . . . , xn ) and g(x1 , . . . , xn ) ∈ K[x1 , . . . , xn ], f · g ∈ A ⇒ f ∈ A and g ∈ A. If A satisfies (P), show that A is integrally closed in K[x1 , . . . , xn ] and F is algebraically closed in K(x1 , . . . , xn ). Solution 2.24. Let f ∈ K[x1 , . . . , xn ] such that f is integral over A, then there exist a positive integer m and a0 , . . . , am−1 ∈ A such that f m + am−1 f m−1 + · · · + a0 = 0.

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This implies f |a0 . Let h ∈ K[x1 , . . . , xn ] such that a0 = f ·h, then f ·h ∈ A. Since A satisfies (P), then f ∈ A and h ∈ A. Therefore A is integrally closed in K[x1 , . . . , xn ]. Let uv ∈ K(x1 , . . . , xn ), where u, v ∈ K[x1 , . . . , xn ], gcd(u, v) = 1. Suppose that uv is algebraic over F , then there exist a positive integer m and (a0 , b0 ), (a1 , b1 ), . . . , (am−1 , bm−1 ) ∈ A2 such that gcd(ai , bi ) = 1 for i = 1, . . . , m − 1 and  u m−1  u m a a0 m−1 + + ··· + = 0. v bm−1 v b0 We deduce that there exist d0 , . . . , dm−1 , dm ∈ A such that dm um + dm−1 um−1 v + · · · + d0 v m = 0.

This implies v|dm . Let v 0 ∈ K[x1 , . . . , xn ] such that dm = vv 0 . Since dm ∈ A and A satisfies (P), we conclude that v ∈ A and v 0 ∈ A. Similarly, we prove that u ∈ A. Therefore uv ∈ F and F is algebraically closed in K(x1 , . . . , xn ). Exercise 2.25. Let K be a field, P (x) and A(x) be polynomials with coefficients in K such that P (x) is monic and non constant. Let φ(A,P ) be the K-endomorphism of K[x]/P (x)K[x] such that φ(A,P ) (f (x)) = A(x)f (x) for any f (x) ∈ K[x]. (1) Show that Det φ(A,P ) = Resx (P (x), A(x)). (2) Let Q(x) ∈ K[x] be monic. Suppose that P (x) and Q(x) are irreducible over K. Let α and β be roots of P (x) and Q(x) respectively in an algebraic closure of K. Suppose that K(β) ⊂ K(α) and let u(x) ∈ K[x] such that β = u(α). (a) Show that for any A(x) ∈ K[x],

Resx (P (x), A(u(x)) = Resx (Q(x), A(x)m ),

where m = [K(α) : K(β)]. (b) Show that for any A(x) ∈ K[x], there exists B(x) ∈ K[x] such that Resx (P (x), A(x)) = Resx (Q(x), B(x)). Solution 2.25. (1) Suppose that the result is true for an algebraically closed field and let Ω be an algebraic closure of K. Clearly the resultant has the same value when we consider P (x) and A(x) as polynomials with coefficients in K or in Ω. On the other hand let ψ(A,P ) be the Ω-endomorphism of Ω[x]/P (x)Ω(x) such that ψ(A,P ) (f (x)) =

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A(x)f (x), then clearly φ(A,P ) and ψ(A, P ) have the same matrix in the basis {¯ 1, x ¯, . . . , xd−1 } of K[x]/P (x)k[x] over K and in the ba¯ x ¯, . . . , xd−1 } of Ω[x]/P (x)Ω[x] over Ω, where d = deg P hence sis {1, Det φ(A,P ) = Det ψ(A,P ) . Therefore, we may suppose that K is algebraically closed. For any A1 (x), A2 (x) ∈ K[x], we have: Resx (P (x), A1 (x)A2 (x)) = Resx (P (x), A1 (x)) · Resx (P (x), A2 (x))

and

Det φ(A1 A2 ,P ) = Det φ(A1 ,P ) · Det φ(A2 ,P ) .

Hence, it is sufficient to prove the formula when A(x) = a ∈ K or A(x) = x − b with b ∈ K.

• If A(x) = a, then Resx (P (x), a) = adeg P . The matrix of φ(a,P ) in the basis {¯ 1, x ¯, . . . , xd−1 }, where d = deg P is the diagonal matrix aId , hence Det φ(a,P ) = ad = adeg P . The formula is proved in this case. • If A(x) = x − b, with b ∈ K, then Resx (P (x), x − b) = (−1)deg P Resx (x − b, P (x)) = (−1)deg P P (b).

On the other hand, let y = x − b and consider the following diagram K[x]/P (x)K[x]

φ(A,P )

K[x]/P (x)K[x]

θ−1

θ

K[y]/Q(y)K[y]

ψ

K[y]/Q(y)K[y]

where Q(y) = P (y + b),

ψ(g(y)) = yg(y)

and θ(f (x)) = f (y + b),

for any g(y) ∈ K[y] and any f (x) ∈ K[x]. Then obviously it is a commutative diagram of K-algebras. Moreover θ is an isomorphism and θ−1 is its inverse. Since φ(A,P ) = θ−1 ◦ ψ ◦ θ, then Det φ(A,P ) = Det ψ. Set P (x) = xd + ad−1 xd−1 + ... + a0 , then Q(y) = P (y + b) = (y + b)d + ad−1 (y + b)d−1 + · · · + a0

= y d + cd−1 y d−1 + · · · + c0 .

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The matrix of ψ in the basis {¯1, y¯, . . . , y d−1 } of K[y]/Q(y)K[y] is given by   0 0 ... −c0 1 0 ... −c1    0 1 ... −c2  .  ... ... ... ...  0 0 ... −cd−1

Therefore

Det φ(A,P ) = Det ψ = −c0 (−1)1+d = (−1)d c0

= (−1)deg P Q(0) = (−1)deg P P (b). We conclude that the formula holds in this case also. (2)(a) Let k = [K(β) : K] and m = [K(α) : K(β)], then B = {1, β, . . . , β k−1 , α, αβ, . . . , αβ k−1 , . . . , αm−1 , . . . , αm−1 β k−1 } is a basis of K(α) over K. Let M be the matrix of the endomorphism, multiplication by A(β) in K(β) relatively to the basis {1, β, . . . , β k−1 } and let N be the matrix of the endomorphism, multiplication by A(β) in K(α) relatively to the basis B, then N is a block matrix, where the blocks are filled by zeros except on the principal diagonal, where M is repeated. Now we have Resx (Q(x), A(x)m ) = Resx (Q(x), A(x))m = (Det M )m = Det N = Det. of the endo. multiplication by A(β) in K(α) = Det. of the endo. multiplication by A(u(α)) in K(α) = Resx (P (x), A(u(x))). (b) For any fields E, F such that F ⊂ E and any γ ∈ E, we denote by E/F E/F mγ , the F -endomorphism of E such that mγ (a) = γa for any a ∈ E. Let B(x) ∈ K[x] such that
NK(α)/K(β) − A(α) = B(β),

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then Resx (P (x), A(x)) = Det φ(A,P ) K(α)/K

= Det mA(α)

= (−1)mk NK(α)/K (A(α))


= NK(β)/K (−1)m NK(α)/K(β) (A(α))
= NK(β)/K NK(α)/K(β) (−A(α))
= NK(β)/K B(β) K(β)/K

= Det mB(β)

= Resx (Q(x), B(x)). Exercise 2.26. Let K be a field, Ω be an algebraic closure of K, d be a positive integer and f (x) be a non constant polynomial with coefficients in K. (1) Show that the following conditions are equivalent. (i) y d − f (x) is absolutely irreducible. (ii) For any c ∈ K ∗ , y d − cf (x) is absolutely irreducible. (iii) Let f (x) = a(x − α1 )d1 · · · (x − αs )ds be the factorization of f (x) in Ω[x] into linear factors, then gcd(d, d1 , . . . , ds ) = 1. (2) Show that the above equivalent conditions hold if gcd(d, deg f ) = 1. Solution 2.26. (1) • (i) ⇒ (ii). Suppose that (ii) does not hold and let c ∈ K ∗ such that y d − cf (x) is reducible over Ω, say y d − cf (x) = A(x)B(x) where A(x), B(x) ∈ Ω[x] and deg A, deg B ≥ 1. Then  y d A(x) − f (x) = B(x), c c1/d hence z d − f (x) is reducible over Ω, contradicting (i). • (ii) ⇒ (iii). Suppose that (iii) does not hold and let e = gcd(d, d1 , . . . , ds ), d0 = d/e and d0i = dei for i = 1, . . . , s. Then e ≥ 2 and 0

0

0

y d − f (x) = y d e − a(x − α1 )ed1 · · · (x − αs )eds 0

0

0

= (y d )e − [a1/e (x − α1 )d1 ...(x − αs )ds ]e . 0

0

Set g(x) = a1/e (x − α1 )d1 · · · (x − αs )ds . Then 0

0

0

y d − f (x) = (y d − g(x))((y d )e−1 + (y d )e−2 g(x) + ... + g(x)e−2 ).

Therefore y d − f (x) is reducible over Ω contradicting (ii).

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• (iii) ⇒ (i). Suppose that y d − f (x) is reducible over Ω and let η be a primitive d-th root of unity in Ω, and α be a root of y d − f (x) in an algebraic closure of K(x), then y d − f (x) = (y − α)(y − ηα) · · · (y − η d−1 α)

and there exist non negative integers i1 , . . . , ir with 1 ≤ r < d such Qr Qr that j=1 (y − αη ij ) ∈ Ω[x, y]. It follows that αr j=1 η ij ∈ Ω[x], hence αr ∈ Ω[x]. Let l be the smallest positive integer such that αl ∈ Ω[x]. We show that for any positive integer m such that αm ∈ Ω[x], we have l|m. Let m be a such integer and let q, t ∈ Z such that m = lq + t with 0 ≤ t < l. We have αm = αlq · αt , and since αm , αl ∈ Ω[x], then αt ∈ Ω[x], hence t = 0. We conclude, in particular, that l|d and l|r. Therefore l < d. Let g(x) ∈ Ω[x] such that αl = g(x). Then f (x) = αd = (αl )d/l = g(x)d/l . Therefore d/l divides di for i = 1, ..., s and then gcd(d, d1 , ..., ds ) > 1, contradicting (iii). (2) Since deg f = d1 + d2 + · · · + ds , then if gcd(deg f, d) = 1, we have gcd(d, d1 , . . . , ds ) = 1. Therefore (iii) holds. Exercise 2.27. Let K be an algebraically closed field of characteristic 0. A k-tuple (x0 , x1 , . . . , xk−1 ) ∈ K k , with xi 6= xj for i 6= j is called a cycle of f (x) ∈ K[x] if f (xi ) = xi+1 for i = 0, . . . , k − 2 and f (xk−1 ) = x0 . The integer k is called the length of the cycle. (1) Let (x0 , x1 , . . . , xk−1 ) be a cycle of f (x) of length k. Show that k is smallest positive integer such that fk (x0 ) = x0 , where fk (x) is the k-th iterate of f (x). Show that for any positive integer n, fn (x0 ) = x0 if and only if k|n. (2) Let g(x) = ax + b with (a, b) ∈ K 2 and a 6= 0. Determine the possible lengths of the cycles of g(x). (3) Let h(x) = x2 − x. Show that h(x) has no cycle of length 2. (4) Let f (x), g(x) ∈ K[x] such that g(ax + b) = af (x) + b with (a, b) ∈ K 2 , a 6= 0 and let k ≥ 2 be an integer. Show that f (x) has a cycle of length k if and only if g(x) has a cycle of the same length. (5) Let d = deg f and suppose that d ≥ 2.

(a) Show that f (x) has cycles of length 1. (b) Fix two integers n > k > 1. Suppose that f (x) has no cycle of length k nor n and let A(x)/B(x) be the irreducible form of the rational function F (x) = (fn (x) − x)/(fn−k (x) − x).

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(i) Show that the number of solutions x ∈ K, counting multiplicities, satisfying one of the equations F (x) = 0 or F (x) = 1 is equal to 2 deg A. (ii) Show that the number of distinct zeros of A(x)/B(x) is at P l most equal to d. l|n l a2 , we have c > |a|, which implies that c + a > 0 and the equation x2 = s = (a + c)/2 is solvable in K. We now have x=

p b 1 . (a + c)/2 and y = p 2 (a + c)/2

(c) Suppose that there exists a strict algebraic extension E of K(i) and let F be the Galois closure of E over K. Let n = [F : K] and G = Gal(F, K). Set n = 2k m, where k is a non negative integer and m is a positive, odd integer. Let H be a 2-Syllow’s subgroup of G, L = K(θ) be its fixed field and f (x) = Irr(θ, K). Since deg f = [L : K] = m and m is odd, then by (6), f (x) has a root in K. Thus m = 1 and then n = 2k . We deduce that | Gal(F, K(i))| = 2k−1 and since F strictly contains K(i), then k ≥ 2. Let H1 be a subgroup of Gal(F, K(i)) of order 2k−2 and L1 be its fixed field. Then K(i) ⊂ L1 ⊂ F and [L1 : K(i)] = 2. This implies that there exists a monic irreducible polynomials of degree 2 with coefficients in K(i), contradicting (10), (b). Exercise 5.3. Let K be a field and G be a group of ring-automorphisms of K[t] such that, for any σ ∈ G, σ(K) = K. Extend in a natural way the action of G on

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K(t) and let K G = {a ∈ K, σ(a) = a},

K[t]G = {f (t) ∈ K[t], σ(f (t)) = f (t)}

and

G

K(t) = {f (t) ∈ K(t), σ(f (t)) = f (t)}. (1) Show that K(t)G = Frac(K[t]G ). (2) Show that there exists β(t) ∈ K[t]G such that K(t)G = K G (β(t)). Solution 5.3. (1) We first prove that if σ ∈ G, then σ(t) = at+b for some a, b ∈ K, a 6= 0. Since σ −1 (σ(t)) = t, then deg σ(t) = deg σ −1 (t) = 1, hence the result. It is obvious that F r(K[t]G ) ⊂ K(t)G . Let u(t) and v(t) ∈ K[t] such that gcd(u(t), v(t)) = 1. Suppose that u(t)/v(t) ∈ K(t)G . Then, we may suppose that v(t) is not constant and v(t)σ(u(t)) = u(t)σ(v(t)), hence u(t) | σ(u(t)). If we set σ(u(t)) = u(t)q(t) with q(t) ∈ K[t], we obtain σ(v(t)) = v(t)q(t). Writing down a Bezout’s identity a(t)u(t) + b(t)v(t) = 1, yields the similar one σ(a(t))σ(u(t)) + σ(b(t))σ((v(t)) = 1, which implies q(t) ∈ K ? . We now have σ(u(t)) = cu(t)

and σ(v(t)) = cv(t)

for some c ∈ K ? . Suppose that we have proved that there exists e ∈ K ? such that c = e/σ(e), then we have u/v = (eu)/(ev), σ(eu) = σ(e)σ(u) = eu

and

σ(ev) = σ(e)σ(v) = ev and the proof is finished. We prove the existence of such an e. By performing euclidean divisions, we obtain: u(t) = v(t)q1 (t) + r1 (t), v(t) = r1 (t)q2 (t) + r2 (t)

and

ri (t) = ri+1 (t)qi+2 (t) + ri+2 (t), ?

for i = 1, . . . , k,

with rk+2 (t) ∈ K and deg ri+1 (t) < deg ri (t) for all i. Apply σ to the first equation and obtain cu = cvσ(q1 ) + σ(r1 ). We subtract from this equation the first one multiplied by c and we obtain:

−cv σ(q1 ) − q1 = σ(r1 ) − cr1 .

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Since v is non constant, deg σ(q1 ) = deg q1 and deg σ(r1 ) = deg r1 , then σ(q1 ) = q1 and σ(r1 ) = cr1 . By induction we may prove that σ(qi ) = qi and σ(ri ) = cri . Now let e = 1/rk+2 , then it satisfies the required properties. (2) By (1), it is sufficient to prove that there exists β(t) ∈ K[t] such that K[t]G = K G [β(t)]. To prove this we distinguish two cases. If K[t]G contains no non constant polynomial, then we may take β(t) = 1. Suppose next that K[t]G contains non constant polynomials. Let β(t) be such a polynomial whose degree is minimal. Clearly K G [β(t)] ⊂ K[t]G . We prove the other inclusion. Let f (t) ∈ K[t]G . Expand this polynomial in the β(t)-basis [Lang (1965), Th. 9, Chap. 5.5]. We Pk obtain f (t) = i=0 ai (t)β i (t), where k is a non negative integer and ai (t) ∈ K[t], with deg ai < deg β(t) for i = 0, . . . , k. For any σ ∈ G, Pk we have f (t) = σ(f (t)) = i=0 σ(ai (t))β i (t), hence σ(ai (t)) = ai (t) for any σ ∈ G and for i = 0, . . . , k. Since the degree of β(t) is minimal among non constant polynomials in K[t]G , then ai (t) ∈ K. Therefore ai (t) ∈ K G and then f (t) ∈ K G [β(t)]. Exercise 5.4. Let f (x) = x4 − 2x2 + 2 and α ∈ C be a root of f (x). (1) Show that f (x) is irreducible over Q. Without the knowledge of Gal(f (x), Q), determine the fields F (if any) such that Q & F & Q(α).

(Eq 1)

(2) Find Gal(f (x), Q) and compute again the fields F satisfying the conditions (Eq 1). Solution 5.4. (1) By Eisenstein’s irreducibility theorem, applied with the prime 2, we conclude that f (x) is irreducible over Q. Let F be a field satisfying the conditions (Eq 1) and let gF (x) be the minimal polynomial of α over F . Since [F : Q] = 2, then gF (x) = (x − α)(x − β) = x2 + ax + b, where a, b ∈ F and β is a root of f (x) distinct from α. Moreover F = Q(a, b). Since the roots of f are given by s√ s√ 2+1 2−1 +i , α2 = −α, α3 = α ¯ and α4 = −¯ α, α1 = α = 2 2 where α ¯ is the complex conjugate of α. We must consider the following three possibilities for gF (x).

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• gF (x) = (x − α)(x + α) = x2 − α2 = x2 − (1 + i).

q√ √ 2+1 x+ 2. • gF (x) = (x−α)(x− α ¯ ) = x2 +(α+ α ¯ )x+|α|2 = x2 −2 2 q√ √ 2−1 • gF (x) = (x−α)(x+ α ¯ ) = x2 +(α ¯ −α)x−|α|2 = x2 −2i 2 x− 2.

In the first case, F = Q(α2 ) = Q(i) and this field satisfies the conditions (Eq 1). q√ √ 2+1 , 2), which is not a quadratic field. In the second case F = Q( 2 q√ √ 2−1 In the third case F = Q(i 2 , 2) and this field is not a quadratic field. We conclude that there exists one and only one field F satisfying the conditions (Eq 1), namely F = Q(i). √ (2) Since√the roots of f (x) are α, −α, α ¯ , −¯ α and since α ¯ = 2/α, then Q(α, 2) is the√splitting field of f (x) over Q. √ We√show that 2 6∈ Q(α). Suppose that 2 ∈ Q(α), then Q(α) = Q( 2, i). we deduce that there exist a, b, c, d ∈ Q such that √ √ α = a + b 2 + ci + di 2. (Eq 2) √ √ Using the complex conjugation, we obtain α ¯ = a +qb 2 − ci − di 2. √ √ 2+1 Adding these two identities, we obtain a + b 2 = 2 . It follows that √ √ a2 + 2b2 + 2ab 2 = 1/2 + 2/2, thus a2 + 2b2 = 1/2

and ab = 1/4.

These equations may be written in the form 2a2 + 4b2 = 1 and (2a2 )(4b2 ) = 1/8. This implies that 2a2 and 4b2 are solutions of the equation y 2 − y + 1/8 = 0. Since this√equation has no rational solu√ tion, then we get a contradiction, thus 2 6∈ Q(α) and Q(α, 2) is the splitting field of f (x) over Q. √ determined by its Any field automorphism σ of Q(α,√ 2) is completely √ √ action on α and 2. We have σ( 2) = ± 2 and σ(α) = ±α or ± α ¯. Since √ |Gal(f (x), Q)| = [Q(α, 2) : Q] = 8, then all the possibilities for σ given above give rise to automorphisms. Let σ1 be the identity. Label the other automorphisms as follows ( ( σ2 (α) = −α σ3 (α) = α ¯ √ √ , √ √ , σ2 ( 2) = 2 σ3 ( 2) = 2

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(

σ4 (α) = −¯ α √ , √ 2 σ4 ( 2) =

( σ6 (α) = −α √ , √ σ6 ( 2) = − 2 and

(

(

σ5 (α) = α √ , √ σ5 ( 2) = − 2

(

σ7 (α) = α ¯ √ √ , σ7 ( 2) = − 2

σ8 (α) = −¯ α √ √ . σ8 ( 2) = − 2

It is easy to verify that Gal(f (x), Q) is generated by σ3 and σ7 . The first generator is of order 2, while the second is of order 4. Moreover, we have σ3 σ7 σ3 = σ7−1 = σ73 = σ8 . It follows that Gal(f (x), Q) is isomorphic to the dihedral group√ of order 8. Let F be a field satisfying (Eq 1) and let H = Gal(Q(α, 2), F ), then |H| = 4 and √ √ Gal(Q(α, 2), Q(α)) ⊆ Gal(Q(α, 2), F ). Therefore H contains σ1 and σ5 . It is seen that the only subgroup of Gal(f (x), Q) of order 4 containing σ5 is H = {σ1 , σ5 , σ2 , σ6 } and Inv(H) = Q(α2 ) = Q(i). We conclude that Q(α) contains one and only one field F satisfying (Eq 1), namely Q(i). √ Remark. We may prove that 2 6∈ Q(α) √ by using number theory. Let A and B be the rings of integers of Q( 2) and Q(α) √ respectively. Since 2 x − 2 is irreducible over F5 , then 5A is prime. If Q( 2) was contained 0 in Q(α), then the splitting of 5B in B would be 5B = P2 P2 or 5A = Q4 , 0 where the indices of the ideals P2 , P2 and Q4 denote their respective inertial degree. On the other hand, the factorization of f (x) over F5 is given by 0 f (x) = (x − 2)(x + 2)(x2 + 2). By Exercise 10.30, √ we have 5B = Q1 Q1 Q2 and then we reach a contradiction. It follows that 2 6∈ Q(α). Exercise 5.5. Let f (x) = x4 − 6x2 + 6, α be a root of f (x) in C, E be a splitting field of f (x) and G = Gal(f (x), Q). √ (1) Show that E = Q(α, 2) and G is isomorphic to the Dihedral group of order 8, D4 =< σ, τ, σ 4 = τ 2 = 1, τ σ = σ 3 τ >.

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(2) Determine a monic irreducible polynomial g(x) ∈ Q[x] satisfying the following conditions. (a) deg g = 4. (b) The splitting field over Q of g(x) is contained in E. (c) The Galois group over Q of g(x) is not isomorphic to G. Solution 5.5. (1) We compute the solutions of the equation f (x) = 0 by setting x2 = y. We √ obtain the quadratic equation y 2 − 6y + 6 = 0, whose are p roots √ 3 ± 3. We conclude that the roots of f (x) are given byp1 3 + 2 3, √ wherep1 and that α √ 2 = ±1. We may √suppose √ √ = 32+ 3. Let 3 − 3. We have αβ = 3 2. Since 3 = α − β √= √ 3, then 2 2 = αβ/(α − 3). We deduce that E = Q(α, β) = Q(α, 2). We √ show that 2 6∈ Q(α). Suppose that the contrary holds, then since √ 3 ∈ Q(α), we have q √ √ √ √ Q( 3, 2) = Q(α) = Q( 3, 3 + 3). We use the following result. Claim. Let F be a field of characteristic not equal to 2 and√let a and √ b be elements of F , not squares. Suppose that F ( a) = F ( b). Then a/b is a square in F . √ √ Proof. We have a = λ + µ b, with λ and µ ∈√F , then a = λ2 + √ √ bµ2 + 2λµ b, hence λ = 0. It follows that ( a)/( b) = µ ∈ F , thus a/b is a square in F . √ √ Applying this result for F = Q( 3), we get √ √ (3 + 3)/2 is a square in Q( 3), which implies that NQ(√3)/Q ((3 + 3)/2) is a square in Q, that √ is 3/2 is a rational square, hence a contradiction. Therefore 2 6∈ Q(α). We now have √ [E : Q] = [Q(α, 2) : Q] √ = [Q(α, 2) : Q(α)][Q(α) : Q] = 8. We deduce that (G : 1) = 8. To define √ any element ρ ∈ G, it is sufficient to determine√explicitly ρ(α) and ρ( 2). Obviously ρ(α) = ±α √ or ρ(α) = ±β and ρ( 2) = ± 2. By taking all possible combinations of signs, we obtain the eight elements of G. Among them consider the automorphisms ( σ and τ of E such that( σ(α) =β τ (α) =α and √ √ √ √ . σ( 2) = 2 τ ( 2) = − 2

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We compute the values of σ(β) and τ (β). We have ! √ √ √ √ 2(α2 − 3) 2(β 2 − 3) σ(β) = σ = = − 2 3/β = −α α β ! √ √ 2(α2 − 3) 2(α2 − 3) τ (β) = τ =− = −β. α α

and

We deduce that σ 2 (α) = σ(β) = −α,

σ 3 (α) = −σ(α) = −β,

σ 4 (α) = −σ(β) = α, √ √ √ √ σ 2 ( 2) = σ 3 ( 2) = σ 4 ( 2) = 2

and

2

τ (α) = α, √ √ τ ( 2) = 2. 2

Thus σ 4 = τ 2 = 1. We also have σ 3 τ (α) = σ 3 (α) = −β,

τ σ(α) = τ (β) = −β, √ √ √ σ 3 τ ( 2) = σ 3 (− 2) = − 2 √ √ √ τ σ( 2) = τ ( 2) = − 2,

and

hence σ 3 τ = τ σ. We conclude that G ' D4 . (2) Since √ √ 3 = α2 − 3 and 2 = αβ/(α2 − 3), √ √ √ √ then 3 ∈ E and 2 ∈ E. Let γ = 3 + 2 and g(x) = Irr(γ, Q, x). 4 2 It is seen that √ g(x) √ = x + 2x − 11 and the splitting field of g(x) is equal to Q( 3, 2). Moreover Gal(γ, Q) ' Z/2Z × Z/2Z, so that the conditions (a), (b) and (c) hold. Exercise 5.6. Let K be a field, f (X) = X n −a1 X n−1 +· · ·+(−1)n an ∈ K[X] be separable and let α1 , . . . , αn be its roots in an algebraic closure of K. (1) Let M = {P (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ]; P (α1 , . . . , αn ) = 0}. Show that M is a maximal ideal of K[x1 , . . . , xn ].

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(2) Let G = Gal(f (x), K) and τ ∈ Sn . Show that the following assertions are equivalent. (i) τ ∈ Gal(f (x), K). (ii) For any polynomial P (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ], we have P (α1 , . . . , αn ) = 0 ⇒ P (τ (α1 ), . . . , τ (αn )) = 0.

(3) For any τ ∈ Sn \ G, let

Sτ = {P (x1 , . . . , xn ) ∈ M; P (τ (α1 ), . . . , τ (αn )) 6= 0} (a) Show that Sτ is a multiplicative semigroup of K[x1 , . . . , xn ] contained in M. (b) If |K| ≥ n! − 1, show that for any τ1 , . . . , τm ∈ Sn \ G, we have Sτ1 ∩ · · · ∩ Sτm 6= ∅.

(4) Let g(x) = x4 − 6x2 + 6. By using (2), show that the Galois group of g(x) over Q is isomorphic to the dihedral group of order 8. Solution 5.6.

(1) Let Φ : K[x1 , . . . , xn ] → K[α1 , . . . , αn ] = K(α1 , . . . , αn ) be the map such that Φ(P (x1 , . . . , xn )) = P (α1 , . . . , αn ), then Φ is a morphism of rings. We have Ker Φ = M and then K[x1 , . . . , xn ]/M ' Im Φ = K(α1 , . . . , αn ). It follows that M is maximal. (2) • (i) ⇒ (ii). Let P (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] such that P (α1 , . . . , αn ) = 0. Then since τ ∈ Gal(f (x), K), we have 0 = τ (P (α1 , . . . , αn )) = P (τ (α1 ), . . . , τ (αn )), thus (ii). • (ii) ⇒ (i). Let E = K(α1 , . . . , αn ) be the splitting field of f (x) over K and let τ be a permutation of α1 , . . . , αn satisfying the condition (ii). We extend the definition of τ to the whole E and obtain an application from E into E. Let x ∈ E = K(α1 , . . . , αn ) = K[α1 , . . . , αn ]. Set x = P (α1 , . . . , αn ), where P is a polynomial with coefficients in K. Define τ (x) by τ (x) = P (τ (α1 ), . . . , τ (αn )). We show that τ is well defined. Suppose that P (α1 , . . . , αn ) = Q(α1 , . . . , αn ),

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then (P − Q)(α1 , . . . , αn ) = 0 and by (ii), (P − Q)(τ (α1 ), . . . , τ (αn )) = 0, thus P (τ (α1 ), . . . , τ (αn )) = Q(τ (α1 ), . . . , τ (αn )). Obviously τ is a morphism of fields. Moreover it is a K-morphism of E into E and since E/K is algebraic, it is an automorphim [Lang (1965), Chap. 7.2 Lemma 1], hence τ ∈ Gal(E, K) (3)(a) From (2) it is seen that, for any τ ∈ Sn \ G, Sτ 6= ∅. Obviously Sτ is a semigroup of K[x1 , . . . , xn ] contained in M. (b) The proof is done by induction on m. For m = 1 the result is proved in (a). Suppose that Sτ1 ∩ · · · ∩ Sτm−1 6= ∅ and let P be an element of this intersection. The proof is finished if P ∈ Sτm . If not let Q ∈ Sτm . We may suppose that Q 6∈ Sτ1 ∩ · · · ∩ Sτm−1 . If for all i ∈ {1, . . . , m − 1}, Q 6∈ Sτi , the result is established since P + Q ∈ Sτ1 ∩ · · · ∩ Sτm . It remains to consider the case where for some non empty subset J of {1, . . . , m − 1}, |J| ≤ m − 2, we have Q ∈ Sτj for any j ∈ J. For any j ∈ {1, m − 1}, consider the equation (in x): Q(τ (α1 ), . . . , τ (αn )) + xP (τ (α1 ), . . . , τ (αn )) = 0. Any of these equations has at most one solution in K. Since m−1 < n! − 1 ≤ |K|, we may choose λ ∈ K distinct from all the roots of the preceding equations. Let R = Q + λP , then one verifies that R ∈ Sτ1 ∩ · · · ∩ Sτm . (4) Eisenstein’s theorem, shows that g(x) is irreducible over Q. The roots of g(x) are given by q q √ √ α1 = 3 + 3, α2 = −α1 , α3 = 3 − 3 and α4 = −α3 .

We will use the following non trivial relations between the roots of g(x): α1 +α2 = 0 and α3 +α4 = 0. Let τ ∈ S4 , then since α1 +α2 +α3 +α4 = 0, we have τ (α1 ) + τ (α2 ) = 0 if and only if τ (α3 ) + τ (α4 ) = 0. Thus, for the determination of the Galois group of g(x), we use the relation α1 + α2 = 0 and forget the second relation. We compute τ (α1 ) + τ (α2 )

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for the 24 permutations of α1 , . . . , α4 . We find that these expressions vanish for the following permutations: τ0 = Id, τ1 = (α1 α3 α2 α4 ), τ2 = (α3 α4 ), τ3 = τ12 = (α1 α2 )(β1 β2 ), τ4 = τ13 = (α1 α4 α2 α3 ), τ5 = τ1 τ2 = (α1 α3 )(α2 α4 ), τ6 = τ12 τ2 = (α1 α2 ), τ7 = τ13 τ2 = (α1 α4 )(α2 α3 ). For all the other permutations, these expressions do not vanish since the value of each of them is one of the followings: α1 + α3 , α1 + α4 , α2 + α4 , α2 + α3 . By (2), it follows that the Galois group of g(x) is contained in the set formed by τ0 , . . . , τ7 . Since the splitting field of g(x) contains Q(α1 ), then the order of its Galois greater than √ group is a multiple of 4 no √ 8. The identity α1 α3 = 6, and the observation that 6 does not belong to Q(α1 ), show that the degree of the splitting field is at least equal to 8. It follows the Galois group is the set whose elements are the permutations τ0 , . . . , τ7 . Now, τ1 is of order 4, τ2 is of order 2, and we have τ13 τ2 = τ2 τ1 , hence Gal(g(x), Q) is isomorphic to the dihedral group of order 8. Exercise 5.7. √ √ √ √ (1) Show that Q( 2)/Q and Q( 4 2)/Q( 2) are Galois but Q( 4 2)/Q is not. (2) Let K, E, F be fields such that K ⊂ E ⊂ F , E/K and F/E are finite Galois extensions with Galois groups M and N respectively. Suppose that the following conditions hold. (a) For any σ ∈ M , there exist automorphisms of F extending σ. ˆ = {zσ , σ ∈ M }. Denote by zσ one fixed such extension and let M (b) For any σ, τ ∈ M , we have zσ zτ = zστ h with h ∈ N . (c) For each σ ∈ M and each h ∈ N , we have zσ hzσ−1 ∈ N . ˆ . Show that Let G be the subgroup of AutK (F ) generated by N and M F/K is Galois and Gal(F, K) = G. Solution 5.7. √ √ √ (1) Since the extensions Q( 2)/Q and Q(√4 2)/Q( 2) are quadratic, then they are Galois. We show that Q( 4 2)/Q is not √normal.√ Let ρ : √ Q( 4 2) → C be the unique embedding such that ρ( 4 2) = i 4 2. Then

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√ √ √ √
ρ Q( 4 2) = Q(i 4 2) 6= Q( 4 2). Hence Q( 4 2)/Q is not normal. Alternatively, we could argue that the irreducible polynomial x4 − 2 over Q has a root in the extension, but not all its roots. (2) We first compute (G : 1). The conditions (a), (b), (c) show that any ˆ. ρ ∈ G may be written in the form ρ = zσ h with h ∈ N and zh ∈ M −1 Suppose that zσ1 h1 = zσ2 h2 , then zσ1 h1 h2 = zσ2 . For any e ∈ E, we have σ2 (e) = zσ2 (e) = zσ2 h2 (e) = zσ1 h1 (e) = zσ1 (e) = σ1 (e), hence σ1 = σ2 and then zσ1 = zσ2 which in turn implies h1 = h2 . It follows that the representation of the elements of G in the form zσ h is ˆ ||N | = |M ||N |. By [Lang (1965), Th. 9, unique, hence (G : 1) = |M Chap. 7.4], F/K is separable. We show that this extension is normal. Let Ω be an algebraic closure of K, then it is clear that the number of Kembeddings ρ : F → Ω is equal to [F : E][E : K] = |N ||M |, hence equal to |G|. Let A be the set of all these K-embeddings and let φ : G → A ρ i be the map such that φ(ρ) is the K-embedding F → F → Ω, where i is the canonical injection. Then φ is one to one, hence φ(G) = A. Therefore, for any K-embedding ρ : F → Ω, we have ρ(F ) = F , that is F/K is normal. Therefore | Gal(F, K)| = [F : K] = |N ||M | = |G|. Since G ⊂ Gal(F, K), then G = Gal(F, K). Exercise 5.8. Let K be a field, E be a Galois extension of K and G = Gal(E, K). Let σ ∈ G. (1) Let n be the smallest positive integer such that σ n = IdE . Show that n = [E : F ], where F is the invariant field of σ. (2) Let β ∈ E and m be the smallest positive integer such that σ m (β) = β. Show that m | n and m is the smallest positive integer such that σ m Gal(E, K(β)) = Gal(E, K(β)). (3) Let f (x) ∈ K[x] be irreducible and separable, β be a root of f and let F be the normal closure of K(β) over K. Show that F = K(β) if and only if for any σ ∈ Gal(F, K) we have n = m, where n and m are the integers defined in (1) and (2) respectively.

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Solution 5.8. (1) By the fundamental theorem of Galois theory, [Lang (1965), Th. 1, Chap. 8.1], E is Galois over F and its Galois group is cyclic generated by σ. Therefore n = | < σ > | = [E : F ]. (2) Obviously m | n. We have

σ m Gal(E, K(β)) = Gal(E, K(β)) ⇔ σ m ∈ Gal(E, K(β)) ⇔ σ m (β) = β,

hence the result. (3) Suppose that F = K(β) and let σ ∈ Gal(F, K). We already know that m ≤ n. Since σ m (β) = β, then σ m = IdF , hence m ≥ n, thus m = n. Suppose now that for any σ ∈ Gal(F, K), the related integers m and n are equal. Let τ ∈ Gal(F, K(β)), then τ (β) = β, hence 1 = m = n = [F : K(β)]. Therefore F = K(β). Exercise 5.9. Let K be a field, f (x) be an irreducible and separable polynomial of degree n with coefficients in K, γ1 , . . . , γn be the roots of f (x) in an algebraic closure of K, E = K(γ1 , . . . , γn ) and G = Gal(E, K). (1) For any i = 1, . . . , n, let Gi = {σ ∈ G, σ(γ1 ) = γi }. Show that G1 is a subgroup of G and explain what is Gi . (2) Let θ ∈ E be an element whose conjugates over K constitute a normal basis of E/K. For any i = 1, . . . , n, choose and fix σi ∈ Gi and let P βi = σ∈G1 σi (σ(θ)). Show that βi 6= βj if i 6= j. (3) Let F (x1 , . . . , xn ) be a non zero polynomial with coefficients in E and suppose that K is infinite. Show that there exist distinct elements α1 , . . . , αn ∈ E such that (a) any σ ∈ G permutes α1 , . . . , αn , (b) F (α1 , . . . , αn ) 6= 0.

Solution 5.9. (1) Since G1 = Gal(E, K(γ1 )), then G1 is a subgroup of G. Let i ∈ {1, . . . , n} and let σi be a fixed element of Gi , then we have • Claim. Gi = σi G1 . • Proof. Let σ ∈ Gi , then σ = σi (σi−1 σ) and clearly σi−1 σ ∈ G1 . We prove the reverse inclusion. Let σ ∈ σi G1 , σ = σi h with h ∈ G1 , then σ(γ1 ) = σi (γ1 ) = γi , thus σ ∈ Gi .

The claim implies that Gi is the left coset of G1 in G with respect to γi .

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P (2) We may write βi in the form βi = τ ∈σi G1 τ (θ). Since the conjugate of θ form a basis of E/K and since ∪i=1,...,n σi G1 is a partition of G, then the βi are all distinct. Moreover, since G operates on G/G1 by translation, then β1 , . . . , βn are conjugate over K. (3) For i = 1, . . . , n, let yi =

n X

βij−1 xj := Li (x1 , . . . , xn )

j=1

and let

G(x1 , . . . , xn ) = F (L1 (x1 , . . . , xn ), . . . , Ln (x1 , . . . , xn )) = F (y1 , . . . , yn ). The determinant of the transition matrix from {x1 , . . . , xn } to Q {y1 , . . . , yn } is equal to i j. By assumption, there exists a positive integer e such that e

e

θ + i = (θ + j)p = θp + j, e

hence θ + (i − j) = θp . Let h = i − j and let k ∈ {1, . . . , p − 1} such that kh ≡ 1 (mod p) then θ and θ + h are conjugate over Fp , hence there exists a unique automorphism σ of Fp (θ) such that σ(θ) = θ + h. We deduce that σ k (θ) = θ + kh = θ + 1. It follows that θ and θ + 1 are conjugate over Fp . This implies that f (θ + 1) = 0, hence f (x + 1) = f (x). (2) Consider the group G of the Fp -automorphisms of Fp (x) generated by the automorphism τ such that τ (x) = x + 1. Then from (1) we conclude that f (x) belongs to the invariant field say E of G. Clearly xp − x + 1 ∈ E, hence Fp (xp − x + 1) ⊂ E ⊂ Fp (x). Obviously [Fp (x) : Fp (xp − x + 1)] = p and by Artin’s theorem [Lang (1965), Th. 2, Chap. 8.1], [Fp (x) : E] = p, hence E = Fp (xp − x + 1) and f (x) = u(xp − x + 1), where u(x) is a polynomial having the properties stated in the exercise.

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Exercise 5.14. Let K be a field, f (x) ∈ K[x] be monic of degree n and separable over K. Let α1 , . . . αn be the roots of f in an algebraic closure of K, E = K(α1 , . . . αn ) and G = Gal(f (x), K). Fix an integer k ∈ {1, . . . , n − 1}. For j = 1, . . . , n, let k X Y f (xi ) . Fj (x1 , . . . , xk ) = σ(αj ) x − σ(αi ) i i=1 σ∈G

(1) Show that for j = 1, . . . , n, Fj (x1 , . . . , xk ) ∈ K[x1 , . . . , xk ]. (2) Suppose that E = K(α1 , . . . αk ) and let i1 , . . . , ik ∈ {1, . . . , n} be distinct integers. Show that the following propositions are equivalent. (i) There exists τ ∈ G such that, for t = 1, . . . , k, τ (αt ) = αit . (ii) For j = 1, . . . , n, Fj (αi1 , . . . , αik ) 6= 0. (iii) There exists j ∈ {1, . . . , n} such that Fj (αi1 , . . . , αik ) 6= 0.

(3) Suppose that the equivalent conditions in (2) hold.

(a) Show that for all j ∈ {1, . . . , n} Fj (αi , . . . , αik ) . τ (αj ) = Qk 1 0 t=1 f (αit ) (b) Deduce that Fj (α1 , . . . , αk ) . αj = Qk 0 t=1 f (αt )
k−1 (4) Show that f (x) | Fj (x, . . . , x). (5) Suppose that K = Q, f (x) = x3 − 3x + 1. Show that G ' Z/3Z and E = K(α1 ). Compute Fj (α1 ) for j = 1, 2, 3 and express α2 and α3 as rational functions of α1 . Solution 5.14. (1) Let τ ∈ G then since the map σ → τ ◦ σ is a bijection of G onto G, then we have k Y X f (xi ) Fjτ (x1 , . . . , xk ) = τ ◦ σ(αj ) = Fj (x1 , . . . , xk ), x − τ ◦ σ(αi ) i i=1 τ ◦σ∈G

hence Fj (x1 , . . . , xk ) ∈ K[x1 , . . . , xk ]. (2) • (i) ⇒ (ii). Fj (αi1 , . . . , αik ) is a sum of terms of the form f (αi1 ) f (αik ) Tj,σ (αi1 , . . . αik ) = σ(αj ) ··· . αi1 − σ(α1 ) αik − σ(αk ) Clearly this term is non zero if and only if σ = τ . It follows that Fj (αi1 , . . . αik ) = Tj,τ (αi1 , . . . αik ) 6= 0.

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• (ii) ⇒ (iii). Obvious. • (iii) ⇒ (i). We have Fj (αi1 , . . . , αik ) =

X

σ∈G

Tj,σ (αi1 , . . . αik ) 6= 0,

hence there exists τ ∈ G such that Tj,τ (αi1 , . . . αik ) 6= 0. This automorphism τ satisfies τ (αt ) = αit for t = 1, . . . , k. (3)(a) We have Fj (αi1 , . . . , αik ) = Tj,τ (αi1 , . . . , αik ) Y Y = τ (αj ) (αi1 − τ (αu )) · · · (αik − τ (αu )) u6=1

u6=k

= τ (αj )f 0 (αi1 ) · · · f 0 (αik )

and this allows to express τ (αj ) as a rational function of αi1 , . . . , αik . (b) Apply (a) when τ = IdE . (4) We continue to use the polynomials Tj,σ defined in (2). We have Y Y Tj,σ (x, . . . , x) = σ(αj ) (x − σ(αu )) · · · (x − σ(αu )) u6=1

≡0

(mod (f (x))

u6=k

k−1

)

¯ in K[x]. Therefore Fj (x, . . . , x) ≡ 0 (mod (f (x))k−1 ). (5) The discriminant of the polynomial f (x) = x3 − 3x + 1 is equal to 92 hence its Galois group is cyclic of order 3 [Lang (1965), Example 2, Chap. 8.2] and its field of decomposition over Q is E = Q(α1 ), where α1 is any of its roots. We have F1 (x1 ) = α1 f (x1 )/(x1 − α1 ) + α2 f (x1 )/(x1 − α2 ) + α3 f (x1 )/(x1 − α3 ).

Denote by s1 , s2 , s3 the elementary symmetric functions of the roots of f of degree 1, 2, 3 respectively. Then F1 (x1 ) = s1 x21 − 2s2 x1 + 3s3 = −6x1 − 3.

Similarly we obtain

F2 (x1 ) = α2 f (x1 )/(x1 − α1 )

+ α3 f (x1 )/(x1 − α2 )

+ α1 f (x1 )/(x1 − α3 )

= s1 x21 − (α12 + α22 + α32 + α1 α2 + α1 α3 + α2 α3 )x1 + α12 α2 + α22 α3 + α32 α1

= s2 x1 + α12 α2 + α22 α3 + α32 α1 = −3x1 + α12 α2 + α22 α3 + α32 α1

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and F3 (x1 ) = α3 (x1 )/(x1 − α1 )

+ α1 f (x1 )/(x1 − α2 )

+ α2 f (x1 )/(x1 − α3 )

= s1 x21 − (α12 + α22 + α32 + α1 α2 + α1 α3 + α2 α3 )x1 + α12 α3 + α22 α1 + α32 α2

= s2 x1 + α12 α3 + α22 α1 + α32 α2 = −3x1 + α12 α3 + α22 α1 + α32 α2 .

We must compute the non symmetric functions A = α12 α2 + α22 α3 + α32 α1 = TrE/Q (α12 α2 ) and B = α12 α3 + α22 α1 + α32 α2 = TrE/Q (α12 α3 ). P We have A + B = i6=j αi2 αj . Since X X X 0= αi2 αj = A + B + αi3 , P P then A + B = − αi3 = − (3αi − 1) = 3. We also have X X AB = αi4 αj αl + 3s23 + αi3 αj3 . i6=j i,j,l distinct We have X X X αi4 αj αl = s3 αi3 = s3 (3αi − 1) = 3 i,j,l distinct and X X αi3 αj3 = (3αi − 1)(3αj − 1) = 9s2 + 3 = −24, i6=j

i6=j

hence AB = −18. It follows that A and B are roots of the equation y 2 − 3y − 18 = 0, hence we may suppose (i.e. there is a labeling of the roots of f ) that A = 6 and B = −3. We have F1 (x1 ) = 6x1 − 3, F2 (x1 ) = −3x1 + 6 and F3 (x1 ) = −3x1 − 3,

hence

α2 = F2 (α1 )/f 0 (α1 ) = (−3α1 + 6)/(3α12 − 3) = (−α1 + 2)/(α12 − 1)

and

α3 = F3 (α1 )/f 0 (α1 ) = (−3α1 − 3)/(3α12 − 3) = (−α1 − 1)/(α12 − 1) = 1/(1 − α).

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Exercise 5.15. Let K be a field, f (x, y) ∈ K[x, y] be irreducible and separable over K(x) and Ω be an algebraic closure of K(x). Let E ⊂ Ω be the splitting field of f over K(x), K0 be the algebraic closure of K in E, G = Gal(f, K(x)) and H = Gal(f, K0 (x)). Show that K0 /K is Galois, H is a normal subgroup of G and G/H ' Gal(K0 , K). Solution 5.15. Since E is separable over K(x), then K0 (x)/K(x) is separable. Let a ∈ K0 , then Irr(a, K) = Irr(a, K(x) and since a is separable over K(x), it is separable over K. We show that K0 /K is normal. Let σ : K0 → Ω be any K-morphism of fields. We must prove that σ(K0 ) = K0 . Extend σ to K0 (x) and obtain σ1 , by setting σ1 (x) = x. Since E is algebraic over K0 , then we may extend σ1 to E and obtain a K(x)-morphism σ2 of fields. Moreover since E is normal over K(x), then σ2 (E) = E. Let µ ∈ K0 and let g(y) = y m + bm−1 y m−1 + · · · + b0 be the minimal polynomial of µ over K. Since g(µ) = 0, then σ2 (µ)m + bm−1 σ2 (µ)m−1 + · · · + b0 = 0.

We conclude that σ2 (µ) ∈ E. Thus σ(µ) is algebraic over K and belongs to E. It follows that σ(µ) ∈ K0 . This implies that σ is a K-morphism of fields from K0 into K0 . Since K0 is algebraic over K, then σ is onto; that is σ(K0 ) = K0 . Let φ : G → Gal(K0 (x), K(x)) be the map such that for any σ ∈ G, φ(σ) = σ|K0 (x) . Clearly φ is a morphism of groups. Since E/K0 (x) is algebraic, then any K(x)-automorphism of K0 (x) may be extended to E, hence φ is surjective. It is clear that H ⊂ Ker φ. Let σ ∈ Ker φ, then σ is an automorphism of E which fixes every element of K0 (x), hence σ ∈ H and then H = Ker φ. The first isomorphism theorem implies that G/H ' Gal(K0 (x), K(x)). Since Gal(K0 (x), K(x)) ' Gal(K0 , K), then G/H ' Gal(K0 , K). Exercise 5.16. Give examples of two polynomials f (x) and g(x) irreducible and separable over K such that deg f < deg g and Gal(f (x), K) = Gal(g(x), K). Solution 5.16. Let f (x) ∈ [K[x] be irreducible and separable over K of degree n and let G be its Galois group. Suppose that |G| > n and let θ be a primitive element of the splitting field of f over K, then the polynomial g(x) = Irr(θ, K) satisfies the required conditions.

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Exercise 5.17. Let K be an field, Ω be an algebraic closure of K, f (x, y) ∈ K[x, y] be separable over K(x) and satisfying the conditions degx f ≥ 1 and degy f ≥ 1. Let G = Gal(f (x, y), K(x)) and G∗ = Gal(f (x, y), Ω(x)). (1) Show that G∗ is a subgroup of G. (2) Considering the case K = Q, and f (x, y) = y 4 − 2x2 , show that we may have the strict inclusion G∗ ( G. Solution 5.17. (1) Let α1 , . . . , αn be the roots of f (x, y) in an algebraic closure of K(x) and E = K(x, α1 , . . . , αn ). Then Ω(x, α1 , . . . , αn ) = Ω.E is a splitting field of f (x, y) over Ω(x). Let σ ∈ G∗ and σ ¯ : K(x, α1 , . . . , αn ) → Ω(x, α1 , . . . , αn )

be its restriction to K(x, α1 , . . . , αn ). Then σ ¯ is a K(x)-isomorphism of E. Clearly the map σ → σ ¯ is an isomorphism of groups from G∗ into G. (2) Let α be a root of f (x, y) in an algebraic closure of K(x), then the roots of f are ±α, ±iα, hence E = Q(x, α, i). Since 2x2 is not a square in Q(x) then f (x, y) is irreducible over Q(x), hence [Q(x, α) : Q(x)] = 4. We show that i 6∈ Q(x, α), and this will imply that√ [E : Q(x)] = 8. Suppose that i ∈ Q(x, α). Since α4 = 2x2 , then x 2 = ±α2 , hence √ 2 ∈ Q(x, α). We have √ √ Q(x, 2)(α) = Q(x, α) = Q(x, 2, i) = Q(x, α)(i). √ √ Since the minimal polynomials of α and i over Q(x, 2) are y 2 ∓ x 2 √ and y 2 +√1 respectively, then αi ∈ √Q(x, 2). Set αi = λ where √ 2 2 λ ∈ Q(x, 2) then √ −α = λ = ∓x 2. This implies that ∓x 2 is a square in Q( 2)(x), which is a contradiction. We conclude that i 6∈ Q(x, α). Therefore [E : Q(x)] = 8. The elements of G are determined by there action on α and on i. Indeed any σ ∈ G satisfies the following conditions: σ(α) = ±α, ±iα , and σ(i) = ±i,

where all the possible combinations of signs are acceptable. Let Q be an algebraic closure of Q. Over this field, we have √ √ f (x, y) = (y 2 − x 2)(y 2 + x 2),

hence

thus G∗ ( G.

¯ G∗ = Gal(f (x, y), Q(x)) ' Z/2Z × Z/2Z,

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Exercise 5.18. Let K be a field, f (x) be irreducible and separable over K, α1 , . . . , αn be the roots of f in an algebraic closure of K. Let G = Gal(f, K), H = Gal(f, K(α)) and φ ∈ G. Consider the natural action of the group generated by φ on the set {α1 , . . . , αn } and let {α1 , φ(α1 ), . . . , φk1 −1 (α1 )}, . . . , {αs , φ(αs ), . . . , φks −1 (αs )},

be the orbits under this action, where ki is the smallest positive integer such that φki (αi ) = αi . Let {Hσ1 , Hσ1 φ, . . . , Hσ1 φm1 −1 }, . . . , {Hσr , Hσr φ, . . . , Hσr φmr −1 }, be the orbits of the elements of (G/H)d under the action of the same group as above, where mi is the smallest positive integer such that Hσi φmi = Hσi . Show that r = s and after changes of numbering if necessary, mi = ki for all i. Solution 5.18. Let σ ∈ G, and m be a positive integer, then we have Hσφm = Hσ ⇔ φm ∈ σ −1 Hσ

⇔ φm ∈ Gal(f, σ −1 (K))

⇔ φm (σ −1 (α)) = σ −1 (α).

This shows that the orbits of Hσ and of σ −1 (α) under the action of the group generated by φ on the sets (G/H)d and {α1 , . . . , αn } respectively, have the same cardinality and the result follows. Exercise 5.19. Let K be a field, f (x) ∈ K[x] be irreducible and separable. Let n = deg f , α1 , . . . , αn be the roots of f in an algebraic closure of K and G = Gal(f (x), K). Let E = Kα1 + · · · + Kαn be the K-vector space spanned by α1 , . . . , αn . Suppose that the αi are labeled in order to satisfy the condition {α1 , . . . , αs } is a basis of E over K with s ≤ n. For any σ ∈ G and any integer j, 1 ≤ j ≤ s, write σ(αj ) in the form σ(αj ) = aσ1j α1 + aσ2j α2 + · · · + aσsj αs ,

where the coefficients aσij belong to K. Consider the s × s matrix: A(σ) = (aσij ). (1) Show that A(σ) ∈ GLs (K) for any σ ∈ G. (2) Show that the map φ : σ → A(σ) is an injective morphism of G into GLs (K).

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(3) If G is the symmetric group and TrK(α)/K (α) 6= 0, show that s = n. Solution 5.19. (1) Suppose that some linear combination of the column of A(σ), with coefficients λ1 , . . . , λs in K, such that, λ1 σ(α1 ) + · · · + λs σ(αs ) = 0. Then σ(λ1 α1 + · · · + λs αs ) = 0. Therefore λα1 + · · · + λs αs = 0 and then λ1 = · · · = λs = 0. It follows that the columns of A(σ) are linearly independent over K and then Det(Aσ ) 6= 0. (2) Let σ, τ ∈ G, then s X σ◦τ aij αi = σ ◦ τ (αj ) i=1

=σ

s X

aτkj αk

k=1

= = =

s X

k=1 s X

k=1 s X k=1

Therefore σ◦τ aij =

s X

k=1

aτkj aσik

!

aτkj σ(αk ) aτkj

s X

aσik αi

i=1

s X

aτkj aσik

k=1

!

!

αi .



 aτ1j aτ2j   = (aσi1 , . . . , aσis )   ...  , aτsj

˙ ). It follows which implies Aσ◦τ = Aσ Aτ and then φ(σ ◦ τ ) = φ(σ)φ(τ that φ is a morphism of groups. Let σ ∈ G. Suppose that σ ∈ Ker φ, then aσij = 0 if i 6= j and aσij = 1 if i = j. Therefore σ(αj ) = αj for j = 1, . . . , s. It follows that σ = IdK(α1 ,...,αs ) . Since K(α1 , . . . , αs ) = K(α1 , . . . , αn ), then σ = IdK(α1 ,...,αn ) . (3) Suppose that G = Sn and let a1 , . . . , an ∈ K such that a1 α1 + · · · + an αn = 0. For any i = 2, . . . , n, apply to this identity the transposition (α1 We get the new identity a1 αi + · · · + ai−1 αi−1 + ai α1 + · · · + an αn = 0.

αi ).

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Subtracting the second identity from the first, we obtain (a1 − ai )(α1 − αi ) = 0, hence a1 = ai . Since α1 + · · · + αn 6= 0, it follows that α1 , . . . , αn are linearly independent over K, that is s = n. Exercise 5.20. Let K be a field of characteristic p > 0, E be a cyclic extension of K of degree pe with e ≥ 2 and let σ be a generator of Gal(E, K). Let F be the unique intermediate field K ⊂ F ⊂ E of degree pe−1 over K. (1) Show that there exists β ∈ E such that E = F (β), the minimal polye−1 nomial of β over F has the form xp − x − δ and σ p (β) = β + 1. (2) Show that E = K(β), σ(β) = β +α, where α ∈ F satisfies the condition σ(δ) − δ = αp − α and TrF/k (α) = 1. Solution 5.20. (1) Since E/F is cyclic and [E : F ] = p, there exists β ∈ E such that E = F (β) and the minimal polynomial of β over F has the form f (x) = xp − x − δ, [Lang (1965), Th. 11, Chap. 7.6]. The Galois group of E e−1 over F is generated by σ p and the roots of f (x) are given by β + i e−1 where i ∈ Z/pZ. Therefore σ p (β) = β + i0 where i0 ∈ Z/pZ. Let k0 ∈ Z/pZ such that k0 i0 = 1, then σp

e−1

(k0 β) = k0 σ p

e−1

(β) = k0 β + 1, e−1

hence replacing β by k0 β, we may suppose, if necessary, that σ p (β) = β + 1. (2) If K(β) 6= E, then K(β)/K is cyclic of degree ≤ pe−1 hence K(β) ⊂ F , which is a contradiction. Therefore E = K(β). Let α = σ(β) − β. We show that α ∈ F . For this it is necessary and sufficient to show that e−1 σ p (α) = α. We have e−1

σp

(α) = σ(σ p

e−1

(β))) − σ p

e−1

(β)

= σ(β + 1) − (β + 1) = σ(β) − β = α,

hence the claim is proved. Since β p − β − δ = 0, then σ(β)p − σ(β) − σ(δ) = 0,

hence (β + 1)p − (β + 1) − σ(δ) = 0. Thus

β p − β + αp − α − σ(δ) = 0.

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Therefore αp − α − σ(δ) = −δ, that is αp − α = σ(δ) − δ. We have TrF/K (α) =

e−1 pX

σ k (α)

k=1

=

e−1 pX

k=1

=

e−1 pX

σ k (σ(β) − β) σ

k+1

k=1

= σp

e−1

+1

= 1.

(β) −

e−1 pX

σ k (β)

k=1

(β) − σ(β) = σ(β + 1) − σ(β)

Exercise 5.21. Let K be a algebraically closed field of characteristic p 6= 2 and m, n be positive integers. Let f (x, y, t) = txm − 2x + 2 + (xm − t)y n ∈ K[x, y, t]. (1) Show that f (x, y, t) is irreducible over K. (2) Let t0 ∈ K. Show that f (x, y, t0 ) is reducible over K if and only if (t20 + 2)m − 2n t0 = 0. If these equivalent conditions hold, show that x − (t20 + 2)/2 is a factor of f (x, y, t0 ). Solution 5.21. (1) We have f (x, y, t) = (xm − y n )t + xm y n − 2x + 2, hence considering this polynomial as an element of k(x, y)[t], it is of degree one, hence irreducible in this ring. Since gcd(xm − y n , xm y n − 2x + 2) = 1 this polynomial is irreducible in k[x, y][t] = k[x, y, t]. (2) Necessity of the condition. Suppose first that, in K(x)[y], we have gcd(t0 xm − 2x0 + 2, xm − t0 ) = 1, then f (x, y, t) is reducible in K(x)[y]. m +2x−2 is reducible over K(x). By [Lang (1965), Therefore y n − −t0xxm −t 0 Theorem 16 VIII 9], there exists a prime number l such that l|n

4|n

and and

(−t0 xm + 2x − 2)/(xm − t0 ) ∈ K(x)l m

m

or 4

(−t0 x + 2x − 2)/(x − t0 ) ∈ −4K(x) .

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In the first case we have −t0 xm + 2x + 2 ∈ K[x]l and xm − t0 ∈ K[x]l .

The second condition implies m = 0 or t0 = 0. But for m = 0 or for t0 = 0 the first condition is not satisfied. In the second case, since K is algebraically closed, then (−t0 xm + 2x − 2)/(xm − t0 ) ∈ K(x)2

and we are moved to the first case. We conclude that if gcd(t0 xm − 2x + 2, xm − t0 ) = 1,

then f (x, y, t0 ) is irreducible over K. We have

gcd(t0 xm − 2x + 2, xm − t0 ) 6= 1

if and only if Resx (t0 xm − 2x + 2, xm − t0 ) = 0. Straightforward computations show that Resx (txm − 2x + 2, xm − t0 ) = (t2 + 2)m − 2m t.

We conclude that if f (x, y, t) is reducible over K, then (t20 + 2)m − 2m t0 = 0.

Conversely this condition implies that there exists d(x) ∈ K[x]\K such that d(x)|t0 xm − 2x + 2 and d(x)|xm − t0 . Therefore   t0 xm − 2x + 2 xm − t0 y m + , f (x, y, t) = d(x) d(x) d(x) which implies that f (x, y, t0 ) is reducible over K. We have seen above that if fm (x, y, t0 ) is reducible over K, then gcd(t0 xm −2x+2, xm −t0 ) 6= 1, hence these univariate polynomials have a common root, say u in K. We have um = t0 , hence t20 − 2u + 2 = 0. Therefore u = (t20 + 2)/2. It follows that x − (t20 + 2)/2 is a factor of f (x, y, t0 ). Exercise 5.22. Let K be a field, L/K be a Galois extension, H be a subgroup of G and E be the invariant field of H. Show that AutK (E) ' (NG (H))/H. Solution 5.22. Consider the map θ : NG (H) → AutK (E), defined by θ(σ) = σ|E . We first show that θ is well defined. Let x ∈ E and h ∈ H, then h(σ(x)) = σ(h1 (x)) = σ(x)

with h1 ∈ H, hence σ(x) ∈ E. Clearly θ is a group homomorphism and is surjective. Let σ ∈ NG (H), then σ ∈ Ker θ if and only if σ|E = IdE if and only if σ ∈ H, hence Ker θ = H and the result follows from the first isomorphism theorem of groups.

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Exercise 5.23. Let K be a field of characteristic p ≥ 0, n ≥ 2, be an integer. If p > 0, suppose that p - n. Let  be a primitive n-th root of unity in an algebraic closure of K. Let f (x, y) ∈ K(x)[y]. Show that f (x, y) ∈ K(xn )[y] if and only if f (x, y) = f (x, y). Solution 5.23. Necessity of the condition: Obvious. Sufficiency of the condition. First proof. Let g(x, y) ∈ K(x, y). We show that g(x, y) ∈ K(xn , y) if and only if g(x, y) = g(x, y). Let G be the group of K()-automorphism of K()(x, y) generated by the unique K()-automorphism σ such that σ(x) = x and σ(y) = y, then obviously G is of order n. Let F ⊂ K()(x, y) be the invariant field of G, then by Artin’s Theorem [Lang (1965), Th. 2, Chap. 8.1], F/K()(x, y) is Galois and its Galois group is equal to G. We have K(xn , y) ⊂ F ⊂ K(x, y) and [K(x, y) : K(xn , y) = n = |G| = [K(x, y) : F ],

hence F = K(xn , y) and the sufficiency of the condition follows. Second proof. Write f (x, y) in the form m X (ai (x)/bi (x))y i , f (x, y) = i=0

where ai (x), bi (x) ∈ K[x] and gcd(ai (x), bi (x)) = 1. Since f (x, y) = (x) (x) = abii(x) . It follows that f (x, y), then abii(x) ai (, x) = λi ai (x) and b1 (, x) = λi bi (x), where λi ∈ K for any i = 0, . . . , m. We deduce that

ai (0) = λi ai (0) and bi (0) = λi bi (0).

Since gcd(ai (x), bi (x)) = 1 then ai (0) 6= 0 or bi (0) 6= 0 thus λi = 1 for i = 0, . . . , m. Fix i ∈ {0, . . . , m} and set ai (x) = cq xq + cq−1 xq−1 + · · · + c0 . Since ai (x) = ai (x) then for any k ∈ {0, . . . , q} we have ck k = ck . It follows that if ck 6= 0, then k = 1, hence n | k. We conclude that ai (x) ∈ K[xn ] for i = 0, . . . , m. The same method works for bi (x) and we may conclude that bi (x) ∈ K[xn ]. Therefore f (x, y) ∈ K(xn )[y]. Exercise 5.24. (1) Let f (x) ∈ Q[x] be irreducible. Suppose that f (x) has a root z = p+iq, where p, q ∈ R, q 6= 0 and q 2 ∈ Q. Show that there exist a ∈ Q and h(x) ∈ Q[x] irreducible such that f (x) = ah(x + iq)h(x − iq) and h(p) = 0.

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(2) Let f (x) ∈ Q[x] of the degree n ≥ 3 such that f (x) has exactly two non real roots z = p + iq and z¯ = p − iq. Show that q 6∈ Q. Solution 5.24. (1) First proof. We may suppose that f (x) is monic. Let n = deg f and Pn f (x) = m=0 am xm , with am ∈ Q for m = 0, . . . , n − 1 and an = 1. We have n X f (x + iq) = am (x + iq)m and (x + iq)

m

=

m=0 m  X k=0

If k is even, say k = 2l, then

 m m−k x (iq)k . k

(iq)k = (iq)2l = (−1)l (q 2 )l ∈ Q. If k is odd, say k = 2l + 1, then (iq)k = (iq)2l · iq = (−1)l (q 2 )l (iq) := bl · iq,

where bl ∈ Q hence

(x + iq)m = Fm (x) + iqGm (x),

where Fm (x) and Gm (x) ∈ Q[x]. Moreover deg Fm = m and deg Gm ≤ m − 1. It follows that f (x + iq) has the form f (x + iq) = F (x) + iqG(x), where F (x) and G(x) ∈ Q[x], F (x) is monic, deg F = n and deg G ≤ n − 1. √ Since 0 = f (p + iq) = F (p) + iqG(p) and since iq ∈ iQ or iq = i d, then F (p) = G(p) = 0, hence F (x) and G(x) have non trivial factor.
Let h(x) = gcd(F (x), G(x)) then f (x + iq) = h(x) F1 (x) + iqG1 (x) , where F1 (x) and G1 (x) ∈ Q[x]. Substitute x − iq for x and obtain f (x) = h(x − iq)F1 (x − iq) + iqG1 (x − iq).

By conjugation, we obtain Hence


f (x) = h(x + iq) F1 (x + iq) − iqG1 (x + iq) .

f 2 (x) = h(x−iq)h(x+iq)(F1 (x−iq)+iqG1 (x−iq))(F1 (x+iq)−iq(x+iq)). Let A(x) = h(x − iq)h(x + iq) and

B(x) = (F1 (x − iq) + iqG1 (x − iq))(F1 (x + iq) − iqG1 (x + iq))

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then obviously, A(x) and B(x) ∈ Q(iq)[x]. Let σ be the unique automorphism of Q(iq) such that σ(iq) = −iq, then Aσ = A and B σ = B, hence A(x) and B(x) ∈ Q[x] and then F 2 (x) = A(x)B(x). Let D(x) = gcd(f (x), A(x)). Since any prime factor of A(x) divides f (x), then D(x) is non-trivial. Moreover since f (x) is irreducible, then D(x) = f (x). Similarly we have gcd(f (x), B(x)) = f (x), hence B(x) 1 = A(X) f (x) f (x) . Therefore f (x) = A(x) = h(x − iq)h(x + iq). Suppose that h(x) is reducible over Q and let h(y) = h1 (y)h2 (y) be non trivial factorization in Q[x, y], then h(x + iq) = h1 (x + iq)h2 (x + iq) hence f (x) = h1 (x − iq)h1 (x + iq)h2 (x − iq)h2 (x + iq), contradicting the irreducibility of f (x) over Q. Second proof. Let h(x) = Irr(p, Q) and let g(x) = h(x + iq)h(x − iq). Let H(x) = h(l(x)), where l(x) = x + iq. We claim that H(x) is irreducible over Q(iq). By contradiction, suppose that H(x) = H1 (x)H2 (x) in Q(iq)[x], then h(x) = H(l−1 (x)) = H1 (l−1 (x))H2 (l−1 (x)), hence h(x) is reducible over Q(iq). Consider the following diagram of fields. Q(iq, p)

Q(iq)

Q(p)

Q(iq) ∩ Q(p)

Q Since Q(p) ⊂ R and iq 6∈ R, then Q(iq) ∩ Q(p) = Q. Since Q(iq)/Q is Galois then Q(iq, p)/Q(p) is Galois and their Galois groups are equal, see [Lang (1965), Chap. 8.1, Th. 4]. It follows that [Q(p) : Q] = [Q(p) : Q(iq) ∩ Q(p)] = [Q(iq, p) : Q(iq)],

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hence H(x) is irreducible over Q(iq). Since g(x) = NQ(iq)/Q (h(x + iq)) and Q(iq)/Q is normal and separable then, by Exercise 4.2, g(x) is irreducible over Q. We have g(p + iq) = h(p + 2iq)h(p) = 0, hence f (x)|g(x) in Q[x]. Therefore f (x) = g(x). (2) Suppose that q ∈ Q, then we may apply (1) and obtain: f (x) = ah(x + iq)h(x − iq), where a ∈ Q, h(x) ∈ Q[x] irreducible and h(p) = 0. Let α1 , . . . , αn−2 be the real root of f (x). Since 0 = f (αj ) = ah(αj + iq)h(αj − iq), then h(αj + iq ) = 0 or h(αj − iq) = 0 and then h(αj + iq) = h(αj − iq) = 0. It follows that n2 = deg h = which is the number of roots of h(x) is ≥ 2(n−2)+1 = 2n−3, that is n ≤ 2, which is excluded by assumptions. Exercise 5.25. Let K be a field, α be algebraic, separable of degree n over K, f (x) = Irr(α, K). Let G be the Galois group of f (x) over K and α1 = α, α2 , . . . , αn Pn be the conjugates of α. Let V = i=1 K(αi ). (1) Show that V is the K-vector space generated by

{1, α1 , . . . , α1n−1 , . . . , αn , . . . , αnn−1 }. Show that DimK V ≥ n and the equality holds if and only if K(α)/K is normal. (2) Show that DimK V ≤ (n − 1)2 + 1. (3) If G is not 2-transitive, show that DimK V < (n − 1)2 + 1. (4) If G is 2-transitive and the characteristic p of K does not divide |G|, show that DimK V = (n − 1)2 + 1. Solution 5.25. (1) The statement about the generators of V is clear. Since K(α) ⊂ V , then DimK V ≥ DimK K(α) = n. We have equality if and only if for any i = 1, . . . , n, αi ∈ K(α), thus if and only if K(α)/K is normal.

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(2) From (1), we know a set of generators of V containing 1 + n(n − 1) elements. Since for any k = 1, . . . , n − 1, α1k + · · · + αnk ∈ K, then αn , . . . , αnn−1 are linear combinations of n−1 1, α1 , . . . , α1n−1 , . . . , αn−1 , . . . , αn−1 . 2 Therefore DimK V ≤ (n − 1) + 1. (3) Suppose that G is not 2-transitive, then the polynomial f (x)/(x − α) = (x − α2 ) · · · (x − αn ) is reducible over K(α). Renumbering if necessary the roots of f (x), we obtain a factorization  over K(α) of the form:  

f (x)/(x − α) = (x − α2 ) · · · (x − αm ) (x − αm+1 ) · · · (x − αn ) for some integer 2 ≤ m ≤ n−1. We deduce that for any k = 1, . . . , n−1, k k α2k + · · · + αm ∈ K(α) and αm+1 + · · · + αnk ∈ K(α), n−1 n−1 hence αm , . . . , αm and αn , . . . , αn are elements of the linear space generated by {1, αij , for j = 1, . . . , n − 1, and i 6= m, n}. Therefore DimK V ≤ 1 + n(n − 1) − 2(n − 1) = 1 + (n − 1)(n − 2) < 1 + (n − 1)2 . n−1 (4) It is sufficient to show that 1, α1 , . . . , α1n−1 , . . . , αn−1 , . . . , αn−1 are linearly independent over K. It is equivalent to show that if the polynomials Pi (x) ∈ K[x], i = 1, . . . , n − 1 satisfy the conditions: Pn−1 deg Pi ≤ n − 1, Pi (0) = 0 and i=1 Pi (αi ) =: a ∈ K, then Pi (x) = 0 for i = 1, . . . , n − 1. For i = 1, . . . , n, let Hi be the subgroup of G fixing each element of K(αi ) and for i 6= j, let Hij be the one which fixes each element of K(αi , αj ). Let e = |Hij | = [K(α1 , . . . , αn ) : K(αi , αj )]. Let E = K(α1 , . . . , αn ). It is easy to verify the degree (resp. index) at each level as it is indicated in the following diagram of fields (resp. groups). E G e

K(αi , αj ) n−1

K(αi ) n

K

n

Hi n−1

Hij e

{id}

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Pn−1 Now suppose that i=1 Pi (αi ) =: a ∈ K, where the polynomials Pi (x) satisfy the conditions deg ≤ n − 1 and Pi (0) = 0 for i = 1, . . . , n − 1. Pn−1 Then for any σ ∈ H1 , we have i=1 Pi (σ(αi )) =: a. We may write Pn−1 this equation in the form P1 (α1 ) + i=2 Pi (σ(αi )) =: a. Given j and l both distinct from 1, since [E : K(α1 , αj )] = e, there exist e elements of H1 which map αj onto αl . Therefore adding all these equations for σ ∈ H1 , we obtain (n − 1)eP1 (α1 ) + e

n n−1 X X j=2 s=2

Ps (αj ) = (n − 1)ea.

since p - |G|, then p - e. Therefore (n − 1)P1 (α1 ) +

n n−1 X X j=2 s=2

Ps (αj ) = (n − 1)a.

We may write this equation in the form (n − 1)P1 (α1 ) + P2 (α2 ) + · · · + P2 (αn ) + · · · + Pn−1 (α2 )

+ · · · + Pn−1 (αn ) = (n − 1)a. Pn Since for any k ∈ {2, . . . , n − 1}, j=1 Pk (αj ) ∈ K, then (n − 1)P1 (α1 ) −

n−1 X k=2

Pk (α1 ) ∈ K.

We deduce that nP1 (α1 ) −

n−1 X k=1

Pk (α1 ) ∈ K.

The assumptions on the degrees of the polynomials Pk shows that ! n−1 X deg nP1 (x) − Pk (x) < n = deg(α1 ). k=1

Therefore nP1 (x) −

n−1 X k=1

Pk (x) = c ∈ K.

Since P (0) = 0 for any k = 1, . . . , n − 1, then c = 0 and nP1 (x) − Pn−1 k k=1 Pk (x) = 0. In a similar way we obtain the following equations

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Pn−1 nPj (x)− k=1 Pk (x) = 0 for j = 1, . . . , n−1. We write these equations in the form (n − 1)P1 (x) − P2 (x) − · · · − Pn−1 (x) = 0

(Eq 1)

−P1 (x) + (n − 1)P2 (x) − · · · − Pn−1 (x) = 0

(Eq 2)

−P1 (x) − P2 (x) − · · · + (n − 1)Pn−1 (x) = 0.

(Eq n-1)

.................................... = 0

We obtain one more equation by adding these n − 1 equations: P1 (x) + P2 (x) + · · · + Pn−1 (x) = 0.

(Eq n)

Adding the k-th equation to this last equation leads to nPk (x) = 0. Since p - n, then Pk (x) = 0 for k = 1, . . . , n − 1. Exercise 5.26. (1) Let L be a cyclic extension of degree 4 over Q, M be its unique quadratic subfield. Show that M ⊂ R. √ (2) Let d be a square free integer such that √ d ≥ 2 and F = Q( d). Let a, b ∈ Z such that the number α = a + b d is not a square in F and √ K = F ( α). Show that K/Q is normal if and only if a2 − db2 ∈ F 2 . (3) Keep the assumptions of (2) and suppose that a2 −db2 ∈ F 2 . Show that a2 −db2 = y 2 or a2 −db2 = dy 2 with y ∈ Z. Show that in the first (resp. second) case Gal(K, Q) ' Z/2Z × Z/2Z (resp. Gal(K, Q) ' Z/4Z). (4) Let E be a number field of degree 4 over Q. Show that E/Q is cyclic if and only if the following conditions hold. (a) There exist a square free integer d ≥ 2 and s, q, r ∈ Z such that d = ( rs )2 + ( rq )2 . p √ (b) E = Q( rd + s d).

Solution 5.26.

(1) If L ⊂ R, the proof is obvious. Suppose next that L 6⊂ R. Let σ be the restriction of the complex conjugation to L, then σ is an automorphism of L, i.e. σ ∈ Gal(L, Q). Since this group is cyclic of order 4, it contains a unique subgroup of order 2, namely H = Gal(L, M ). Moreover since σ is of order √ 2, then H = hσi. Therefore M = Inv(hσi) ⊂ R. √ (2) Let β = a−b d. It is clear that the conjugates of α are the followings: √ √ √ √ α, − α, β, − β. Suppose that a2 − db2 = γ 2 ∈ F 2 . We have √ √ √ √ αβ = γ 2 , hence α β = ±γ. It follows that β = ±γ/ α ∈ K, which shows that K/Q is normal. Since χ(Q) = 0, then K/Q is Galois.

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√ √ √ Suppose now that K/Q is Galois. Then β ∈ K. Set β = A + B α, where A, B ∈ F . We deduce that √ √ A2 + B 2 α + 2AB α = a − b d ∈ F. √ Therefore A = 0 or B = 0. If B = 0, then A2 = √ a − b d. Applying √ the automorphism τ : F → F defined by τ (x + y d) = x − y d, we √ obtain a + b d = τ (A)2 ∈ F 2 , which is a contradiction. Suppose next that A = 0. We have √ √ a − b d = B 2 α = B 2 (a + b d), hence

√ a−b d a2 − db2 √ = √ . B = a+b d (a + b d)2 2

This proves a2 − db2 ∈ F 2 . √ (3) Since a2 − db2 ∈ F 2 , set√a2 − db2 = (c + e d)2 , where c, d ∈ Q. Then a2 − db2 = c2 + de2 + 2ce d. We conclude that c = 0 or e = 0. If e = 0, then a2 − db2 = c2 , which represents the first case mentioned in the conclusion of (3). Suppose next that c = 0. We have a2 − db2 = de2 , which represents the second case. We now prove the assertion relative to the Galois group. Let σ1 , σ2 , σ3 , σ4 be the elements of the Galois group of K over Q such that p p √ √ √ √ σ1 = IdK , σ2 ( α) = − α, σ3 ( α) = β and σ3 ( α) = − β.

It is clear that σ2 is of order 2, hence K/Q is cyclic if and only if √ σ3 √ is of order (α) = (σ3 ( α))2 = β. We deduce that √ 4. We have√σ3√ σ3 ( d) = − d. We have α β = γ ∈ F , hence p √ σ32 ( α) = σ3 ( β) √ = σ3 (γ/ α) √ = σ3 (γ)/σ3 ( α) p = σ3 (γ)/ β √ = (σ3 (γ)/γ) α. Therefore σ3 is of order 4 or 2 according to σ3 (γ)/γ = −1 or σ3 (γ)/γ = 1. We deduce that Gal(K, Q) √ ' Z/4Z) if and only if σ3 (γ) = −γ, which is equivalent to γ = y d with y ∈ Z and then equivalent to a2 − db2 = dy 2 . Similarly we have Gal(K, Q) ' Z/2Z × Z/2Z if and only if σ3 (γ) = γ, which is equivalent to γ = y with y ∈ Z and then equivalent to a2 − db2 = y 2 .

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(4) • Necessity of the conditions. p √ By (2) and (3), there exist a, b ∈ Z such that E = Q( a√+ b d),√for some integer d ≥ 2 and some integers, a, b such that a+b d 6∈ Q( d) and a2 − db2 = du2 with u ∈ Z. This last relation implies that d | a. Set a = da1 , then da21 = b2 + u2 . We deduce that d = (b/a1 )2 +(u/a1 )2 . If we replace b by s, u by q and a by rd, we obtain the proof of (b) and (c). Notice that the identity d = ( rs )2 + ( rq )2 shows that d is a sum of two squares of integers. Therefore any prime divisor of d is equal to 2 or congruent to 1 modulo 4. • Sufficiency of the conditions. √ √ We must show that rd + s d 6∈ Q( d)2 and r2 d2 − s2 d = dy 2 with y ∈ Q. Since d = ( rs )2 + ( rq )2 , then r2 d − s2 = q 2 and √ we obtain √ the second claim√by multiplying by d. Suppose that rd+s d ∈ Q( d)2 , √ 2 2 2 2 2 then (rd + s d)(rd − s d) ∈ Q , hence r d − s d = y with y ∈ Q which contradicts the equation r2 d − s2 = q 2 . Exercise 5.27. Let p be a prime number, K be a field, K(α) and K(β) be Galois extensions of K and N be a normal p-subgroup of Gal(K(α, β), K) such that every element of order p in N fixes α or β. Show that the order of N divides [K(αβ) : K]. Solution 5.27. Let G = Gal(K(α, β) : K), γ = αβ and f (x) = Irr(γ, K). Obviously N 0 acts in the set of roots of f (x). Let γ be any root of f (x). We show that 0 the orbits, under the action of N , of γ and γ have the same cardinal. Since 0 0 γ and γ are conjugate over K, there exists σ ∈ G such that σ(γ) = γ . For any n ∈ N , we have 0

0

0

σ(n(γ)) = n (σ(γ)) = n (γ ) 0

for some n ∈ N . Therefore, σ induces an injective map from Orb(γ) 0 0 into Orb(γ ). Clearly, σ −1 induces an injective map form Orb(γ ) into Orb(γ). We conclude that all the orbits under the action of N have the same cardinality. Thus, to obtain the stated result, it is sufficient to prove |Orb(γ)| = (N : 1). This will follow if no non trivial element of N fixes γ. Suppose that there exists n ∈ N , n 6= Id such that n(γ) = γ. Let pe , e−1 with e ≥ 1, be the order of n. We may suppose that e = 1, otherwise np has order p and fixes γ. Our assumptions on the elements of order p of N implies n(α) = α or n(β) = β. If n(α) = α, then αβ = γ = n(γ) =

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n(α)n(β) = αn(β), hence n(β) = β, thus n = Id which is a contradiction. We get the same contradiction if we suppose that n(β) = β. Remark. We give a second proof of the claim that all the orbits under the action of N have the same cardinality. Let H = N ∩Gal(K(α, β), K(γ)). We first show that Orb(γ) = {n1 (γ), . . . , ns (γ)}, where {n1 , . . . , ns } is a complete set of representatives of the elements of N modulo H. Let n1 , n2 ∈ N , then n1 (γ) = n2 (γ) ⇔ n−1 2 n1 (γ) = γ ⇔ n−1 2 n1 ∈ H

⇔ n1 H = n2 H.

We conclude that |Orb(γ)| = (N : H). 0 0 Let γ = σ(γ) be a conjugate of γ, then as above, we have |Orbγ | = (N : σHσ −1 ) = (N : H). Therefore all the orbits have the same cardinality. Exercise 5.28. Let K be a field. (1) Let G be a finite group. Show that there exist fields E and F such that K ⊂ F ⊂ E, E/F is Galois and Gal(E, F ) ' G. (2) Let n be a positive integer. Show that there exists an extension J of K which has no finite extension of degree k with 1 ≤ k ≤ n. Solution 5.28. (1) Let m = |G| and let x1 , . . . , xm be algebraically independent variables over K and let σi for i = 1, . . . , m be the elementary symmetric function of degree i of the xj . Then K(x1 , . . . , xm ) is a Galois extension of k(σ1 , . . . , σm ) whose Galois group is isomorphic to Sm . Embedding G in Sm , we may consider G as a subgroup of Sm . Let F be the invariant field of G, then K ⊂ F ⊂ K(x1 , . . . , xm ), K(x1 , . . . , xm )/F is Galois and Gal(K(x1 , . . . , xm ), F ) ' G. The proof is compete if we let E = K(x1 , . . . , xm ). (2) Let m be an integer such that n! < m!/2. Let G = Am and let E and F be the fields determined by (1) and satisfying the conditions: K ⊂ F ⊂ E, E/F is Galois and Gal(E, F ) ' Am . Let Ω be an algebraic closure of F containing E. Consider the set L of subfields of Ω containing F such that L ∩ E = F . These conditions on L are equivalent to LE/L is Galois and Gal(LE, L) ' ∩Am . The following diagram may be useful.

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LE

L

E

F

The set L is non empty since it contains F . Moreover L is ordered by the inclusion. Any family of totally ordered elements of L has a greatest element (which is the union of the sets in the family). It follows that L is an inductive set and then Zorn’s lemma implies that it has a maximal element say J. Suppose now that J has an extension of degree k with 1 ≤ k ≤ n and let N ⊂ Ω be its normal closure over J. We may visualize the lattice of fields trough the following diagram. N JE = N E

N

J

JE

N ∩ JE

E

F Since J ∈ L, then Gal(JE, J) ' Am . By [Lang (1965), Th. 4, Chap. 8.1], JE/(N ∩ JE) is Galois, N E/N is also Galois and Gal(JE, N ∩ JE) ' Gal(N E, N ). Since J is maximal in L then N ∈ L.

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Since Gal(JE, N ∩ JE) ( Gal(JE, J) = Am ,

then J ( N ∩JE. Since (N ∩JE)/J is normal, then Gal(JE, N ∩JE) is distinguished in Gal(JE, J) = Am . Therefore Gal(JE, N ∩JE) = {Id}. It follows that Gal(N E, N ) = {Id}, hence N E = N which implies E ⊂ N . Now we have m! > n! [N : J] ≥ [EJ : J] = 2 and this contradicts the fact that N is the normal closure over J of some extension L/J with [L : J] = k and 1 ≤ k ≤ n. Exercise 5.29. Let K be a field, Ω be an algebraic closure of K, α ∈ Ω be separable of degree n over K. Let α1 , α2 , . . . , αn be the conjugates of α, where α1 = α. (1) Let a ∈ K ? and β be a root of f (x). If aα is a root of f (x), show that aβ is a root of f (x). (2) Let s ≥ 1 be an integer. Define in A = {1, . . . , n} the relation R by i R j if αis = αjs . Show that R is an equivalence relation. (3) Denote by i(s) the equivalence class of i for the relation R. Show that |i(s) | = [K(αi ) : K(αis )]. (4) Show that for any i, j ∈ A, |i(s) | = |j(s) |. (5) Let r and s be coprime positive integers. Show that for any i ∈ A, |i(s) ∩ i(r) | = 1. (6) Let r and s be coprime positive integers. Show that for any i ∈ A, |i(rs) | ≥ |i(s) ||i(r) |. (7) Suppose that there exists a prime p such that [K(αp ) : K] < n. Show that p | n, f (x) ∈ K[xp ] and 1(p) = {ζ1 α1 , . . . , ζp α1 }, where the ζi are the p-th roots of unity and these roots are distinct and belong to the Galois closure of K(α). (8) Let P be the set of prime numbers p such that [K(αp ) : K] < n. Show that |P| ≤ logn/log2. (9) Suppose that [K(αm ) : K] < n, show that there exists a prime p | m such that [K(αp ) : K] < n. (10) Suppose that K = Q and let α be the positive real root of f (x) = x6 −2 and p be a prime number. (a) Show that [Q(αp ) : Q] < 6 if and only if p = 2 or p = 3. Deduce that for any positive integer m, [Q(αm ) : Q] < 6 if and only if m ≡ 0 (mod 2) or m ≡ 0 (mod 3).

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P5 (b) Let β = i=0 ai αi be a primitive element of Q(α) over Q. Show that [Q(β p ) : Q] < 6 if and only if p = 2 (resp.p = 3) and the ai satisfy the following equations a0 a1 + 2a2 a5 + 2a3 a4 = 0 a0 a3 + a1 a2 + 2a4 a5 = 0 a0 a5 + a1 a4 + a2 a3 = 0

(resp.

b1 a0 + b0 a1 + 2b5 a2 + 2b4 a3 + 2b3 a4 + 2b2 a5 = 0 b2 a0 + b1 a1 + b0 a2 + 2b5 a3 + 2b4 a4 + 2b3 a5 = 0 b4 a0 + b3 a1 + b2 a2 + b1 a3 + b0 a4 + 2b5 a5 = 0 b5 a0 + b4 a1 + b3 a2 + b2 a3 + b1 a4 + b0 a5 = 0). Here the bi are given by b0 = a20 + 2a23 + 4a1 a5 + 4a2 a4 b1 = 2a0 a1 + 4a2 a5 + 4a3 a4 b2 = a21 + 2a24 + 2a0 a2 + 4a3 a5 b3 = 2a0 a3 + 2a1 a2 + 4a4 a5 b4 = a22 + 2a25 + 2a0 a4 + 2a1 a3 b5 = 2a0 a5 + 2a1 a4 + 2a2 a3 . Solution 5.29. (1) Let q(x) and r(x) ∈ K[x] such that deg r(x) < n and f (ax) = f (x)q(x) + r(x), then obviously q(x) = an . Substituting α for x in the preceding identity, leads to r(α) = 0, hence r(x) = 0. Thus f (ax) = an f (x). It follows that f (aβ) = 0. (2) Easy. (3) Let Σ = {σ : K(αi ) → K, σ is a K(αis )-embedding} and consider the map φ : i(s) → Σ such that for any j ∈ i(s) , we have φ(j) is the unique K-embedding of K(αi ) into K satisfying φ(j)(αi ) = αj . We verify that φ(j) fixes αis . We have φ(j)(αis ) = αjs = αis , hence the result. We show that φ is one to one. Suppose that φ(j) = φ(k), then αj = φ(j)(αi ) = φ(k)(αi ) = αk ,

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hence j = k. We show that φ is onto. Let σ : K(αi ) → K be a K(αis )-embedding such that σ(αi ) = αj , then αis = σ(αis ) = αjs , hence j ∈ i(s) . (4) We define an injective map from i(s) into j(s) . Let f (x) = Irr(α, K), G = Gal(f (x), K). Since G is transitive, then there exists σ ∈ G such that σ(αi ) = αj . Let k ∈ i(s) , then αks = αis , hence σ(αk )s = αjs . Therefore σ induces an injective map from i(s) into j(s) . By symmetry there exists an injective map from the later into the former. Therefore the two sets have the same cardinality. (5) Obviously i ∈ i(s) ∩ i(r) . Let k and l ∈ i(s) ∩ i(r) , thenαkr = αlr and αks = αls . By Bezout’s identity, there exist u and v such that ur + vs = 1. We deduce that αk = αkur αkvs = αlur αlvs = αl , hence the result. (6) By (2), it is sufficient to show that ∪j∈i(r) j(s) ⊂ i(rs) and that this

union is disjoint. Let j ∈ i(r) and k ∈ j(s) , then αjr = αir and αks = αjs , hence αkrs = αjrs = αirs , which means k ∈ i(rs) . Suppose that there exist j1 and j2 ∈ i(r) such that j1(s) = j2(s) , then j1 and j2 ∈ j1(r) ∩ j1(s) , hence by (5), j1 = j2 . (7) By (3), we conclude that [K(αp ) : K] < n if and only if there exists a root β of f (x) such that β 6= α and β p = αp . We use this β in our proof. We have β = ζα, where ζ is a primitive p-th root of unity. Pn As in (1), we conclude that f (ζx) = ζ n f (x). Let f (x) = i=0 ai xi , where ai ∈ K and an = 1, then n n X X ζn ζ i−n ai xi = ζ n ai xi . i=0

i=0

It follows that for any i for which ai 6= 0, we have ζ i−n = 1, that is i ≡ n (mod p). In particular 0 ≡ n (mod p) and then i ≡ 0 (mod p) for all the indices i such that ai 6= 0. We conclude that f (x) ∈ K[xp ]. For any i = 1, . . . , p, we have (ζi α1 )p = α1p hence ζi α1 ∈ 1(p) . Conversely if αi ∈ 1(p) , then αip = α1p , thus αi = ηα1 for some p-th root of unity η. Since for any i = 1, . . . , p, ζi = α1 /αj for some j ∈ {1, . . . , n}, then the Galois closure of K(α) contains the p-th roots of unity and these roots are distinct since f (x) is separable over K. (8) Notice that by the definition of P, for any p ∈ P, we have |i(p) | ≥ 2. Let B be any finite subset of P, then by (4) and (6), we have Y 2|B| ≤ |i(p) | ≤ |i(Qp∈B p) | ≤ n. p∈B

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We deduce that |B| ≤ log n/ log 2, hence P is finite and |P| ≤ (log n)/(log 2). Remark. Since any element of P is a divisor of n, then |P| ≤ r, where r is the number of prime factors of n. Write n in the form Qr n = i=1 pei i . Then log n =

r X i=1

ei log pi ≥

r X i=1

ei log 2 ≥ r log 2,

hence the result. (9) Let k = |1(m) |, then the assumptions implies that k ≥ 2. We may suppose that 1(m) = {1, . . . , k}. For any i ∈ 1(m) , let ζi = α1 /αi , then obviously ζi is a m-th root of unity. Let d be the least common multiple of the orders of these ζi , then d | m. Let Um be the set of m-th roots of unity. We show that H := {ζ1 , . . . , ζk } is a subgroup of Um . Let ζi and ζj be elements of H, then ζi−1 α1 and ζj−1 α1 are roots of f (x). By (1), we conclude that ζj−1 ζi−1 α1 is a root say αt of f (x). It follows that α1 /αt = ζj ζi . Thus t ∈ 1(m) and then ζj ζi ∈ H. Now H being a subgroup of the cyclic group Um is itself cyclic of order k. By definition of d it follows that k = d. Since k ≥ 2, then k has at least a prime divisor p. Therefore H contains an element of order p, say ζi0 with i0 ∈ {1, . . . , k}. We deduce that α1 = ζi0 αi0 . Thus α1p = αip0 . Since α1 6= αi0 , then [K(αp ) : K] < n. (10)(a) Suppose that [Q(αp ) : Q] < 6, then by (7), p | 6 so that p = 2 or p = 3. The roots of f (x) are given by α, −α, jα, −jα, j 2 α and −j 2 α, where j is a primitive cube root of unity. Therefore we have the following relations between α and some of its conjugates α2 = (−α)2

and α3 = (jα)3 = (j 2 α)3 .

This shows that [Q(α2 ) : Q] = 3 and [Q(α3 ) : Q] = 2. We deduce that [Q(αm ) : Q] < 6 ⇐⇒ There exists a prime ⇐⇒ m ≡ 0

(mod 2)

or

p | m [Q(αp ) : Q] < 6 m≡0

(mod 3).

(b) Let σ1 , . . . , σ6 : Q(α) → C be the distinct embeddings. We suppose that they are labeled in the following way: σ1 (α) = α,

σ2 (α) = −α,

σ4 (α) = −jα,

σ3 (α) = jα, 2

σ5 (α) = j α,

and σ6 (α) = −j 2 α.

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P5 Let γ = i=0 ci αi be an element of Q(α). We look at the conditions σi (γ) = γ for i = 2, . . . , 6. We have σ2 (γ) = γ ⇐⇒ c0 − c1 α + c2 α2 − c3 α3 + c4 α4 − c5 α5 = ⇐⇒ c1 = c3 = c5 = 0

5 X

ci α i

i=0

⇐⇒ γ ∈ Q(α2 ),

σ3 (γ) = γ ⇐⇒ c0 + c1 jα + c2 j 2 α2 + c3 α3 + c4 jα4 + c5 j 2 α5 = 2

5 X

4

ci αi

i=0

⇐⇒ c0 + c1 jα + c2 (−1 − j)α + c3 α + c4 jα + c5 (−1 − j)α5 =

3

5 X

ci α i

i=0

⇐⇒ c0 − c2 α2 + c3 α3 − c5 α5 + c1 jα − c2 jα2 + c4 jα4 − c5 jα5 =

5 X

ci α i

i=0

⇐⇒ c1 = c2 = c4 = c5 = 0 ⇐⇒ γ ∈ Q(α3 ),

2 2

3

4

2 5

σ4 (γ) = γ ⇐⇒ c0 − c1 jα + c2 j α − c3 α + c4 jα − c5 j α =

5 X

ci αi

i=0

⇐⇒ c0 − c1 jα + c2 (−1 − j)α2 − c3 α3 + c4 jα4 − c5 (−1 − j)α5 =

5 X

ci α i

i=0

⇐⇒ c0 − c2 α2 − c3 α3 + c5 α5 − c1 jα − c2 jα2 + c4 jα4 + c5 jα5 =

5 X

ci α i

i=0

⇐⇒ c1 = c2 = c3 = c4 = c5 = 0 ⇐⇒ γ ∈ Q,

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σ5 (γ) = γ ⇐⇒ c0 + c1 j 2 α + c2 jα2 + c3 α3 + c4 j 2 α4 + c5 jα5 = 2

5 X

ci αi

i=0 4

⇐⇒ c0 + c1 (−1 − j)α + c2 jα + c3 α + c4 (−1 − j)α + c5 jα5 =

5 X

3

139

ci α i

i=0

⇐⇒ c0 − c1 α + c3 α3 − c4 α4 − c1 jα + c2 jα2 − c4 jα4 + c5 jα5 =

5 X

ci α i

i=0

⇐⇒ c1 = c2 = c4 = c5 = 0 ⇐⇒ γ ∈ Q(α3 ),

and

σ6 (γ) = γ ⇐⇒ c0 − c1 j 2 α + c2 jα2 − c3 α3 + c4 j 2 α4 − c5 jα5 =

5 X

ci αi

i=0

⇐⇒ c0 − c1 (−1 − j)α + c2 jα2 − c3 α3 + c4 (−1 − j)α4 − c5 jα5 =

5 X

ci αi

i=0

⇐⇒ c0 − c1 α − c3 α3 − c4 α4 + c1 jα + c2 jα2 − c4 jα4 − c5 jα5 =

5 X

ci αi

i=0

⇐⇒ c1 = c2 = c3 = c4 = c5 = 0 ⇐⇒ γ ∈ Q.

In these equivalences, we have used the fact that {1, α, . . . , α5 } (resp. {1, α, . . . , α5 , j, jα, . . . , jα5 } is a basis of Q(α) (resp. Q(α, j)) over Q. By (9), we conclude that [Q(β p ) : Q] < 6 if and only if p = 2 and there exists i ∈ {2, . . . , 6} such that σi (β)2 = β 2 or p = 3 and there exists i ∈ {2, . . . , 6} such that σi (β)3 = β 3 . Using the above equivalences, we conclude that σi (β)2 = β 2 ⇐⇒ σi (β 2 ) = β 2 ⇐⇒ i = 2

⇐⇒ β 2 ∈ Q(α2 ).

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Since β 2 = a20 + 2a23 + 4a1 a5 + 4a2 a4 + α(2a0 a1 + 4a2 a5 + 4a3 a4 ) + α2 (a21 + 2a24 + 2a0 a2 ) + α3 (2a0 a3 + 2a1 a2 + 4a4 a5 ) + α4 (2a25 + 2a0 a4 + 2a1 a3 ) + α5 (2a0 a5 + 2a1 a4 + 2a2 a3 ), then σi (β)2 = β 2 if and only if β satisfies the system of equations in the aj given in the statement, which is obtained by equating to 0 the coefficients of α, α3 and α5 in this expression of β 2 . Similarly we have σi (β)3 = β 3 ⇐⇒ σi (β 3 ) = β 3 ⇐⇒ i = 3

or i = 5 ⇐⇒ β 3 ∈ Q(α3 ).

We may conclude the proof on expressing β 3 in the form β 3 = P5 i i=0 ci α and then equating to 0 the coefficients c1 , c2 , c4 and c5 . Remark. For example, any primitive element of Q(α) of the form β = a3 α3 + a5 α5 (resp. β = a1 α + a4 α4 ) satisfies the condition [Q(β 2 ) : Q] = 3 (resp. [Q(β 3 ) : Q] = 2). Exercise 5.30. Two complex numbers α and β are said to be equivalent if there exists (A, B, C, D) ∈ Z4 such that β = (Aα + B)/(Cα + D) with AD − BC = ±1. (1) Let (a, b, c) ∈ Z3 such that a 6= 0 and the polynomial f (x) = ax2 + bx + c is irreducible over Z. Show that the roots of this polynomial are equivalent if and only if, there exists (x, y) ∈ Z2 such that a2 x2 + (b2 − 2ac)xy + c2 y 2 = ±b2 .

(Eq 1)

(2) Show that the equivalent conditions in (1) hold in the following cases. (i) a = ±c. (ii) a | b. (iii) c | b.

(3) Let (a, b, c, d) ∈ Z4 such that a 6= 0 and the polynomial g(x) = ax3 + bx2 + cx + d is irreducible over Z. (a) Suppose that g(x) has two distinct equivalent roots. Show that Disc(g) is a perfect square.

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(b) Let h(x) = 8x3 − 24x2 + 18x − 1 and γ be a root of h(x). Show that Disc(h) = 34 · 26 and the list of the roots of h(x) is given 00 6γ−7 , γ = 4γ−7 by γ, γ 0 = 4γ−4 4γ−6 . Conclude that the converse of (a) is false. (c) Suppose that Disc(g) is a perfect square and let α, β be two distinct roots of g(x). Show that there exists (A, B, C, D) ∈ Z4 such that β = (Aα + B)/(Cα + D) with AD − BC = (A + D)2 and AD − BC 6= 0.

(4) Let (a, b, c, d) ∈ Z4 such that a 6= 0 and the polynomial g(x) = ax3 + bx2 + cx + d is irreducible over Z and let α be a root of g(x). Suppose that Disc(g) is a perfect square. Show that there exists (u, v, w) ∈ Z3 such that the roots of the polynomial gˆ(x) := Irr((uα + v)/w, Q) are equivalent. (5) Consider the polynomial h(x) defined in (3), (b). Determine a polyˆ nomial h(x) whose roots are equivalent and generate the field Q(γ). (6) Let (a, b, c, d) ∈ Z4 such that a 6= 0 and the polynomial g(x) = ax3 + bx2 + cx + d is irreducible over Z and let α be a root of g(x). Suppose that Dis(g) is a perfect square. Let β = (Aα + B)/(Cα + D), with (A, B, C, D) ∈ Z4 , gcd(A, B, C, D) = 1, be a root of g(x) distinct from α. Show that any prime factor of AD − BC is a divisor of a Disc(g). Solution 5.30. √ √ (1) Let α = −(b + d)/2a and β = −(b − d)/2a be the roots of f (x), where d = b2 − 4ac. Suppose that β = (Aα + B)/(Cα + D) with (A, B, C, D) ∈ Z4 and AD − BC = ±1. Then Aα + B = Cαβ + Dβ = Cc/a + Dβ, hence (A + D)α + B = Cc/a + D(α + β). Therefore a(A + D)α + aB = Cc − bD.

Since 1, α is a basis of Q(α) over Q, then A + D = 0 and aB = Cc − bD. We have A2 + BC = A(−D) + BC = ∓1, hence ∓b2 = b2 BC+b2 A2 = b2 BC+(aB−cC)2 = a2 B 2 +(b2 −2ac)BC+c2 C 2 . This implies that x = B, y = C is an integral solution of the equation a2 x2 + (b2 − 2ac)xy + c2 y 2 = ∓b2 .

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Conversely, suppose that a2 x2 + (b2 − 2ac)xy + c2 y 2 = ±b2 with (x, y) ∈ Z2 . If b = 0, then β = −α which implies that α and β are equivalent. Suppose that b 6= 0, then b2 | (ax − cy)2 . Let z ∈ Z such that ax − cy = bz, then ±b2 = b2 xy + b2 z 2 , hence z 2 + xy = ±1. We have x − ac y = ab z, hence x − αβy = −(α + β)z. Therefore β = zα+x yα−z . We have proved that z 2 + xy = ±1, hence α and β are equivalent. (2) (i) If a = c, then (Eq 1) becomes: c2 x2 + (b2 − 2c2 )xy + c2 y 2 = ±b2 or b2 xy + (cx − cy)2 = ±b2 . (ii) (iii) (3)(a)

(b)

Clearly x = 1, y =  is a solution of this equation. If a | b, then (b/a, 0) is a solution of (Eq 1). If c | b, then (0, b/c) is a solution of (Eq 1). Let α and β be two distinct equivalent roots of g(x). Let (A, B, C, D) ∈ Z4 such that β = (Aα + B)/(Cα + D) with AD−BC = ±1. The third root γ of g(x) is given by γ = −b/a−α−β, hence Q(α, β, γ) = Q(α). Therefore Gal(g(x), Q) ' Z/3Z ' A3 . It follows that Disc(g) is a square in Z. The computation of the discriminant of h(x) may be done by using the formula given in [Lang (1965), Exercise 11, Chap. 5]. It is possible to obtain the same result by using the following general formula for discriminants. Let F (x) be a polynomial of degree n and let a be its leading coefficient, then Disc(F ) = (−1)n(n−1)/2 an−2

n Y

0

F (θi ),

i=1

where θ1 , . . . , θn denote the roots of F (x). Moreover, if F (x) is irreducible over K, then 0

Disc(F ) = (−1)n(n−1)/2 an−2 NQ(θ)/Q (F (θ)). 0

Since g (x) = 6(4x2 − 8x + 3), then Disc(g) = −26 .33 NQ(γ)/Q (4γ 2 − 8γ + 3).

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Set δ = 4γ 2 − 8γ + 3. We compute the characteristic polynomial p(x) of δ. We have δ = 4γ 2 − 8γ + 3 1 γδ = 4γ 2 − 6γ + 2 17 1 2 2 γ δ = 6γ − γ + , 2 2 hence

3 − x p(x) = 1/2 1/2

−8 −(6 + γ) −17/2

4 4 = x3 + 19x2 + 3. 6−γ

It follows that NQ(γ)/Q (δ) = −3 and then Disc(g) = 26 .34 . On the other hand, let u(x) = 6x−7 4x−4 , then it is easy to verify the following identities: 4x − 7 u(2) (x) := u(u(x)) = , 4x − 6

γ 0 = u(γ) and γ 00 = u(2) (γ). Moreover straightforward computations lead to the identity: (4x − 4)3 h(u(x)) = −8h(x). This shows that u(γ) and u(2) (γ) are roots of h(x). Therefore γ, γ 0 , γ 00 are equivalent but here, although the discriminant is a perfect square, we have AD − BC = 4 6= ±1 for u(x) and for u(2) (x). This shows that the converse of (a) is false. (c) Let α, β, γ be the roots of g(x). Since Disc(g) is a square, then Q(α)/Q is Galois, cyclic. Therefore β ∈ Q(α). Let a0 , a1 , a2 ∈ Q such that β = a0 + a1 α + a2 α2 . We distinguish the cases a2 = 0 and a2 6= 0. • a2 6= 0. The equation satisfied by α may be written in the form α3 = b0 + b1 α + b2 α2 . Using this equation, it is easy to show the equivalence of the following propositions (P1 ): There exists (A, B, C, D) ∈ Z4 such that (C, D) 6= (0, 0) and β = (Aα + B)/(Cα + D). (P2 ): The linear system of equations −B + a2 b0 C + a0 D = 0

−A + (a0 + a2 b1 )C + a1 D = 0

(a1 + a2 b2 )C + a2 D = 0,

has a solution (A, B, C, D) ∈ Q4 such that (C, D) 6= (0, 0).

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We will prove (P2 ). The matrix of the system is given by   0 −1 a2 b0 a0 M =  −1 0 a0 + a2 b1 a1  . 0 0 a1 + a2 b2 a2

The determinant of the matrix extracted from M by suppressing its last column is equal to −(a2 b2 + a1 ). • a1 6= −a2 b2 . In this case we may compute A, B, C in terms of D. If we choose D ∈ Q \ {0}, then we find A, B, C ∈ Q such that (A, B, C, D) is a solution of the system. • a1 = −a2 b2 . The third equation of the system shows that a2 D = 0. If a2 = 0, then a1 = 0 and then β = a0 ∈ Q, which is a contradiction. Therefore D = 0 and then solving the first two equations gives A = (a0 + a2 b1 )C and B = a2 b0 C. Therefore β = (a0 +a2 b1 )α+a2 b0 . If we multiply the numerator and denominator of α this fraction by a same integer, we obtain a similar relation with integral coefficients. To end the proof of (c), we must show that AD − BC = (A + D)2 and AD − BC 6= 0. Since g(β) = 0, then g(x) | (Cx + D)3 g(u(x)) Ax+B . It follows that there exists r ∈ Q such in Q[x], where u(x) = Cx+D that (Cx + D)3 g(u(x)) = rg(x). From this identity it follows that u(β) is a root of g(x). We discuss the possible values of u(β). If u(β) = β, we find that u(x) = x, which is a contradiction. Suppose that u(β) = α. Since u(2) (x) = (A2 +BC)x+B(A+D) C(A+D)x+D 2 +BC , then C(A + D) = D2 − A2 = B(A + D) = 0, hence A + D = 0 or B = C = A − D = 0. Clearly, the second possibility is impossible, hence A + D = 0. This implies that u(2) (x) = x. Consider the value of u(γ). If u(γ) = α, then u(2) (γ) = u(α) = β, which is a contradiction. Obviously me may reject the possibility u(γ) = γ. If u(γ) = β, then u(2) (γ) = u(β) = α, which is a contradiction. We conclude that the only possibility for u(β) is u(β) = γ. Therefore u(3) (α) = α. Since u(3) (x) =

(A3 + 2ABC + BCD)x + B(A2 + BC + D2 + AD) , C(A2 + AD + D2 + BC)x + D3 + 2BCD + ABC

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then B(A2 + BC + D2 + AD) = 0, C(A2 + AD + D2 + BC) = 0 2

and

2

(A − D)(A + AD + D + BC) = 0.

If B = C = A − D = 0, then u(2) (x) = x, which is a contradiction. Hence (A + D)2 − (AD − BC) = A2 + BC + D2 + AD = 0.

It follows that u(3) (x) = x and (A+D)2 = (AD −BC). This implies that AD − BC 6= 0. (4) Let β be a root of g(x) distinct from α. According to (3), (c), there exists (A, B, C, D) ∈ Z4 such that β = (Aα + B)/(Cα + D) with AD − BC = (A + D)2 and AD − BC 6= 0. We may suppose that gcd(A, B, C, D) = 1. If AD − BC = ±1, then the result is trivial. If not, we make a finite number of transformations on the generator of the field Q(α). First kind of transformations. Suppose that |AD − BC| ≥ 2 and p | gcd(A, D), where p is a prime number. Since BC ≡ AD (mod p2 ), then p2 | B or p2 | C. If p2 | B, then β/p =

A1 (α/p) + B1 (A/p)(α/p) + (B/p2 ) = , C(α/p) + (D/p) C1 (α/p) + D1

where A1 = A/p,

B1 = B/p2 ,

C1 = C

and D1 = D/p.

We have |A1 D1 − B1 C1 | = |(AD − BC)/p2 | < |AD − BC|,

hence replacing α by α0 = α/p, we reduce AD − BC. If p2 | C, then pβ =

(A/p)(pα) + (B) A1 (pα) + B1 = , 2 (C/p )(pα) + (D/p) C1 (pα) + D1

where A1 = A/p,

B1 = B,

C1 = C/p2

and D1 = D/p.

We have |A1 D1 − B1 C1 | = |(AD − BC)/p2 | < |AD − BC|,

hence replacing α by α0 = pα, we reduce AD − BC. We repeat these transformations of the first kind until |AD − BC| = 1 or |AD − BC| ≥

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2 and gcd(A, D) = 1. If |AD − BC| = 1, then we have found our generating element of Q(α). If |AD − BC| ≥ 2 and gcd(A, D) = 1, we make a second kind transformation, hereafter described. Second kind of transformations. Here we suppose that |AD − BC| ≥ 2 and gcd(A, D) = 1. For any k ∈ Z, we have β+k =

(A + Ck)(α + k) + B + kD − kA − k 2 C A1 α + B1 = , C(α + k) + D − kC C1 α + D1

where A1 = A + Ck, C1 = C

B1 = B + kD − kA − k 2 C,

and D1 = D − kC.

One may verify that A1 D1 − B1 C1 = AD − BC. Let p be a prime number such that p | AD − BC. Since AD − BC = (A + D)2 and gcd(A, D) = 1, then p - A, p - D and p - BC. Choose k ∈ Z such that p | A1 , that is k ≡ −A/C (mod p). Since −A/C ≡ D/C (mod p), then p | D1 . At this stage we have |A1 D1 − B1 C1 | ≥ 2 and gcd(A1 , D1 ) ≡ 0 (mod p), so that we can use the transformations of the first kind to reduce |A1 D1 − B1 C1 |. It is clear that after a finite number of transformations, we reach a stage for which AD − BC| = 1. The composition of all the transformations used in the process will produce a transformation on α of the form T (α) = (uα + v)/w. 6γ−7 . Here (5) We have γ 0 = 4γ−4 AD − BC = 4, A = 6, D = −4, gcd(A, D) = 2, B = −7 and C = 4, hence we can apply a transformation of the first kind on replacing γ 3ˆ γ −7 by γˆ = 2γ. We obtain the following identity: 2γ 0 = 3(2γ)−7 (2γ)−2 = γ ˆ −2 . Here AD −BC = 1 so that we have reached the final primitive element, ˆ ˆ namely 2γ. The polynomial h(x) is given by h(x) = x3 − 6x2 + 9x − 1. (6) Let p be a prime factor of AD − BC. Suppose that p - a. We will show that p | Disc(g), that is g(x) is inseparable over Q. Let α and β be two roots of g(x) related by the identity β = (Aα + B)/(Cα + D) with (A, B, C, D) ∈ Z4 , gcd(A, B, C, D) = 1, AD − BC = (A + D)2 and AD − BC 6= 0. Set u(x) = (Ax + B)/(Cx + D), then u(2) (x) = (A2 x + B2 )/(C2 x + D2 ), where A2 = A2 + BC, B2 = B(A + D), C2 = C(A + D) and D2 = D2 + BC.

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We first claim that (C, D) 6= (0, 0) (mod p) and (C2 , D2 ) 6= (0, 0) (mod p). We prove the claim only for (C, D). The proof for the other is similar. Since g(x) | (Cx + D)3 g(u(x)), then (Cx + D)3 g(u(x)) = (r/s)g(x),

where r, s ∈ Z, gcd(r, s) = 1 and rs 6= 0. We may write this identity in the form  s a(Ax + B)3 + b(Ax + B)2 (Cx + D)  + c(Ax + B)(Cx + D)2 + d(Cx + D)3 = rg(x).

If p | s, then g(x) ≡ 0 (mod p), which is a contradiction. Therefore s 6≡ 0 (mod p). Suppose by contradiction that C ≡ 0 (mod p) and D ≡ 0 (mod p), then A ≡ 0 (mod p) and by the preceding identity, we conclude that saB 3 ≡ rg(x) (mod p). Since none of the integers s, a, B is 0 modulo p, then r 6≡ 0 (mod p). We look at this congruence as an identity in Fp [x]. While saB 3 is a constant, rg(x) is a polynomial of degree 3. Therefore, we have reached a contradiction and our claim is established. From the relations α + u(α) + u(2) (α) = −b/a,

αu(α) + αu(2) (α) + u(α)u(2) (α) = c/a (2)

αu(α)u

and

(α) = −d/a,

we deduce the following identities (Cx + D)(C2 x + D2 )x + (C2 x + D2 )(Ax + B) + (Cx + D)(A2 x + B2 ) = −b/a(Cx + D)(C2 x + D2 ) + λ1 g(x), x(C2 x + D2 )(Ax + B) + x(Cx + D)(A2 x + B2 ) + (Ax + B))(A2 x + B2 ) = (c/a)(Cx + D)(C2 x + D2 ) + λ2 g(x), and x(Ax + B))(A2 x + B2 ) = −(d/a)(Cx + D)(C2 x + D2 ) + λ3 g(x),

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where λ1 , λ2 , λ3 ∈ Q. Indeed λ1 has the formλ1 = u1 /ak , where u1 ∈ Z and k is a non negative integer. To prove this let S = {am , m ∈ N}, then S is a multiplicative subset of Z. Let R = S −1 Z be the ring of fractions of Z for this multiplicative subset. Since a is invertible in R, we can perform the Euclidean division in R[x] of (Cx + D)(C2 x + D2 )x + (C2 x + D2 )(Ax + B) + (Cx + D)(A2 x + B2 ) − b/a(Cx + D)(C2 x + D2 ) by g(x) and obtain a quotient q1 (x) and a remainder r1 (x) both in R[x]. The quotient and the remainder being unique in Q[x], we conclude that r1 (x) = 0 and λ1 ∈ R. The proof for λ2 and λ3 is similar and will be omitted. This result allows us to reduce modulo p the three identities and obtain identities in Fp [x]. Let g¯(x) be the reduction of g(x) modulo p and let ρ be one of its roots in an algebraic closure of Fp . In each of these identities, we substitute ρ for x. This is possible because a 6≡ 0 (mod p), (C, D) 6= (0, 0) (mod p) and (C2 , D2 ) 6= (0, 0) (mod p). For example the first identity becomes: ρ + u(ρ) + u(2) (ρ) = −b/a. This implies that ρ, u(ρ), u(2) (ρ) is the complete list of the roots of g¯(x) in an algebraic closure of Fp . Since AD − BC ≡ 0 (mod p), then the ¯ A¯ B determinant ¯ ¯ is 0 in Fp . It follows that there exist µ1 , µ2 ∈ Fp , CD ¯ B) ¯ + µ2 (C, ¯ D) ¯ = (0, 0). Since (C, ¯ D) ¯ 6= not both 0, such that µ1 (A, ¯ = (µ2 /µ1 )D. ¯ It (0, 0), then µ1 6= 0. Therefore A¯ = (µ2 /µ1 )C¯ and B follows that u(x) = µ2 /µ1 and then u(ρ) = u(2) (ρ) = µ2 /µ1 . This shows that g¯(x) is not separable over Fp . Exercise 5.31. Let p be a prime number, r be a positive integer, q = pr , E = Fq (t) and F = Fq (tq − t). (1) For any α ∈ Fq , let σα be the unique Fq -automorphism of Fq (t) such that σα (t) = t + α. Let G = {σα , α ∈ Fq }. Show that G is a subgroup of AutFq (Fq (t)) isomorphic to (Fq , +). Show that Inv(G) = F . Deduce that E is a Galois extension of F of degree q and Gal(E, F ) = G ' (Z/pZ)r . (2) Determine all of the fields M such that F ⊂ M ⊂ Fq (t) and [Fq (t) : M ] = p. For any of these fields M , show that M = Fq (tp − tαp−1 ) for some α ∈ F∗q .

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(3) Let α ∈ F∗q be the element determined in (2). Show that we may express explicitly tq − t in the form: tq − t =

r X i=1

ai tp

r+1−i

− (tαp−1 )p

r−i




,

(Eq 1)

with ar = 1. (4)(a) For any k ∈ {1, ..., r} show that there exist ( kr ) fields Lk such that Fq (tq − t) ⊂ Lk ⊂ Fq (t) and [Fq (t) : Lk ] = pk . (b) Let α1 , . . . , αk be linearly independent elements of Fq over F1 (t). Let Hk be the subgroup of G generated by σα1 , . . . , σαk . Show Pk that Hk = {σβ , for β i=1 bi αi , bi ∈ Fp }. Deduce that [Fq (t) : Inv(Hk )] = pk . (c) Show that there exists a unique polynomial gk (t) ∈ Fq [t], of the Pk i form gk (t) = i=0 ai tp , where at ∈ Fq , ak = 1, and satisfying the condition Inv(Hk ) = Fq (gk (t)). (5) Suppose that r = 3 ie q = p3 . Find the list of fields L such that Fq (tq − t) ⊂ L ⊂ Fq , by giving for each L a polynomial g(t) ∈ Fq [t] such that L = Fq (g(t)). Draw the diagram of these fields. Solution 5.31. (1) Let φ : Fq → AutFq (Fq (t)) be the map defined by φ(α) = σα . Clearly φ is a morphism of the group (Fq , +) into the group AutFq (Fq (t), ◦). We obviously have Ker φ = {0} and Im φ = G. Therefore φ induces an isomorphism of Fq onto G. Let K be the invariant field of G. For any α ∈ Fq , we have σα (tq − t) = (t + α)q − (t + α) = tq + αq − t − α = tq − t, hence F ⊂ K. Consider the following inclusion: F ⊂ K ⊂ Fq (t). Clearly we have [Fq (t) : F ] = q. By Artin’s Theorem [Lang (1965), Th. 2, Chap. 8.1], Fq (t)/K is Galois with Galois group equal to G and [Fq (t) : K] = |G| = q. Therefore K = F = Fq (tq −t). We conclude that E is a Galois extension of F of degree q and its Galois group is equal to G. Since G is isomorphic to Fq , then G is isomorphic to (Z/pZ)r . (2) By the fundamental theorem of Galois [Lang (1965), Th. 1 and Cor., Chap. 8.1], M is the invariant field of a subgroup H of G of order p. Since G contains r such subgroups, there are exactly r fields M . Let α ∈ F∗q and let σα be the Fq -automorphism of Fq (t) defined in (1), then σα generates a subgroup H of order p of G. Moreover any subgroup of G of order p arises in this way. Let β ∈ F∗q , then β generates the

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same subgroup as α if and only if β = λα with λ ∈ F∗p . We show that Inv(H) = Fq (tp − tαp−1 ). We have σα (tp −tαp−1 ) = (t+α)p −(t+α)αp−1 = tp +αp −tαp−1 −αp = tp −tαp−1 , hence tp − tαp−1 ∈ Inv(H). We have Fq (tp − tαp−1 ) ⊂ Inv(H) ⊂ Fq (t). By Artin’s Theorem [Lang (1965), Th. 2, Chap. 8.1], H = Gal(Fq (t), Inv H), hence p = (H : 1) = [Fq (t) : Inv(H)]. On the other hand, [Fq (t) : Fq (tp − αtp−1 )] = p. We deduce that M = Inv(H) = Fq (tp − tαp−1 ). (3) We may write (Eq 1) in the form: r

tq − t = tp − tp

r−1

(αp−1 )p

p−k+2

−t

+ ak−1 (t + ak (tp

r−k+1

− tp

r−1

+ a2 (tp

pr−k+1

r−k

(α

r−1

− tp

r−2

p−1 pr−k+1

)

(αp−1 )p

r−k

(αp−1 )p

r−2

) + ···

)

) + · · · + ar (tp − tαp−1 ),

hence (1) is equivalent to a2 = (αp−1 )p ar α

ak p−1

= (α

p−1

r−1

= βp

r−k+1

)p

r−1

,

ak−1 ,

for k = 2, . . . , r

and

= ar β = 1,

where β = αp−1 . Using the first r − 1 equation, it is easy to prove r−1 r−2 r−(k−1) by induction that ak = β p +p +···+p . In particular, we have r−1 r−2 ar = β p +p +···+p and then ar β = β p

r−1

+···+p+1

= β (p

r

−1)/(p−1)

= αp

r

−1

=1

and the last equation is satisfied. (4)(a) Since G ' (Z/pZ)r , then any subgroup of G is of order pk with k = 0, 1, . . . , r. The number of subgroups Hk of order pk is equal to ( kr ). By Galois theorem relative to the correspondence between subgroups and subfields, there are exactly ( kr ) fields Lk such that F ⊂ Lk ⊂ Fq (t) and [Fq (t) : Lk ] = pk .

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(b) Let σβ ∈ G, then σβ ∈ Hk ⇔ there exist b1 , . . . , bk ∈ N, σβ = σαb11 ◦ σαb22 ◦ · · · ◦ σαbkk ⇔ there exist b1 , . . . , bk ∈ N, σβ = σb1 α1 +···+bk αk

⇔ there exist b1 , . . . , bk ∈ N, t + β = t + b1 α1 + · · · + bk αk ⇔ there exist b1 , . . . , bk ∈ N, β = b1 α1 + · · · + bk αk .

Since Fq (t) is Galois over Inv Hk , then [Fq (t) : Inv Hk ] = |Hk | = pk . Pk i (c) Let gk (t) = i=0 ai tp ∈ Fq [t] with ak = 1. Then gk (t) ∈ Inv(H) ⇔ σαi (gk (t)) = gk (t) ⇔ ⇔

k X

i

i=0

k X

i

ai (tp + αjp ) =

for j = 1, . . . , k

k X

i

ai (tp )

for j = 1, . . . , k

i=0

i

ai αjp = 0

for j = 1, . . . , k

i=0

⇔ a0 α1 + a1 α1p + · · · + ak−1 α1p

k−1

= −α1p

k

⇔ .................................... k−1

a0 αk + a1 αkp + · · · + ak−1 αkp

k

= −αkp .

Therefore the polynomial gk (t) has the given form and satisfies the condition gk (t) ∈ Inv(Hk ) if and only if the above system has a solution (a0 , a1 , ..., ak−1 ) ∈ (Fq )k . The determinant of this system is given by k−1 α1 αp ... α1p 1 p pk−1 α2 α2 ... α2 D= . .............. ... .................. k−1 α αp ... αp k

k

k

By [Lang (1965), Coro. 2, Chap. 8.5], D 6= 0, hence a0 , a1 , ..., ak−1 are uniquely determined. This implies that there exists one and only one polynomial gk (t) ∈ Fq [t] having the given shape and belonging to Inv(Hk ). (5) We begin with the fields Fq (tq − t) and Fq (t) of degree 1 and q respectively over F . The degree of any other intermediate field L is a divisor of q = p3 , hence equal to p or p2 . (a) Intermediate fields L such that [L : F ] = p. Let α be a primitive element of Fq over Fp (i.e. a root of a monic polynomial f (x) ∈ Fp [x] of

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degree 3). The group G contains 3 subgroups of order p, namely the subgroups generated by σ1 , σα , σα2 respectively. Therefore there are 3 fields L such that Fq (tq − t) ⊂ L ⊂ Fq (t) and [Fq (t) : L] = p, namely L = Fq (tp − t) or L = Fq (tp − tαp−1 ) or L = Fq (tp − tα2(p−1) ). (b) Intermediate fields L such that [L : F ] = p2 . Such fields are invariant fields of subgroups of G of order p2 . There are ( 32 ) = 3 such subgroups namely the subgroups generated by {σ1 , σα }, {σ1 , σα2 } and {σα , σα2 } respectively. • Invariant field of the subgroup generated by {σ1 , σβ } where β = α or β = α2 . We compute the coefficients a1 , a2 ∈ Fq of g(t) = 2 tp + a1 tp + a2 t such that g(t + 1) = g(t) and g(t + β) = g(t). These condition lead to the equations 2

1 + a1 + a2 = 0 and β p + a1 β p + a2 β = 0, that is 2

a1 + a2 = −1, and a1 β p + a2 β = −β p . The determinant of this system of linear equations is equal to 1 1 D = p = β − β p , hence non zero. We deduce that β β 2 −1 1 β p −1 − 1 a1 = p2 /(β − β p ) = − p−1 −β β β −1 = −((β p−1 )p + · · · + β p−1 + 1),

a2 = −1 − a1 = (β p−1 )p + · · · + β p−1 . Therefore 2

L = Fq (tp + a1 tp − (1 + a1 )t) with a1 = −(β p−1 )p − · · · − β p−1 − 1.

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153

• Invariant field of the subgroup generated by {σα , σα2 }. We com2 pute the coefficients b1 , b2 ∈ Fq of g(t) = tp + b1 tp + b2 t such that g(t + α) = g(t) and g(t + α2 ) = g(t). We find the equations: 2 2 b1 αp + b2 α = −αp and b1 α2p + b2 α2 = −α2p . The determinant p α α of this system is given by D = 2p 2 = αp+2 − α2p+1 , hence α α 2 −αp α b1 = 2p2 2 /(αp+2 − α2p+1 ) −α α 2

2

α2p +1 − αp +2 = − 2p+1 α − αp+2 2 p2 +2 α (αp −1 − 1) = − p+2 p−1 α (α − 1)

= −αp

2

−p

((αp−1 )p + · · · + (αp−1 ) + 1)

= −(αp−1 )p [(αp−1 )p + · · · + αp−1 + 1]

= −((αp−1 )2p + · · · + (αp−1 )p+1 + (αp−1 )p )

and

2 p α −αp p+2 − α2p+1 ) b2 = 2p 2 /(α α −α2p 2

2

αp +2p − α2p +p = αp+2 − α2p+1 2 2 α2p +p − αp +2p = α2p+1 − αp+2 2 2 αp +2p (αp −p − 1) = αp+2 (αp−1 − 1)

= αp

2

+p−2

((αp−1 )p−1 + · · · + (αp−1 ) + 1)

= (αp−1 )p+2 ((αp−1 )p−1 + · · · + αp−1 + 1)

= (αp−1 )2p+1 + · · · + (αp−1 )p+3 + (αp−1 )p+2 . We now draw the diagram of intermediate fields.

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α −α 2 2 α2p +p − αp +2p = α2p+1 − αp+2 2 2 αp +2p (αp −p − 1) = αp+2 (αp−1 − 1) 2

p−1 p−1 Theory((α and Applications: =Galois αp +p−2 )p−1 + · · Solved · + (αExercises ) + and 1) Problems

= (α

p−1 p+2

)

((α

p−1 p−1

)

+ ··· + α

p−1

10939-main

+ 1)

= (αp−1 )2p+1 + · · · + (αp−1 )p+3 + (αp−1 )p+2 .

154

We now draw the diagram of intermediate fields.

Galois Theory and Applications: Solved Exercises and Problems

Fq (t)

  Fq tp − tα2(p−1)




Fq (tp − t)

Fq tp − tαp−1

 2  Fq tp + a1 tp + a2 t

 2  Fq tp + c1 tp + c2 t

 2  Fq tp + b1 tp + b2 t

Fq (tq − t)

Here a1 = − c1 = − b1 = −

p X

αp−1

i=0 p  X


i

i=p

α


p−1 i

a2 =

p X

αp−1

i=1

i

,

,

b2 =

α2(p−1)

i=0

2p X

,

c2 =

p  X


i

,

α2(p−1)

i=1

2p+1 X

i=p+2

αp−1


i

i

and

.

Exercise 5.32. Let k and n be positive integers such that 1 ≤ k ≤ n. A group G acting on a set Ω of cardinality n is said to be k-transitive if given two ordered subsets of cardinality k of Ω, {a1 , . . . , ak } and {b1 , . . . , bk }, there exists σ ∈ G such that σ(ai ) = bi for i = 1, . . . , k. Denote by G{a1 ,...,ak } be the subgroup of G whose elements are the σ ∈ G such that σ(ai ) = ai for i = 1, . . . , k. (1) Show that the following propositions are equivalent. (i) G is k-transitive. (ii) There exists a subset {a1 , . . . , ak } of cardinality k of Ω satisfying the condition: for any subset {b1 , . . . , bk } of cardinality k of Ω, there exists σ ∈ G such that σ(ai ) = bi for i = 1, . . . , k.

(2) Let 1 ≤ l < k. Show that the following propositions are equivalent.

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(a) G is k-transitive. (b) G is l-transitive and for any {a1 , . . . , al } ⊂ Ω, G{a1 ,...,al } is (k − l)transitive on Ω \ {a1 , . . . , al }. (c) G is l-transitive and there exists {a1 , . . . , al } ⊂ Ω such that G{a1 ,...,al } is (k − l)-transitive on Ω \ {a1 , . . . , al }.

(3) Let K be a field, f (x) be a separable polynomial of degree n with coefficients in K. Let Ω be the set of roots of f (x) in an algebraic closure of K and G = Gal(f (x), K). Let k be an integer such that 2 ≤ k ≤ n. Show that the following propositions are equivalent.

(d) G is k-transitive. (e) f (x) is irreducible over K and for any integer 2 ≤ l ≤ k and any subset {α1 , . . . , αl−1 } of cardinality l − 1 of Ω, the polynomial Ql−1 fl (x) = f (x)/ i=1 (x − αi ) is irreducible over K(α1 , . . . , αl−1 ). (h) f (x) is irreducible over K and and there exists a subset of Ω, {α1 , . . . , αk−1 } of cardinality k − 1 such that for any integer Ql−1 2 ≤ l ≤ k, the polynomial fl (x) = f (x)/ i=1 (x − αi ) is irreducible over K(α1 , . . . , αl−1 ).

(4) A subset A of Ω is called a non trivial block of G if 2 ≤ |A| < |Ω| and for any σ ∈ G, we have σ(A) = A or σ(A) ∩ A = ∅. Show that if G 6= 1 and G is intransitive, then it has a non trivial block. (5) The group G is said to be imprimitive if there exists a partition of Ω of the form Ω = Ω1 ∪ · · · ∪ Ωr , satisfying the following conditions. • 2 ≤ |Ωi | < |Ω| for i = 1, . . . , r. • For any i ∈ {1, . . . , r}, there exists j ∈ {1, . . . , r} such that σ(Ωi ) = Ωj .

Show that if G is imprimitive, then it has a non trivial block. Show that the converse holds if G is transitive. (6) Let f (x) ∈ K[x] be irreducible and separable. Show that the following propositions are equivalent. (u) G is imprimitive. (v) For any root α of f (x) there exists a field F such that K K(α). (w) There exists a root α of f (x) and a field F such that K F

F K(α)

(7) Let G be an imprimitive group. If G is k-transitive, show that k = 1. Solution 5.32. (1) • (i) ⇒ (ii). Obvious.

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(2)

(3)

(4)

(5)

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• (ii) ⇒ (i). Let {c1 , . . . , ck } and {b1 , . . . , bk } be two subsets of Ω of cardinality k. By (ii), there exist σ1 and σ2 ∈ G such that σ1 (ai ) = ci and σ2 (ai ) = bi for i = 1, . . . , k. Let σ = σ1 σ2−1 . Then σ(bi ) = ci for i = 1, . . . , k. • (a) ⇒ (b). That G is l-transitive is obvious. Fix a subset {a1 , . . . , al } of Ω of cardinality l and let {b1 , . . . , bk−l } and {c1 , . . . , ck−l } be two subsets of Ω \ {a1 , . . . , al }. Consider the subsets of Ω of cardinality k, B = {a1 , . . . , al , b1 , . . . , bk−l } and C = {a1 , . . . , al , b1 , . . . , bk−l }. By (a), there exists σ ∈ G such that σ(ai ) = ai and σ(bj ) = cj for i = 1, . . . , l and j = 1, . . . , k − l. This σ belongs to G{a1 ,...,al } and maps bj onto cj for j = 1, . . . , k − l. • (b) ⇒ (c). Obvious. • (c) ⇒ (a). Let {b1 , . . . , bk } and {c1 , . . . , ck } be two subsets of Ω of cardinality k. Let σ1 and σ2 ∈ G such that σ1 (bi ) = ai and σ2 (ci ) = ai for i = 1, . . . , l. Let τ ∈ G{a1 ,...,al } such that τ (σ1 (bi )) = σ2 (ci ) for i = l + 1, . . . , k. Then σ := σ2−1 τ σ1 maps bi onto ci for i = 1, . . . , k. • (d) ⇒ (e). We use the well known result that a separable polynomial over a given field is irreducible if and only if its Galois group acts transitively on its roots. Set f1 (x) = f (x). Fix l ∈ {2, . . . , k} and {α1 , . . . , αl−1 } be a subset of Ω. Suppose that fl−1 (x) is irreducible over K(α1 , . . . , αl−2 ). To show that fl (x) is irreducible over K(α1 , . . . , αl−1 ), it is sufficient to prove that Gal(fl (x), K(α1 , . . . , αl−1 )) is transitive on Ω \ {α1 , . . . , αl−1 }. This assertion is true by (2). • (e) ⇒ (h). Obvious. • (h) ⇒ (d). Since f (x) is irreducible, then G is transitive. Since f (x)/(x−α1 ) is irreducible over K(α1 ), then Gα1 is transitive on Ω\ {α1 }. It follows, from (2), that G is 2-transitive on Ω. Suppose that G is (l−1)-transitive. Since fl (x) is irreducible over K(α1 , . . . , αl−1 ), then G{α1 ,...,αl−1 } is transitive over Ω \ {α1 , . . . , αl−1 }. By (2), G is l-transitive. Since G 6= 1, there exists α ∈ Ω and σ ∈ G such that σ(α) = α. Let A = {σ(α), σ ∈ G}, then |A| ≥ 2. On the other hand, since G is transitive, there exists θ ∈ Ω such that θ 6∈ A. Thus |A| < |Ω|. Obviously σ(A) = A for any σ ∈ G. Therefore A is a non trivial block of G. Suppose that G is imprimitive and let A = Ω1 . Since for any σ ∈ G, σ(Ω1 ) = Ωj , then σ(A) = A or σ(A) ∩ A = ∅. To prove the reverse

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implication, when G is transitive, let A satisfying one of the preceding conditions. Let A1 = A, . . . , Ar be the distinct values of σ(A) for σ ∈ G. Let b ∈ Ω. Let a ∈ A, since G is transitive, let σ ∈ G such that σ(a) = b, then b ∈ σ(A), thus ∪Ai = Ω. It is easy to verify that Ai ∩ Aj = ∅ for i 6= j. (6) • (u) ⇒ (v). Let Ω = Ω1 ∪ · · · ∪ Ωr be a partition of Ω satisfying the condition of imprimitivity. Let α be a root of f (x), which we may suppose that it belongs to Ω1 . Let L = K(α1 , . . . , αn ), H = Gal(L, K(α)), J = {σ ∈ G, σ(Ω1 ) = Ω1 }. Clearly J is a subgroup of G containing H. Let F = Inv(J), then K ⊂ F ⊂ K(α). It remains to show that these inclusions are strict. Let β ∈ Ω1 such that β 6= α and let σ ∈ G such that σ(α) = β. Then σ ∈ J and σ 6∈ H, thus J 6= H. Let α2 ∈ Ω2 and let σ ∈ G such that σ(α) = α2 , then σ ∈ G \ J. • (v) ⇒ (w). Obvious. ¯ be an algebraic closure of K. Let m = [F : K] • (w) ⇒ (u). Let K ¯ For and σ1 , . . . , σm be the distinct K-embeddings of F into K. any i ∈ {1, . . . , m} let σij , j = 1, . . . , n/m be the extensions of σi to K(α). Let Ωi := {σij (α), j = 1, . . . , n/m}. Since the σij for i ∈ {1, . . . , m} and j ∈ {1, . . . , n/m} represent the list of all ¯ then Ω = ∪n/m Ωi . Since the the K-embeddings of K(α) into K, i=1 inclusions of fields are strict, then 2 ≤ |Ωi | < n for i = 1, . . . , n/m. If Ωi ∪ Ωh 6= ∅, then σij (α) = σhl (α) for some j and l. It follows that σij = σhl , hence i = h, which implies Ωi = Ωh . (7) Suppose that G is k-transitive and by contradiction that k ≥ 2. In particular G is 2-transitive. Let Ω = Ω1 ∪ · · · ∪ Ωr be a partition of Ω satisfying the condition of imprimitivity. Let a1 , b1 ∈ Ω1 and a2 ∈ Ω2 , then there exists σ ∈ G such that σ(a1 ) = a1 and σ(b1 ) = a2 . This is a contradiction since we have σ(Ω1 ) 6= Ω1 and Ω1 ∩ σ(Ω1 ) 6= ∅.

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Exercise 6.1. Let p be a prime number and n be a positive integer. Show that in Z[x], we have (1 + x + · · · + xp−1 )n ≡ (1 − x)n(p−1) (mod p). Solution 6.1. We do computations in Fp (x). We have n  n  (1 − x)p 1 − xp = = (1 − x)n(p−1) , (1 + x + · · · + xp−1 )n = 1−x 1−x hence the result. Exercise 6.2. Let q be a prime power. Let f (x) be a polynomial of degree 3 with coefficients in Fq . (1) If q > 3, show that the equation y 2 = f (x) has at least one solution (x, y) ∈ F2q . (2) Show that the above conclusion does not hold for q = 3. Solution 6.2. (1) If f has a root in Fq , then the claim is obvious, so may suppose that f has no root in Fq . Suppose by contradiction that for any a ∈ Fq , (q−1)/2 f (a) 6∈ F?2 = −1 for any a ∈ Fq , hence f (a)r + 1 = 0 q then f (a) for any a ∈ Fq , where r = (q − 1)/2. It follows that f (x)r + 1 = (xq − x)g(x),

(Eq 1)

rf 0 (x)f (x)r−1 = g 0 (x)(xq − x) − g(x).

(Eq 2)

where g(x) ∈ Fq [x] and deg g = 3r − q = 3r − (2r + 1) = r − 1. Differentiating this equation, we obtain:

159

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Multiply equations (Eq 1) and (Eq 2) by rf 0 (x) and f (x) respectively and obtain equations (10 ) and (20 ) respectively. Subtract (20 ) from (10 ) and obtain: rf 0 (x) = rf 0 (x)(xq − x)g(x) − f (x)g 0 (x)(xq − x) + f (x)g(x), hence rf 0 (x) − f (x)g(x) = (xq − x)(rf 0 (x)g(x) − f (x)g 0 (x)). Equating the degrees of both sides of this equation, we obtain 3 + deg g = r + 2 ≥ q = 2r + 1

that is r ≤ 1 thus q ≤ 3, which is a contradiction. (2) Let f (x) = x3 − x − 1, then for any a ∈ F3 , we have f (a) = −1, so that f (a) is never a square in F3 . Exercise 6.3. Let K be a field of characteristic p > 0 and d be a positive integer. (1) Let Kd = {xd1 + · · · + xdn , n ≥ 1, xi ∈ K}. Show that Kd is a subfield of K. (2) Let K d = {xd , x ∈ K}. Show that Kd is the smallest subfield of K containing K d . (3) Determine Kd for any d ≥ 1 if K = F34 . (4) Suppose that K = Fq . Let F be a subfield of K. Show that there exists d ≥ 1 such that F = Kd . Solution 6.3. (1) Let y = xd1 + · · · + xdn and z = ud1 + · · · + udm be elements of Kd , then clearly y +z and yz ∈ K d . Moreover, we have −1 = p−1 = 1d +· · ·+1d (p − 1 times), hence −1 ∈ Kd and then −y = −1 · y ∈ Kd . Let y ∈ Kd such that y 6= 0, then y −1 = y d−1 · (y −1 )d = y · · · y · (y −1 )d ∈ Kd .

We conclude that Kd is a subfield of K. (2) By (1), Kd is a subfield of K. Let F be a subfield of K containing K d , then it contains any finite sum of elements of K d . Therefore F ⊃ Kd . We conclude that Kd is the smallest subfield of K containing K d . (3) The subfields of K are: F3 , F32 and F34 . Let δ = gcd(d, 34 − 1) = gcd(d, 24 .5). We examine several cases. 4

• δ = 1. In this case K d \{0} is a subgroup of K ∗ of order 3 δ−1 = 80. Therefore, according to (2) and the list of subfields of K, Kd = F34 .

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• δ = 2. Here K d \{0} is a subgroup of K ∗ of order 3 2−1 = 40. Therefore, for the same reason as above, Kd = F34 . 4 • δ = 4. Here K d \{0} is a subgroup of K ∗ of order 3 4−1 = 20, hence Kd = F34 . 4 • δ = 5. Here K d \{0} is a subgroup of K ∗ of order 3 5−1 = 16, hence Kd = F34 . 4 −1 • δ = 10. Here K d \{0} is a subgroup of K ∗ of order 3 10 = 8, hence Kd = F32 . 4 −1 = 5, hence • δ = 16. Here K d \{0} is a subgroup of K ∗ of order 3 16 d Kd = F34 , because if a a subfield F of K contains K , then K d \{0} is a subgroup of F ∗ . In this case we may exclude the possibility F = F32 . 4 • δ = 8. Here K d \{0} has order 3 8−1 = 10, hence Kd = F34 . 4 −1 • δ = 20. Here K d \{0} has order 3 20 = 4, hence Kd = F32 . 34 −1 d • δ = 40. Here K \{0} has order 40 = 2, hence Kd = F3 . • δ = 80. Here K d \{0} has order 1, hence Kd = F3 . q−1

(4) Let e be the order of F ∗ , then F ∗ = {x e , x ∈ F∗q }. Therefore F = Fqd , with d = q−1 e . Since F is a field then F = Kd . Exercise 6.4. Let p be a prime number, m and n be positive integers such that m ≥ 2.

(1) If m ≡ 0 (mod p) or m ≡ 1 (mod p), show that the polynomial f (x) = xm + x + 1 is separable over Fp . n (2) Let g(x) = xp + x + 1 and let g1 (x), . . . , gr (x) be its irreducible factors in Fp [x] of degree d1 , . . . , dr respectively. Show that lcm(d1 , . . . , dr ) = 2n. Deduce that if (p = 2 and n ≥ 3) or p ≥ 3, then g(x) is reducible over Fp . n (3) Let h(x) = xp +1 + x + 1 and let h1 (x), . . . , hs (x) be its irreducible factors in Fp [x] of degree e1 , . . . , es respectively. Show that lcm(e1 , . . . , es ) = 3n. Deduce that if (p = 2 and n 6∈ {1, 3}) or p ≥ 3, then h(x) is reducible over Fp . (4) Show that r ≥ pn /(2n) and s ≥ (pn + 1)/(3n). Solution 6.4. (1) We have f 0 (x) = mxm−1 + 1. If m ≡ 0 (mod p), then f 0 (x) = 1 which implies that f (x) is separable over Fp . If m ≡ 1 (mod p), then f 0 (x) = xm−1 + 1. Suppose that f (x) and f 0 (x) have a common root say α, then αm−1 = −1 and αm = −α − 1. It follows that αm = −α

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and αm = −α −1, which is a contradiction. Therefore f (x) is separable over Fp . (2) Let I be the ideal generated by g(x) and let Ij , for j = 1, . . . , r, be the ideal generated by gj (x) in Fp [x]. Since by (1), g(x) is separable over Fp , we may consider the following sequence of rings isomorphisms ψ

Fp [x]/I −→ defined by

r Y

j=1

ˆ φ

(Fp [x]/Ij ) −→

r Y

j=1

ψ −1

(Fp [x]/Ij ) −→ Fp [x]/I,

ψ(a(x) + I) = (a(x) + I1 , . . . , a(x) + Ir ), ψ −1 is the inverse of ψ and ˆ 1 (x) + I1 , . . . , ar (x) + Ir ) = (a1 (xp ) + I1 , . . . , ar (xp ) + Ir ). φ(a Let φ = ψ −1 ◦ φˆ ◦ ψ, then φ(a(x) + I) = a(xp ) + I. Let k be a positive integer, then we have φk (x + I) = x + I ⇐⇒ φˆk (x + I1 , . . . , x + Ir ) = (x + I1 , . . . , x + Ir ) k

⇐⇒ xp + Ij = x + Ij ⇐⇒ deg gj | k

for

j = 1, . . . , r

for j = 1, . . . , r

⇐⇒ lcm(d1 , . . . , dr ) | k. Let d = lcm(d1 , . . . , dr ). We have n

φn (x + I) = xp + I = −x − 1 + I

and φ2n (x + I) = x + I,

hence by the above equivalences, d | 2n. Since φn (x + I) = −x − 1 + I, d then d 6= n. Suppose that d < n. From the identity xp + I = x + I, d we conclude that xp − x ≡ 0 (mod g(x)), which is a contradiction. Suppose now that n < d < 2n. Then d = 2n0 with n0 | n and 0 < n0 < d n. Set d = n + t, where 0 < t < n. From the identity xp + I = x + I, we conclude that d

x + I = xp + I = xp n

n+t

t

+I t

= (xp )p + I = (−x − 1)p + I t

= −xp − 1 + I.

t

We deduce that xp + x + 1 ≡ 0 (mod g(x)), which is a contradiction. Therefore d = 2n. Suppose that g(x) is irreducible over Fp , then r = 1, d = d1 = pn = 2n. From this it is obvious that p = 2 and n = 2j . We deduce that 2j =

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j + 1. Using arguments from analysis one may show that j ∈ {0, 1} and then n ∈ {1, 2}. It is possible to get the same conclusion by arguing that 2j is equal to the number of subsets of a set, say E, containing j elements. Since the empty set and the sets {e} for e ∈ E are such subsets and their number is equal to j + 1, then j < 2. (3) We use the same proof as in (2). Making the necessary changes, mutadis mutandis, we keep the same notations as in (2). Here we have n

φn (x + I) = xp + I = −1 − 1/x + I,

φ2n (x + I) = −1/(x + 1) + I

and

3n

φ (x + I) = x + I,

hence by the above equivalences, lcm(e1 , . . . , es ) | 3n. Since φn (x+I) = −x − 1 + I, then d 6= n. Suppose that d < n. From the identity d d xp + I = x + I, we conclude that xp − x ≡ 0 (mod h(x)), which is a contradiction. Suppose now that n < d < 2n. Then d = 3n0 with n0 | n and 0 < n0 < n. Set d = n + t, where 0 < t < n. From the identity d xp + I = x + I, we conclude that d

x + I = xp + I = xp n

n+t

+I

t

t

= (xp )p + I = (−1 − 1/x)p + I t

= −1 − (1/x)p + I. t

We deduce that xp +1 +x+1 ≡ 0 (mod h(x)), which is a contradiction. From the computation of φ2n (x + I), it is seen that d 6= 2n. Suppose that 2n < d < 3n. Set d = 2n + u, where 0 < u < n, then d

x + I = xp + I = xp 2n

= (xp )p = u

u

2n+u

+I −1 pu ) +I +I =( x+1

−1 + I. x pu + 1

We deduce that xp +1 +x+1 ≡ 0 (mod h(x)), which is a contradiction. Therefore d = 3n. Suppose that h(x) is irreducible over Fp , then r = 1, d = d1 = pn + 1 = 3n. Using the function defined by F (x) = px − 3x + 1, it is seen that this equality holds only if p = 2, n = 1 or n = 3. (4) Since d = lcm(d1 , . . . , dr ), then di ≤ d for any i ∈ {1, . . . , r}. It follows Pr from (2), that pn = i=1 di ≤ 2nr, thus r ≥ pn /(2n). The proof for s is similar and will be omitted.

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Exercise 6.5. Pm i Let p be a prime number, q = pn and L(x) = i=0 ai xq ∈ Fq [x]. Show that for any α ∈ Fqn , L(TrFqn /Fq (α)) = TrFqn /Fq (L(α)). Solution 6.5. We have L(TrFqn /Fq (α)) = L(α + αq + · · · + αq =

m X i=0

= =

m X

i=0 m X

=

i=0

)

ai (α + αq + · · · + αq i

ai (αq + αq i

ai αq +

i=0

m X

n−1

i+1

m X

i+1

i=0

qi

ai α +

i=0

)q

+ · · · + αq

ai αq

m X

n−1

ai α

qi

i

i+n−1

+ ··· + !q

m X

n−1

ai α q

i+n−1

i=0

+ ··· +

= L(α) + L(α)q + · · · + L(α)q

)

m X i=0

ai α

qi

!qn−1

= TrFqn /Fq (L(α)), hence the result.

Exercise 6.6. Let q be a prime power, n be a positive integer and x ∈ Fqn . Show that x ∈ F2qn if and only if NFqn /Fq (x) ∈ F2q . Solution 6.6. If x = 0, the equivalence is obvious. From now on, we suppose that x 6= 0. Since the norm map from Fqn into Fq is surjective, then the proposition ∗2 (P) For x ∈ F∗qn , x ∈ F∗2 q n if and only if NFqn /Fq (x) ∈ Fq is equivalent to the proposition ∗2 (Q) NFqn /Fq (F∗2 q n ) = Fq , so we prove (Q). In the sequel, we omit the reference to the index Fqn /Fq in the norm and will write N (x) for the norm of x. Obviously we have NFqn /Fq (F∗2 qn ) ⊂ F∗2 . To get equality of these two sets, we show that they have the same q cardinality. It is easy to show that |F∗2 | = (q − 1)/2. For the other set q ? ∗ ?2 ∗2 consider the norm maps N1 : Fqn → Fq and N2 : Fqn → Fq which are morphisms of groups. Since N1 is surjective then | Ker N1 | = (q n −1)/(q−1).

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We have ∗2 ∗2 ∗2 Im N2 = N2 (F∗2 q n ) ' Fq n / Ker N2 = Fq n /(Ker N1 ∩ Fq n ),

hence ∗2 |N (F∗2 q n )| = |N2 (Fq n )|

∗2 = |F∗2 q n |/| Ker N1 ∩ Fq n |

≥ |F∗2 q n |/| Ker N1 | n q − 1 qn − 1 / = 2 q−1

= (q − 1)/2. ∗2 We conclude that N (F∗2 q n ) = Fq .

Exercise 6.7. Let f (x) = x2 + ax + b ∈ Fq [x]. Suppose that f (x) is irreducible over Fq and that a 6= 0 if q is odd. Let α be a root of f (x) in an algebraic closure of Fq . Show that {α, αq } is a basis of Fq (α) over Fq . Solution 6.7. Suppose that λα + µαq = 0, where λ, µ ∈ Fq . Suppose that µ 6= 0, then 2 λ 6= 0 and αq−1 = −λ/µ ∈ F?q , hence α(q−1) = 1. It follows that the order of α in the multiplicative group Fq (α)? divides gcd((q − 1)2 , q 2 − 1). We have ( q−1 if q is even 2 2 gcd((q − 1) , q − 1) = . 2(q − 1) if q is odd If q is odd then α2(q−1) = 1, hence α2 ∈ Fq , which is excluded by our assumptions. If q is even, then αq−1 = 1. Therefore α ∈ Fq which is a contradiction. It follows that µ = 0, then λ = 0, which implies that {α, αq } is a basis of Fq (α) over Fq . Exercise 6.8. Let p be a prime number, q be a positive power of p, n be a positive integer and α ∈ Fqn be an element generating a normal basis of Fqn over Fq . Suppose that there exists a ∈ Fq such that α + a does not generate a normal basis of Fqn over Fq . Show that p - n and a = − TrFqn /Fq (α)/n.

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Solution 6.8. Since α + a does not generate a normal basis of Fqn over Fq , then there exists a positive integer m, 1 ≤ m ≤ n − 1 and a0 , a1 , . . . , am−1 ∈ Fq such that a0 (α + a) + a1 (α + a)q + · · · + am−1 (α + a)q

m−1

+ (α + a)q

m

= 0.

We deduce that

a0 α + a1 αq + · · · + am−1 αq

m−1

m

+ αq + a(a0 + · · · + am−1 + 1) = 0.

We write this equation in the form: a0 α + a1 αq + · · · + am−1 αq

m−1

+ αq

Using the trace map, we obtain

m

= −a(a0 + · · · + am−1 + 1).

a0 TrFqn /Fq (α) + a1 TrFqn /Fq (αq )  m−1  + · · · + am−1 TrFqn /Fq αq  m + TrFqn /Fq αq = −na(a0 + · · · + am−1 + 1).

Since conjugate elements in a given extension have the same trace, then TrFqn /Fq (α)(a0 + · · · + am−1 + 1) = −na(a0 + · · · + am−1 + 1).

Since α generates a normal basis, then a0 + · · · + am−1 + 1 6= 0 and TrFqn /Fq (α) 6= 0. It follows that p - n and a = − TrFqn /Fq (α)/n. Exercise 6.9. Let σ : Fqn → Fqn be the Frobenius automorphism and let f (x) (resp. g(x)) be the characteristic (resp. minimal) polynomial of the Fq -linear map σ. Show that f (x) = g(x) = xn − 1. Solution 6.9. n For any α ∈ Fqn , we have σ n (α) = αq = α, hence σ n = IdFqn . It follows that f (x) = xn −1. It is known that g(x) is monic and g(x) | f (x). Suppose that g(x) 6= xn − 1, then the degree m of g(x) is at most equal to n − 1. Set g(x) = xm + am−1 xm−1 + · · · + a0 , where a0 , . . . , am−1 ∈ Fq . For any α ∈ Fqn , we have g(σ)(α) = 0, hence σ m + am−1 σ m−1 + · · · + a0 IdFqn = 0.

It follows that any α ∈ Fqn is a root of the polynomial m

h(x) = xq + am−1 xq

m−1

+ · · · + a0 x.

This fact is impossible since the number of roots of this polynomial, in an algebraic closure of Fq , is at most equal to q m . We conclude that g(x) = xn − 1.

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Exercise 6.10. (1) Let E be a vector space of finite dimension over the field K and let T : E → E be a linear map. Let v ∈ E and I = {f (x) ∈ K[x], f (T )(v) = 0}. Show that I is a principal ideal of K[x] generated by some monic polynomial which will be denoted by M(T,v) (x). Show that this polynomial divides the characteristic polynomial and the minimal polynomial of T over K. (2) Let p be a prime number, r be a positive integer and q = pr . Let n be a positive integer, α be an element of the field Fqn and σ be the Frobenius automorphism of this field. Show that α generates a normal basis of Fqn over Fq if and only if M(σ,α) (x) = xn − 1. Hint. One may use Exercise 6.9. (3) Write n in the form n = pe t, where e and t are positive integers and e gcd(p, t) = 1. Let xn − 1 = (f1 (x) · · · fr (x))p be the factorization of xn − 1 into a product of irreducible factors over Fq . Let φi (x) = (xn − 1)/fi (x). Show that α generates a normal basis over Fq if and only if φi (σ)(α) 6= 0 for i = 1, . . . , r. (4) If moreover t = 1, show that α generates a normal basis over Fq if and only if TrFqn /Fq (α) 6= 0. Solution 6.10. (1) Clearly I is an ideal of the ring K[x]. Since this ring is principal, then I is principal generated by some polynomial, which may be supposed to be monic. Since the characteristic polynomial and the minimal polynomial of T over K belong to I, then they are divisible by M(T,v) (x). (2) Suppose that M(σ,α) (x) 6= xn −1. By (1) and by Exercise 6.9, xn −1 ∈ I, hence M(σ,α) (x) | xn −1. Therefore the degree m of g(x) := M(σ,α) (x) satisfies the condition m < n. Set g(x) = xm + am−1 xm−1 + · · · + a0 , where a0 , . . . , am−1 ∈ Fq , then σ m + am−1 σ m−1 · · · + a0 IdFqn (α) = 0. Therefore m

αq + am−1 αq

m−1

· · · + a0 α = 0.

We conclude that α does not generate a normal basis of Fqn over Fq .

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Suppose that α does not generate a normal basis of Fqn over Fq , then there exist a positive integer m, m ≤ n − 1 and a0 , . . . , am ∈ Fq , with m am 6= 0 such that a0 α + a1 αq + · · · + am αq = 0. Let g(x) = a0 + a1 x + · · · + am xm ,

then g(σ)(α) = 0. Therefore M(σ,α) (x) | g(x) and then M(σ,α) (x) 6= xn − 1. (3) Since, by Exercise 6.9, M(σ,α) (x) | xn −1, then the equivalence follows from (2). (4) If t = 1, that is e

e

xn − 1 = xp − 1 = (x − 1)p , then r = 1, f1 (x) = x − 1 and

φ1 (x) = (xn − 1)/(x − 1) = xn−1 + · · · + x + 1,

hence by (3), α generates a normal basis over Fq if and only if TrFqn /Fq (α) 6= 0. Exercise 6.11. For any finite family {u1 (X1 , . . . , Xn ), . . . , ur (X1 , . . . , Xn )} of polynomials with coefficients in Fq , we denote by V (u1 , . . . , un ) the subset of Fnq , V (u1 , . . . , un ) = {(x1 , . . . , xn ) ∈ Fnq , ui (x1 , . . . , xn ) = 0, for i = 1, . . . , r}.

Let f1 (X1 , . . . , Xn ), . . . , fr (X1 , . . . , Xn ) ∈ Fq [X1 , . . . , Xn ] and let F (X1 , . . . , Xn ) = (1 − f1q−1 ) · · · (1 − frq−1 ) − 1.

Show that V (f1 , . . . , fr ) = V (F ). Solution 6.11. ~ = (X1 , . . . , Xn ) and ~x = (x1 , . . . , xn ). Let ~x ∈ V (f1 , . . . , fn ), then Set X clearly F (~x) = 0, hence ~x ∈ V (F ). Let ~x ∈ V (F ). Suppose that there exists j ∈ {1, . . . , r} such that fj (~x) 6= 0, then fj (~x)q−1 = 1, hence F (~x) = −1. Therefore we reach a contradiction. We conclude that ~x ∈ V (f1 , . . . , fr ). Exercise 6.12. Let q be a prime power, N = q − 1 and a ∈ Fq . Let ( 1 − XN if a = 0 . δa (X) = PN k − k=1 (X/a) if a 6= 0 ( 1 if x = a (1) Let x ∈ Fq . Show that δa (x) = . 0 if x 6= a

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(2) Let (a11 , . . . , an1 ), . . . , (a1M , . . . , anM ) be M distinct tuples of elements of Fq and let b1 , . . . , bM ∈ Fq . Consider the polynomial f (X1 , . . . , Xn ) =

M X i=1

bi

n Y

δaj (Xj ). i

j=1

Show that f (X1 , . . . , Xn ) ∈ Fq [X1 , . . . , Xn ] and for any k ∈ {1, . . . , M }, f (a1k , . . . , ank ) = bk . Solution 6.12. (1) If a = 0, then δa (X) = 1 − X N , hence δa (0) = 1 and for x 6= 0, PN δa (x) = 1 − xq−1 = 0. If a 6= 0, then δa (X) = − k=1 (X/a)k , hence δa (a) = −

N X

k=1

1k = −(q − 1) = 1

and for x 6= a, δa (x) = −

N X

k=1

y k = −y(y q−1 − 1)/(y − 1),

where y = x/a. Therefore δa (x) = 0. (2) Clearly, f (X1 , . . . , Xn ) ∈ Fq [X1 , . . . , Xn ]. Let k ∈ {1, . . . , M }. To get Qn Qn the result, we show that j=1 δaj (ajk ) = 1 and j=1 δaj (ajk ) = 0 for i k i 6= k. These claims are true by (1) and the proof is complete. Exercise 6.13. Let n be a positive integer, p be a prime number and q = pr , where r is positive integer. Let n ¯ ∈ {1, . . . , q − 1} be the unique integer such that n≡n ¯ (mod q − 1). (1) Let (a, b) ∈ F2q such that a 6= 0. Show that (aX + b)n ≡ (aX + b)n¯

(mod (X q − X)Fq [X]).

(2) Deduce that for any j ∈ {1, . . . , q − 1} we have     X n n ¯ ≡ (mod p). m j j≤m≤n m≡j

(mod q−1)

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Solution 6.13. (1) It is equivalent to prove that the functions f (x) = (ax + b)n and g(x) = (ax + b)n¯ take the same values point wise in Fq . This assertion is trivial for x = −b/a. Set n = (q − 1)m + n ¯ , then for any x 6= −b/a, we have f (x) = (ax + b)(q−1)m (ax + b)n¯ = (ax + b)n¯ = g(x). (2) On one hand, by (1), we have (X + 1)n ≡ (X + 1)n¯

(mod (X q − X)Fq [X]).

On the other hand, we have

   q−1 n X X  n  Xm = 1 + (X + 1)n =  m m=0 j=1

X

j≤m≤n

m≡j

hence

(X +1)n ≡ 1+



q−1 X    j=1

m≡j

X

j≤m≤n

(mod q−1)

(mod q−1)

   n   Xj m 

    n X m , m

(mod (X q −X)Fq [X]).

By equating the coefficients of X j in the right sides of the two congruence equations, we obtain the result. Exercise 6.14. Pq−1 Let a and b ∈ F?q . Show that j=0 aq−1−j bj = 1.

Solution 6.14. If b = a, then

q−1 X

aq−1−j bj =

j=0

If b 6= a, then

q−1 X j=0

aq−1−j bj = bq−1

q−1 X j=0

aq−1 =

X

1 = 0.

q−1 X 1 − (a/b)q = 1. (a/b)j = bq−1 1 − a/b j=0

Exercise 6.15. Let f (x) ∈ Fq [x]. Show that the following propositions are equivalent.

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(i) For any a ∈ Fq , f (x + a) = f (x). Pm (ii) f (x) has the form f (x) = k=0 ck (xq − x)k with ck ∈ Fq .

Solution 6.15.

• (i) ⇒ (ii). The representation of f (x) in base xq −x [Lang (1965), Th. 9, P Chap. 5.5] has the form f (x) = k ck (x)(xq − x)k , where ck (x) ∈ Fq [x] and deg ck < q. From the hypothesis and from the uniqueness of this representation, we conclude that for any k = 0, . . . , m, and any a ∈ Fq , ck (x + a) = ck (x), hence ck (x + a) − ck (x) = 0 for any k and any a. It follows that ck (x + y) = ck (x), which implies ck (x) ∈ Fq . • (ii) ⇒ (i). Easy. Exercise 6.16. Let p be a prime number, q = pr , f (x) ∈ Fq [x] be a monic polynomial of degree d and let E = Fq [x]/(f (x)). Recall that the trace map, Tr : E → Fq , d−1 defined by Tr(α) = α + αq + · · · + αq is linear over Fq . (1) If f is irreducible, show that the trace map is surjective. (2) Suppose that f is reducible over Fq . (a) If f = g k , where g is irreducible over Fq and k ≥ 2, show that there exists e ∈ E \ {0}, such that e2 = 0 and Tr(e) = e. (b) If f = f1e1 · · · frer , where r ≥ 2 and the fi are irreducible over Fq , show that there exists e ∈ E \ {0, 1}, such that e2 = e and Tr(e) = de. (c) Suppose that p - d. Deduce that Fq ⊂ Im(Tr) and that this inclusion is strict. Solution 6.16. (1) It is sufficient to prove that the image of the map is not trivial. By d−1 contradiction suppose that for any α ∈ E, α + αq + · · · + αq = 0, d−1 then the equation x + xq + · · · + xq = 0 has q d solutions, which is impossible. (2)(a) Let e = (g(x))h , where  k/2 if k is even h= , (k + 1)/2 if k is odd then e 6= 0 and e2 = 0. We deduce that for any j ≥ 2, ej = 0, hence Tr(e) = e.

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(b) Let u(x) ∈ Fq [x] such that u(x) ≡ 1

(mod f2e2 (x) · · · frer (x)),

(mod f1e1 (x)) and u(x) ≡ 0

then u2 − u = u(u − 1) ≡ 0

(mod f (x)).

Let e = u(x), then clearly e 6= 0, e 6= 1 and ek = e for any integer k ≥ 1. Therefore Tr(e) = de. (c) Since p - d, then 1/d ∈ Fq . We deduce that 1 = Tr(1/d) ∈ Im(T r), hence Fq ⊂ Im(Tr). In the case (a), we have Tr(e) = e, e 6= 0 and e2 = 0, hence e ∈ / Fq . In the case (b), we have e 6= 0 and e2 = e, hence e is a 0 divisor and then e ∈ / Fq . But in both cases we have proved that e ∈ Im(Tr), hence the inclusion is strict. Exercise 6.17. Show that the polynomial f (x) = xn + x + 1 is reducible over F2 if and only if there exists a non constant polynomial g(x) ∈ F2 [x] of degree at most n − 1 such that f (x) | g(x2 ) − g(x). Solution 6.17. We prove the necessity of the condition. Let f (x) = f1 (x)f2 (x) be a non trivial factorization of f (x) in F2 [x] with gcd(f1 (x), f2 (x)) = 1. Let g(x) ∈ F2 [x] be the unique polynomial of degree k < n (determined by the Chinese remainder theorem) such that g(x) ≡ 0 (mod f1 (x)) and g(x) ≡ 1 (mod f2 (x)). Then we have g(x)2 − g(x) ≡ 0 (mod f (x)), hence g(x2 ) − g(x) ≡ 0

(mod f (x)).

The sufficiency of the condition is obvious. Exercise 6.18. Let m be a positive integer, q be a prime power and a ∈ F∗q . Show that there exists f (x) ∈ Fq [x] monic, irreducible of degree m such that f (0) = a. Solution 6.18. Since the result is obvious if m = 1, then we may suppose that m ≥ 2. Let α be a generator of the group F∗qm . We have NFqm /Fq (α) = ααq · · · αq

m−1

= α1+q+···+q

m−1

= α(q

m

−1)/(q−1)

.

Let c = NFqm /Fq (α), then c ∈ Fq∗ and c is a generator of this group. It follows that there exists k ∈ {0, . . . , q − 2} such that a = ck . Let Fqd be the field generated by αk over Fq , then d|m. Suppose that d < m

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d

then (αk )q −1 = 1, hence α(q −1)k = 1, thus k(q d − 1) ≡ 0 mod q m − 1 contradicting the fact that k ≤ q − 2. We conclude that d = m, that is αk is a primitive element of Fqm over Fq . Clearly (−1)m αk is also a primitive element. Let g(x) = Irr((−1)m αk , Fq ), then deg g = m and we have g(0) = (−1)m NFqm /Fq ((−1)m αk ) 2

= (−1)m (−1)m (NFqm /Fq (α))k = (−1)m

2

+m k

c = ck

= a. Exercise 6.19. Let m, n be positive integers and q be a prime power. Let α ∈ Fqmn be an element generating a normal basis over Fq . (1) Show that TrFqmn /Fqm (α) generates a normal basis of Fqm over Fq . (2) Show that the preceding result no longer holds if the trace is replaced by the norm. Solution 6.19. (1) Let γ = TrFqmn /Fqm (α). Then n−1 n−1 X X mj m j γ= α(q ) = αq . j=0

j=0

Pn−1 qmj+k For k = 0, . . . , m − 1, we have γ = . Suppose that j=0 α Pm−1 qk = 0, where the coefficients ak belong to Fq . Then k=0 ak γ Pm−1 Pn−1 q mj+k a = 0. The left side of this identity is a linear comk=0 j=0 k α qk

i

bination, with coefficients in Fq of the family {αq , i = 0, . . . , mn − 1} i and each αq appears one and only one time. Hence ak = 0 for m−1 k = 0, . . . , m − 1. Therefore, γ, γ q , . . . , γ q are linearly independent over Fq , and then they constitute a normal basis of Fqm over Fq . (2) Let α be a root of f (x) = x4 + x3 + x2 + x + 1 in an algebraic closure of F2 . It is easy to verify the irreducibility of this polynomial over F2 , therefore F2 (α) = F24 . The conjugates of α over F2 are α, α2 , α4 = α3 + α2 + α + 1 and α8 = α3 . It is easy to verify that these elements are linearly independent over F2 , hence they form a normal basis of F24 over F2 . On the other hand, we have NF24 /F22 (α) = αα4 = α5 = 1, hence this norm does not generate a normal basis of F22 over F2 .

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Exercise 6.20. Let p be a prime number, q be a power of p, f (x) = xm − ax − b be an irreducible polynomial with coefficients in Fq and let α be a root of f (x) in an algebraic closure of Fq . (1) Show that for any 1 ≤ k ≤ m − 2, TrFq (α)/Fq (αk ) = 0. (2) Deduce that if g(α) generates a normal basis of Fq (α) over Fq , where g(x) ∈ Fq [x], 1 ≤ deg g ≤ m − 2, then m 6= 0 (mod p) and g(0) 6= 0. Solution 6.20. (1) Let α1 = α, α2 , . . . , αm be the roots of f (x) in an algebraic closure of Fq . For any k ∈ {1, . . . , m} let σk be the elementary symmetric Pm function of the αi of degree k and let sk = i=1 αik . Newton’s formula [Small (1991), Prop. 2.3, Chap. 2] reads for k = 1, . . . , m: sk − σ1 sk−1 + · · · + (−1)k−1 σk−1 s1 + (−1)k kσk = 0.

By induction it is seen that sk = 0 for k = 1, . . . , m − 2. Therefore TrFq (α)/Fq (αk ) = 0. (2) Let g(x) = a0 + a1 x + · · · + am−2 xm−2 be a polynomial of degree at most m − 2 with coefficients in Fq . Suppose that m ≡ 0 (mod p) or g(0) = a0 = 0, then we have rFq (α)/Fq (g(α))

= ma0 + a1rFq (α)/Fq (α) + · · · + am−2rFq (α)/Fq (αm−2 ) m

hence g(α), g(α)q , ..., g(α)q −1 are linearly dependent over Fq . It follows that g(α) does not generate a normal basis of Fq (α) over Fq . Exercise 6.21. Let q be a prime power, E be the set of maps f : Fnq → Fq . Let

I = {g ∈ Fq [x1 , . . . , xn ], g(a1 , . . . , an ) = 0 for any (a1 , . . . , an ) ∈ Fnq }.

(1) Let ψ : Fq [x1 , . . . , xn ] → E be the map such that ψ(g(x1 , . . . , xn )) is the map of Fqn into Fq induced by g. Show that ψ is morphism of Fq -algebras of Fq [x1 , . . . , xn ] onto E. Deduce Fq [x1 , . . . , xn ]/I ' E. (2) Let F1 (x1 , . . . , xn ), . . . , Fr (x1 , . . . , xn ) ∈ Fq [x1 , . . . , xn ], V = {(a1 , . . . , an ) ∈ Fnq , Fi (a1 , . . . , an ) = 0 for any i = 1, . . . , r}

and J = {F ∈ Fq [x1 , . . . , xn ], F (a1 , . . . , an ) = 0, for any (a1 , . . . , an ) ∈ V }. Show that J = I + F1 Fq [x1 , . . . , xn ] + · · · + Fr Fq [x1 , . . . , xn ].

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Solution 6.21. (1) We omit the proof of every thing except that ( ψ is onto. 1 if x = a For any a = (a1 , . . . , an ) ∈ Fnq , let fa (x) = . We first show 0 if not that the set {fa , a ∈ Fnq } is a basis of E. Suppose that λ1 fa(1) + · · · + λm fa(m) = 0,

(6.1)

where λ1 , . . . , λm ∈ Fq and a(1) , . . . , a(m) are distinct elements of Fnq , then for any j ∈ {1, . . . , m}, λj = (λ1 fa(1) + · · · + λm fa(m) )(a(j) ) = 0, hence the set is free over Fq . P Let f ∈ E, then one easily verifies that f = f (a)fa , which proves that the given set generates E over Fq . Now to prove the surjectivity of ψ, it is sufficient to prove that for any a ∈ Fnq , fa is a image of some polynomial, i.e., fa is a polynomial function. Let a = (a1 , . . . , an ) ∈ Fnq and Fa (x1 , . . . , xn ) = (1 − (x1 − a1 )q−1 ) · · · (1 − (xn − an )q−1 ), then clearly ψ(Fa ) = fa . P (2) Obviously we have I + Fi Fq [x1 , . . . , xn ] ⊂ J. Let H ∈ J and let

F (x1 , . . . , xn ) = 1 − (1 − F1 (x1 , . . . , xn )q−1 ) · · · (1 − Fr (x1 , . . . , xn )q−1 ). One verifies that F (a1 , . . . , an ) = P

( 0

1

if (a1 , · · · , an ) ∈ V if not

and F (x1 , . . . , xn ) ∈ Fi Fq [x1 , . . . , xn ]. Let G = H − HF . If (a1 , . . . , an ) ∈ V , then G(a1 , . . . , an ) = 0 and if (a1 , . . . , an ) 6∈ V , then G(a1 , . . . , an ) = H(a1 , . . . , an ) − H(a1 , . . . , an ) = 0, hence G(a1 , . . . , an ) = 0. It follows that G ∈ I and H = G + HF ∈ I +

r X i=1

Fi Fq [x1 , . . . , xn ].

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Exercise 7.1. Let r ≥ 2 be an integer and f (x) be a polynomial of degree n with coefficients in Fqr . Suppose that f (Fq ) ⊆ Fq . (1) If n ≤ q − 1, show that f (x) ∈ Fq [x]. (2) If n ≥ q, let g(x) ∈ Fqr [x] be the unique polynomial such that deg g ≤ q − 1 and f (x) ≡ g(x) (mod xq − x). Show that g(x) ∈ Fq [x]. Solution 7.1. (1) Set f (x) = a0 + a1 x + · · · + aq−1 xq−1 , where ai ∈ Fqr for i = 0, . . . , q − 1 and let ξ0 , . . . , ξq−1 be the distinct elements of Fq with ξ0 = 0. We may write the equations f (ξi ) = a0 + a1 ξi + · · · + aq−1 ξiq−1 in the form     f (ξ0 ) a0  f (ξ1 )   a1       ···  = A ··· , f (ξq−1 ) aq−1 where   1 0 ··· 0 1 ξ1 · · · ξ1q−1  . A= · · · ··· ··· ···  1

We have

ξq−1

···

1 Det A = ξ1 · · · ξq−1 · · · 1

q−1 ξq−1

··· ··· ···

ξ1q−2 · · · , q−2 ξq−1

which is non zero, hence A is invertible and A−1 has its coefficients in Fq . It follows that ai ∈ Fq for i = 0, . . . , q − 1. 177

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(2) Since f (x) ≡ g(x) (mod xq − x), let h(x) ∈ Fqr [x] such that f (x) = g(x) + (xq − x)h(x).

We have g(Fq ) = f (Fq ) ⊆ Fq . Since deg g ≤ q − 1, then we may apply (1) and conclude that g(x) ∈ Fq [x]. Exercise 7.2. Let q be a prime power, H be a subgroup of F?q of order d and s = (q −1)/d. Let f (x) be a polynomial with coefficients in Fq . Show that f (H) ⊂ H if and only if there exist a non negative integer r and g(x) ∈ Fq [x] such that f (x) ≡ xr g(x)s (mod xd − 1). Solution 7.2. • Sufficiency of the condition. Notice that H = {bs , b ∈ Fq }. Let a ∈ H. Since ar ∈ H and g(a)s ∈ H then f (a) = ar g(a)s ∈ H, thus f (H) ⊂ H. • Necessity of the condition. We recall Lagrange’s interpolation theorem [Bourbaki (1950), Application: Formule d’interpollation de Lagrange, Chap. 4.2] or [Ayad (1997), Exercice 1.14]. Let K be a field, a1 , . . . , an be distinct elements of K and b1 , . . . , bn be elements of K distinct or not. Then there exists one and only one polynomial h(x) with coefficients in K of degree at most n − 1 such that h(ai ) = bi for i = 1, . . . , n. This polynomial may be written explicitly. Moreover if no condition on the degree of h(x), is prescribed, then any two polynomials h1 (x) and h2 (x) such that h1 (ai ) = h2 (ai ) = bi , satisfy the condition ! n Y h2 (x) ≡ h1 (x) mod (x − ai ) . i=1

To get this result it is equivalent to prove that h2 (x) ≡ h1 (x) (mod x − ai ) for i = 1, . . . , n. But this last claim is true since h2 (ai ) − h1 (ai ) = 0 for i = 1, . . . , n. This theorem of Lagrange in mind, write f (x) in the form f (x) = xr f1 (x), where r is a non negative integer and f1 (x) is a polynomial with coefficients in Fq such that f1 (0) 6= 0. For any a ∈ H, we have f (a) = ar f1 (a). since ar and f (a) ∈ H, then f1 (a) ∈ H. According to the remark made above on H, there exists ba ∈ Fq such that f1 (a) = bsa . Let g(x) ∈ Fq [x] such that g(a) = ba for any a ∈ H. Since, for any a ∈ H, f (x) and xr g(x)s take the same value on a, then by Lagrange’s theorem, these polynomials are congruent modulo Q d a∈H (x − a) that is modulo x − 1.

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Exercise 7.3. Let q be a prime power and n ≥ 1 be an integer. Let f (X) ∈ Fq [X] such that deg f < q n and f (X) permutes Fqn . Let σ be the permutation induced by f (X) and let g(X) ∈ Fqn [X] be the unique polynomial of degree at most q n − 1 such that g(x) = σ −1 (x) for any x ∈ Fqn . Show that g(X) ∈ Fq [X]. Solution 7.3. • First proof. Let d be the order of σ, then for any x ∈ Fqn , we have σ d (x) = f d (x) = x hence σ −1 (x) = f d−1 (x) for any x ∈ Fqn . It follows n that g(X) ≡ f d−1 (X) (mod X q − X) and the proof is complete. • Second proof. Since, for any x ∈ Fqn we have f (g(x)) = x, then there exists h(X) ∈ Fqn [X] such that n

f (g(x)) = X + (X q − X)h(X).

For any Fq -automorphism σ of Fqn , denote by uσ (X) the unique polynomial obtained from u(x) by applying σ to the coefficients. Operating σ on the above identity leads to: n

f σ (g σ (X)) = X + (X q − X)hσ (X), hence n

f (g σ (X)) = X + (X q − X)hσ (X).

This shows that g σ (X), as a permutation of Fqn , coincides with σ −1 . Since g(X) and g σ (X) have their degrees smaller than q n , then g σ (X) = g(X). It follows that g(X) ∈ Fq [X]. Exercise 7.4. Let n and m be positive integers, a and b ∈ F?q and f (X) = aX n + bX m . Let d = gcd(n − m, q − 1). Suppose that (−b/a)(q−1)/d = 1. Show that f (X) does not permute Fq . Solution 7.4. According to Bezout’s Identity, there exist u and v ∈ Z such that u(n − m) + v(q − 1) = d.

Let ξ be a generator of F?q , then ξ d generates the unique subgroup of order (q − 1)/d of F?q and −b/a = ξ id for some positive integer i. We deduce that
n−m −b/a = ξ id = ξ iu(n−m) = ξ iu .

This shows that f (ξ iu ) = 0 which means that the map induced by f (X) is not injective.

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Exercise 7.5. Let q be a prime power. Recall that if σ : Fq → Fq is a map, then the unique polynomial f (x) with coefficients in Fq and of degree at most q − 1 such that f (a) = σ(a) for any a ∈ Fq is given by X σ(c)(1 − (x − c)q−1 ). (Eq 1) f (x) = c∈Fq

(1) Let a and b be distinct elements of Fq and τ the transposition (ab). By using (Eq 1), show that the unique polynomial f (x) ∈ Fq representing τ with deg f (x) ≤ q − 1 is given by f (x) = x + (a − b)(x − a)q−1 + (b − a)(x − b)q−1 .

Show that deg f (x) = q − 2. (2) Let a, b and c be distinct elements of Fq and ρ the 3-cycle (abc). Show that the unique polynomial g(x) ∈ Fq representing ρ with deg g(x) ≤ q − 1 is given by g(x) = x + (a − b)(x − a)q−1 + (b − c)(x − b)q−1 + (c − a)(x − c)q−1 .

Show that deg g(x) = q − 2 if and only if −3 is not a square in Fq . Solution 7.5.

(1) Since f (x) represents τ , then f (x)−x represents τ −IdFq . Using (Eq 1), we get f (x) − x = (a − b)(x − a)q−1 + (b − a)(x − b)q−1 , hence f (x) = x + (a − b)(x − a)q−1 + (b − a)(x − b)q−1 .

It is clear from this identity that the coefficient of xq−1 is zero. This could be also proved by arguing that f (x) is a permutation polynomial and thus its degree is not a divisor of q − 1. The coefficient α of xq−2 is given by α = −a(a − b)(q − 1) + −b(b − a)(q − 1),

that is α = (a − b)2 , hence non zero. It follows that deg f (x) = q − 2. (2) Here g(x) − x represents ρ − IdFq . By (Eq 1), we obtain

g(x) − x = (a − b)(x − a)q−1 + (b − c)(x − b)q−1 + (c − a)(x − c)q−1 ,

hence g(x) = x + (a − b)(x − a)q−1 + (b − c)(x − b)q−1 + (c − a)(x − c)q−1 .

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As in (1), the degree of g(x) is not equal to q − 1. The coefficient α of xq−2 is given by α = a(a − b) + b(b − c) + c(c − a).

We have α = 0 if and only if a is a root of the polynomial x2 − (b + c)x + b2 + c2 − bc. The discriminant of this polynomial is given by D = −3(b − c)2 . Thus α 6= 0 if and only if −3 is not a square in Fq . Exercise 7.6. Let K be a field of characteristic p ≥ 0, f (x) and g(y) be non-constant polynomials with coefficients in K. If p > 0, suppose that f (x) 6∈ K[xp ] or g(x) 6∈ K[xp ]. (1) Show that f (x) − g(y) is square free. (2) Suppose that K = Fq and f (x) is a permutation polynomial over Fq . Show that (f (x) − f (y))/(x − y) has no factor of the form ax + by + c in Fq [x, y] with a 6= 0 or b 6= 0. Solution 7.6. (1) Suppose that h2 (x, y)|f (x)−g(y), where h(x, y) is a non-constant polynomial with coefficient in K, then h(x, y)|f 0 (x) and h(x, y)|g 0 (y). From the assumptions, one of the polynomials f 0 (x) and g 0 (y) is non zero. Without loss of generality, suppose that f 0 (x) 6= 0, then h(x, y) ∈ K[x]. It follows that h(x, y) divides the content in K[x] of f (x) − g(y), which is equal to 1, hence a contradiction. We conclude that f (x) − g(y) is square free. (2) Suppose that (ax+by+c) | (f (x)−f (y))/(x−y), with a, b, c ∈ K, a 6= 0 or b 6= 0, then by (1), ax + by + c 6= λ(x − y) for any λ ∈ K. Suppose that b 6= 0, then for any x ∈ Fq , we have f (x) − f ((−c − ax)/b) = 0, which implies that f (x) is not a permutation polynomial. Exercise 7.7. Let σ : F2n → F2n be a map such that σ(0) = σ(e) = 0 for some e ∈ F?2n and σ(i) 6= σ(j) for all i, j ∈ F?2n with i 6= j. Let f (X) ∈ F2n [X] be the unique polynomial of degree at most 2n − 1 such that f (x) = σ(x) for any x ∈ F2n . Show that deg f = 2n − 1. Solution 7.7. Since the map σ is not injective, then it is not surjective, hence there exists α 6∈ σ(F2n ). Let P (X) = f (X) + α[(X + e)2

n

−1

+ 1],

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then clearly P (e) = α and for any x 6= e, P (x) = f (x), hence P (X) is a permutation polynomial. Obviously, by the very definition of P (X), deg P ≤ 2n − 1. Since it is a permutation polynomial its degree must be at most 2n − 2. Since α 6= 0, then the leading term of the polynomial f (X) is n αX 2 −1 and its degree is 2n − 1. Second proof. We will use the following formula, which is Lagrange interpolation formula [Small (1991), Remarks, Chap. 2] X σ(c)(1 − (x − c)q−1 ). f (x) = c∈Fq

We obtain: X

f (x) =

c∈Fq ,c6∈{0,e}

σ(c)(1 − (x − c)q−1 ).

P The coefficient λ of x is given by λ = c6∈{0,e} σ(c). The assumptions shows that there exists d ∈ F?2n such that d 6= σ(c) for any c ∈ F2n . We may write λ in the form λ = d1 + d2 + · · · + dq−2 , where the di are distinct and di 6∈ {0, d}. There exists i0 such that di0 = −d. Since the opposite of any di for i 6= i0 is equal to some dj , then it follows that λ = di0 6= 0, that is the coefficient xq−1 is nonzero. q−1

Exercise 7.8. Let p ≥ 3 be a prime number and σ = (01) be the transposition of {0, . . . , p − 1}. Let f (X) be the unique polynomial with coefficients in Fp of degree at most p − 1 such that for any x ∈ Fp , f (x) = σ(x). Show that f (X) = X +

p−1 Y

(X − k).

k=2

Solution 7.8. We have f (x) − x = σ(x) − x = Since deg f ≤ p − 1, then

   1

−1   0

f (X) − X = (aX + b)

p−1 Y

k=2

if x = 0 if x = 1 if x 6= 0, 1

(X − k),

.

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where a, b ∈ Fp . We have

1 = f (0) = (−1)p−2 (p − 1)!b and

−1 = f (1) − 1 = (a + b)

p−1 Y

(k − 1) = −(a + b)(p − 2)!,

k=2

hence b = 1 and a = 0. Second proof. We will use the following formula, which is Lagrange interpolation formula [Small (1991), Remarks, Chap. 2] X (σ(c) − c)(1 − (X − c)p−1 ). f (X) − X = c∈Fp

We have

σ(c) − c = hence

   1

if c = 0

−1   0

if c = 1

,

if c 6= 0, 1

f (X) − X = (X − 1)p−1 − X p−1 . The degree of this polynomial is equal to p−2, and for any a ∈ {2, . . . , p−1}, Qp−1 we have f (a) − a = 0. It follows that f (X) − X = k=2 (X − k) and then f (X) = X +

p−1 Y

(X − k).

k=2

Exercise 7.9. Let p be a prime number and n ≥ 2 be an integer. Let f : Z/pn Z → Z/pn Z be the map such that f (0) = 0 and f (x) = 1 for x 6= 0. Show that there exists no polynomial F (X) ∈ Z/pn Z[X] such that f (x) = F (x) for any x ∈ Z/pn Z. Solution 7.9. Suppose the contrary. Since f (0) = 0, then F (x) = am xm + · · · + a1 x, where ai ∈ Z/pn Z. Let Fˆ (x) ∈ Z[x] be a lift of F (x). We have Fˆ (p) ≡ 0 (mod p), which contradicts the assumption F (p) = 1. Exercise 7.10. Let n ≥ 2 be an integer. Show that n is prime if and only if there exists a polynomial f (x) with integral coefficients such that f (0) ≡ 1 (mod n), f (1) ≡ 0 (mod n) and f (k) ≡ k (mod n) for k = 2, . . . , n−1, i.e. f induces the permutation (01) on the set Z/nZ.

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Solution 7.10. The necessity of the condition is well known. We prove the sufficiency. By contradiction, suppose that n is not prime and let p be a prime factor of n. Let f (x) ∈ Z[x] satisfying the prescribed conditions. Then f (p) = a0 + · · · + am pm ≡ 0 (mod p), hence p | a0 = f (0) ≡ 1 (mod p), which is a contradiction. Exercise 7.11. Let p be a prime number and let f (x) ∈ Z[x] be a polynomial permuting the elements of Z/pZ. Show that f 0 (x) permutes the elements of Z/pZ or there exists a ∈ Z such that f (x) + a(xp − x) permutes the elements of Z/p2 Z. Solution 7.11. Suppose that f 0 (x) does not permute the elements of Z/pZ, then there 0 exists a ∈ {0, . . . , p − 1} such that for any x ∈ Z f (x) − a 6= 0 (mod p). Let g(x) = f (x) + a(xp − x). It is clear that g(x) induces the same function over Z/pZ as f (x), hence it permutes the elements of this ring. Notice that g 0 (x) = f 0 (x) − a. Let b ∈ Z and consider the equation g(y) ≡ b (mod p2 ). Let x0 such that g(x0 ) ≡ b (mod p). We show that there exists a solution of the form y0 = x0 + λp, where λ is an integer to be determined. We have g(y0 ) ≡ g(x0 ) + λpg 0 (x0 ) ≡ b (mod p2 )

≡ b + µp + λpg 0 (x0 )

(mod p2 ) (mod p2 ).

We deduce that µ + λg 0 (x0 ) ≡ 0 (mod p) and this equation determines λ and then y0 . Exercise 7.12. Let p be a prime number, e be a positive integer and q = pe . Suppose that q ≥ 3. (1) Let a ∈ F?q and fa (x) = −a2

h

(x − a)q−2 + a−1

q−2

iq−2 −a .

Show that the polynomial fa (x) induces the transposition (0 a) of Fq . (2) Deduce that any permutation of Fq is the composition of permutations induced by xq−2 and linear polynomials with coefficients in Fq .

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Solution 7.12. (1) We have fa (0) = −a2

h

(−a)q−2 + a−1

q−2

iq−2 −a

= −a2 (−a)q−2 = −a2 (−a−1 ) = a, fa (a) = −a2

h

a−1

q−2

iq−2 −a =0

and for b ∈ Fq , b 6= 0, b 6= a, q−2 h iq−2 −a fa (b) = −a2 (b − a)q−2 + a−1 h q−2 iq−2 = −a2 (b − a)−1 + a−1 −a h q−2 iq−2 b = −a2 −a a(b − a) iq−2 h a(b − a) −a = −a2 b h −a2 iq−2 = b. = −a2 b This implies that fa (x) as a (polynomial) function of Fq into itself represents the transposition (0 a). (2) Clearly fa (x) as a permutation is the composition of permutations induced by xq−2 and by permutations induced by linear polynomials. Since the transpositions of the form (0 a), when a runs in F?q generates the set of permutations of Fq , then the permutation induced by xq−2 together with those generated by linear polynomials generate the full symmetric group operating on Fq . Exercise 7.13. Let f (x) be a non constant polynomial with integral coefficients, p be a prime number and e be a positive integer. Let a ∈ Z such that f (a) ≡ 0 (mod pe ). (1) Show that the number of solutions b distinct modulo pe+1 of the equation f (x) ≡ 0 (mod pe+1 ) is equal to (i) 0 if f 0 (a) ≡ 0 (mod p) and f (a) 6≡ 0 (mod pe+1 ). (ii) 1 if f 0 (a) 6≡ 0 (mod p). (iii) p if f 0 (a) ≡ 0 (mod p) and f (a) ≡ 0 (mod pe+1 ).

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(2) Let n ≥ 2 be an integer. Deduce that f (x) permutes the elements of Z/pn Z if and only if f (x) permutes the elements of Z/pZ and f 0 (a) 6≡ 0 (mod p) for any a ∈ Z. Solution 7.13. (1) Let b ∈ Z such that b = a + cpe with c ∈ Z, then the equation f (x) ≡ 0 (mod pe+1 ) is equivalent to f (a)/pe + cf 0 (a) ≡ 0 (mod p). If f 0 (a) 6≡ 0 (mod p), then this condition determines c in one and only one way. Therefore (ii) is proved. If f 0 (a) ≡ 0 (mod p), then the same equation is equivalent to f (a)/pe ≡ 0 (mod p), i.e. f (a) ≡ 0 (mod pe+1 ). Thus (i) and (iii) are proved. (2) Suppose that f (x) permutes the elements of Z/pn Z. Let b ∈ Z. Consider the equation f (x) ≡ b (mod p). Let a ∈ Z such that f (a) ≡ b (mod pn ), then f (a) ≡ b (mod p), hence the equation modulo p has a solution and f (x) permutes the elements of Z/pZ. Let a ∈ Z, b = f (a) and g(x) = f (x) − b, then a is a solution of the equations g(x) ≡ 0 (mod p) and g(x) ≡ 0 (mod pn ). Applying (1) to g(x), we conclude that we have the case (ii), hence f 0 (a) = g 0 (a) 6≡ 0 (mod p). We prove the converse. Let b ∈ Z. Consider the equation f (x) ≡ b (mod pn ). Let a ∈ Z such that f (a) ≡ b (mod p). Applying (ii) to g(x) = f (x) − b, we obtain that there exists a2 ∈ Z such that a2 ≡ a (mod p) and f (a2 ) ≡ b (mod p2 ). By induction, we may prove that there exists an ∈ Z such that f (an ) ≡ b (mod pn ), hence f permutes Z/pn Z. Exercise 7.14. Let q be a prime power and n be a positive integer. Let Gn = {f (x) ∈ Fqn [x], deg f < q n , f permutes Fqn } and let Hn = {f (x) ∈ Fq [x], deg f < q n , f ∈ Gn }. (1) Show that Hn is a subgroup of Gn . (2) Let φ : Fqn → Fqn , be the Frobenius automorphism and let f ∈ Gn . Show that f ∈ Hn if and only if f ◦ φ = φ ◦ f . (3) Let d be a positive divisor of n and let Md = {α ∈ Fqn , deg α (over Fq ) = d}. Let α0 ∈ Md . Show that Md = {f (α0 ), f ∈ Hn }.

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Solution 7.14. (1) Hn 6= ∅ since x belongs to this set. Obviously the composition of two elements of Hn belongs to Hn . According to Exercise 7.3, the symmetric of an element of Hn is an element of Hn . Therefore Hn is a subgroup of Gn . (2) Suppose that f ∈ Hn , then φ(f (x)) = (f (x))q = f (xq ) = f (φ(x)). P Conversely let f (x) = ai xi ∈ Gn . Suppose that f (φ(x)) = φ(f (x)), P P q q i q q then f (x ) = f (x) for any x ∈ Fqn , i.e. ai (xq )i = ai (x ) for P P any x ∈ Fqn . Therefore ai y i = aqi y i for any y ∈ Fqn . Now the P q i polynomials f (y) and g(y) = ai y both have their degrees smaller than q n and are equal as functions over Fqn , hence they are equal as polynomials, which implies f ∈ Hn . (3) Let Nd = {f (α0 ), f ∈ Hn }. We first show that Nd ⊂ Md , that is f (α0 ) ∈ Md for any f ∈ Hn . For this purpose we prove that for any field F such that Fq ⊂ F ⊂ Fqn , and any f ∈ Hn , we have f (F ) = F . Let f and F be such objects and let γ ∈ F , then since f (x) ∈ Fq [x], f (γ) ∈ F . Therefore f (F ) ⊂ F . Since F is finite and f is injective, then f (F ) = F . We have Fqd = Md ∪ e|d Fqe , hence f (α0 ) ∈ Md , thus e deg v. We have R(x) = θ(α) = hence

(a0 un + a1 un−1 v + · · · + an v n )v m , (b0 um + b1 um−1 v + · · · + bm v m )v n

deg R = n deg u + mdegv − m deg u − n deg v = (n − m)(deg u − deg v) > 0,

where the degree of a rational function over K is the difference between the degree of the numerator and the degree of the denominator. We deduce that n > m and R(x) =

a0 un + a1 un−1 v + · · · + an v n . (b0 um + b1 um−1 v + · · · + bm v m )v n−m

It follows that any prime factor of v(x) divides u(x), thus α(x) ∈ K[x].

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? deg u < deg v. Let µ(x) = θ( x1 ). Then R(x) = θ(α(x)) = µ



1 α(x)



=µ



u(x) v(x)



215

.

Therefore we are moved to the preceding case. We conclude that v(x) 1 u(x) = α(x) ∈ K[x]. ? deg u = deg v. Let c be the quotient of the leading coefficient of u by that of v. We have deg(u − cv) < deg(u) = deg v and R(x) = γ(α − c) = γ



u − cv v



,

where γ(x) = θ(x + c). In this case, we are moved to the second case. We conclude that 1 v(x) = ∈ K[x]. u(x) − cv(x) α(x) − c (2) • (a) ⇒ (b). Since f (x) = g(h(x)), then K(f ) ⊂ K(h) ⊂ K(x). Since [K(x) : K(f )] = deg f, [K(x) : K(h)] = deg h and deg h < deg f , then F = K(h) is a strict intermediate field between K(f ) and K(x). • (b) ⇒ (a). Let F be a field such that K(f ) ( F ( K(x). Since F contains the non-constant polynomial f (x) ∈ K[x], then there exists h(x) ∈ K[x], non-constant, such that F = K(h). We have f (x) = g(h(x)), where g(x) = u(x) v(x) ∈ K(x) and gcd(u(x), v(x)) = 1, hence f (x) = u(h(x)) v(h(x)) . Since f (x) is a polynomial then v(h(x)) | u(h(x)). By Bezout’s identity there exist A(x), B(x) ∈ K[x] such that A(x)u(x) + B(x)v(x) = 1. We deduce that A(h(x))u(h(x)) + B(h(x))V (h(x)) = 1, hence v(h(x)) is a constant. Therefore v(x) is constant (recall that h(x) is not constant). It follows that g(x) ∈ K[x] and f (x) = g(h(x)). The assertion on the degrees of g and h is easy to prove and will be omitted. (3)(u) Obvious.

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(v) For any j ∈ {1, . . . , n} we have 0 = f (α1 ) − g(βj ) = A(α1 , βj )B(α1 , βj ). From (1), it follows that there exists j ∈ {1, . . . , n} such that A(α1 , βj ) = 0. We deduce that [K(α, βj ) : K(α1 )] < n = [K(βj ) : K(x)], hence K(βj ) and K(α1 , . . . , αm ) are not linearly disjoint over K(x). (w) Since K(α1 , . . . , αm ) is a Galois extension of K(x), by [Lang (1965), Th. 4, Chap. 8.1], it is linearly disjoint over K(x) from any extension F of K(x) such that F ∩ K(α1 , . . . , αm ) = K(x). We deduce that K(x) ( K(α1 , . . . , αm ) ∩ K(βj ). We now have K(x) ( K(α1 , . . . , αm ) ∩ K(βj ) ⊂ K(βj ), x = g(βj ) and g(x) is indecomposable over K, then by (1), there is no strict intermediate field between K(x) and K(βj ). Therefore K(α1 , . . . , αm ) ∩ K(βj ) = K(βj ). This implies K(βj ) ⊂ K(α1 , . . . , αm ). By conjugating, we conclude that K(β1 , . . . , βn ) ⊂ K(α1 , . . . , αm ). By symmetry, we obtain the reverse inclusion. Therefore f (y) − x and g(z) − x have the same splitting field over K(x). Exercise 8.11. Let K be a field, f (x) and g(x) be non constant polynomials with coefficients in K. (1) Show that K(f ) and K(g) are not linearly disjoint over K. (2) Suppose that K = C. (a) Show that C(f ) ∩ C(g) = C if and only if C[f ] ∩ C[g] = C. (b) Show that C(x2 +x)∩C(x2 ) = C and C(xn )∩C(xm ) = C(xk ), where m and n are given positive integers and k is their least common multiple. (3) Give an example of two fields E and F contained in some field Ω such that there exists α ∈ Ω, algebraic over E and over F but transcendental over E ∩ F .

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Solution 8.11. (1) Since f (x) and g(x) are elements of K(x) and Trdeg K(x)/K = 1, then f (x) and g(x) are algebraically dependent over K. It follows that there exist a0 (g), . . . , am (g) ∈ K[g], with am (g) 6= 0 such that a0 (g) + a1 (g)f + · · · + am (g)f m = 0. Therefore 1, f, . . . , f m are linearly dependent over K(g). Since f is transcendental over K, then these elements are linearly independent over K, hence K(f ) and K(g) are not linearly disjoint over K. (2)(a) The necessity of the condition is obvious. We prove its sufficiency. Suppose by contradiction that there exists h(x) ∈ (C(f ) ∩ C(g)) \ C. Write h(x) in the forms h(x) = A(f (x))/B((f (x)) = C(g(x))/D(g(x)), where A(x), B(x), C(x) and D(x) are polynomials with complex coefficients such that gcd(A(x), B(x)) = gcd(C(x), D(x)) = 1. We may suppose that h(x) is chosen in order to satisfy the condition deg B + deg D is minimal. Then A(f (x))D(g(x)) = B((f (x))C(g(x)). Since gcd(A(f (x)), B(f (x))) = gcd(C(g(x)), Dg((x))) = 1, then A(f (x)) | C(g(x)) and C(g(x)) | A(f (x)), hence there exists λ ∈ C such that C(g(x)) = λA(f (x)). It follows that D(g(x)) = λB(f (x)). By differentiating the preceding identities, we get C 0 (g(x))g 0 (x) = λA0 (f (x))f 0 (x). It follows that D0 (g(x))g 0 (x) = λB 0 (f (x))f 0 (x). If B 0 (f (x)) = 0, then B 0 (x) = 0 and D0 (g(x)) = D0 (x) = 0, so that h(x) ∈ (C[f ] ∩ C[g]) \ C, which is a contradiction. We now have A0 (f (x))/B 0 ((f (x)) = C 0 (g(x))/D0 (g(x)),

which contradicts our assumption on the minimality of deg B + deg D.

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(b) To show that C(x2 + x) ∩ C(x2 ) = C, it is sufficient according to (a), to prove that C[x2 + x] ∩ C[x2 ] = C. Let h(x) be an element of C[x2 + x] ∩ C[x2 ], then h(x) = an (x2 + x)n + an−1 (x2 + x)n−1 + · · · + a0 = bn (x2 )n + bn−1 (x2 )n−1 + · · · + b0 ,

where a0 , . . . , an and b0 , . . . , bn are complex numbers. Suppose that n ≥ 1 and an bn 6= 0. Equating the leading coefficients (resp. the coefficients of x2n−1 ) in these two expressions of h(x), we obtain: an = bn and nan = 0, which is a contradiction. Thus n = 0 and h(x) ∈ C. We now consider the other intersection. Obviously C(xk ) ⊆ C(xn ) ∩ C(xm ). We prove the reverse inclusion. Let h(x) be an element of the intersection then h(x) = A(xm )/B((xm ) = C(xn )/D(xn ), where A(x), B(x), C(x) and D(x) are polynomials with complex coefficients such that gcd(A(x), B(x)) = gcd(C(x), D(x)) = 1. As in (a), we obtain C(xn ) = λA(xm ) and D(xn ) = λB(xm ). P Let U (x) = ai xi be the polynomial appearing on the left (and on the right) side of the first identity, then if ai 6= 0, we have i 6≡ 0 (mod m) and i 6≡ 0 (mod n). It follows that i 6≡ 0 (mod k), whenever ai 6= 0, that is U (x) ∈ C[xk ]. We may obtain the same conclusion for V (x) = D(xn )) = λB(xm ). We conclude that h(x) ∈ C[xk ]. (3) Consider E = C(x2 ), F = C(x2 + x), Ω = E = C(x) and α = x. Exercise 8.12. Let K be a field, f (x, y), g(x, y) and h(x, y) be polynomials with coefficients in K such that f (x, y) is not constant. (1) If f is irreducible over K, show that there exist R(x, y) and q(x, y) ∈ K[x, y] such that R is irreducible and R(g, h) = f (x, y)q(x, y). (2) If f is irreducible over K, let I = {A(x, y) ∈ K[x, y], such that A(g, h) ≡ 0

(mod f (x, y))}.

Show that I is a principal prime ideal of K[x, y] generated by R(x, y). Deduce that K[g, h] ∩ f K[x, y] 6= (0) if and only if g and h are algebraically independent over K.

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(3) If f is arbitrary (irreducible or not), show that there exists S(x, y) ∈ K[x, y] \ K such that S(g, h) ≡ 0 (mod f (x, y)). Solution 8.12. (1) Since f is non constant, then degx f ≥ 1 or degy f ≥ 1. Suppose that degy f ≥ 1, the proof is similar if degx f ≥ 1 and let α be a root of f (x, y) in an algebraic closure of K(x). Let E = K(x, α), then E is a transcendental extension of K and Trdeg E/K(x) = 1. Therefore there exists R(u, v) ∈ K[u, v] such that R(g(x, α), h(x, α)) = 0 and we may suppose that R is irreducible over K. It follows that there exists q(x, y) ∈ K(x)[y] such that R(g(x, y), h(x, y)) = f (x, y)q(x, y). Using the content in K(x), we obtain cont(R(g(x, y), h(x, y))) = cont(f (x, y)) cont(q(x, y)) = cont(q(x, y)). We conclude that cont(q(x, y)) ∈ K[x], which implies q(x, y) ∈ K[x, y]. (2) Let ψ : K[x, y] → K[x, y]/f K[x, y] be the morphism of rings such that ψ(A(x, y)) = A(g, h), then Ker ψ = I so that I is a prime ideal of K[x, y] and we have R(x, y)K[x, y] ⊆ I. Since the image of ψ is not a field, then Ker ψ is not maximal. Let M be a maximal ideal of K[x, y] containing Ker ψ. Consider the chain of prime ideals {0} ⊂ R(x, y)K[x, y ⊂ Ker ψ ⊂ M. Since the Krull dimension of K[x, y] is equal to 2, then R(x, y)K[x, y] = Ker ψ = I. Suppose that g(x, y) and h(x, y) are algebraically dependent over K and let A(u, v) be an irreducible polynomial with coefficients in K such that A(g, h) = 0. Since A ∈ I, then up to a multiplicative constant, the polynomials A(x, y) and R(x, y) are equal. Let u(x, y) ∈ K[g, h] ∩ f K[x, y], then u(x, y) = v(g, h) ≡ 0 (mod f ), hence v(x, y) ∈ I and then Ax, y) | v(x, y). Since A(g, h) = 0, then u(x, y) = v(g, h) = 0. Suppose that K[g, h] ∩ f K[x, y] = (0). Since R(g, h) belongs to this intersection, then R(g, h) = 0, which implies g and h are algebraically dependent over K. (3) Let f (x, y) = f1 (x, y) · · · fr (x, y)

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be the factorization of f into irreducible factors over K, distinct or not. For each i ∈ {1, . . . , r} let Ri (x, y) be the polynomial defined in (1) and satisfying the condition Ri (g, h) ≡ 0 (mod fi (x, y)), then Qr Qr i=1 Ri (g, h) ≡ 0 (mod f (x, y)), so that S(x, y) = i=1 Ri (x, y) satisfies the required conditions. Exercise 8.13. Let m, n be coprime positive integers and q be a prime power. Let α ∈ Fqn , β ∈ Fqm and γ = αβ. Show that γ generates a normal basis of Fqmn if and only if α (resp. β) generates a normal basis of Fqn (resp. Fqm ) over Fq . Solution 8.13. • Necessity of the conditions. We prove that β generates a normal basis of Fqm . The same proof works for α and will be omitted. Since Fqm and Fqn are linearly disjoint over Fq , we have TrFqmn /Fqm (γ) = TrFqmn /Fqm (αβ) = β TrFqmn /Fqm (α) = β TrFqn /Fq (α). Since γ generates a normal basis of Fqmn , then the left side of the above identity is non zero. Therefore TrFqn /Fq (α) 6= 0. It follows that β generates a normal basis of Fqm over Fq if and only if β TrFqn /Fq (α) does, that is TrFqmn /Fqm (γ) does. This last claim is true by Exercise 6.19. So β generates a normal basis of Fqm over Fq . • Sufficiency of the conditions. Consider the following diagram of fields. Fqmn

Fq m

Fqn

Fq Since gcd(m, n) = 1, then the fields Fqm and Fqn are linearly disjoint over Fq and Gal(Fqmn , Fq ) ' Gal(Fqm , Fq ) × Gal(Fqn , Fq ). i

j

The list of conjugates of γ is given by αq β q for i = 0, . . . , n − 1 and n−1 j = 0, . . . , m − 1. Now {α, . . . , αq } is a basis of Fqn over Fq and m−1 {β, . . . , β q } is a basis of Fqmn over Fqn , hence i

j

{αq β q , for i = 0, . . . , n − 1 and j = 0, . . . , m − 1}

is a basis of Fqmn over Fq .

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Exercise 8.14. Let q be a prime power, {α1 , . . . , αn } be a normal basis of Fqn over Fq and {β1 , . . . , βn } be its dual basis. Show that this second basis is a normal basis of Fqn over Fq . One may use the following matrices    n−1  β1 β2 ... βn α1 α1q . . . α1q  βq  q q n−1  β2q ... βnq      A =  α2 α2 . . . α2  and B =  1 . ... ... ...   ... . . . . . . . . . ...  n−1 n−1 n−1 n−1 β1q β2q . . . βnq αn αnq . . . αnq Solution 8.14. For any matrix C, we denote by C t its transpose. We may suppose that i−1 α1 , . . . , αn are labeled such that αi = α1q . Our assumption implies that AB = In , hence BA = In . Since A is symmetric, we deduce that (AB)t = B t At = B t A = In . Therefore B t A = In = BA, which implies B t = B and then βi = β q This means that {β1 , . . . , βn } is a normal basis of Fqn over Fq .

i−1

.

Exercise 8.15. Let q be a prime power, m and n be coprime positive integers, α1 , . . . , αm be elements of Fqm . Show that they form a basis of Fqm over Fq if and only if they form a basis of Fqmn over Fqn . Solution 8.15. Consider the following diagram of fields: Fqmn

Fqm

Fqn

Fq Since gcd(m, n) = 1, then the fields Fqm and Fqn are linearly disjoint over Fq . Therefore α1 , . . . , αm are linearly independent over Fq if and only if they are linearly independent over Fqn . To complete the proof, we observe that [Fqmn : Fqn ] = [Fqm : Fq ] = m.

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Exercise 8.16. Let K be field, E and F linearly disjoint extensions of K. Show that E ∩ F = K. Show that the converse is false. Solution 8.16. Suppose that E ∩ F 6= K and let α ∈ E ∩ F \ K, then 1, α are linearly independent over K, but linearly dependent over F , a contradiction. Therefore E ∩ F = K. For the converse, let E = Q(21/3 ) and F = Q(j21/3 ), where j is a primitive cube root of unity, then E ∩ F = Q and E, F are not linearly disjoint over Q. Exercise 8.17. Let K be a field, f (x) and g(y) be irreducible polynomials over K. Let α (resp. β) be a root of f (x) (rep. g(y) in an algebraic closure of K. Show that the ideal generated by f (x) and g(y) in K[x, y] is maximal if and only K(α) and K(β) are linearly disjoint over K. Solution 8.17. We first show that K[x, y]/(f (x), g(y)) ' K(α)[y]/(g(y)). Let φ : K[x, y] → K(α)[y]/(g(y)) be the unique K-morphism of rings such that φ(x) = α and φ(y) = y + (g(y)). Obviously φ is surjective. Let m = deg g and let F (x, y) ∈ K[x, y] such that φ(F ) = 0. Performing an Euclidean division of F (x, y) by g(y) in K[x][y], we obtain F (x, y) = g(y)q(x, y) + r(x, y), where q and r ∈ K[x, y] and Pm−1 degy r < m. Since φ(F ) = 0, then φ(r) = 0. Set r(x, y) = i=1 ai (x)y i , Pm−1 then i=1 ai (α)¯ y i = 0. Since {1, y¯, . . . , y¯m−1 } is a basis of K(α)[y]/(g(y)) over K(α), then ai (α) = 0 for i = 0, . . . , m − 1. It follows that f (x) | ai (x) for i = 0, . . . , m − 1. Set ai (x) = f (x)bi (x) for i = 0, . . . , m − 1, then F (x, y) = g(y)q(x, y) + f (x)

m−1 X

ai (x)y i .

i=1

Thus F belongs to the ideal generated by f (x) and g(y), that is Ker φ is contained in this ideal. The proof of the reverse inclusion is obvious, so that

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Ker φ = (f (x), g(y)). The first isomorphism theorem implies the mentioned isomorphism. We now have (f (x), g(y)) is maximal in K[x, y] if and only if K(α)[y]/(g(y)) is a field, thus if and only if g(y) is irreducible over K(α). Therefore if and only K(α) and K(β) are linearly disjoint over K. Exercise 8.18. Let K be a field, f (x, y) and g(u, z) be irreducible polynomials with coefficients in K such that degy f = d1 and degz g = d2 . Let α ∈ K(x) and β ∈ K(u) such that f (x, α) = 0 and g(u, β) = 0. Suppose that f is absolutely irreducible. (1) Show that [K(x, u, β) : K(x, u)] = d2 and [K(x, u, β, α) : K(x, u, β)] = d1 . Deduce that K(x, u, β) and K(x, u, α) are linearly disjoint over K(x, u). (2) By considering the example K = Q, f (x, y) = y 4 − 2x2 and g(u, z) = z 2 − 2u2 , show that the conclusion of (1) does not hold if we remove the condition of absolute irreducibility. Solution 8.18. (1) Consider the following diagram of fields: K(x, u, α, β) K(x, u, β) K(x, u, α) K(x, α)

K(u, β) K(x, u)

K(x)

K(u)

We first prove that [K(x, u, β) : K(x, u)] = d2 . We have K(u, x)/K(u) is purely transcendental while K(u, β)/K(u) is algebraic, hence K(u, x) and K(u, β) are linearly disjoint over K(u) [Lang (1965), Pro. 3, Chap. 10.5]. It follows that [K(x, u, β) : K(x, u)] = [K(u, β) : K(u)] = d2 . We now prove that [K(x, u, β, α) : K(x, u, β)] = d1 . It is sufficient to prove that f (x, y) is irreducible over K(u, β) = K(u)[β]. Suppose that

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f (x, y = f1 (x, y)f2 (x, y), where f1 and f2 ∈ K(u)[β][x, y] with degy f1 and degy f2 ≥ 1. Set X X f1 (x, y) = ajk xj y k and f2 (x, y) = bjk xj y k , where ajk and bjk ∈ K(u)[β]. We may write ajk and bjk in the form X  X  h ajk = Ahjk (u)β h /D(u) and bjk = Bjk (u)β h /D(u).

¯ such that D(u0 ) 6= 0 and such that, when the polyChoose u0 ∈ K nomial g(u, z) is expressed through the decreasing powers of z, then ¯ its leading coefficient does not vanish at u = u0 . Choose z0 ∈ K such that g(u0 , z0 ) 6= 0. If we substitute u0 for u and z0 for z, we obtain the factorization f (x, y) = f 1 (x, y)f 2 (x, y), where f 1 and f 2 are the polynomials obtained by the substitution from f1 and f2 respectively. This contradicts the absolute irreducibility of f (x, y), hence [K(u, x, β, α) : K(u, x, β)] = d1 . It follows that [K(u, x, β, α) : K(u, x)] = d1 d2 = [K(u, x, α) : K(u, x)][K(u, x, β) : K(u, x)] which means that K(u, x, α) and K(u, x, β) are linearly disjoint over K(u, x). √ √ (2) Here we have β = ±u 2 and α2 = ±x 2, d1 = 4 and d2 = 2. We have the following diagram: K(x, u, α, β) √ K(u, x, 2) K(u, x) hence 8 = d1 d2 6= [K(u, x, β, α) : K(u, x)] = 4. Exercise 8.19. Let K be a field, A be a graded commutative K-algebra without divisors of 0, L be its fraction field and d be a non negative integer. A rational function f = g/h ∈ L is said to be homogeneous of degree d if g and h are homogeneous and deg g − deg h = d. Let L0 = {f ∈ L, f homogeneous, deg f = 0} ∪ {0}.

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(1) Show that L0 is a field and K ⊂ L0 ⊂ L. (2) Let g ∈ L, homogeneous, of degree e ≥ 1. Show that g is transcendental over L0 . If moreover g is chosen to satisfy the condition that its degree e ≥ 1 is minimal, show that L = L0 (g). (3) Suppose that there exist f1 , . . . , fr ∈ L be homogeneous such that L = K(f1 , . . . , fr ). Show that there exist g1 , . . . , gr−1 ∈ L0 such that L0 = K(g1 , . . . , gr−1 ). Prove that gcd(deg f1 , . . . , deg fr ) = e, where e is defined in (2). Hint. One may use Exercise 1.5 and Exercise 10.26. (4) Let x1 , . . . , xn be a set of algebraically independent variables over K. Suppose that A = K[x1 , . . . , xn ]. Determine in this case L0 . Solution 8.19. (1) Clear. (2) Set g = g1 /g2 where g1 , g2 ∈ A, homogeneous with deg g1 > deg g2 . Suppose that g is algebraic over L0 , then g satisfies an equation of the form (an /bn )g n + · · · + (ak /bk )g k + · · · + (a0 /b0 ) = 0 with n ≥ 1 and deg ai = deg bi . Let D = lcm(b0 , . . . , bn ). Replacing g by its value and multiplying by D, this equation becomes (Dan /bn )g1n + · · · + (Dak /bk )g1k g2n−k + · · · + (Da0 /b0 )g2n = 0. We have deg(Dan /bn )g1 = deg D + n deg g1 and for any k < n, deg(Dak /bk )g1k g2n−k = deg D + k deg g1 + (n − k) deg g2

< deg D + k deg g1 + (n − k) deg g1 = deg D + n deg g1 ,

hence a contradiction. We conclude that g is transcendental over L0 . Suppose now that e is minimal. To prove that L ⊂ L0 (g) it is sufficient to show that any f ∈ A, homogeneous, belongs to L0 (g). The inclusion in the opposite direction is obvious and will be omitted. Let f ∈ A homogeneous, then deg f = eq + r, where q and r are integers and 0 ≤ r < e. We have deg(f /g q ) = deg f − qe = r and f /g q is homogeneous. Since e is minimal, then r = 0, hence f /g q ∈ L0 and f ∈ L0 (g). (3) Let di = deg fi for i = 1, . . . , r and d = gcd(d1 , . . . , dr ). By Bezout’s identity, there exist integers u1 , . . . , ur such that u1 d1 + · · · + ur dr = 1. The integers u1 , . . . , ur are coprime (see Exercise 1.5), hence by Exercise 10.26, there exists a matrix with integral coefficients M =

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(cji ) ∈ Zr×r such that cj1 = uj and det M = 1. Set M −1 = (bji ). Let Qr cj φi = j=1 fj i for i = 1, . . . , r, which means (φ1 , . . . , φr ) = (f1 , . . . , fr )M .

We deduce that (f1 , . . . , fr ) = (φ1 , . . . , φr )M hence fi =

Qr

,

bji

j=1

deg φ1 =

−1

X

φi . This shows that L = K(φ1 , . . . , φr ). We have X j cji dj = d and deg φi = ci dj ≡ 0 (mod d) deg φ

/d

for i ≥ 2. For i = 1, . . . , r − 1, let gi = φi+1 /φ1 i+1 . Then deg gi = 0, hence gi ∈ L0 . We have proved that L = K(φ1 , . . . , φr ). It is easy to see that L = K(g1 , . . . , gr−1 , φ1 ). Consider the chain of fields: K(g1 , . . . , gr−1 ) ⊂ L0 ⊂ K(g1 , . . . , gr−1 , φ1 ) = L. By (2), L/L0 is a purely transcendental extension, hence φ1 is transcendental over L0 , thus also over K(g1 , . . . , gr−1 ). It follows that L/K(g1 , . . . , gr−1 ) is purely transcendental. Since L is algebraic over any field F such that K(g1 , . . . , gr−1 ) ( F ⊂ L, then L0 = K(g1 , . . . , gr−1 ), L = L0 (φ1 ) and that the degree of φ1 is minimal. Hence e = deg φ1 = d = gcd(deg f1 , . . . , deg fr ). (4) In this case L = K(x1 , . . . , xn ). Let α ∈ L0 . We may write α in the form P ai1 ···in xi11 · · · xinn . α = P i1 +···+in =d jn j1 j1 +···+jn =d bj1 ···jn x1 · · · xn Dividing numerator and denominator of this fraction by xdn , we get P ai1 ···in (x1 /xn )i1 · · · (xn−1 /xn )in−1 α = P i1 +···+in =d , jn−1 j1 j1 +···+jn =d bj1 ···jn (x1 /xn ) · · · (xn−1 /xn )

hence α ∈ K(x1 /xn , . . . , xn−1 /xn ), thus L0 ⊂ K(x1 /xn , . . . , xn−1 /xn ). The reverse inclusion is obvious. We claim and we omit the prof that L = K(x1 /xn , . . . , xn−1 /xn , xn ).

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Exercise 8.20. Let K be a field of characteristic 0, x1 , . . . , xn be algebraically independent − variables over K. Denote by → x the n-tuple (x1 , . . . , xn ). Let Ω be an algebraic closure of K(x1 , . . . , xn ). − − (1) Let f1 (→ x ), . . . , fn (→ x ) be elements of Ω and let J be the matrix, − − called the Jacobian matrix of f1 (→ x ), . . . , fn (→ x ), and given by J =  ∂f1 ∂fn  · · · ∂x1 ∂x1  .. . . ..   . . .  . Show that the following conditions are equivalent. ∂f1 ∂fn ∂xn · · · ∂xn

− − (i) The elements f1 (→ x ), . . . , fn (→ x ) are algebraically independent over K. (ii) The matrix J is non singular.

Show that the condition on the characteristic of K cannot be omitted. − − (2) Deduce that if g1 (→ x ), . . . , gn (→ x ) are rational functions with coefficients → − in K such that gi ( x ) ∈ K(x1 , . . . , xi ) and degxi gi ≥ 1 for i = 1, . . . , n, − − then g1 (→ x ), . . . , gn (→ x ) are algebraically independent over K. Here degxi gi denotes the maximum among the degrees in xi of the numerator − and the denominator of gi (→ x ). − − (3) Let r be a positive integer such that r ≤ n and h1 (→ x ), . . . , hr (→ x ) be  ∂h1 ∂hr  · · · ∂x1 ∂x1   elements of Ω. Let Jr =  ... . . . ...  . Show that the following ∂h1 ∂hr ∂xn · · · ∂xn assertions are equivalent. − − (i) The elements h1 (→ x ), . . . , hr (→ x ) are algebraically independent over K. (ii) The rank of the matrix Jr is equal to r. − − (4) Let r, h1 (→ x ), . . . , hr (→ x ) and Jr as in (3). Show that − − Rank(Jr ) = Trdeg(K(h1 (→ x ), . . . , hr (→ x ))/K). Solution 8.20. − − (1) • (ii) ⇒ (i). Suppose that f1 (→ x ), . . . , fn (→ x ) are algebraically dependent over K and let φ(y1 , . . . , yn ) ∈ K[y1 , . . . , yn ], irreducible such that φ(f1 , . . . , fn ) = 0. Then differentiating this identity relatively

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to xk , for any k ∈ {1, . . . , n}, we get n X ∂fi ∂φ (f1 , . . . , fn ). = 0. ∂y ∂x i k i=1

This shows that ∂φ ∂φ (f1 , . . . , fn ), . . . , (f1 , . . . , fn ) ∂y1 ∂yn is a solution of a homogeneous system of linear equations, where the coefficients appearing in the i-th equation are the ones in the i-th row of J. Since the characteristic of K is 0, then degyi If

∂φ (y1 , . . . , yn ) = degyi φ(y1 , . . . , yn ) − 1. ∂yi

∂φ ∂yi (f1 , . . . , fn )

= 0 for some i ∈ {1, . . . , n}, then φ(y1 , . . . , yn ) |

thus

∂φ ∂yi (y1 , . . . , yn )

∂φ (y1 , . . . , yn ), ∂yi

= 0 which implies that φ does not depend on yi .

∂φ It follows that there exists i ∈ {1, . . . , n} such that ∂y (f1 , . . . , fn ) 6= i 0, hence the mentioned solution of the homogeneous system is non trivial. We conclude that the determinant of the system, that is Det(J), is zero. • (i) ⇒ (ii). Let E = K(x1 , . . . , xn , f1 , . . . , fn ). Since f1 , . . . , fn are algebraic over K(x1 , . . . , xn ), then Trdeg(E/k) = n. The assumptions show that {f1 , . . . , fn } is a transcendence basis of E over K. It follows that for any i ∈ {1, . . . , n}, xi , f1 , . . . , fn are algebraically dependent over K. Hence there exist

φi (y0 , y1 , . . . , yn ) ∈ K[y0 , y1 , . . . , yn ] such that φi (xi , f1 , . . . , fn ) = 0, i = 1, . . . , n. Moreover for any i ∈ {1, . . . , n}, we have degy0 φi ≥ 1. Differentiate each of the preceding equations relatively to xk for k = 1, . . . , n and obtain δik

n X ∂fj ∂φi ∂φi (xi , f1 , . . . , fn ) + (xi , f1 , . . . , fn ) = 0, ∂y0 ∂x k ∂yj j=1

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where δik = 1 if i = k and 0 otherwise. We may write these equations in the form ! ∂φi n X ∂fj − ∂yj (xi , f1 , . . . , fn ) = δik , i = 1, . . . , n, k = 1, . . . , n. ∂φi ∂xk ∂y0 (xi , f1 , . . . , fn ) j=1 This set of equations shows that there exists an n × n matrix H such that JH = In , where In is the identity matrix. This implies that J is non singular. Here we look at the condition on the characteristic of K. Let K be any field of characteristic p > 0, n = 2, f1 (x1 , x2 ) = xp1 and f2 (x1 , x2 ) be any polynomial with coefficients in K such that degx2 f2 ≥ 1, then while these polynomials are algebraically independent over K, their Jacobian matrix is 0. Therefore the equivalence of (i) with (ii) cannot be generalized for arbitrary characteristic. (2) Here the Jacobian matrix of g1 , . . . , gn is triangular and has no zero on its main diagonal, hence Det(J) 6= 0 and then by (1), g1 , . . . , gn are algebraically independent over K. (3) • (i) ⇒ (ii). Since Trdeg(Ω/K) = n and h1 , . . . , hr are algebraically independent over K, then we may complete this set of algebraic function by say hr+1 , . . . , hn and obtain a transcendence basis of Ω  ∂hi over K. The rank of the Jacobian matrix ∂xj 1≤i≤n of h1 , . . . , hn is 1≤j≤n

equal to n, hence the extracted matrix formed by the r first columns has rank r. Therefore Rank(Jr ) = r.  ∂hi • (ii) ⇒ (i). We may suppose that the matrix H = ∂x 1≤i≤r exj 1≤j≤r

tracted from Jr is non singular. Let hi = xi for i = r + 1, . . . , n and let J be the Jacobian matrix of h1 , . . . , hn , then it is seen that Det(J) = Det(H) 6= 0. Thus, by (1), h1 , . . . , hn are algebraically independent over K. It follows that h1 , . . . , hr are algebraically independent over K. − − (4) Let s = Rank(Jr ) and t = Trdeg(K(h1 (→ x ), . . . , hr (→ x ))/K). We may suppose that the s first column of Jr are linearly independent. The matrix extracted from Jr and constituted by these scolumn has rank s, hence by (3), h1 , . . . , hs are algebraically independent over K. Thus s ≤ t. On the other hand, since t ≤ r, we may suppose that h1 , . . . , ht are algebraically independent over K. The Jacobian matrix H of h1 , . . . , ht is then of rank t, by (3). Since H is extracted from Jr , then s = Rank(Jr ) ≥ t. We conclude that s = t.

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Exercise 8.21. c(y) Let K be a field, f (x) = a(x) b(x) and g(x) = d(y) be non constant rational functions with coefficients in K such that gcd(a(x), b(x)) = gcd(c(x), d(x)) = 1. (1) Show that the following propositions are equivalent. (i) There exists h(x) ∈ K(x)\K such that f = h ◦ g. (ii) The polynomial d(x)d(y)(g(x)−g(y)) divides b(x)b(y)(f (x)−f (y)) in K[x, y]. (2) Let J(x, y) ∈ K[x, y]. Show that if there exist u(x), v(x) ∈ K[x] such that J(x, y) = u(x)v(y) − v(x)u(y), then (a) J(y, x) = −J(x, y) (b) J(w, x)J(y, z) + J(w, y)J(z, x) + J(w, z)J(x, y) = 0.

Show that the converse holds if K is infinite. (3) Let f (x) = (x4 − x3 − 8x − 1)/(2x4 + x3 − 16x + 1) ∈ C(x). Decompose f (x) in the form f (x) = h(g(x)), where g(x) and h(x) ∈ C(x) with deg g = deg h = 2. (4) Deduce that there exist exactly three fields Fi for i = 1, 2, 3 such that C(f ) ( Fi ( C(x). Moreover they are given by Fi = k(gi ), where g1 (x) = (x2 − 2x)/(2x2 − 3x + 1),

g2 (x) = (x2 − 2jx)/(2x2 − 3jx + j 2 ) 2

2

2

and

2

g3 (x) = (x − 2j x)/(2x − 3j x + j).

(5) Show that C(gi , gj ) = C(x) for i, j ∈ {1, 2, 3}, i 6= j. Express x as a rational function in g1 , g2 with complex coefficients. Solution 8.21. (1) • (i) ⇒ (ii). We have d(x)d(y)(g(x) − g(y)) = c(x)d(y) − c(y)d(x)

and

b(x)b(y)(f (x) − f (y)) = a(x)b(y) − a(y)b(x).

The assumption (i) implies that K(f ) ⊂ K(g) ⊂ K(x). Moreover x is a root of a(z) − b(z)f ∈ K(f )[z] and of c(z) − d(z)g ∈ K(g)[z]. This last polynomial is irreducible in K[g, z], hence also in K(g)[z], then it is equal to the minimal polynomial of x over K(g) up to a multiplication by some element of the form λ − µg ∈ K[g]. Since x is also a root of a(z) − b(z)f = a(z) − b(z)h(g),

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then c(z)−d(z)g divides a(z)−h(g)b(z) in K(g)[z], hence c(z)−d(z)g divides a(z) − h(g)b(z) in K[g, z]. Set a(z) − h(g)b(z) = (c(z) − gd(x))q(g, z), where q(g, z) ∈ K[g][z]. We deduce that     a(x) c(x) c(x) a(z) − b(z) = c(z) − d(z) q ,z . b(x) d(x) d(x) Multiply by b(x)d(x)e+1 and obtain: d(x)

e+1

e

(a(z)b(x) − a(x)b(z)) = (c(z)d(x) − c(x)d(z))d(x) q



 c(x) ,z . d(x)

• (ii) ⇒ (i). Since c(x)d(y) − c(y)d(x) divides a(x)b(y) − a(y)b(x) in c(x) K[x, y], then c(y) − d(x) d(y) divides a(y) − a(x) b(x) b(y) in K(x)[y], that is c(y) − g(x)d(y) divides a(y) − f (x)b(y) in K(x)[y]. Performing Euclidean divisions in K(g(x))[y], we obtain a(y) = (c(y) − g(x)d(y))q1 (g, y) + r1 (g, y)

and

b(y) = (c(y) − g(x)d(y))q2 (g, y) + r2 (g, y),

where q1 , q2 , r1 , r2 ∈ K(g(x))[y] and deg r1 < deg g. degy r2 < deg g. Moreover r1 6= 0 and r2 6= 0. We deduce that a(y) − f (x)b(y) = (c(y) − g(x)d(y))(q1 − f (x)q2 ) + r1 − f (x)r2 . Our assumption implies r1 − f (x)r2 = 0. Let p(g) (resp. q(g)) be the leading coefficient of r1 (resp. r2 ) as a polynomial in y with p(x) coefficients in K(g), then f (x) = p(g) q(g) = h(g) with h(x) = q(x) . (2) Let u(x), v(x) ∈ K[x] such that J(x, y) = u(x)v(y) − v(x)u(y), then clearly (a) holds. The identity (b) is a particular case of the following one: (AB−CD)(F G−HE)+(AE−F D)(HB−CG)+(AG−HD)(CE−F B) = 0 after setting A = u(w), B = v(x), C = u(x), D = v(w), E = v(y), F = u(y), G = v(z), and H = u(z). This last identity may be verified by writing its left side in the form Aγ + Dµ. It is readily seen that γ = µ = 0.

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We now prove the converse. Suppose that J(x, y) satisfies (a) and (b). Let w = α and z = β in (b), where α and β are elements of K to be determined. Then J(α, x)J(y, β) + J(α, y)J(β, x) + J(α, β)J(x, y) = 0. Since K is infinite and J 6= 0, then there exists (α, β) ∈ K 2 such that J(α, β) 6= 0. Therefore, we get: J(x, y) = −J(α, x)

J(y, β) J(β, x) − J(α, y) . J(α, β) J(α, β)

Using the identity (a), we obtain J(x, y) = J(x, α)

J(x, β) J(y, β) − J(y, α) , J(α, β) J(α, β)

so that J(x, y) = u(x)v(y) − u(y)v(x)

J(x,β) . The assumption J(α, β) 6= 0, where u(x) = J(x, α) and v(x) = J(α,β) made on (α, β) implies that J(x, β) 6= 0, hence v(x) 6= 0. (3) We begin with a claim. Let J(x, y) ∈ K[x, y] such that J(y, x) = −J(x, y), then J(x, y) = (x − y)H(x, y), where H(x, y) ∈ K[x, y] is symmetric. For the proof, let

J(x, y) = (x − y)H(x, y) + R(x) be the Euclidean division of J by x − y in K[x][y], then R ∈ K[x]. We have 0 = J(x, x) = R(x). We deduce that H(x, y) is symmetric. Let g(x) ∈ K(x), deg g = 2 such that f (x) = h(g(x)) for some h(x) ∈ K(x). Let a(x), b(x) ∈ K[x] such that gcd(a(x), b(x)) = 1 and g(x) = a(x)/b(x). By (1) the polynomial a(x)b(y) − a(y)b(x) divides u(x)v(y) − u(y)v(x), where u(x) (resp. v(x)) is the numerator (resp. denominator) of f (x). Easy computations show that F (x, y) :=

u(x)v(y) − u(y)v(x) = 3x3 y 3 + 3(x + y)(x2 + y 2 + 8xy) − 24. x−y

Let G(x, y) = a(x)b(y)−a(y)b(x) . Write this polynomial in the form of a x−y sum of its homogeneous components. Its leading homogeneous component say G+ (x, y) divides 3x3 y 3 and is symmetric. Suppose first that

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v(x) ∈ C∗ , that is g(x) ∈ C[x], then we may express the numerator and the denominator of f (x) in the form: x4 − x3 − 8x − 1 = a(g(x)) 4

and

3

2x + x − 16x + 1 = b(g(x)), where a(x), b(x) ∈ C[x] and gcd(a(x), b(x)) = 1. Differentiating these equations, we obtain 4x3 − 3x2 − 8 = a0 (g(x))g 0 (x) 3

2

0

and

0

8x + 3x − 16 = b (g(x))g (x). It follows that g 0 (x) divides the polynomials 4x3 − 3x2 − 8 and 8x3 + 3x2 − 16. But is easy to see that these two polynomials have no common root, hence they are relatively prime which implies g 0 (x) ∈ C and then deg g = 1, which contradicts our assumption. We conclude that g(x) is not polynomial. Now g(x) = u(x) v(x) with u, v non constant and M ax(deg u, deg v) = 2. In this case it is seen that deg G+ (x, y) ≤ 2. Since G+ is symmetric and divides 3x3 y 3 , then G+ (x, y) = axy, where a ∈ C? . Moreover, replacing u(x) by u(x)/a, if necessary, we may suppose that G+ (x, y) = xy. Let H(x, y) ∈ C[x, y] such that F (x, y) = G(x, y)H(x, y). Set G(x, y) = G2 + G1 + G0

and

H(x, y) = H4 + H3 + H2 + H1 + H0 , where G2 , G1 , G0 (resp. H4 , H3 , H2 , H1 , H0 ) represent the homogeneous components of G (resp. H). Here the indices of the components are equal to their degrees. By identification in the identity F = G · H, we obtain the following equations G2 H4 = 3x3 y 3

(Eq 1)

G2 H3 + G1 H4 = 0

(Eq 2)

G2 H2 + G1 H3 + G0 H4 = 0

(Eq 3) 2

2

G2 H1 + G1 H2 + G0 H3 = 3(x + y)(x + y + 8xy)

(Eq 4)

G2 H0 + G1 H1 + G0 H2 = 0

(Eq 5)

G1 H0 + G0 H1 = 0

(Eq 6)

G0 H0 = −24.

(Eq 7)

Since G2 = xy, then by (Eq 1), H4 = 3x2 y 2 . Replacing in (Eq 2), G2 and H4 by their values, we obtain: H3 = −3xyG1 . Since G1 is

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homogeneous of degree 1 and symmetric, then G1 = a(x + y) with a ∈ C? , hence H3 = −3axy(x + y). Set G0 = b with b ∈ C, then replacing in (Eq 3), G1 , G2 , H3 , H4 and G0 by their values, we get: H2 = 3a2 (x+y)2 −3bxy. From (Eq 6), we deduce that H1 = 24a b2 (x+y). We put in (Eq 4) the values of G2 , H1 , G1 , H2 , G0 and H3 and we obtain 3a3 (x2 + y 2 ) + (

24a + 6a3 − 6ab)xy = 3(x2 + 8xy + y 2 ), b2 3

2 2

hence a3 = 1 and a(4−bb2+a b ) = 4. Therefore a and b are determined by the following equations a3 = 1 and ab3 + 3b2 − 4a = 0. Now a is a cube root of unity and b is a root of the polynomial c(x) = ax3 + 3x − 4a = a(x − a2 )(x2 + 4a2 x + 4a) = a(x − a2 )(x + 2a2 )2 . We deduce that b = a2 or b = −2a2 . We examine each of these values of b. • Case b = a2 . Recall the preceding computations of the Hi , Gj . We have: G0 = a2 ,

G1 = a(x + y),

H1 = 24(x + y),

G2 = xy, 2

2

2

H0 = −24a,

H2 = 3a (x + y + xy),

H3 = −3axy(x + y)

and

H4 = 3x2 y 2 .

Replace them in (Eq 5), and obtain x2 + y 2 + 2xy = 0, which is a contradiction. Therefore, this value of b must be rejected. • Case b = −2a2 . Here we have G0 = −2a2 ,

G1 = a(x + y)

and

G2 = xy.

It follows that the polynomial J(x, y) = a(x)b(y) − a(y)b(x) is given by: J(x, y) = (x − y)G(x, y) = (x − y)(−2a2 + a(x + y) + xy) = x2 y − xy 2 + ax2 − ay 2 − 2a2 x + 2a2 y.

We verify if the conditions (a) and (b) of (2) are satisfied.

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Obviously (a) is fulfilled. We have J(w, x)J(y, z) + J(w, y)J(z, x) + J(w, z)J(x, y) = (w2 x − wx2 + aw2 − ax2 − 2a2 w + 2a2 x) .(y 2 z − zy 2 + ay 2 − az 2 − 2a2 y + 2a2 z)

+ (w2 y − wy 2 + aw2 − ay 2 − 2a2 w + 2a2 y)

.(z 2 x − xz 2 + az 2 − ax2 − 2a2 z + 2a2 x)

+ (w2 z − z 2 w + aw2 − az 2 − 2a2 w + 2a2 z)

.(x2 y − xy 2 + ax2 − ay 2 − 2a2 x + 2a2 y) = 0. So that condition (b) is satisfied. Since C is infinite, we may choose (α, β) ∈ C2 such that J(α, β) 6= 0. For example α = 0 and β = a works. We have J(0, a) = 1 and then a(x) = G(x, 0) = ax2 − 2a2 x and b(x) = G(x, a) = 2ax2 − 3a2 x + 1. Therefore g(x) =

x2 − 2ax ax2 − 2a2 x = . 2ax2 − 3ax + 1 2x2 − 3ax + a2

We now compute the rational function h(x) ∈ C(x) such that f (x) = c(x) h(g(x)). Set h(x) = d(x) , where c(x), d(x) ∈ C[x], gcd(c(x), d(x)) = 1 and M ax(deg c, deg d) = 2. Suppose first that deg c = 2 and deg d = 1, then   b2 (x)c a(x) b(x) x4 − x3 − 8x − 1   = f (x) = 4 , 2x + x3 − 16x + 1 b2 (x)d a(x) b(x)

hence b(x)|2x4 +x3 −16x+1, that is 2x2 −3ax+a2 |2x4 +x3 −16x+1. This is a contradiction since b(a) = 0 and v(a) 6= 0. Similarly we eliminate the case deg c = 1 and deg d = 2. It remains to consider the case deg c = deg d = 2. We may suppose that c(x) is monic. Set c(x) = x2 + λx + µ and d(x) = αx2 + βx + γ. Using the identity h(g(x)) = f (x), we obtain (x2 − 2ax)2 + λ(x2 − 2ax)(2x2 − 3ax + a2 ) + µ(2x2 − 3ax + a2 )2 α(x2 − ax)2 + β(x2 − ax)(2x2 − 3ax + a2 ) + γ(2x2 − 3ax + a2 )2 x4 − x3 − 8x − 1 = 4 . 2x + x3 − 16x + 1

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This identity may be written in the form A(x)/B(x) = where

x4 − x3 − 8x − 1 , 2x4 + x3 − 16x + 1

A(x) = (1 + 2λ + 4µ)x4 − a(4 + 7λ + 12µ)x3

+ a2 (4 + 7λ + 13µ)x2 − (2λ + 6µ)x + aµ

and B(x) = (α + 2β + 4γ)x4 − a(4α + 7β + 12γ)x3

+ a2 (4α + 7β + 13γ)x2 − (2β + 6γ)x + aγ

It follows that: −a(4 + 7λ + 12µ)/(1 + 2λ + 4µ) = −1 a2 (4 + 7λ + 12µ)/(1 + 2λ + 4µ) = 0

−(4 + 7λ + 12µ)/(1 + 2λ + 4µ) = −8 aµ/(1 + 2λ + 4µ) = −1

and (4α + 7β + 12γ)/(1 + 2λ + 4µ) = 2 −a(4α + 7β + 12γ)/(1 + 2λ + 4µ) = 1 a2 (4α + 7β + 12γ)/(1 + 2λ + 4µ) = 0 −(2β + 6γ)/(1 + 2λ + 4µ) = −16 αγ/(1 + 2λ + 4µ) = 1.

3+4a 1 Solving the first four equations, we obtain λ = − 2+7a and µ = 2+7a . We deduce that 1 + 2λ + 4µ = −a/(2 + 7a). The five last equations lead to

α = (−2+14a)/(2+7a), β = (3−8a)/(2+7a) and γ = −1/(2+7a). We conclude that h(x) =

(2 + 7a)x2 − (3 + 4a)x + 1 . (−2 + 14a)x2 + (3 − 8a)x − 1

Conversely, it is easy to verify that this h(x) satisfies the identity g(h(x)) = f (x). Indeed there are three decompositions of f (x) in the form f (x) = h(g(x)), where deg g = deg h = 2. Each one corresponds to the values of g(x) and h(x) computed above and to the choice a ∈ {1, j, j 2 }, where j is a primitive cube root of unity in C.

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(4) Let F be a field such that C(f ) ⊂6= F ⊂6= C(x). By Luroth’s Theorem [Schinzel (2000), Th. 2, Chap. 1.1], there exists g(x) ∈ C(x) such that F = C(g). Moreover since [C(x) : C(f )] = d = 2, then [C(x) : F ] = 2, then deg g = 2. Since f ∈ F , then f = h(g) where h(x) ∈ C(x) and 2 −2ax deg h = 2. It follows by (3) that g(x) = 2x2x−3ax+a 2 , where a is a cube root of unity in C. Fix a primitive cube root of unity j and let F1 , F2 , F3 be the fields corresponding to a = 1 or a = j or a = j 2 respectively. Denote by g1 (x), g2 (x) and g3 (x) the corresponding rational functions. We show that the fields F1 , F2 , F3 are pairwise distinct. Suppose that F2 = F1 , then there exist p, q, r, s ∈ C such that 2

−2x p 2xx2 −3x+1 +q x2 − 2jx = 2 −2x 2 2 x 2x − 3jx + j r 2x2 −3x+1 + s

and ps − rq 6= 0. Hence

(p + 2q)x2 − (2p + 3q)x + q x2 − 2jx = . (r + 2s)x2 − (2r + 3s)x + s 2x2 − 3jx + j 2 We conclude immediately that q = 0, and then x2 − 2x − 2r+3s p x+

r+2s 2 p x

s p

=

x2 − 2jx , 2x2 − 3jx + j 2

which is a contradiction. Similarly we may prove that F1 6= F3 and F2 6= F3 . (5) The assertion on the composition of the fields Fi is obvious. We write x in the form: x = φ(g1 , g2 )/(ψ(g1 , g2 ), with g1 (x) = (x2 − 2x)/(2x2 − 3x + 1) 2

2

and 2

g2 (x) = (x − 2jx)/(2x − 3jx + j ).

From the first equation, we deduce that x2 = (2x + g1 (1 − 3x))/(1 − 2g1 ).

Replace this value of x2 in the second equation and obtain: x(2 − 2j + g1 (4j − 3)) + g1 . g2 = x(4 − 3j + g1 (6j − 6)) + j 2 + g1 (2 − 2j 2 ) Therefore j 2 − g1 + (2 − 2j 2 )g1 g2 x= . 2 − 2j + (4j − 3)g1 + (3j − 4)g2 + (6 − 6j)g1 g2 Exercise 8.22. Let K be a field, Ω be a algebraic closure of K, u(t) and v(t) be non constant polynomials with coefficients in K.

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(1) Show that there exists P (x, y) ∈ K[x, y] irreducible such that P (u(t), v(t)) = 0. Show that, for any f (x, y) ∈ K[x, y], if f (u(t), v(t)) = 0 then P (x, y) | f (x, y). (2)(a) Show that there exist u1 (t) and v1 (t) ∈ K[t] such that P (u1 (t), v1 (t)) = 0, degx P = deg v1 and degy P = deg u1 . (b) Considered as a polynomial in x (resp. y) with coefficients in K[y] (resp. K[x]), show that the leading coefficients of P (x, y) is in K ∗ . (3) Show that P (x, y) is irreducible in Ω[x, y]. (4) Show that P (x, y) is irreducible in Ω(( x1 ))[y] and in Ω(( y1 ))[x]. Solution 8.22. (1) Consider the chain of fields K(u) ⊂ K(u, v) ⊂ K(t) = K(u, t). Since t is algebraic over K(u), then v is algebraic over K(u). Let F (y) = y m +

a0 (u) am−1 (u) m−1 y + ··· + D(u) D(u)

be the minimal polynomial of v over K(u), where D(x), ai (x) ∈ K[x] and gcd(D(x), a0 (x), . . . , am−1 (x)) = 1. Let P (x, y) = D(x)y m + am−1 (x)y m−1 + · · · + a0 (x), then P (x, y) is irreducible over K and P (u(t), v(t)) = 0. Let f (x, y) ∈ K[x, y] such that f (u(t), v(t)) = 0. Let q(x, y), r(x, y) ∈ K(x)[y] such that f = P q + r and degy r < degy P . We have r(u(t), v(t)) = 0, hence P (x, y) divides r(x, y) in K(x)[y]. It follows that r(x, y) = 0 and then f = P q. We compare the contents in K(x) of the polynomials appearing in the two sides of this identity. We obtain cont(f ) = cont(P ) cont(q) = cont(q). It follows that cont(q) ∈ K[x]. Therefore q ∈ K[x, y] and then P (x, y) | f (x, y) in K[x, y]. Remark. We may prove this property of divisibility in the following way. Let ψ : K[x, y] → K[t] be the unique K-morphism of algebras such that ψ(x) = u(t) and ψ(y) = v(t), then K[x, y]/ Ker ψ ' Im ψ ⊂ K[t]. Therefore Ker ψ is a non zero prime ideal of K[x, y] containing P (x, y). We prove that Ker ψ = P (x, y)K[x, y]. Since the image of ψ is not a field, then Ker ψ is not maximal. Let M be a maximal ideal of K[x, y] containing Ker ψ. Consider the chain of prime ideals {0} ⊂ P (x, y)K[x, y ⊂ Ker ψ ⊂ M.

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Since the Krull dimension of K[x, y] is equal to 2, then Ker ψ = (P ). Hence P (x, y) | f (x, y) for any f (x, y) ∈ K[x, y] satisfying the condition f (u(t), v(t)) = 0. (2)(a) Luroth’s Theorem [Schinzel (2000), Th. 2, Chap. 1.1], shows that there exists φ(t) ∈ K(t) such that K(u, v) = K(φ). Moreover Exercise 8.10 implies that we may choose φ(t) ∈ K[t]. This choice being made, let u1 (x) = A(x)/B(x) ∈ K(x) such that gcd(A, B) = 1 and u(t) = u1 (φ(t)) = A(φ(t))/B(φ(t)). Bezout’s Theorem implies that there exist λ(x) and µ(x) ∈ K[x] such that λ(x)A(x) + µ(x)B(x) = 1. We deduce that λ(φ(t))A(φ(t)) + µ(φ(t))B(φ(t)) = 1. Since B(φ(t)) | A(φ(t)), then B(φ(t)) ∈ K ∗ . Therefore B(t) ∈ K ∗ , ie u1 (x) ∈ K[x]. Similarly if we set v(t) = v1 (φ(t)), then v1 (x) ∈ K[x]. We have 0 = P (u(t), v(t)) = P (u1 (φ(t)), v1 (φ(t))), hence P (u1 (t), v1 (t)) = 0. Since K(u1 (φ(t)), v1 (φ(t))) = K(φ(t)), then K(u1 (z), v1 (z)) = K(z). Let A(x) (resp. B(y)) be the leading coefficient of P (x, y) considered as a polynomial in y (resp. in x), (u1 (z),y) is the minimal polynomial of v1 (z) over K(u1 (z))], then PA(u 1 (z)) hence degy P = [K(u1 , v1 ) : K(u1 )] = [K(z) : K(u1 (z))] = deg u1 . A similar reasoning leads to degx P (x, y) = deg v1 . (b) Clearly z is integral over K[u1 (z)]. Therefore v1 (z) in integral over K[u1 (z)] and then the leading coefficient of P (x, y) as a polynomial in x is in K ∗ . Therefore we may suppose that P (x, y) is monic relatively to y. (3) We apply the results proved in (1), (2)(a) and (2)(b) for u(t), v(t) when K is replaced by Ω. We obtain the polynomials P (x, y) ∈ Ω[x, y], φ(t), u1 (t), v 1 (t) ∈ Ω[t] such that P (x, y) is irreducible, monic in x with moreover the following properties P¯ (u(t), v(t)) = P¯ (¯ u1 (t), v¯1 (t)) = 0, ¯ degx P¯ = deg v¯1 , degy P¯ (x, y) = deg u ¯1 and Ω(u(t), v(t)) = Ω(φ(t)).

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By (1) we conclude that P¯ (x, y) | P (x, y) in Ω[x, y]. From K(u(t), v(t)) = K(φ(t)) we deduce that Ω(u(t), v(t)) = Ω(φ(t)), hence we may suppose that ¯ = φ(t) and then u1 (t) = u1 (t) and v1 (t) = v1 (t). φ(t) Now since degy P¯ (x, y) = deg u1 (t) = degy P (x, y), P¯ (x, y)|P (x, y) and these two polynomials are monic in y, then P¯ (x, y) = P (x, y). We conclude that P (x, y) is irreducible over Ω. (4) Let n = deg u1 (t) = degy P (x, y). Set P (x, y) = y n + an−1 (x)y n−1 + · · · + a0 (x),

ˆ be an algebraic closure where ai (x) ∈ K[x] for i = 1, . . . , n − 1. Let Ω of Ω((x)) and let β : ω(u1 (t)) → Ω((x)) be the unique Ω-embedding such that β(u1 ) = x1 . Since v1 is algebraic over K(u1 ), there exists ˆ Since P (u1 , v1 ) = 0, then an extension βˆ : Ω(u1 , v1 ) = Ω(t) → Ω. 1 ˆ (u1 , v1 )) = 0, hence P ( , β(v ˆ 1 )) = 0. Therefore β(v ˆ 1 ) is a root of β(P x the monic polynomial P ( x1 , y) ∈ Ω((x))[y] of degree n. Since   1 Ω(u1 )(v1 ) = Ω(t) = Ω(u1 ) , t then ˆ 1 )) = Ω((x)) Ω((x))(β(v

1 ˆ β(t)

hence

" ˆ [Ω((x))(β(v1 )) : Ω((x))] = Ω((x))

1 ˆ β(t)

!

!

,

: Ω((x)) .

Set u1 (t) = a0 tn + a1 tn−1 + · · · + an , where a0 , . . . , an ∈ K and a0 6= 0, then a1 an u1 tn + tn−1 + · · · + − = 0. a0 a0 a0 Set c =

1 a0

and ci =

ai a0

#

for i ∈ {1, . . . , n}, then

tn + c1 tn−1 + · · · + cn − cu1 = 0.

(Eq 1)

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Divide this equation by u1 tn and obtain  n−1  n 1 1 1 −1 −1 + cn−1 u1 + · · · + c1 u−1 + u−1 (cn u1 − c) 1 1 = 0. t t t The element w = cn u−1 1 − c is non zero and we obtain:  n−1  n 1 1 −1 −1 −1 1 −1 + cn−1 u1 w + · · · + c1 u−1 + u−1 = 0. 1 w 1 w t t t We deduce that   n   n−1 1 ˆ −1 βˆ 1 + cn−1 xβ(w) βˆ t t   ˆ −1 βˆ 1 + xβ(w) ˆ −1 = 0. + · · · + c1 xβ(w) t We have ˆ ˆ n u−1 − c) = cn x − c, β(w) = β(c

ˆ ˆ −1 ∈ Ω[[x]] is also a unit. hence β(w) is a unit of Ω[[x]] and then β(w) 1 ˆ Now β( t ) is a root of the polynomial h(x, y) ∈ Ω((x))[y], ˆ −1 y n−1 + · · · + c1 xβ(w) ˆ −1 + xβ(w) ˆ −1 . h(x, y) = y n + cn−1 xβ(w) By application of Eisenstein’s irreducibility criterion we conclude that h(x, y) is irreducible over Ω((x)). It follows that      1 ˆ Ω(x) β : Ω((x)) = degy h = n t and then, by (Eq 1), ˆ 1 )) : Ω((x))] = n = deg P [Ω((x))(β(v y



 1 ,y . x

We conclude that P ( x1 , y) is irreducible over Ω((x)).

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Exercise 9.1. Let f (x, y) and g(x, y) be polynomials with integral coefficients. Suppose δg δf δg that C[f, g] = C[x, y] and δf δx (0, 0) δy (0, 0) − δy (0, 0) δx (0, 0) = ±1. (1) Show that Z[f, g] = Z[x, y]. (2) Show that the condition on the partial derivatives of of f (x) and g(x) cannot be omitted. Solution 9.1. (1) Let f1 (x, y) = f (x, y) − f (0, 0) and g1 (x, y) = g(x, y) − g(0, 0), then f1 (x, y), g1 (x, y) ∈ Z[x, y],

C[f1 , g1 ] = C[f, g] = C[x, y] and δg1 δf1 δg1 δf1 (0, 0) (0, 0) − (0, 0) (0, 0) = ±1. δx δy δy δx Clearly if Z[f1 , g1 ] = Z[x, y], then Z[f, g] = Z[x, y], hence we may suppose that f (0, 0) = 0 and g(0, 0) = 0. Set f (x, y) = f10 x + f01 y + higher degree terms

and

g(x, y) = g10 x + g01 y + higher degree terms, where f10 g01 − f01 g10 = ±1. Making the following change of variables, x0 = f10 x + f01 y,

y 0 = g10 x + g01 y,

allows us to write f and g in the form: f = x0 + higher degree terms and y = y 0 + higher degree terms, hence we may suppose that f = x + higher degree terms 243

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and g = y + higher degree terms. P Since C[f, g] = C[x, y], then x = (i,j) cij f i g j , where cij ∈ C for any (i, j). We are going to prove that cij ∈ Z, that is x ∈ Z[f, g]. It will appear from the proof that the same reasoning works for y. Define the following order relation in Z2 . We say that (i, j) > (i0 , j 0 ) if i + j > i0 + j 0 or i + j = i0 + j 0 and i > i0 . It is a total order in Z2 . We prove by induction that cij ∈ Z. Since f (0, 0) = g(0, 0), then c00 = 0. Let (i, j) ∈ Z2 such that (i, j) > (0, 0) and suppose that cij ∈ Z for any (i0 , j 0 ) satisfying the condition (i0 , j 0 ) < (i, j). The P 0 0 polynomial x − (i0 ,j 0 ) d. Show that there exist c2 , . . . , cn ∈ K such that

g(y1 , . . . , yn ) = ay1d + Ad−1 (y2 , . . . , yn )y1d−1 + · · · + A0 (y2 , . . . , yn ),

where a ∈ K ? and the polynomials Ai have their coefficients in K. Solution 9.3. Write f (x1 , . . . , xn ) in the form

f (x1 , . . . , xn ) = fd (x1 , . . . , xn ) + fd−1 (x1 , . . . , xn ) + · · · + f0 (x1 , . . . , xn ),

where fj (x1 , . . . , xn ) is 0 or homogeneous of degree j and fd (x1 , . . . , xn ) 6= 0. Then g(y1 , . . . , yn ) = fd (y1 , y2 − c2 y1 , . . . , yn − cn y1 )

+ fd−1 (y1 , y2 − c2 y1 , . . . , yn − cn y1 )

+ · · · + f0 (y1 , y2 − c2 y1 , . . . , yn − cn y1 ).

It is clear that degy1 fj (y1 , y2 − c2 y1 , . . . , yn − cn y1 ) < d for j = 0, . . . , d − 1. Set X fd (x1 , . . . , xn ) = ai1 ···in xi11 · · · xinn , i1 +···+in =d

then

gd (y1 , . . . , yn ) =

X

i1 +···+in =d

ai1 ···in y1i1 (y2 − c2 y1 )i2 · · · (yn − cn y1 )in .

From this we conclude that the coefficient, say a, of y1d belongs to K and is given by X a= ai1 ···in (−c2 )i2 · · · (−cn )in = fd (1, −c2 , . . . , −cn ). i1 +···+in =d

The polynomial fd (1, y2 , . . . , yn ) is non zero. Otherwise the non zero homogeneous polynomial fd (y1 , . . . , yn ) would be divisible by y1 − 1, which is a contradiction. (A divisor of an non zero homogeneous polynomial is itself homogeneous!) The degree d0 of the polynomial in n − 1 variables, fd (1, y2 , . . . , yn ) satisfies the condition d0 ≤ d < |K|, hence there exists (−c2 , . . . , −cn ) ∈ K n−1 such that fd (1, −c2 , . . . , −cn ) 6= 0, that is a 6= 0. Exercise 9.4. Let m and n be positive integers such that 1 ≤ m ≤ n. Let K be a field and x1 , . . . , xn be algebraically independent variables over K. Denote by ~x the n-tuple (x1 , . . . , xn ). For any i = 1, . . . , m, let gi (xi ) be a non constant polynomial with coefficients in K.

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(1) Show that any f (~x) ∈ K[~x] may be written in one and only one way in the form X f (~x) = aj1 ···jm (~x)g1 (x1 )j1 · · · gm (xm )jm ,

where the sum is finite, aj1 ···jm (~x) ∈ K[~x], degxi aj1 ···jm (~x) < deg gi for i = 1, . . . , m and degxi aj1 ···jm (~x) < degxi f for m < i ≤ n. (2) Suppose that for any i = 1, . . . , m, deg gi ≥ 2. Show that for any i = 1, . . . , m, xi 6∈ K[g1 (x1 ), . . . , gm (xm )]. Solution 9.4. (1) We proceed by induction on m. Suppose first that m = 1. The proof follows the one given in [Lang (1965), Th. 9, Chap. 5.5]. In this theorem, the involved polynomials are polynomials in one variable with coefficients in a field K, which is not the case in this exercise. The Euclidean division in K[x2 , . . . , xn ][x1 ] of f (~x) by g1 (x1 ) leads to f (~x) = g1 (x1 )q(~x) + b0 (~x), where q(~x) and b0 (~x) ∈ K[~x] and degx1 b0 < deg g1 . We show that degxi b0 ≤ degxi f for i = 2, . . . , n. We fix i and we distinguish three cases. • degxi q = degxi b0 . Let B(x1 , . . . , xˆi , . . . , xn ) and C(x1 , . . . , xˆi , . . . , xn ) be the leading coefficients of q(~x) and b0 (~x) respectively, when these polynomials are written as decreasing powers of xi . Here the hated variable xi in the given polynomials means that this variable is not present in the polynomials. If Bg1 + C = 0, then g1 | C, hence degx1 b0 ≥ degx1 C ≥ deg g1 , which is a contradiction. Therefore, Bg1 + C 6= 0 and then degxi q = degxi b0 = degxi f. • degxi q < degxi b0 . In this case, we have degxi b0 = degxi f. • degxi q > degxi b0 . Here we have degxi b0 < degxi q = degxi f. In any case, we have proved that degxi b0 ≤ degxi f. By induction, suppose that q(~x) = b1 (~x) + b2 (~x)g1 (x1 ) + · · · + bd (~x)g1 (x1 )d , where degx1 bi < deg g1 and degxi bi ≤ degxi q for i ≥ 2, then f (~x) = b0 (~x) + b1 (~x)g1 (x1 ) + · · · + bd+1 (~x)g1 (x1 )d+1 .

(Eq 1)

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Here the conditions on the degrees of the bi , relatively to x1 and relatively to xi are satisfied. Since bi (~x) is the remainder of the Euclidean division of   f (~x) − b0 (~x) − b1 (~x)g1 (x1 ) − · · · − bi−1 (~x)g1 (x1 )i−1 /g1 (x1 )i−1

by g1 (x1 ), then this representation of f (~x) is unique. Suppose that the representation of any polynomial, with the conditions on the degrees, exists and is unique for any system of m − 1 polynomials. Write each polynomial bi (~x) appearing above as a polynomial in g2 (~x), . . . , gm (~x) with coefficients in K[(~x)] and with the degrees conditions, that is X (i) bi (~x) = ch1 ···hm (~x)g2 (x2 )h2 (i) · · · gm (~x)hm (i) , (i)

(i)

with degxj ch1 ···hm (~x) < deg gj for j = 2, . . . , m and degxj ch1 ···hm (~x) ≤ degxj bi (~x). Then X (i) f (~x) = ch1 ···hm (~x)g1 (x1 )i g2 (x2 )h2 (i) · · · gm (~x)hm (i) , so that the representation of f (~x) exists. We prove the uniqueness of the representation. Suppose that beside the representation X f (~x) = aj1 ···jm (~x)g1 (x1 )j1 · · · gm (xm )jm , we have the second one X f (~x) = bi1 ···im (~x)g1 (x1 )i1 · · · gm (xm )im .

Rewriting each of these expressions as sums of linear combinations of ascending powers of g1 (x1 ), we conclude by the uniqueness of the representation in the unique polynomial g1 (x1 ), that for fixed k, X aj1 ···jm (~x)g1 (x1 )j1 · · · gm (xm )jm j1 =k

=

X

j1 =k

bi1 ···im (~x)g1 (x1 )i1 · · · gm (xm )im .

Here we have two representations of the same polynomial in the system {g2 (x1 ), . . . , gm (xm )}. By the inductive hypothesis we conclude that the coefficients in the representations are equal. (2) The unique representation of xi in the system {g1 , . . . , gm } is given by X xi = aj1 ···jm (~x)g1 (x1 )j1 · · · gm (xm )jm ,

where aj1 ···jm (~x) = 0 if (j1 , . . . , jm ) 6= (0, . . . , 0) and aj1 ···jm (~x) = xi if (j1 , . . . , jm ) = (0, . . . , 0). Therefore xi cannot be expressed in the form xi = u(g1 (x1 ), . . . , gm (xm )), where u(y1 , . . . , ym ) is a polynomial with coefficients in K because such an expression is a representation in the system {g1 , . . . , gm }.

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Exercise 9.5. (1) Let f (x) and h(x) be polynomials with integral coefficients of degree m and n respectively and let p be a prime number. Suppose that h(x) is monic, irreducible over Fp , m and n ≥ 2, gcd(m, n) 6= 1 and p does not divide the leading coefficient of f (x). Let g(x) = h(x)/p. Show that f (Q) ∩ g(Q) = ∅. (2) Given a polynomial F (x) with rational coefficients of degree m ≥ 2 and a positive integer n such that gcd(m, n) 6= 1, show that there exists a polynomial G(x) of degree n for which the equation F (x) = G(y) has no solution (x, y) ∈ Q2 . Solution 9.5. (1) Suppose that f (Q) ∩ g(Q) 6= ∅ and let (a, b) ∈ Q2 such that pf (a) = h(b). • Case νp (b) ≥ 0. Since h(x) is irreducible modulo p, then νp (h(b)) = 0. From the above equation, we conclude that νp (f (a)) = −1. It follows that νp (a) < 0 and then νp (f (a)) = mνp (a) = −1, which is a contradiction since m ≥ 2. • Case νp (b) < 0. In this case, we have νp (h(b)) < 0, and by the above equation satisfied by (a, b), we get νp (f (a)) < 0 and then νp (a) < 0. We now have νp (f (a)) = mνp (a)

and νp (h(b)) = nνp (b).

The equation relating a and b implies that 1 + mνp (a) = nνp (b), which in turn shows that gcd(m, n) = 1, which contradicts our assumptions. Thus the intersection is empty. (2) We may write F (x) in the form F (x) = ad f (x), where a and d are relatively prime integers and f (x) is a polynomial with integral coefficients. Let p be prime number which is not a divisor of the leading coefficient of f (x). Since m = deg f ≥ 2, we may find a polynomial h(x) of degree n such that according to (1), the equation f (x) = h(x)/p has no rational solution. Put G(x) = ah(x)/p, then deg G = n and the equation F (x) = G(y) has no rational solution.

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Exercise 9.6. Let K be a field, n ≥ 2 be an integer and g, h ∈ K[x1 , . . . , xn ] be non constant. Suppose that the leading homogeneous form of g(x)h(x) is square free. Show that gK[x1 , . . . , xn ] + hK[x1 , . . . , xn ] 6= K[x1 , . . . , xn ]. Hint. Use Exercise 3 of this chapter. Solution 9.6. • We may suppose that K is infinite. For if K were finite, let K1 = K(t), where t is a new variable, then if gK1 [x1 , . . . , xn ] + hK1 [x1 , . . . , xn ] 6= K1 [x1 , . . . , xn ], we conclude that 1∈ / gK1 [x1 , . . . , xn ] + hK1 [x1 , . . . , xn ], hence 1∈ / gK[x1 , . . . , xn ] + hK[x1 , . . . , xn ], so that the same property will hold for K. • Let f (x) = g(x)h(x) and d = deg f (x). According to Exercise 9.3, we may suppose that f has the form f (x1 , . . . , xn ) = xd1 + Ad−1 (x2 , . . . , xn )x1d−1 + · · · + A0 (x2 , . . . , xn ). It follows that the factors g and h of f , considered as polynomials in x1 , with coefficients in K[x2 , x3 , . . . , xn ], may be supposed to be monic. In particular, degx1 f = deg f + . In the sequel, we denote by W + the leading homogeneous component of the arbitrary polynomial W . Since the result asked for is true if the polynomials g and h are not coprime, we may suppose that gcd(g, h) = 1. Suppose by contradiction that there exist u(x1 , . . . , xn ) and v(x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] such that ug + vh = 1. Performing the Euclidean division of v by g in K[x2 , . . . , xn ][x1 ], we obtain v = gq +r, where r 6= 0 and degx1 r < degx1 g. Since (u+qh)g +rh = 1, then g + | r+ h+ . Since degx1 r+ = degx1 r < degx1 g = degx1 g + , then some non constant factor of g + divides h+ . Therefore f + is not square free, contradicting the hypotheses. Exercise 9.7. Let a, b ∈ Z such that (a, b) 6= (0, 0). Show that the polynomial f (x, y) = y 3 − x2 y 2 − 3x2 y + x4 y − ax2 + b is irreducible over Q.

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Solution 9.7. Suppose that f (x, y) = g(x, y)h(x, y) is a non trivial factorization of f in Q[x, y]. Since f is monic in y, we must have degy g ≥ 1 and degy h ≥ 1. Without loss of generality, we may suppose that degy g = 1, degy h = 2 and g(x, y) = cy − P (x), where c ∈ Q and P (x) ∈ Q[x], deg P ≥ 1, then we may suppose that c = 1. It follows that f (x, P (x)) = 0 and therefore P (x) | ax2 − b in Q[x]. Since this condition is impossible if a = 0, we suppose that a 6= 0. If deg P = 1, set P (x) = λx + µ with λ 6= 0. The identity f (x, P (x)) = 0, immediately implies λ = 0, a contradiction. If deg P = 2, then P (x) = λ(ax2 − b) with λ ∈ Q? . Using again the same identity, we get the equation (λa)3 − (λa)2 + (λa) = 0, which has no non zero rational solution λa. In particular, such a factorization f = gh cannot exist, and therefore f must be irreducible. Exercise 9.8. Let K be a field, α1 , . . . , αm be non zero, algebraic and separable over K of degree d1 , . . . , dm respectively and u1 , . . . , um be algebraically independent − − variables. Set → u = (u1 , . . . , um ) and → α = (α1 , . . . , αm ). − (1) Show that α1 , . . . , αm are algebraic, separable over K(→ u ) and → − → − → − → − [K( u , α ) : K( u )] = [K( α ) : K]. − − (2) Show that γ = u1 α1 + · · · + um αm is a primitive element of K(→ u,→ α) → − over K( u ). − − − (3) Let P (→ u , x) ∈ K(→ u )[x] be the minimal polynomial of γ over K(→ u ). → − Show that P ∈ K[ u , x]. em (4) Let d = d1 · · · dm . We denote the elements α1e1 · · · αm with 0 ≤ ei ≤ di − 1 as follows: θ1 = 1, θ2 = α1 , . . . , θd1 = α1d1 −1 , θd1 +1 = α2 , . . . ,

dm −1 θ2d1 = α1d1 −1 α2 , . . . , θd = α1d1 −1 · · · αm .

(a) Show that for any j ∈ {1, . . . , d}, γθj may be written in the form γθj =

d X

− lji (→ u )θi ,

i=1

− where lji ∈ K[→ u ] and deg lji = 1. → − (b) Show that P ( u , x) divides the polynomial l11 − x l12 ... l l − x ... 22 − D(→ u , x) = 21 ... ... ... l ld2 ... d1

l1d l2d . . . . ldd − x

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(5) Show that for any i ∈ {1, . . . , m},

∂P → (− u , γ) i αi = − ∂u . − ∂P → ∂x ( u , γ)

− →? ? − → ? (6) Let u? = (u?1 , . . . , u?m ) ∈ K m such that ∂P ∂x (u , γ ) 6= 0, where γ = → − u?1 α1 + · · · + u?m αm . Show that γ ? is a primitive element of K( α ) over K. (7) Let   →  − →? ? ? m ∂P − u ,γ = 0 . F = u ∈K , ∂x (a) Show that   →  − →? →? ?  ∂P − m ∂P − ? ? F = u ∈K , u , γ = 0, u , γ = 0, i = 1, . . . , m . ∂x ∂ui − → − → (b) Let U = K m \ F. Show that for any u? ∈ U, P (u? , x) is irreducible over K. (8) Let K = Q, α1 = 21/2 and α2 = 21/4 . Compute D(u1 , u2 , x) and deduce P (u1 , u2 , x). Solution 9.8. (1) It is trivial that for any i ∈ {1, . . . , m}, αi is algebraic and separable − over K(→ u ). Consider the diagram − − K(→ u,→ α)

− K(→ u)

− K(→ α)

K − − The extensions K(→ u ) and K(→ α ) are linearly disjoint over K, hence the result on the degrees. − (2) Let Ω be an algebraic closure of K(→ u ). Suppose that there exist two → − → − → − K( u )-isomorphisms σ, τ : K( u , α ) → Ω such that σ(γ) = τ (γ), then u1 σ(α1 ) + · · · + um σ(αm ) = u1 τ (α1 ) + · · · + um τ (αm ), hence σ(αi ) = τ (αi ) for i = 1, . . . , m. Therefore σ = τ and γ is a − − − primitive element of K(→ u,→ α ) over K(→ u ).

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− (3) The polynomial P (→ u , x) is given by ! m Y X → − P ( u , x) = x− ui σ(αi ) , σ

i=1

− − − where the product runs over the set of K(→ u )-embeddings of K(→ u,→ α) into Ω, hence the result. Notice that any of these embeddings is an − extension of one and only one embedding of K(→ α ) into Ω fixing the elements of K. em (4) (a) Each θj has the form θj = α1e1 · · · αm with 0 ≤ ei ≤ di − 1, hence e1 +1 e1 e2 +1 em em em +1 · · · αm γθj = u1 α1 · · · αm + u2 α1 α2 + um α1e1 · · · αm . If ei + 1 ≤ mi − 1, then there exists k such that em = θik . If ei + 1 = mi , then α1e1 · · · αiei +1 · · · αm e1 ei +1 em em α1 · · · αi · · · αm = α1e1 · · · (−adi −1 αidi −1 − · · · − a1 αi − a0 ) · · · αm em em = −adi −1 α1e1 · · · αidi −1 · · · αm − a1 α1e1 · · · αi · · · αm e

e

i−1 i+1 em − a0 α1e1 · · · αi−1 αi+1 . · · · αm This shows that γθj is a linear combination of the θk , k = 1, . . . , d, where the coefficients are linear polynomials in u1 , . . . , um with coefficients in K. (b) The preceding equations may be written in the form: (l11 − γ)θ1 + l12 θ2 + · · · + l1d θd = 0

l21 θ1 + (l22 − γ)θ2 + · · · + l2d θd = 0 ......................................................

ld1 θ1 + ld2 θ2 + · · · + (ldd − γ)θd = 0. This implies that the homogeneous system (l11 − γ)x1 + l12 x2 + · · · + l1d xd = 0 l21 x1 + (l22 − γ)x2 + · · · + l2d xd = 0

.........................................................

ld1 x1 + ld2 x2 + · · · + (ldd − γ)xd = 0 has a non trivial solution, namely θ1 , . . . , θd . Therefore the determinant D of this system is zero, i.e. D(u1 , . . . , um , γ) = 0. − − Since P (→ u , x) is the minimal polynomial of γ over K(→ u ), then → − → − → − P ( u , x) | D( u , x). It is possible to compute D( u , x) without using the above determinant. Indeed we have Y − D(→ u , x) = (x − u1 σ1 (α1 ) − . . . − um σm (αm )) , (σ1 ,...,σm )

where the product runs over all the m-tuples (σ1 , . . . , σm ) and σi is a K isomorphism of K(αi ) into Ω.

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(5) We fix j ∈ {1, . . . , m} and we differentiate ! the identity m X − P → u, ui αi = 0 i=1

relatively to uj . We obtain ∂P → ∂P → (− u , γ) + (− u , γ) = 0. (Eq 1) αj ∂x ∂uj → − − − − Since K(→ u,→ α )/K(→ u ) is separable then ∂P ∂x ( u , γ) 6= 0, hence − ∂P → ∂u ( u , γ) αj = − ∂Pj → . − ∂x ( u , γ) − →? ? (6) Since ∂P ∂x (u , γ ) 6= 0, then the preceding identity shows that for any − j = 1, . . . , m, αj ∈ K(γ ? ), hence K(→ α ) = K(γ ? ) and γ ? is a primitive → − element of K( α ). (7) (a) Easy application of equations (Eq 1). − → − → (b) Since P (u? , γ ? ) = 0, then Irr(γ ? , K) | P (u? , x). These two polynomials being monic in x, to get their equality it is sufficient to show that they have the same degree. We have − → − degx P (u? , x) = degx P (→ u , x) → − − = [K( u , γ) : K(→ u )] − − − = [K(→ u,→ α ) : K(→ u )] → − = [K( α ) : K]

= [K(γ ? ) : K], hence the result. Here are some explanations of the preceding equalities. The first one is due to the fact that P is monic in x. Since γ is a root of P , we obtain the second one. The third equality − − comes from the fact that γ is a primitive element of K(→ u,→ α ) over → − K( u ). From (1), we deduce the fourth equality. The last equality − is due to the fact that γ ? is a primitive element of K(→ α ) over K. (8) We have √ √ D(u1 , u2 , x) = (x + u1 2 + u2 21/4 )(x + u1 2 − u2 21/4 ) √ √ .(x − u1 2 + u2 i21/4 )(x − u1 2 − u2 i21/4 ) √ √ .(x − u1 2 − u2 21/4 )(x − u1 2 + u2 21/4 ) √ √ .(x + u1 2 − u2 i21/4 )(x + u1 2 + u2 i21/4 ) = (x2 + 2u21 )2 − 2(2u1 x − u22 )2 .(x2 + 2u21 )2 − 2(2u1 x + u22 )2

= P (u1 , u2 , x)Q(u1 , u2 , x)

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with P (u1 , u2 , x) = (x2 + 2u21 )2 − 2(2u1 x − u22 )2 and Q(u1 , u2 , x) = (x2 + 2u21 )2 − 2(2u1 x + u22 )2 .

Question. May it happen that there exists (u?1 , . . . , u?m , x? ) ∈ K m+1 such that → → ∂P − ∂P − (u? , x? ) = 0, (u? , x? ) = 0 ∂x ∂ui for i = 1, . . . , m and such that one of the following conditions holds? (i) There exists i such that u?m 6∈ K. Pm (ii) For any i, u?m ∈ K but x? 6= i=1 u?i αi .

Exercise 9.9. Let K be a field, fi (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] for i = 1, . . . , n be algebraically independent over K. (1) Suppose that xi , for i = 1, . . . , n, is integral over K[f1 , . . . , fn ], show that K(f1 , . . . , fn ) ∩ K[x1 , . . . , xn ] = K[f1 , . . . , fn ].

(2) Let f (x, y) = x2 y 2 and g(x, y) = xy(x + y) and suppose that the characteristic of the field K is not equal to 2. (a) Show that f and g are algebraically independent over K. (b) Show that x is algebraic over K(f, g), but not integral over K[f, g]. (c) Show that K(f, g) ∩ K[x, y] 6= K[f, g]. Solution 9.9. (1) Let h(x1 , . . . , xn ) be an element of the intersection of the two rings, then h is integral over K[f1 , . . . , fn ] and belongs to the fraction field of this last ring. This ring is isomorphic to K[x1 , . . . , xn ], hence factorial. Therefore K[f1 , . . . , fn ] is integrally closed and then h ∈ K[f1 , . . . , fn ]. The other inclusion is trivial. (2) (a) Suppose that there exists φ(u, v) ∈ K[u, v]\K such that φ(f, g) = 0. Let φ+ (u, v) be the the leading form of φ(u, v). We may write φ+ (u, v) in the form +

e k

φ (u, v) = u v

m Y

(u − θi v),

i=1

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where e, k are non negative integers, the product may be empty and the θi are algebraic over K. Since f and g are homogeneous, then φ+ (f, g) is the leading form of φ(f, g), hence φ+ (f, g) = 0. We deduce that f = 0 or g = 0 or f = θi g for some i ∈ {1, . . . , m} which is impossible. √ √ (b) We have xy =  f , with  = ±1 and x + y = g/ f . Hence x and √ √ y are roots of the polynomial u2 − g/( f )u +  f . Therefore q p p 2x = g/ f ± g 2 /f − 4 f . We deduce that



p p 2 2x − g/ f = g 2 /f − 4 f ,

√ hence  f x2 = xg − f and then

f x4 − g 2 x2 + 2f gx − f 2 = 0. This means that x is algebraic over K(f, g) and is a root of the polynomial φ(z) = f z 4 − g 2 z 2 + 2f gz − f 2 . We can use another method to find again this polynomial. Since x is a common root of the polynomials φ1 (z) = x2 z 2 − f

and φ2 (z) = xz 2 + x2 z − g,

then Resz (φ1 (z), φ2 (z)) = 0. Using standard formulas on the resultant, we obtain √ √    f f 2 f 2 f 4 −g x 2 −x −g Resz (φ1 (z), φ2 (z)) = x x 2 + x x x x x
= x2 f 2 + g 2 x2 − 2f gx − f x4 .

Since x 6= 0 then φ(x) = 0. It remains to show that x is not integral over K[f, g]. It is enough to show that φ(z) is irreducible over K[f, g]. This equivalent to prove that the polynomial ψ(u, v, z) = uz 4 − v 2 z 2 + 2uvz − u2 is irreducible over K. We consider this polynomial as a polynomial in v of degree 2 and we find that its discriminant is equal to 4uz 6 . Therefore ψ is irreducible over K.

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(c) We have g 2 /f = (x + y)2 ∈ K(f, g) ∩ K[x, y]. Suppose that g 2 /f ∈ K[f, g], then there exists ρ(u, v) ∈ K[u, v] such that g 2 = f ρ(f, g). Since f and g are algebraically independent over K, we may compare the degrees in f of the two sides of this identity and get a contradiction. Therefore K[f, g] is strictly contained in K(f, g) ∩ K[x, y]. Exercise 9.10. Let K be field, d ≥ 2 be an integer and x1 , . . . , xn be algebraically independent variables. Set ~x = (x1 , . . . , xn ). For any F (~x) ∈ K[~x], denote by n−1 Sd (F ) the univariate polynomial defined by Sd (F ) = F (y, y d , . . . , y d ). (1) Show that the map Sd : F → Sd (F ) is a surjective morphism of Kalgebras from K[~x] into K[y]. (2) Show that the restriction of Sd to the set of polynomials F (~x), such that degxi < d for i = 1, . . . , n, is injective. (3) Let Ω be an algebraically closed field containing K and F (~x) ∈ K[~x]. Suppose that F is reducible over Ω. Show that F is reducible over K, where K is the algebraic closure of K contained in Ω. (4) Factorize F (x1 , x2 ) = 6x31 + 3x2 x21 + x22 + 2x2 x1 + 6x1 + 3x2 into irreducible factors in Q[x1 , x2 ]. Solution 9.10. (1) Clearly Sd is a morphism of algebras. Let f (y) ∈ K[y], then f (y) = Sd (f (x1 )), thus Sd is surjective. (2) Since the image of a monomial by Sd is a monomial, it is sufficient to prove that any monomial ay i is the image of one monomial M = ax1 · · · xn ∈ K[~x]. n−1 Suppose that Sd (M ) = ay i , then ay i = ay i1 +di2 +···+d in , hence i = i1 + di2 + · · · + dn−1 in . This implies that i1 , . . . , in are the digits in base d of i. Since the representation in base d exists and is unique, then the restriction of Sd is injective and also bijective by (1). (3) Choose a total order (for example the lexicographic order) in the set of monomials ax1 · · · xn ∈ K[~x]. In order to prove the result, we may suppose that F is monic relatively to this order. Suppose that F (~x) = F1 (~x)F2 (~x) is a non trivial factorization of F (~x) in Ω[~x]. We may suppose that F1 and F2 are monic. Let d be a positive integer such

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that d > degxi F for i = 1, . . . , n, then       n−1 n−1 n−1 F y, y d , . . . , y d = F1 y, y d , . . . , y d F2 y, y d , . . . , y d .

This is a factorization in Ω[y] of a polynomial in one variable with coefficients in K. Moreover, by (2), it is a non trivial factorization. n−1 The coefficients of F1 (y, y d , . . . , y d ) are symmetric functions of a set n−1 ¯ the of roots of F (y, y d , . . . , y d ). Since these roots are elements of K, d dn−1 ¯ coefficients of F1 (y, y , . . . , y ) belong to K. The same claim is true n−1 for F2 (y, y d , . . . , y d ). It follows that F1 (~x) and F2 (~x) ∈ K[~x]. (4) Here we let d = 3 and we get F (y, y 3 ) = y 6 + 3y 5 + 2y 4 + 9y 3 + 6y = y(y 5 + 3y 4 + 2y 3 + 9y 2 + 6). We factorize φ(y) = y 5 + 3y 4 + 2y 3 + 9y 2 + 6 over Q. It is easy to verify that φ does not vanish for ±1, ±2, ±3, ±6, so that φ(y) has no factor of degree 1. It follows that if φ(y) is reducible over Q, then φ(y) = g(y)h(y), with g(y) and h(y) ∈ Z[y], deg g = 2 and deg h = 3. Since φ(y) ≡ y 2 (y 3 + y 2 + 1) (mod 2), then g(y) = y 2 + 2ay + 2b and h(y) = y 3 + (2c + 1)y 2 + 2dy + 2e + 1,

where a, b, c, d, e ∈ Z. Identifying the coefficients of xi , for i = 0, . . . , 4, in the identity φ(y) = g(y)h(y), we obtain the following equations b(2e + 1) = 3, a(2e + 1) + 2bd = 0, e + 2ad + b(2c + 1) = 4, d + b + a(2c + 1) = 1, a + c = 1. The first equation shows that 2e + 1 = , b = 3 or 2e + 1 = 3, b =  with  = ±1.

• If 2e + 1 =  and b = 3. The second equation implies a = −6d. The last equation gives c = 1 + 6d. Substituting these values of a, b, c, e, in terms of d and e in the fourth equation leads to 72d2 − 17d + 1 − 3 = 0. The ˙ discriminant of this quadratic equation is equal to ∆ = 1 + 2883. If  = −1, then ∆ < 0 and the quadratic equation has no integral (nor real) solution. We conclude that φ(y) is irreducible over Q. If  = 1, then ∆ = 5 · 173. It follows that ∆ ≡ 0 (mod 5) but ∆ 6≡ 0 (mod 52 ). We are led to the same conclusion as before.

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• If 2e + 1 = 3 and b = . By the same method as above, we find that d satisfies the following equation 8d2 − 8d + 27(1 − ) = 0. Its discriminant is equal to 64 if  = 1 and −52 if  = −1. If  = −1, then φ(y) is irreducible over Q. If  = 1, then the roots of the quadratic equation are given by d = 0 or d = 1. If d = 0, the second equation implies that a = 0. The last equation gives c = 1. The fourth one implies b = 1. It follows that g(y) = y 2 + 2 and h(y) = y 3 + 3y 2 + 3. One verifies that these polynomials satisfy the relation φ(y) = g(y)h(y). Notice that g(y) and h(y) are irreducible over Q, so that we have found the factorization of φ(y) into irreducible factors over Q. We now have F (y, y 3 ) = y(y 2 + 2)(y 3 + 3y 2 + 3). Suppose that F (x1 , x2 ) = F1 (x1 , x2 )F2 (x1 , x2 ) is a non trivial factorization of F in Q[x1 , x2 ], then F (y, y 3 ) = F1 (y, y 3 )F2 (y, y 3 ) = y(y 2 + 2)(y 3 + 3y 2 + 3). Without loss of generality, we must consider the following cases (i) F1 (y, y 3 ) = y, F2 (y, y 3 ) = (y 2 + 2)(y 3 + 3y 2 + 3). (ii) F1 (y, y 3 ) = y 2 + 2, F2 (y, y 3 ) = y(y 3 + 3y 2 + 3). (iii) F1 (y, y 3 ) = y(y 2 + 2), F2 (y, y 3 ) = (y 3 + 3y 2 + 3). In the first case, F1 (x1 , x2 ) = x1 . Since x1 - F (x1 , x2 ), then we must reject this case. In the second case F1 (x1 , x2 ) = x21 + 2. Since x21 + 2 F (x1 , x2 ), this case does not hold. In the third case, we get F1 (x1 , x2 ) = x2 + 2x1

and F2 (x1 , x2 ) = x2 + 3x21 + 3.

It is now easy to verify that these polynomials satisfy the relation F (x1 , x2 ) = F1 (x1 , x2 )F2 (x1 , x2 ). Exercise 9.11. Let K be a field. (1) Let g(x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] and t a new variable. Suppose that g(tx1 , . . . , txn ) = tm u(x1 , . . . , xn ), where m is a positive integer and u is a polynomial with coefficients in K. Show that m = deg g, u = g and g is homogeneous. (2) Let f (x1 , . . . , xn ) an homogeneous polynomial with coefficients in K. Suppose that f = gh is a non trivial factorization of f in K[x1 , . . . , xn ]. Show that g and h are homogeneous.

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Solution 9.11. (1) Set g(x1 , . . . , xn ) = u(x1 , . . . , xn ) = Then g(tx1 , . . . , txn ) =

X

t

P

ij



X

X

ai1 ,...,in xi11 · · · xinn

and

bi1 ,...,in xi11 · · · xinn .

ai1 ,...,in xi11 · · · xinn = tm

X

bi1 ,...,in xi11 · · · xinn .

It follows that for any (i1 , . . . , in ), i1 + · · · + in = m and ai1 ,...,in = bi1 ,...,in , which implies that deg g = m, u = g and g is homogeneous. (2) • First proof. Suppose that g is not homogeneous. Let g + and g − (resp. h+ and h− ) be the homogeneous parts of g (resp. h) of highest and lowest degree. We have g + 6= g − but we may have h+ = h− . We also have f = g + h+ and g − h− = 0, therefore h− = 0. This implies h = 0 which is a contradiction. • Second proof. Let t a new variable then, from the identity f = gh, we deduce that td f (x1 , . . . , xn ) = g(tx1 , . . . , txn )h(tx1 , . . . , txn ), where d = deg f . We consider this identity as a factorization of td f in K[x1 , . . . , xn ][t]. We deduce that g(tx1 , . . . , txn ) = tk u(x1 , . . . , xn )

and

l

h(tx1 , . . . , txn ) = t v(x1 , . . . , xn ), where k + l = d and u, v are polynomials in x1 , . . . , xn . We conclude from (1) that g and h are homogeneous. Exercise 9.12. Let f (X, Y ) ∈ Q[X, Y ] be irreducible and let α be a root of f in an algebraic closure of Q(X). Let u(X, Y ) = un (X)Y n + · · · + u0 (X) ∈ Q[X, Y ], β = u(X, α) and let g(X, Y ) ∈ Q[X, Y ] be the unique irreducible polynomial(up to a multiplication by a constant) satisfying g(X, β) = 0. Suppose that n ≥ 1 and that un (X) has no rational root. For any h(X, Y ) denote by V (h) the set V (h) = {(a, b) ∈ C2 , h(a, b) = 0}. If V (g) ∩ Q2 , (resp. V (g) ∩ Z2 ) is finite, show that V (f ) ∩ Q2 , (resp. V (f ) ∩ Z2 ) is finite. Solution 9.12. We have g(X, u(X, α)) = g(X, β) = 0, hence g(X, u(X, Y )) = f (X, Y )q(X, Y )/D(X),

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where D(X) ∈ Q[X], q(X, Y ) ∈ Q[X, Y ]. Moreover, we may suppose that D is monic and gcd(D(X), q0 (X), . . . , qm (X)) = 1, where q0 (X), . . . , qm (X) denote the coefficients of q as a polynomial in Y . Let P (X) be an irreducible factor of D(X), then since g(X, u(X, Y )) ∈ Q[X, Y ], we conclude that P (X) divides the content (in Q[X]) of f (X, Y ), hence D(X) = 1 and g(X, u(X, Y )) = f (X, Y )q(X, Y ). Let (a, b) ∈ Q2 (resp. (a, b) ∈ Z2 ) such that f (a, b) = 0, then g(a, u(a, b)) = 0. Set c = u(a, b), then (a, c) ∈ V (g) ∩ Q2 (resp. (a, c) ∈ V (g) ∩ Z2 ). The hypotheses implies that a and c have a finite number of values. To determine b, we must solve the equation u(a, b) = un (a)bn + · · · + u0 (a) = c. For a given c and a given a not a root of un (X), this equation has a finite number of solutions b in Q and in Z and the proof is complete. Exercise 9.13. Let f (x) and g(x) ∈ Q[x] be non constant polynomials. (1) Suppose that g(Q) ⊂ f (Q). Show that there exists P (x) ∈ Q[x] such that g(x) = f (P (x)). (2) If g(Q) = f (Q), deduce that there exist a and b ∈ Q, a 6= 0 such that g(x) = f (ax + b). (3) Show that a similar result to (1) does not hold over a finite field. Solution 9.13. (1) Let F (X, Y ) = f (X) − g(Y ) and let F (X, Y ) = F1 (X, Y ) · · · Fr (X, Y ) be its factorization into irreducible factors over Q. Clearly the leading coefficient of Fi (X, Y ) for i = 1, . . . , r as a polynomial in X and also as a polynomial in Y is a constant. Let A be the set of elements y of Q such that for any i ∈ {1, . . . , r}, Fi (X, y) is irreducible over Q, then by Hilbert’s irreducibility theorem [Schinzel (2000), Chap. 4.4], A is infinite. Let y ∈ A, then g(y) ∈ f (Q), hence there exists x ∈ Q such that f (x) − g(y) = 0. Therefore there exists i0 ∈ {1, . . . , r} such that for infinitely y ∈ Q, the polynomial Fi0 (X, y) is irreducible over Q Pn and has a root x ∈ Q. Set Fi0 (X, Y ) = i=0 ai (Y )X i , where n ≥ 1, ai (Y ) ∈ Q[Y ] for i = 1, . . . , r and an (Y ) = c ∈ Q. For y ∈ A, Fi0 (X, y) is irreducible and has a rational root, hence n = 1 and Fi0 (X, Y ) = cX + a0 (Y ). We now have f (X) − g(Y ) = (cX + a0 (Y ))G(X, Y ),

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with G(X, Y )
∈ Q[X, Y ]. Substituting for X the value −a0 (Y )/c yields f − a0 (Y )/c = g(Y ) and the proof is complete. (2) Easy. (3) Consider the polynomials f (X) = X q and g(X) = X q − X over Fq , then clearly {0} = g(Fq ) ⊂ f (Fq ) = Fq . If there exists P (X) ∈ Fq [X] such that g(X) = f (P (X)), then deg P = 1. Set P (X) = aX + b, then f (aX + b) = aX q + b 6= g(X). Exercise 9.14. Let K be a field, f (x) ∈ K[x] of degree d ≥ 1. We say that f is functionally decomposable over K if there exist two non constant polynomials u(x), v(x) ∈ K[x] such that f (x) = u(v(x)). In this case u(x) (resp. v(x)) is called the left composition factor (resp. right composition factor) of f (x). (1) Show that the following propositions are equivalent. (i) f is functionally decomposable over K. (ii) There exists h(x) ∈ K[x]\K, deg h < deg f such that (h(x)−h(y)) | (f (x) − f (y)) in K[x, y]. (2) Let h(x) ∈ K[x] be non constant. Suppose that deg h < deg f . Show that the following propositions are equivalent. (i) h(x) is a right composition factor of f (x). (ii) There exists g(x) ∈ K[x] such that h(x) − t divides f (x) − g(t) in K[x, t]. (iii) For any a ∈ K, there exists b ∈ K such that (h(x) − a) | (f (x) − b). (iv) There exist a positive integer r > deg f / deg h, distinct elements of K, a1 , . . . , ar and elements of K, b1 , . . . , br not necessarily distinct such that for any i ∈ {1, . . . , r}, h(x) − ai | f (x) − bi . (v) There exist a ∈ K, a positive integer k and a polynomial g(x) ∈ K[x] such that Resx (f (x) − z, h(x) − t) = a(z − g(t))k . (vi) There exists a polynomial g(x) ∈ K[x] such that the remainder of the Euclidean division of f (x) − z by h(x) − t in K[t, z][x] is equal to g(t) − z.

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Solution 9.14. (1) • (i) ⇒ (ii). If f (x) = u(v(x)), where u and v are non constant, then v(x) − v(y) | f (x) − f (y) in K[x, y]. • (ii) ⇒ (i). Let d = deg f , e = deg h and D = bd/ec. The h(x)-adic expansion of f (x) [Lang (1965), Th. 9, Chap. 5.5] has the form: f (x) =

D X

fi (x)(h(x))i ,

i=0

where fi (x) is a polynomial with coefficients in K of degree at most e − 1 for i = 0, . . . , D. We have D X i=0

fi (x)(h(x))i −

D X i=0

fi (y)(h(y))i ≡ 0

(mod h(x) − h(y)).

Since h(x) ≡ h(y)

(mod h(x) − h(y)),

then D X i=0


fi (x) − fi (h) (h(y))i ≡ 0

(mod h(x) − h(y)).

Therefore, there exists A(x, y) ∈ K[x, y] such that D X i=0



fi (x) − fi (h) (h(y))i = A(x, y) h(x) − h(y) .

Comparing the degrees of the two sides of this identity leads to A(x, y) = 0 and then to D X

fi (x)(h(y))i =

i=0

D X

fi (y)(h(y))i .

i=0

Substituting 0 for x in this identity, we obtain two h(y)-adic expansions of the same polynomial in K[y]. It follows that fi (y) ∈ K for PD i = 0, . . . , D. Set fi (y) = bi and gi (y) = i=0 bi xi . Then f (y) =

D X

bi (h(c))i = g(h(x),

i=0

which means that h(x) is a right composition factor of f (x).

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(2) We prove the following implications. (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i) ⇔ (v) and (ii) ⇔ (vi). The implications (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are easy and will be omitted. • (iv) ⇒ (i). The conditions contained in (iv) express the fact that f (x) is a solution of the following system of congruence equations: f (x) ≡ bi

(mod h(x) − ai ),

i = 1, . . . , r.

By the Chinese Remainder Theorem, a solution of this system is Pr given by f0 (x) = i=1 bi Yi (x), where Yi (x) is a polynomial satisfying the conditions:   Y Yi (x) ≡ 1 (mod h(x) − ai ) and Yi (x) ≡ 0 mod (h(x) − aj ) . j6=i

We may take

Yi (x) =

Y Y (h(x) − aj )/ (ai − aj ). j6=i

We thus have

j6=i

f (x) = f0 (x) + u(x)

r Y

i=1

(h(x) − ai ),

where u(x) is a polynomial. We clearly have deg f0 < r deg h and by (iv), deg f < r deg h, hence f (x) = f0 (x) =

r X i=1

Set bi , (a j6=i i − aj )

ci = Q

Y bi (h(x) − aj ). j6=i (ai − aj )

Q

gi (x) = ci

Y (x − aj )

j6=i

and

j6=i

then

f (x) =

r X

gi (h(x)) = g(h(x)),

i=1

hence h(x) is a right composition factor of f (x).

g(x) =

r X i=1

gi (x),

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• (i) ⇒ (v). Using Exercise 8.6(4), we obtain


k Resx (f (x) − z, h(x) − t) = a H(z, t) ,

where a ∈ K, k is a positive integer and H(z, t) is an irreducible polynomial over K satisfying H(f (x), h(x)) = 0. This polynomial H satisfying this property is unique up to multiplication by a constant in K. Since h(x) is a right composition factor of f (x), there exists g(x) ∈ K[x] such thatf (x) − g(h(x)) = 0, hence H(z, t) = z − g(t) and the proof is complete. • (v) ⇒ (i). Substitute f (u) for z and h(u) for t, where u is a new variable, in the identity and obtain

Resx (f (x) − z, h(x) − t) = a(z − g(t))k

Resx (f (x) − f (u), h(x) − h(u)) = a(f (u) − g(h(u)))k .

Since the polynomials for which we compute the resultant have the common root x = u, then the resultant vanishes, hence f (u) = g(h(u)) and h(x) is a right composition factor of f (x). • (ii) ⇒ (vi). Since h(x) − t | f (x) − g(t), there exists a polynomial q(x, t) such that f (x) − g(t) = (h(x) − t)q(t, x), hence • (vi) ⇒ (ii). Clear.

f (x) − z = (h(x) − t)q(t, x) + g(t) − z.

Exercise 9.15. Let u(x), v(x) be two non constant polynomials with coefficients in a factorial ring A, p be a prime of A and K the fraction field of A. (1) Suppose that u(x) and v(x) satisfy the Eisenstein’s irreducibility criteria for the prime p. Show that the same property holds for u(v(x)). (2) Deduce that if the polynomial f (x) with coefficients in A is p-Eisenstein, then for any positive integer k, f [k] is irreducible over K, where f [k] is the k-th iterate (for the composition of functions) of f (x). (3) Let u1 , . . . , un , x be algebraically independent variables over K. Let F (x) = xn + u1 xn−1 + · · · + un ∈ K(u1 , . . . , un )[x]. Show that for any positive integer k, F [k] (x) is irreducible over K, where F [k] is the k-th iterate of F .

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Solution 9.15. (1) Set u(x) =

n X

ui xi

i=0

and v(x) =

m X

vi x i ,

i=0

where ui and vi ∈ A. Then u(x) ≡ un xn

(mod p)

and v(x) ≡ vm xm

(mod p).

n mn Therefore u(v(x)) ≡ un vm x (mod p). It follows that all the coefficients of u(v(x)) except the leading one are congruent to 0 modulo p. We have

u(v(0)) = un v0n + · · · + u1 v0 + u0 ≡ u0

(mod p2 ),

hence the result. (2) The conclusion for the iterates of a polynomial is trivial. (3) Consider the unique morphism of rings Φ : K[u1 , . . . , un , x] → K[T, x] such that Φ(ui ) = T , for i = 1, . . . , n and Φ(x) = x, then Φ(F (x)) = xn + T xn−1 + · · · + T, hence Φ(F (x)) is T -Eisenstein. It follows that Φ(F )[k] is T -Eisenstein, hence irreducible over K. Since Φ(F [k] ) = Φ(F )[k] , then F (x)[k] is irreducible. Exercise 9.16. (1) Let K be field, E be an extension of K, f (x), g(x) and h(x) ∈ K(x). If K(f, g) = K(h), show that E(f, g) = E(h). (2) Suppose that f nor g is constant. Show that there exists P (z, w) ∈ K[z, w], irreducible such that P (f, g) = 0. Show that P (z, w) is absolutely irreducible. Solution 9.16. (1) Since E(f, g) is the smallest field containing E, f , g and since E(h) contains E, f , g, then E(f, g) ⊂ E(h). On the other hand, we have h ∈ K(f, g) ⊂ E(f, g), hence E(h) ⊂ E(f, g). Therefore E(f, g) = E(h).

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(2) We have K ⊂ K(f ) ⊂ K(f, g) ⊂ K(x). Clearly the transcendence degree over K of K(x) and of K(f ) is equal to 1. Therefore the same property is true for K(f, g). It follows that this last field is an algebraic extension of K(f ). Let Q(f, w) ∈ K(f )[w] be the minimal polynomial of g over K(f ). Consider the set of the denominators of all the coefficients of this polynomial and let d(f ) be their least common multiple. Then clearly P (z, w) := d(z)Q(z, w) is ¯ be an algebraic cloirreducible in K[z, w] and P (f, g) = 0. Let K sure of K. By Luroth’s Theorem [Schinzel (2000), Th. 2, Chap. 1.1], there exists h(x) in K[x] such that K(f, g) = K(h). By (1), we ¯ ¯ ¯ w], irreducible such that have K(f, g) = K(h). Let P¯ (z, w) ∈ K[z, ¯ w]. P¯ (f, g) = 0. Clearly P¯ (z, w) | P (z, w) in K[z, Consider the following diagram and the similar one when K is replaced ¯ by K. K(x) K(h)

K(f )

K(g)

K We have degz P (z, w) = [K(f, g) : K(g)] = deg h/ deg g ¯ ¯ = [K(f, g) : K(g)] = degz P¯ (z, w). Therefore P (z, w) and P¯ (z, w) are equal up to a multiplication by a ¯ and this implies that P (z, w) is absolutely irreducible. constant in K

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Exercise 9.17. Let K be a field, Ω be an algebraic closure of K and F (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ]. Let A be a Cartesian subset of Ωn , i.e. a set of the form A = A1 ×A2 ×· · ·×An . Suppose that F (a1 , . . . , an ) = 0 for any (a1 , . . . , an ) ∈ A. (1) If |Ai | > degxi F , show that F = 0. (2) Deduce that there exist h1 , . . . , hn ∈ K[x1 , . . . , xn ] such that Pn F (x1 , . . . , xn ) = i=1 hi gi , deg hi ≤ deg F − deg gi , where gi ∈ K[xi ] Q and gi (xi ) = b∈Ai (xi − b) for i = 1, . . . , n. Solution 9.17.

(1) We proceed by induction on n. If n = 1, the statement is trivial. Suppose that the result holds for n−1 indeterminates. Write F (x1 , . . . , xn ) in the form F (x1 , . . . , xn ) = F0 + F1 xn + · · · + Fm xm n,

where Fi (x1 , . . . , xn−1 ) ∈ K[x1 , . . . , xn−1 ] for i = 0, . . . , m. Let (a1 , . . . , an−1 ) ∈ A1 × A2 × · · · × An−1 . Consider the univariate polynomial f (xn ) = F0 (a1 , . . . , an−1 ) + F1 (a1 , . . . , an−1 )xn + · · · + Fm (a1 , . . . , an−1 )xm n.

It satisfies the condition f (an ) = 0 for any an ∈ An , hence f (xn ) = 0. It follows that for any i = 0, . . . , m, we have Fi (a1 , . . . , an−1 ) = 0 for any (a1 , . . . , an−1 ) ∈ A1 × A2 × · · · × An−1 . Therefore, by the inductive hypothesis, Fi (x1 , . . . , xn−1 ) = 0 for i = 0, . . . , m. (2) We first show that F (x1 , . . . , xn ) may be written in the form n X F (x1 , . . . , xn ) = hi gi + F¯ (x1 , . . . , xn ), i=1

where hi ∈ K[x1 , . . . , xn ], deg hi ≤ deg F − deg gi , Y gi (xi ) = (xi − b) ∈ K[xi ] b∈Ai

for i = 1, . . . , n and degxi F¯ < deg gi . If n = 1, the result follows from the euclidean division of F by g1 . Suppose that n ≥ 2. Divide F by g1 in K[x2 , . . . , xn ][x1 ]. We have F = g1 h1 + F¯1 where F¯1 and h1 ∈ K[x2 , . . . , xn ][x1 ] with degx1 F¯1 < deg g1 , degxi F¯1 ≤ degxi F for i 6= 1 and degxi h1 ≤ degxi F − deg g1 . We do the same operation with F¯1 (x1 , . . . , xn ) instead of F and g2 instead of g1 . We obtain F¯1 = g2 h2 + F¯2 with similar conditions on the degrees of the polynomials. We iterate the process and obtain the announced identity with F¯ = F¯n . We apply Pn (1) to F¯ and conclude that F¯ = 0 and F (x1 , . . . , xn ) = i=1 hi gi .

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Exercise 9.18. Let K be a field, f (x, y) ∈ K[x, y) and f + (x, y) be the leading homogeneous form of f . Suppose that f + is square free. We say that two non trivial factorizations f + (x, y) = u1 (x, y)u2 (x, y)

and f + (x, y) = v1 (x, y)v2 (x, y)

of f in K[x, y) are essentially the same if vi = λi ui for i = 1, 2 or v1 = λ1 u2 and v2 = λ2 u1 , where λ1 and λ2 ∈ K ? . In the opposite case we say that the factorizations are essentially different. Let d be the number of essentially different factorizations of f + in K[x, y) (1) Let V0 (f ) = {λ ∈ K, f (x, y) − λ is reducible over K}. Show that |V0 (f )| ≤ d. (2) Determine these values of λ ∈ V0 (f ) when K = Q and f (x, y) = x4 − y 4 − x2 − 5y 2 . (3) Show that the result in (1) does not hold if we remove the square free condition on f + . (4) Suppose that f + has no factor of degree 1 over K. Let V1 (f ) = {(a, b, c) ∈ K 3 , f (x, y) − (ax + by + c) is reducible over K}. Show that |V1 (f )| ≤ d. Solution 9.18. (1) We define a one to one map φ from V0 (f ) into the set of the distinct factorizations of f + , which will be denoted by F(f + ). Let λ ∈ V0 (f ) and f (x, y) − λ = gλ (x, y)hλ (x, y) a non trivial factorization of f − λ over K. We may suppose that deg g = k ≥ l = deg h. Set f (x, y) =

n X

fi (x, y),

i=0

gλ (x, y) =

k X

gi (x, y)

i=0

hλ (x, y) =

l X i=0

hi (x, y),

and

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where k + l = n, fi (x, y), (resp. gj (x, y)) (resp. ht (x, y)) is 0 or homogeneous of degree i (resp. j), (resp. t). The existence of the factorization is equivalent to the solubility of the following system of equations: fn = gk hl fn−1 = gk hl−1 + gk−1 hl ........................................................... fk = gk h0 + gk−1 h1 + · · · + gk−l hl

fk−1 = gk−1 h0 + gk−2 h1 + · · · + gk−l−1 hl

................................................................. fk = gk h0 + gk−1 h1 + · · · + gk−l hl

fk−1 = gk−1 h0 + gk−2 h1 + · · · + gk−l−1 hl

.................................................................... fl = gl h0 + gl−1 h1 + · · · + g0 hl

ft = gt h0 + gt−1 h1 + · · · + g0 ht

for

f0 − λ = g0 h0 .

1≤t≤l−2

These equations may be written in the form: fn = gk hl hl−1 gk−1 fn−1 = + fn hl gk ...... = ................................................ fk h0 gk−1 h1 gk−l = + + ··· fn hl gk hl gk fk−1 gk−1 h0 gk−2 h1 gk−l−1 = + + ··· + fn gk hl gk hl gk ...... = .................................................... fl gl h0 gl−1 h1 g0 = + + ··· + fn gk hl gk hl gk ft gt h0 gt−1 h1 g0 ht = + + ··· + for fn gk hl gk hl gk hl f0 − λ = g0 h0 .

(Eq 0) (Eq 1)

(Eq n-k) (Eq n-(k-1))

(Eq n-l) 0 d and let r = e − d. Then r ≥ 1 and since e

e

d

f2 (xp ) − f2 (y p )

d

divides f3 (xp ) − f3 (y p ),

we get r

r

f2 (xp ) − f2 (y p )

divides f3 (x) − f3 (y).

0

Since r ≥ 1, then (x − y)2 | f3 (x) − f3 (y). We deduce that f3 (x) = 0 f3 (y) = 0, which is a contradiction. We conclude that e ≤ d. Since e

e

f2 (xp ) − f2 (y p )

divides f3 (xp

s+e

) − f3 (y p

s+e

),

then

0

f2 (x) − f2 (y)

s

s

divides f3 (xp ) − f3 (y p ).

(b) Since f2 (t) 6= 0, then, by (1), f2 (t) is a right composition factor of s f3 (tp ). s (c) According to the result of (b), let F (t) ∈ K[t] such that f3 (tp ) = e F (f2 (t)). Then substituting tp for t, we get f (t) = F (f1 (t)), that is f1 (t) is a right composition factor of f (t). (4) Set f (t) = F (f1 (t)), then f (x) − f (y) = F (f1 (x)) − F (f1 (y)). This shows that f1 (x) − f1 (y) is a divisor of f (x) − f (y) in K[x, y].

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Chapter 10

Integral elements, Algebraic number theory

Exercise 10.1. Let K be a field, x be an indeterminate and Ω be an algebraic closure of K(x). For any positive integer n let x1/n denotes one fixed root of the polynomial y n − x in Ω. Denote by Q+ the set of non negative rational numbers. Let A be the subring of Ω whose elements f (x) have the form P f (x) = r∈Q+ ar xr , where ar ∈ K, ar = 0 for all r except finitely many of them. P (1) For any f (x) ∈ A, f (x) = r∈Q+ ar xr , let S(f ) = {r ∈ Q+ , ar 6= 0}. Define the degree of f (x) as follows. If f (x) = 0 set deg f = −∞. If f (x) 6= 0, set deg f = maxr∈S(f ) r. Let g(x) and h(x) be elements of A. Show that deg gh = deg g + deg h and deg(g + h) ≤ max(deg g, deg h). (2) Show that A is an integral domain and A? = K ? . (3) Show that A is integrally closed. (4) Show that A is integral over K[x]. (5) Let f (x) ∈ K[x]. If f (x) is irreducible in A, show that f (x) is irreducible in K[x]. Show that the converse is false. (6) Let f (x) ∈ A such that deg f > 0. If f (x) is irreducible in A, show that it is prime. (7) Let a ∈ K ? and r be a positive rational number. Show that any divisor (in A) of axr has the form bxs , where b ∈ K ? , s is a non negative rational number and s ≤ r. (8) Show that x cannot be written in the form x = up1 · · · pk , where u ∈ A? and p1 , . . . , pk are irreducible elements of A. (9) Let B be a Noetherian ring. Show that any b ∈ B \ B ? , b 6= 0, may be expressed as a product of irreducible elements of B. (10) Deduce from (8) and (9) that A is not Noetherian. 279

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(11) Show that A is not local. Hint. One may use the following result. A commutative ring B is local if and only if B \ B ? is an ideal. Solution 10.1. (1) If f (x) = 0 or g(x) = 0 then the assertion on the degrees of f (x)g(x) and f (x) + g(x) are satisfied. If f (x) 6= 0 and g(x) 6= 0, set f (x) = P P r and g(x) = s≤s0 bs xs with ar0 6= 0 and bs0 6= 0. Then r≤r0 ar x P f (x)g(x) = ar bs xr+s , with ar0 bs0 6= 0 and r + s < r0 + s0 if (r, s) 6= (r0 , s0 ), hence deg f g = deg f + deg g. We omit the proof for the degree of f (x) + g(x). (2) Obvious. (3) Let f (x) ∈ Frac(A). Suppose that f (x) is integral over A, that is, there exist a positive integer n and a0 (x), . . . , an−1 (x) ∈ A such that f (x)n + an−1 (x)f (x)n−1 + · · · + a0 (x) = 0. There exists a positive integer N such that f (x) ∈ K(x1/N ) and ai (x) ∈ K[x1/N ]. Set f (x) = g(x1/N ) = u(x1/N )/v(x1/N ) with u(y), v(y) ∈ K[y] and gcd(u(y), v(y)) = 1. Substituting xN for x in the integral equation satisfied by f (x), we obtain: g(x)n +an−1 (xN )g(x)n−1 +· · ·+a0 (xN ) = 0. It follows that g(x) is integral over K[x]. Since g(x) ∈ K(x) and K[x] is integrally closed, then g(x) ∈ K[x], hence v(x) | u(x). We deduce that v(x) ∈ K ? and then f (x) ∈ A. P (4) Let f (x) = r≤r0 ar xr be an element of A. To obtain that f (x) is integral over K[x], it is sufficient to prove that xr is integral over K[x]. Set r = p/q, where p and q are positive integers, then it is sufficient to prove that x1/q is integral over K[x]. This claim is true since x1/q is a root of the monic polynomial y q − x with coefficients in K(x). (5) The first part of the statements of (5) is obvious and will be omitted. We prove the second part by finding a counterexample. Let f (x) = x, then f (x) is irreducible in K[x] but reducible in A since f (x) = (x1/2 )2 . (6) Suppose that f (x) divides a(x)b(x) in A, then a(x)b(x) = f (x)g(x), where g(x) ∈ A. Let N be a positive integer such that f (x), g(x), a(x), b(x) ∈ K[x1/N ], then f (xN ), g(xN ), a(xN ), b(xN ) ∈ K[x] and we have a(xN )b(xN ) = f (xN )g(xN ). Since f (x) is irreducible in A, then f (xN )) is irreducible in A, hence irreducible in K[x] by (5). The preceding identity implies that f (xN ) divides one, say a(xN ), among the polynomials a(xN ) and b(xN ) in K[x]. Let c(x) ∈ K[x] such that a(xN ) = c(x)f (xN ). Substituting x1/N for x, we get a(x) =

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c(x1/N )f (x), which proves that f (x) divides a(x) in A, establishing that f (x) is prime in A. (7) Let f (x) ∈ A be a divisor of axr and let g(x) ∈ A such that axr = f (x)g(x). Let N be a positive integer such that axr , f (x), g(x) ∈ K[x1/N ]. Then we get a factorization of axrN in K[x]: axrN = f (xN )g(xN ). It follows that f (xN ) = bxt , where b ∈ K ? and t is a positive integer such that t ≤ rN . Let s = t/N , then f (x) = bxs . (8) Suppose that x = up1 · · · pk , where u ∈ A? and p1 , . . . , pk are irreducible elements of A. Then by (7) pi = bi xsi , where si ≤ r. But bi xsi = bi (xsi /2 )2 is reducible in A. Thus x cannot be expressed as a product of irreducible elements of A. (9) Let T = {t ∈ B \ B ? , t 6= 0, t is not a product of irreducibles of A}.

Suppose that T 6= ∅. Consider the set I of ideals tB for t ∈ T . By the ascending chain condition, this set of ideals contains a maximal element, say t0 B. Since t0 B ∈ T , then t0 is not irreducible, thus t0 = t1 t2 , where t1 nor t2 is a unit and t1 t2 6= 0. Clearly t1 ∈ T or t2 ∈ T . Without loss of generality, we may suppose that t1 ∈ T . We now have t0 B ( t1 B, contradicting the maximality of t0 B in I. We conclude that T = ∅, that is every non zero element of B, not a unit, may be expressed as a product of irreducible elements of B. (10) By (9), if one finds an element f (x) ∈ A \ A? , f (x) 6= 0, which cannot expressed as a product of irreducibles, then one may conclude that A is not Noetherian. By (8), x is such an element. Thus A is not Noetherian. Remark. We may prove that A is not Noetherian in the following way. Consider the chain of ideals k

xA ⊂ x1/2 A ⊂ · · · ⊂ x1/2 A ⊂ x1/2

k+1

A ⊂ ··· . k

k+1

It is seen that this chain never terminates. For if x1/2 A = x1/2 k k+1 then x1/2 | x1/2 , which is a contradiction to (7). (11) We use the hint. Here A? = K ? and then

A,

A \ A? = {0} ∪ {f (x) ∈ A, deg f > 0}.

Since x, x + 1 ∈ A \ A? and x + 1 − x 6∈ A \ A? , then A is not local. Exercise 10.2. Define the polynomials for n ≥ 1.

x n




for n ∈ N by

x 0




= 1 and

x n




=

x(x−1)···(x−(n−1)) n!

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(1) Show that these polynomials constitute a basis of Q[x] over Q. (2) Let

(3) (4) (5) (6) (7) (8)

A = {f (x) ∈ Q[x], f (Z) ⊂ Z}.
Show that A is a free Z-module and B = { nx , n ∈ N} is one of its bases. Show that A is a subring of Q[x] and A? = {−1, 1}. Show that Frac(A) = Q(x). Show that A is integrally closed. Show that A is not Noetherian.
Show that for n ≥ 1, nx is irreducible in A. Give an example of an element of A which has at least two distinct factorizations. For any f (x) ∈ A\{0}, denote by d(f ), the integer d(f ) = gcda∈Z f (a). Let f (x) and g(x) ∈ A \ {0}. Show that (i) d(f )d(g) | d(f g). (ii) d(f m ) = d(f )m for any m ≥ 1.

Show that we may have d(f )d(g)
6= d(f g). (9) Let f (x) = a0 + a1 x + · · · + an nx with ai ∈ Z and an 6= 0. Show that d(f ) = gcd(f (0), f (1), . . . , f (n)) = gcd(a0 , a1 , . . . , an ).

Solution 10.2.
(1) Suppose that a0 + a1 x + · · ·
+ an nx = 0, with a0 , a1 , . . . , an ∈ Q. Set f (x) = a0 + a1 x + · · · + an nx . From f (0) = 0, we get a0 = 0. Suppose that aj = 0 for any j ≤ k − 1, then 0 = f (k) = ak . To prove that B generates the vector space, it is sufficient to show that xj , for j ≥ 0 is a linear combination of the elements of B with rational coefficients. Since   x n! = x(x − 1) · · · (x − (n − 1)) = xn + lower degree terms, n the conclusion follows by induction. (2) Obviously A is a Z-module. A similar reasoning as in (1), shows that B is a basis of A over Z. (3) It is clear that the product of two elements of A is an element of A. Let f (x) ∈ A? and let g(x) ∈ A such that f (x)g(x) = 1, then deg f = deg g = 0 and then f (x) = ±1. (4) From Z[x] ⊂ A ⊂ Q[x], we conclude that Q(x) = Frac(Z[x]) ⊂ Frac(A) ⊂ Frac(Q[x]) = Q(x),

thus Frac(A) = Q(x).

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(5) Let f (x) ∈ Q(x) be integral over A. Write f (x) in the form f (x) = u(x)/v(x) with u(x) and v(x) ∈ Z[x] and gcd(u(x), v(x)) = 1. Representing u(x) and v(x) in the form u(x) = cont(u)u1 (x) and v(x) = cont(v)v1 (x), where u1 (x) and v1 (x) ∈ Z[x], it is seen that gcd(cont(u), cont(v)) = 1 and gcd(u1 (x), v1 (x)) = 1. Since f (x) is integral over A, there exist a0 (x), . . . , an−1 (x) such that f (x)n + an−1 (x)f (x)n−1 + · · · + a0 (x) = 0. Since Z is integrally closed, this equation shows that for any m ∈ Z such that v(m) 6= 0, we have f (m) ∈ Z. Let C be a positive real number greater than all the absolute values of the roots of v(x). We distinguish two cases. • cont(v) = 1. Here cont(v) | cont(u) and v(m) | u(m) for any integer m > C. Exercise 1.29 shows that v(x) | u(x) in Z[x], which in turn implies that f (x) ∈ Z[x] and then f (x) ∈ A. • cont(v) 6= 1. We have cont(v1 ) | cont(u) and v1 (m) | u(m) for any integer m ≥ C. Exercise 1.29 shows that v1 (x) | u(x) in Z[x]. Since cont(v1 ) = 1 and Z[x] is a unique factorization domain, then v1 (x) | u1 (x) in Z[x]. Our assumptions on u1 and v1 implies that v1 (x) = ±1. We may suppose that v1 (x) = 1 and then f (x) = cont(u)u1 (x) , where d is a positive integer such that gcd(cont(u), d) = d 1. Since f (m) ∈ Z for any m > C,
then d | u1 (m) for any m > C. Set u1 (x) = b0 + b1 x + · · · + bk xk with bj ∈ Z for j = 0, . . . , k. We claim and we omit the proof that, since u1 (x) ∈ Z[x], bj ≡ 0 (mod j!). We may write u1 (x) in the form u1 (x) = c0 + c1 x + · · · + ck x(x − 1) · · · (x − (k − 1)) with cj ∈ Z for j = 0, . . . , k. Let q be a positive integer such that dq > C and let j ∈ {0, . . . , k}. We have 0 ≡ u1 (dq) (mod d) ≡ c0 (mod d), that is c0 ≡ 0 (mod d). Suppose by induction that i!ci ≡ 0 (mod d) for any i = 0, . . . , j − 1. We have u1 (dq + j) = ck (dq + j)(dq + j − 1) · · · (dq + j − (k − 1)) + · · · + cj (dq + j)(dq + j − 1) · · · (dq + 1) + · · · + c1 (dq + j) + c0 ≡ 0

(mod d),

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hence u1 (dq + j) ≡ c0 + c1 j + c2 j(j − 1)

+ · · · + cj−1 j(j − 1) · · · 2 + cj j!

(mod d)

j(j − 1) j + 2!c2 1! 2 j(j − 1) · · · 2 + · · · + (j − 1)!cj−1 + cj j! (j − 1)!

≡ c0 + 1!c1

(mod d).

The inductive hypothesis implies that j!cj ≡ 0 (mod d). We now have cont(u)u1 (x) d c0 + c1 x + · · · + ck x(x − 1) · · · (x − (k − 1)) = cont(u)   d   c0 c1 2!c2 x k!ck x = cont(u) + x+ + ··· + . d d d 2 d k

f (x) =

This shows that f (x) ∈ A.


(6) For any non negative integer j let Pj (x) = xj . Let I1 be the ideal of A generated by P1 (x). For any prime number p let Ip be the ideal generated by P1 (x), P2 (x), . . . , Pp−1 (x), Pp (x), then I1 ⊂ I2 ⊂ . . . ⊂ Ip ⊂ . . . . We show that this sequence of ideals is not stationary. Let p1 and p2 be two consecutive primes with p1 < p2 . We are going to show that P2 (x) 6∈ Ip1 , which will imply that Ip1 ( Ip2 . Suppose that P2 (x) ∈ Ip1 , then Pp2 (x) = a1 (x)P1 (x) + · · · + ap1 (x)Pp1 (x) with aj (x) ∈ A for j = 1, . . . , p1 . Put x = p2 in this identity. We get 1 = a1 (p2 )P1 (p2 ) + a2 (p2 )P2 (p2 ) + · · · + ap1 (p2 )Pp1 (p2 ).

(Eq 1)

2 −(j−1)) . Since For any j = 1, . . . , p1 , we have Pj (p2 ) = p2 (p2 −1)···(p j! j ≤ p1 < p2 , then Pj (p2 ) ≡ 0 (mod p2 ). Now (Eq 1) reads 1 ≡ 0 (mod p1 ), which is a contradiction.
(7) Let n be a positive integer. Suppose that nx = g(x)h(x) with g(x) and h(x) ∈ A. Let k = deg g and m = deg h, then k + m = n. We
x may suppose
that k ≤ m. Let bk (resp. cm ) be the coefficient of k x (resp. m ) in the representation of g(x) (resp. h(x)) in the basis B. Equating the leading coefficients of the two sides of the above identity,

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we get 1/n! = (bk /k!)(cm /m!). If k = 0 and m = n, then b0 cn = 1, hence b0 = ±1, thus g(x) ∈ A? . Suppose that k 6= 0, then (n − k)!k! = n!bk cm = n(n − 1) · · · (n − (k − 1))(n − k)!,

hence k! = n(n − 1) · · · (n − (k − 1))bk cm . Since |bk | ≥ 1, |cm | ≥ 1 and n > k, we deduce that
|n(n − 1) · · · (n − (k − 1))bk cm | > k!, which is a contradiction. Thus nx is irreducible in
A.
x For any positive integer n, we have n nx = (x − (n − 1)) n−1 . From what was just been proved the element of A in the right side of this identity has exactly 2 irreducible factors, while the element on the left side has 1 + r irreducible factors, where r is the number of prime numbers distinct or not dividing n, so that in general 1 + r > 2, which means that these factorizations are different. Even if n is a prime number, that is r = 1, these factorizations are different since n is not associate to neither of the factors appearing on the right side of the identity. (8) (i) For any a ∈ Z, d(f ) | f (a) and d(g) | g(a), hence d(f )d(g) | f (a)g(a), thus d(f )d(g) | d(f g). (ii) By (i), d(f )n | d(f n ). Let p be a prime number such that pe || d(f )m , then e = rm and pr || d(f ). It follows that for any a ∈ Z, pr | f (a) and then prm | f (a)m . Therefore pe | d(f )m . For the last statement Let f (x) = x and g(x) = x − 1, then obviously d(f ) = d(g) = 1 and d(f g) = 2, so that d(f )d(g) 6= d(f g) in this case. (9) We have f (0) = a0 f (1) = a0 + a1 f (2) = a0 + 2a1 + a2 ··· = ················································   n f (n) = a0 + na1 + n(n − 1)/2a2 + · · · + an−1 + an . n−1 These equations show that a given positive d divides f (0), f (1), . . . , f (n) if and only if d divides a0 , a1 , . . . , an , thus gcd(f (0), f (1), . . . , f (n)) = gcd(a0 , a1 , . . . , an ). In Exercise 1.8, the following result is proved. Let K be a field of characteristic 0, f (X) be a polynomial of degree n and m > n be an integer. Then    n X m m−j−1 f (m) = (−1)n−j f (j). j n−j j=0

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This shows that d(f ) = gcd(f (0), f (1), . . . , f (n)). Remark. In Exercise 10.1(9) it is proved that in a Noetherian ring any element, except 0 and a unit, may be expressed as a product of irreducibles of the ring. For the ring subject of the present exercise, we may see that any f (x) ∈ A, f (x) 6∈ {0, −1, 1} is a product of irreducible elements of A in the following way. Let p1 , . . . , pr be the list (may be empty) of distinct or not of prime numbers dividing f (x) in A. Let f1 (x) ∈ A such that f (x) = p1 · · · pr f1 (x). If f1 (x) is irreducible in A, the polynomial f (x) is decomposed into a product of irreducible elements of A. If not, let f2 (x) be an irreducible divisor of f1 (x) in A and let f3 (x) ∈ A such that f1 (x) = f2 (x)f3 (x). We use the same reasoning for f3 (x) instead of f1 (x). Since deg f3 (x) < deg f1 (x), this process must terminates after a finite number of steps. At the end of this process, we get a factorization of f (x) into a product of irreducibles of A. This implies that the converse of the property stated in Exercise 10.1(9) does not hold, that is we may have a decomposition into irreducibles even if the ring is not Noetherian. Exercise 10.3. (1) Let K be a field. An element F (x) ∈ K[[x]] is said to be rational if there exist u(x) and v(x) in K[x] such that v(0) 6= 0 and F (x) = u(x)/v(x). P∞ Let F (x) = n=0 an xn ∈ K[[x]]. Show that F (x) is rational if and only if there exist two non negative integers s and n0 and elements b0 , . . . , bs of K such that s ≥ 1, b0 6= 0 and b0 an + b1 an−1 + · · · + bs an−s = 0, for any n ≥ n0 . (2) Let g(x) and h(x) be coprime polynomials with rational coefficients such that h(0) = 1. Suppose that the power series g(x)/h(x) has integral coefficients. Show that g(x) and h(x) ∈ Z[x]. (3) Let g(x) and h(x) ∈ Z[x] be coprime such that h(0) 6= 0. Suppose that g(x)/h(x) ∈ Z[[x]]. Show that h(0) = ±1. (4) Let α be an algebraic number of degree n such that TrQ(α)/Q (αm ) ∈ Z for any positive integer m. Show that α is an algebraic integer. Solution 10.3. (1) • Necessity of the condition. Let u(x) and v(x) ∈ K[x], with v(0) 6= 0, such that F (x) = u(x)/v(x). Set v(x) = v0 +v1 x+· · ·+vs xs and let r = deg u. Let n0 = M ax(s, r + 1), then for n ≥ n0 , the coefficient of xn in the series v(x)F (x), which coincide with u(x),

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will be 0. Since this coefficient is equal to v0 an + · · · + vs an−s , then the result follows by putting bi = vi . • Sufficiency of the condition. Let v(x) = b0 +b1 x+· · ·+bs xs . The relations on the ai and the bj implies that v(x)F (x) is a polynomial, say u(x) with coefficients in K and of degree no greater than n0 − 1. Hence F (x) = u(x)/v(x) is rational. (2) Clearing denominators and dividing out by the greatest common divisor of the set of coefficients of g(x) and h(x), we may write F (x) in the form F (x) = g(x)/h(x) = u(x)/v(x), where u(x) and v(x) are coprime polynomials with integral coefficients. Further, we may suppose that v(0) > 0. We may write a Bezout’s Identity relating u(x) and v(x) in the form a(x)u(x) + b(x)v(x) = c, where c is a positive integer and a(x) and b(x) are polynomials with integral coefficients. It follows that a(x)F (x) + b(x) = c/v(x). The left side of this identity is a series, say P∞ G(x) = n=0 qn xn , with integral coefficients. Thus we get an identity P∞ of the form n=0 qn xn = c/(v0 + v1 x + · · · + vs xs ). We may suppose that in this identity, gcdn≥0 (qn ) = 1. Otherwise, we get an analogue identity satisfying this condition by dividing c and all the coefficients qn by this greatest common divisor. We show that v0 = 1. Suppose that the contrary holds and let p be a prime factor of v0 . Since c = v0 q0 , then p | c. Let r be the smallest non negative integer such that p - qr . Write the above identity in the form r−1 ∞ X X (v0 + · · · + vs xs ) qn xn + (v0 + · · · + vs xs ) qn xn = c. n=r

n=0

Since

(v0 + · · · + vs xs ) then

r−1 X

n=0

S := (v0 + · · · + vs xs )

qn xn ≡ 0

∞ X

n=r

(mod p),

qn xn ≡ 0

(mod p).

We have S = v0 qr xr + (v0 qr+1 + v1 qr )xr+1 + (v0 qr+2 + v1 qr+1 + v2 qr )xr+2 + · · · , hence v0 qr+1 + v1 qr ≡ 0 (mod p), v0 qr+2 + v1 qr+1 + v2 qr ≡ 0 (mod p), and so on. The first congruence shows that p | v1 . By induction we get p | vi for i = 0, . . . , s, contradicting the assumptions. We conclude that v0 = 1. We now have g(x)/h(x) = u(x)/v(x), with gcd(g(x), h(x)) = gcd(u(x), v(x)) = 1 and h(0) = v(0) = 1, hence g(x) = u(x) and h(x) = v(x). Thus g(x) and h(x) have integral coefficients.

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g(x)/h(0) ∈ Z[[x]], then g(x)/h(0) and h(x)/h(0) ∈ Z[x], hence (3) Since h(x)/h(0) h(0) divides gcd(g(x), h(x)). We conclude that h(0) = ±1. Qn (4) • First Proof. Let f (x) = Irr(α, Q) = i=1 (x − αi ) and let n

g(x) = x f (1/x) =

n Y

(1 − αi x),

i=1

then g 0 (x)/g(x) =

n X i=1

=− =−

n

∞

X X −αi = −αi (αi x)k 1 − αi x i=1

n X ∞ X

(αi )

i=1 k=0

∞ X

k+1 k

k=0 ∞ X

x =−

k=0

n X i=1

αik+1

!

xk

TrQ(α)/Q (αk+1 )xk .

k=0

From the assumptions we deduce that g 0 (x)/g(x) ∈ Z[[x]]. Obviously g(0) = 1, hence we may apply (2) and obtain g(x) ∈ Z[x]. Since f (x) has the same coefficients as g(x), except that they are written in the inverse order, then f (x) ∈ Z[x]. • Second Proof. Let K = Q(α) and consider the lattice M = Z + Zα + · · · Zαn−1 . Let L be its dual lattice, that is L = {γ ∈ K, TrK/Q (γM ) ⊂ Z}. By assumptions, we have Z[α] ⊂ L. Since L is a free Z-module of rank n, then Z[α] is a free Z-module. In particular, it is finitely generated over Z. By [Lang (1965), Int. 2, Chap. 9.1], α is integral over Z. Exercise 10.4. Let K be a field, f (t), g(t), f1 (t) and g1 (t) be non constant polynomials with coefficients in K. Show that the following propositions are equivalent. (i) There exists F (t) ∈ K[t], such that f (t) = F (f1 (t)) and g(t) = F (g1 (t)). (ii) The polynomial f1 (x) − g1 (y) divides f (x) − g(y) in K[x, y]. Hint. One may use Exercise 9.20 and Exercise 9.21.

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Solution 10.4. • (i) ⇒ (ii). Obvious. • (ii) ⇒ (i). Let φ(x, y) ∈ K[x, y] be an irreducible factor of f1 (x)−g1 (y) and let u, v be variables such that φ(u, v) = 0. Let w = f (u) and w1 = f1 (u), then w = g(v) and w1 = g1 (v). Consider the following diagram of fields. K(u, v) K(u)

K(v) K(u) ∩ K(v)

K(w1

K(w)

K Since K(u) ∩ K(v) is an intermediate field between K and K(u) and since this last field is a purely transcendental extension of the first, then, by Luroth’s Theorem [Schinzel (2000), Th. 2, Chap. 1.1], K(u) ∩ K(v) is purely transcendental over K. Let w2 ∈ K(u) ∩ K(v) such that K(u) ∩ K(v) = K(w2 ). Moreover since this intersection contains w1 = f1 (u) ∈ K[u], then we may suppose that w2 = A(u) ∈ K[u] see Exercise 8.10(1). Using similar argument as above, we may conclude that K(u) ∩ K(v) = K(w3 ) and w3 = B(v) ∈ K[v]. We have w3 = (aw2 + b)/(cw2 + d), where a, b, c, d ∈ K and ad − bc 6= 0. It follows that (aw2 + b)w3 − (cw + d) = 0, hence (aA(u)) + b)B(v) − (cA(u) + d) = 0. But it is easy to see that the leading coefficient of φ(x, y), considered as a polynomial in x or in y, is a constant, hence v is integral over K[u]. We deduce, by transitivity, that B(v) is integral over K[A(u)]. This implies that, in the preceding equation satisfied by B(v), we have a = 0. Thus B(v) = λA(u) + µ and then w3 = λw2 + µ with λ ∈ K ? and µ ∈ K. Therefore we may

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suppose that K(u) ∩ K(v) = K(w2 ), where w2 = f2 (u) = g2 (v) and f2 , g2 are polynomials with coefficients in K. Set w1 = H1 (w2 ) and w = H(w2 ), where H1 and H are rational functions with coefficients in K. Using the relations f1 (u) = H1 (f2 (u)) and f (u) = H(f2 (u)), one may prove easily that H1 and H are polynomials. Since u is transcendental over K, then we have f1 (x) = H1 (f2 (x)), f (x) = H(f2 (x)), g1 (y) = H1 (g2 (y)) and g(y) = H(g2 (y)). Our assumptions read as follows H1 (f2 (x)) − H1 (g2 (y)) divides H(f2 (x)) − H(g2 (y)). Exercise 9.20 implies that H1 (x) − H1 (y) divides H(x) − H(y). The preceding exercise shows that H1 (t) is a right composition factor of H(t), that is H(t) = F (H1 (t)) with F (t) ∈ K[t]. We deduce that f (t) = H(f2 (t)) = F (H1 (f2 (t))) = F (f1 (t))

and

g(t) = H(g2 (t)) = F (H1 (g2 (t))) = F (g1 (t)) and the proof is complete. Exercise 10.5. Let K be a number field of degree n, A be its ring of integers and α ∈ A. Show that α | NK/Q (α). Solution 10.5. Let α1 = α, α2 , . . . , αn be the conjugates (distinct or not) of α over Q. We Qn Qn have NK/Q (α) = α i=2 αi . Obviously i=2 αi ∈ K. Since the αi are Qn Qn algebraic integers, then so is i=2 αi . Therefore i=2 αi ∈ A. It follows that α | NK/Q (α). Exercise 10.6. Let θ ∈ C be an algebraic integer. If θ is a unit, show that Z[θ] = Z[1/θ]. Solution 10.6. It is sufficient to prove that Z[1/θ] ⊂ Z[θ]. Let

f (x) = xn + an−1 xn−1 + · · · a1 x + ,

where  = ±1 and a1 , . . . , an−1 ∈ Z, be the minimal polynomial of θ over Q. Then 1/θ = 2 /θ = −(θn−1 + an−1 θn−2 + · · · a1 ) ∈ Z[θ].

The stated inclusion, now follows easily.

Exercise 10.7. Let a √ and b be non zero, distinct square free rational integers, K = √ Q( a, b) and A be the ring of integers of A. Suppose that a ≡ b ≡ 1 (mod 3). Show that there exists no α ∈ A such that A = Z[α].

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Solution 10.7. Suppose that A = Z[α] for some α ∈ A and let f (x) be the minimal √ √ polynomial of α over√Q. Consider the elements√of A, α1 = (1+ a)(1+ b), √ √ √ α √2 = (1 + a)(1 − b), α3 = (1 − a)(1 + b) and α4 = (1 − a)(1 − b). For i = 1, . . . , 4, let fi (x) ∈ Z[x] such that αi = fi (α). For any g(x) ∈ Z[x], denote by g(x), the reduced polynomial modulo 3 of g(x). Fix n i ∈ {1, . . . , 4}. We show that f (x) | fi (x)fj (x) for j 6= i and f (x) - fi (x) for any positive integer n. By Exercise 2.3, it is equivalent to prove the following congruences in Z[α]: fi (α)fj (α) ≡ 0 (mod p) for i 6= j and fi (α)n 6≡ 0 (mod p) for n ≥ 1, that is αi αj ≡ 0 (mod p) for i 6= j and αin 6≡ 0 (mod p). In any case, for i 6= j, αi αj may be written in form √ √ αi αj = (1 + a)(1 − a)γ or √ √ αi αj = (1 + b)(1 − b)γ, √ √ where γ ∈ √A, thus√it is sufficient to prove that (1+ a)(1− a) ≡ 0 (mod p) and (1 + b)(1 − b) ≡ 0 (mod p). This claim is true since by assumptions a ≡ b ≡ 1 (mod 3). To show that αin 6≡ 0 (mod p), it is sufficient to verify that TrK/Q (αin ) 6≡ 0 (mod p). Since α1 , . . . , α4 are conjugate over Q, then TrK/Q (αin ) = TrK/Q (α1n ) and we may suppose that i = 1. Since a ≡ b ≡ 1 (mod 3), then we have α12 − α1 = α1 (α1 − 1) √ √ √ √ √ √ = (1 + a + b + ab( a + b + ab √ √ √ = a + b + ab + a(1 + 2b) + b(1 + 2a) + 3 ab ≡0

(mod 3).

We deduce that α1n ≡ α1 (mod 3) and by conjugation, αin ≡ αi (mod 3). It follows that TrK/Q (α1n ) = α1n + · · · + α4n ≡ α1 + · · · + α4 ≡ 1

(mod 3).

We conclude that TrK/Q (αin ) 6≡ 0 (mod 3). n Fix now i ∈ {1, . . . , 4}. Since f (x) - fi (x) for any positive integer n, then there is an irreducible and monic factor gi (x) of f (x) in F3 [x] such that gi (x) - fi (x). Moreover, for any j 6= i, gi (x) | fj (x). It follows that g1 (x), . . . , g4 (x) are four distinct monic irreducible factors of f (x), which is of degree 4. We conclude that deg gi (x) = 1 for i = 1, . . . , 4, contradicting the fact that F3 [x] contains exactly three monic polynomials of degree 1. We deduce that A 6= Z[α] for any α ∈ A.

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Exercise 10.8. Let n ≥ 3 be an integer, ξn be a primitive n-th root of unity in C and Φn (x) be the minimal polynomial of ξn over Q. (1) If n is odd show that Φn (−1) = 1. (2) Let Ψn (x) be the minimal polynomial of ξn + ξn−1 over Q. (a) Show that deg Ψn = φ(n)/2, where φ is the Euler’s function. (b) Show that Φn (x) = xφ(n)/2 Ψn (x + x−1 ). (c) If n 6≡ 0 (mod 4), show that ξn +

ξn−1

(Eq 1)

is an algebraic unit.

k

(3) Let k ≥ 3 be an integer and n = 2 .

k−1

(a) Let R(y) = Resx (x2 − yx + 1, x2 + 1). Show that R(y) is a monic polynomial with integral coefficients and R(0) = 4. (b) Using the result of Exercise 3.6, show that Ψn (0) = ±2. Conclude that ξn + ξn−1 is not an algebraic unit.

Solution 10.8. Q n (1) We have: − 1)/(x − 1). Since n is odd, then d|n Φd (x) = (x d6=1 Y Φd (−1) = ((−1)n − 1)/(−1 − 1) = 1. d|n

d6=1

Using an inductive reasoning and the fact that Φ3 (−1) = 1, we conclude that Φn (−1) = 1. (2)(a) Let θn = ξn + ξn−1 , then the degree of Ψn is equal to the number of distinct conjugates of θn over Q. Any embedding σ of Q(ξn ) into C is completely determined by the value it takes on ξn . Since σ(ξn ) = ξns with s ∈ {1, . . . , n − 1} and gcd(s, n) = 1, then σ(θn ) = ξns + ξn−s = ξns +ξnn−s . Let σ and τ be two embeddings Q(ξn ) into C and suppose that σ(θn ) = τ (θn ). Let s and t such that σ(θn ) = ξns + ξnn−s and τ (θn ) = ξnt + ξnn−t , then σ(θn ) = τ (θn ) ⇔ ξns + ξnn−s = ξnt + ξnn−t

⇔ (ξns−t − 1)(ξnt − ξnn−s ) = 0 ⇔ ξns−t = 1

or ξnt = ξnn−s

⇔ t = s or t = n − s.

We deduce that the number of distinct conjugates of θn over Q is equal to φ(n)/2 and they are given by ξns + ξnn−s with gcd(n, s) = 1 and 1 ≤ s ≤ (n − 1)/2. We conclude that degΨn = φ(n)/2.

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(b) To prove the identity, notice that the polynomials on both sides are monic and are of the same degree φ(n). Since Φ(x) is the minimal polynomial of ξn and also the polynomial on the right hand side vanishes for x = ξ, the two polynomials are identical. (c) Let m = n if n is odd and m = n/2 if n ≡ 2 (mod 4), then, in any case, m is odd and m ≥ 3. This implies that ξn2 is a root of Φm (x). It follows that ξn2 + 1 is a root of the polynomial Φm (x − 1), which is monic. Moreover, by (1), the constant term of this polynomial is equal to Φm (−1) = 1. Hence ξn2 + 1 is an algebraic unit. Thus ξn + ξn−1 = (ξn2 + 1)/ξn is an algebraic unit. (3)(a) To simplify the forthcoming computations,pwe set λ = 2k−1 . Since the roots of x2 − yx + 1 are given by (y ± y 2 − 4)/2, then   k−1 R(y) = Resx x2 − yx + 1, x2 +1
= Resx x2 − yx + 1, xλ + 1    !λ !λ p p 2−4 2−4 y y − y y + + 1  + 1 = 2 2 =2+

y+

!λ p y2 − 4 + 2

y−

p

y2 − 4 2

!λ

  λ  p X j λ−j λ  /2λ y2 − 4 y =2+ j j=0   λ   X p j λ (−1)j y 2 − 4 y λ−j  /2λ + j j=0    p λ X j λ   = 2 + 2 y 2 − 4 y λ−j  /2λ j j=0 j even   λ/2   X λ = 2 + 2 (y 2 − 4)i y λ−2i  /2λ . 2i i=0

Obviously, the coefficients of R(y) are
rational integers. The leading Pλ/2 λ coefficient of R(y) is equal to i=0 2i /2λ−1 , hence equal to 1. The constant term of R(y) corresponds to i = λ/2 and then R(0) = 2 + 2(−4)λ/2 /2λ = 4.

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(b) We have k

Φ2k (x) = (x2 − 1)/(x2

k−1

k−1

− 1) = x2

+ 1,

k−1

hence Irr(ξn , Q, x) = x2 + 1. Let T (y) be the characteristic polynomial of ξ + ξ −1 , then by (2), (a), we have T (y) = Ψn (y)2 . Using Exercise 3.6, we get k−1

R(y) = Resx (x2

+ 1, yx − x2 − 1) = NQ(ξ)/Q (ξ)T (y) = Ψn (y)2 .

Using (3), (a) and identifying the constant terms in this identity, we obtain Ψn (y) = ±2. We conclude that ξn + ξn−1 is not an algebraic unit. Exercise 10.9. Let α be an algebraic integer of degree n and α1 = α, α2 , . . . , αn be its conjugates over Q. (1) If |αi | = 1 for i = 1, . . . , n, show that α is a root of unity. (2) If |αi | ≤ 1 for i = 1, . . . , n, show that α = 0 or α is a root of unity. (3) Let A(x) and B(x) be relatively prime polynomials in Z[x] with B(0) 6= 0. Suppose that the series F (x) = A(x)/B(x) belongs to Z[[x]] and that its radius of convergence is equal to 1. Show that the poles of F (x) are roots of unity. (4) Let F (x) ∈ Z[[x]] \ Z[x] and let R be its radius of convergence. Suppose that F (x) is algebraic over Q[x]. Show that R = 1 and the poles of F are roots of unity or R < 1. Solution 10.9. Qn (1) • First proof. For any integer m ≥ 1, let fm (x) = i=1 (x − αim ). Since the αim are conjugate algebraic integers, then fm (x) ∈ Z[x]. Let A be the set of these polynomial fm (x) and B the set whose elements are the roots of any of these polynomials. We show that A is finite, which in turn will imply that B is finite. For any 1 ≤ h ≤ n, let sh (x1 , . . . , xn ) be the elementary symmetric function of degree h

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of x1 , . . . , xn . Then

X αim1 · · · αimh |sh (α1m , . . . , αnm )| = 1≤i1 0. Clearly v1 , v2 + v1 , v3 , . . . , vn still generate N . Here we have (a1 − a2 )v1 + a2 (v2 + v1 ) + a3 v3 · · · + an vn = 0. The sum of the absolute values of the coefficients is given by |a1 − a2 | + |a2 | + · · · + |an | = |a1 | + |a3 | + · · · + |an |, so it is smaller than the preceding one which is minimal. In this case also we have reached a contradiction. Thus v1 , . . . , vn are linearly independent over Z. We prove the last point. M/Zα is torsion free by the first point. Since it is finitely generated, then it is free over Z. Let s ≤ r be its rank. We may suppose that {e1 , . . . , es } is a basis. To show that the short exact sequence i

s

0→Zα → M → M/Zα → 0 splits, we prove that there exists a morphism of Z-modules f : M/Zα → M such that sof = IdM/Zα . The unique morphism of modules f such that f (ei ) = ei for i = 1, . . . , s satisfies the conditions. Therefore the short exact sequence splits and M ' L Zα M/Zα. It follows that α may be completed for obtaining a basis of M and the case d = 1 is finished. Suppose now that d ≥ 2. Let b1j = c1j /d. Then by what have been proved it is possible to find a matrix A = (aij ) such that Det A = 1 and a1j = b1j . Multiply the first row of A by d and obtain a matrix B with Det B = d Det A = d and the matrix B has the required properties. Remark. Indeed the result proved in the second point is valid if we replace Z by any principal domain, [Ribenboim (2001), Th. 1, Chap. 6.2]. The third point may be proved by using a general result: If 0 → E → F → G → 0 is a short exact sequence of modules and if G is free, then the sequence splits, [Ribenboim (2001), Lemma 1, Chap. 6.2].

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(2) Let {γ1 , . . . , γn } be an integral basis of K and let c11 , . . . , c1n ∈ Z such that 1 = c11 γ1 + · · · + c1n γn . Since gcd(c11 , . . . , c1n ) = 1, then by (1), there exists a n × n matrix B = (bij ) such that b1j = c1j for Pn j = 1, . . . , n and Det B = 1. For i = 1, . . . , n, let ωi = j=1 bij γj ,     γ1 ω1 γ  ω  2  2       . .   then   = B   . Since B in invertible and B −1 has integral .  .      .  .  γn ωn coefficients, then ω1 , . . . , ωn is an integral basis. Since ω1 = 1, the proof is complete. Exercise 10.27. Let (α1 , . . . , αm ) ∈ Cm such that for any (i, j) ∈ {1, . . . , m}2 , we have Pm αi αj = k=1 aijk αk , where aijk ∈ Z for all ijk. Show that α1 , . . . , αm are algebraic integers. Solution 10.27. We have α12 = a111 α1 + · · · + a11m αm

α1 α2 = a121 α1 + · · · + a12m αm

...................................................... α1 αm = a1m1 α1 + · · · + a1mm αm , where the coefficients aijk are integers. We may express these equations in the form:     α1 α1  α2   α2     α1   ···  = M  ··· , αm αm where M is the matrix with integral coefficients given by   a111 a112 · · · a11m  a121 a122 · · · a12m  . M =  ··· ··· ··· ··· 

a1m1 a1m2 · · · a1mm This implies that α1 is an eigenvalue of the matrix M . It follows that α1 is a root of the polynomial f (x) = Det(M − xI). Since this polynomial has integral coefficients and since the leading coefficient is equal to (−1)m , then α1 is an algebraic integer. The same proof is valid for α2 , . . . , αn .

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Exercise 10.28. Let K be a number field, A be its ring of integers and θ ∈ A such that A 6= Z[θ]. Let p be a prime number and S = {α + pA, α ∈ Z[θ]}. (1) Show that S is a subring of A/pA. (2) Consider the following identities (i) S = Z[θ]/pA. (ii) S = Z[θ]/pZ[θ]. Show that (i) has no meaning and (ii) is false. Show that S is a quotient of some ring by an ideal. Solution 10.28. (1) Obvious. (2) Since pA is not contained in Z[θ], then Z[θ]/pA has no meaning. The zero elements of S and of Z[θ]/pZ[θ] are equal to pA and pZ[θ] respectively. Since Z[θ] ( A, then pZ[θ]p ( A, thus (ii) is false. We claim and we omit the proof that S = (Z[θ] + pA)/pA. Exercise 10.29. Let p be a prime number, K be a number field of degree n over Q and A be its ring of integers. Let θ be an element of A. It is known that Disc(θ) may be written in the form Disc(θ) = I(θ)2 Disc(K), where I(θ) is a non negative integer called the index of θ. Moreover I(θ) 6= 0 if and only if θ is primitive over Q. In this case I(θ) = (A : Z[θ]), see [Marcus (1977), Exercise 27, Chap. 1]. Show that the following assertions are equivalent. (i) p | I(θ). (ii) There exist an integer m, 1 ≤ m ≤ n − 1 and a0 , a1 , . . . , am−1 ∈ Z such that θm + am−1 θm−1 + · · · + a0 ≡ 0 (mod pA). (iii) Fp [θ] ( A/pA. Solution 10.29. ? (i) ⇔ (ii). • First proof. Let {ω1 , . . . , ωn } be an integral basis. For j = Pn 0, . . . , n − 1, write θj in the form θj = i=1 cji ωj . These equations may be formulated as follows

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 ω1 ω     2        .    j 1≤i≤n  = M   , where M is the matrix M = (ci )0≤j≤n−1 .   .         .    ωn θn−1 Let σj , for j = 1, . . . , n, be the distinct embeddings of K into C. Then from the above identity of matrices, we get     σj (ω1 ) 1  σ (ω )   σ (θ)   j 2    j     .  .    . =M   .    .      .    . σj (ωn ) σj (θ)n−1 

1 θ . . .





All these n equations may be summarized in the following single one
1≤j≤n
1≤j≤n σj (θi ) 0≤i≤n−1 = M σj (ωi ) 1≤i≤n . Taking determinants of both sides and squaring, we obtain Disc(1, θ, . . . , θn−1 ) = Det(M )2 Disc K,

hence I(θ) = | Det(M )|. It follows that p | I(θ) if and only if Det(M ) ≡ 0 (mod p), if and only if the columns of M are linearly dependent over Fp . This is equivalent to the existence of a0 , . . . , an−1 ∈ Z, not all 0 such that an−1 θn−1 + a1 θ + · · · + a0 ≡ 0

(mod pA),

that is if and only if there exist an integer m, 1 ≤ m ≤ n − 1 and a0 , a1 , . . . , am−1 such that θm + am−1 θm−1 + · · · + a0 ≡ 0 (mod pA). • Second proof. We have p | I(θ) ⇔ p | (A : Z[θ])

⇔ There exists γ ∈ A whose order in A/Z[θ] is equal to p

⇔ There exists γ ∈ A \ Z[θ], ⇔ There exists γ ∈ A \ Z[θ],

pγ ∈ Z[θ]

γ=

n−1 X i=0

ai θ

i

!

/p with ai ∈ Z

⇔ There exist integers m, a0 , a1 , . . . , am−1 , with 1 ≤ m ≤ n − 1

such that θm + am−1 θm−1 + · · · + a0 ≡ 0

(mod pA).

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? (ii) ⇔ (iii). (ii) means that 1, θ, . . . , θn−1 are linearly dependent over Fp . Since the dimension of A/pA over Fp is equal to n, then (iii) has the same meaning. We thus get the equivalence of (ii) and (iii). Exercise 10.30. Let K be a number field of degree n, A be its ring of integers, θ ∈ A and f (x) be its characteristic polynomial over Q. (1) Let p be a prime number. For any u(x) ∈ Z[x], denote by u(x) the polynomial with coefficients in Fp obtained from u(x) by reducing modulo p its coefficients. Let I = {u(x)) ∈ Fp [x], u(θ) ≡ 0 (mod p)}. Show that I is a non zero principal ideal of Fp [x]. Let u0 (x) be a monic generator of I and let Mθ (x) be any monic polynomial with integral coefficients such that Mθ (x) = u0 (x). Call this polynomial the minimal polynomial modulo p of θ. Show that Mθ (x) | f (x) in Fp [x]. Recall from Exercise 10.29 that p | I(θ) if and only if deg Mθ (x) < n. This equivalence may be used to answer the questions hereafter. (2) Let pA = P1e1 · · · Prer be the splitting of p in A. Write f (x) in the form f (x) =

s Y

fj (x)hj + pg(x),

(Eq 1)

j=1

where g(x), f1 (x), . . . , fs (x) are polynomials with integral coefficients, deg g < n, f1 (x), . . . , fs (x) are monic, distinct and irreducible over Fp . (a) For any i ∈ {1, . . . , r}, show that there exists a unique j = j(i) ∈ {1, . . . , s} such that fj (θ) ≡ 0 (mod Pi ). Qs 0 (b) Show that Mθ (x) ≡ j=1 fj (x)hj (x) (mod p), where 1 ≤ h0j ≤ hj for j = 1, . . . , s. (c) Deduce that the map φ : {1, . . . , r} → {1, . . . , s} defined by φ(i) = j(i), where j(i) is the unique integer in {1, . . . , s} such that fj(i) (θ) ≡ 0 (mod Pi ), is surjective. (d) Using Exercise 10.29, show that p | I(θ) if and only if there exists j ∈ {1, . . . , s} such that h0j < hj . (3) Let j ∈ {1, . . . , s} and let J = {i ∈ {1, . . . , r} such that fj (θ) ≡ 0 (mod Pi )}. Show that h0j = maxi∈J dei /νPi (fj (θ))e. (4) Let j ∈ {1, . . . , s} such that hj ≥ 2. Show that h0j < hj if and only if fj (x) | g(x) in Fp [x]. (5) Show that the following conditions are equivalent. (i) p | I(θ).

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(ii) There exists j ∈ {1, . . . , s} such that hj ≥ 2 and fj (x) | g(x) in Fp [x]. (6) In each of the following cases, determine I(θ), Disc(K), an integral basis and for any prime p dividing Disc(θ), its splitting in K. (i) f (x) = x3 − 4. (ii) f (x) = x3 + 4x + 8. Solution 10.30. (1) Obviously I is an ideal of Fp [x], hence principal generated by some polynomial u0 (x). Since f (x) ∈ I, then I = 6 (0). Therefore u0 (x) 6= 0 and we may suppose that u0 (x) is monic. Since Mθ (x) = u0 (x) and f (x) ∈ I, then Mθ (x) | f (x) in Fp [x]. Qs hj (2)(a) Since Mθ (x) | f (x) in Fp [x], then Mθ (x) | j=1 fj (x) . Let i ∈ {1, . . . , r} and let ψ(x) ∈ Fp [x] be the minimal polynomial of θ + Pi Qs hj over Fp . Since ψ(x) | Mθ (x), then ψ(x) divides j=1 fj (x) , hence there exists j = j(i) ∈ {1, . . . , s} such that ψ(x) = fj (x). From ψ(θ + Pi ) = 0, we conclude that fj (θ) ≡ 0 (mod Pi ). We show that this index j = j(i) is unique. Suppose that fj (θ) ≡ fj 0 (θ) ≡ 0 (mod Pi ) for j 6= j 0 . Since fj (x) and fj 0 (x) are coprime, then there exist u(x) and v(x) ∈ Fp [x] such that u(x)fj (x) + v(x)fj 0 (x) = 1. Substituting θ for x, we get 0 ≡ 1 (mod p), which is a contradiction. Thus the index j(i) is unique. Qs Qs hj h0j (b) Since Mθ (x) | j=1 fj (x) (x), then Mθ (x) ≡ j=1 fj (x) (x) (mod p), where 0 ≤ h0j ≤ hj for j = 1, . . . , s. We show that, in fact, h0j ≥ 1 for j = 1, . . . , s. Let γ = Mθ (θ). Then γ ∈ A and γ ≡ 0 (mod p), hence γ satisfies an equation of the form: γ n − pan−1 γ n−1 + p2 an−2 γ n−2 + · · · + (−1)n pn a0 = 0, where ai ∈ Z for i = 0, . . . , n − 1. It follows that Mθ (θ)n − pan−1 Mθ (θ)n−1 + p2 an−2 Mθ (θ)n−2 + · · · + (−1)n pn a0 = 0. We conclude that there exists q(x) ∈ Z[x] such that Mθ (x)n − pan−1 Mθ (x)n−1 + · · · + (−1)n pn a0 = f x)q(x). We deduce that Mθ (x)n ≡ f x)q(x) (mod p). From this, it is seen that any prime factor of f (x) in Fp [x] is a divisor of Mθ (x). Thus h0j ≥ 1 for j = 1, . . . , s.

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(c) Suppose that there exists j0 ∈ {1, . . . , s} such that for all i ∈ Q h0 {1, . . . , r}, fj0 (θ) 6≡ 0 (mod Pi ). Then j=1,...,s fj j (θ) ≡ 0 (mod p), j6=j0

contradicting the property established in (b), that h0j0 ≥ 1. Thus the map φ is surjective. (d) In Exercise 10.29, it is proved that p | I(θ) if and only if deg Mθ (x) < n, hence by (b) if and only if there exists j ∈ {1, . . . , s} such that h0j < hj . 00 (3) Let hj = maxi∈J dei /νPi (fj (θ))e. Let i ∈ J then 0

0

hence hj ≥ 0

00

0

νPi (fj (θ))hj = hj νPi (fj (θ) ≥ ei ,

ei νPi (fj (θ)) .

It follows that h0j ≥ dei /νPi (fj (θ))e and then

hj ≥ hj . We prove the reverse inequality. For any i ∈ J, we have 00 ei ei νP (fj (θ)) ≤ dei /νPi (fj (θ))e, hence νP (fj (θ)) ≤ hj . It follows that i

i

00

00

for any i ∈ J, ei ≤ hj νPi (fj (θ)). We conclude that fj (θ)hj ≡ 0 (mod Piei ). Therefore hj 0 ≤ hj 00 . (4) Substituting θ for x in (Eq 1), we obtain the following identity. Qs s h0j Y 0 j=1 fj (θ) fj (θ)hj −hj = −g(θ). p j=1

If hk > h0k for some k ∈ {1, . . . , s}, then hk ≥ 2 and the left side of this identity is an integer, hence fk (θ) | g(θ) in A. By (2)(a), there exists i ∈ {1, . . . , r} such that fk (θ) ≡ 0 (mod Pi ). It follows that g(θ) ≡ 0 (mod Pi ). Since fk (x) is the minimal polynomial of θ + Pi over Fp , then fk (x) | g(x) in Fp [x]. Conversely suppose that there exists k ∈ {1, . . . , s} such that hk ≥ 2 and fk (x) | g(x) in Fp [x]. Let J = {i ∈ {1, . . . , r} such that fk (θ) ≡ 0 (mod Pi )}. By (3), we have h0k = maxi∈J dei /νPi (fk (θ))e. Let t ∈ J such that this maximum is reached, that is h0k = det /νPt (fk (θ))e. • If νPt (fk (θ)) ≥ et , then h0k = 1. Since hk ≥ 2, then, by (2)(d), the result is proved in this case. • If νPt (fk (θ)) < et , let q(x) and r(x) ∈ Z[x] such that g(x) = fk (x)q(x) + pr(x). Using (Eq 1), we obtain s Y fj (θ)hj = −pfk (θ)q(θ) − p2 r(θ). j=1

Let A and B be the left and right side of this identity respectively. We have h0k = det /νPt (fk (θ))e < 1 + et /νPt (fk (θ)),

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hence h0k νPt (fk (θ)) < et + νPt (fk (θ)). Thus νPt (A) = h0k νPt (fk (θ)) + (hk − h0k )νPt (fk (θ))

< et + νPt (fk (θ)) + (hk − h0k )νPt (fk (θ)).

On the other hand, we have νPt (B) ≥ et + νPt (fk (θ)). It follows that et + νPt (fk (θ)) < et + νPt (fk (θ)) + (hk − h0k )νPt (fk (θ)),

which implies that h0k < hk .

(5) ? (i) ⇒ (ii). By (2), (d), there exists j ∈ {1, . . . , s} such that h0j < hj . It follows that hj ≥ 2 and then by (4), fj (x) | g(x) in Fp [x]. 0 ? (ii) ⇒ (i). By (4), hj < hj , hence by (2), (d) p | I(θ). (6) (i) Here f (x) = x3 − 4, hence Disc(f ) = Disc(θ) = −33 24 . This shows that 3 is ramified in A. In this case, (Eq 1) reads: f (x) = (x − 1)3 + 3(x2 − x − 1)

and its is seen that x − 1 - x2 − x − 1 in F3 [x], hence 3 - I(θ). Therefore, for the prime 3, Mθ (x) = f (x). Dedekind’s Theorem (see [Marcus (1977), Th. 27, Chap. 3]) shows that 3 is totally ramified. For the prime 2, we have f (x) = x3 − 2.2 and x | 2 in F2 [x], hence 2 | I(θ). Obviously θ/2 is not integral since its minimal polynomial is given by f1 (x) = x3 − 1/2. Let µ = θ2 , then the minimal polynomial of µ is given by f2 (x) = x3 − 42 . From this it is seen that θ2 /2 is integral, (thus Mθ (x) = x2 for the prime 2) but θ2 /4 is not. Recall that for a number field E and for any algebraic integer γ ∈ E which is primitive over Q, there exists an integral basis of the form {1, f1 (γ)/d1 , . . . , fn−1 (γ)/dn−1 }, where fi (x) is a monic polynomial of degree i with integral coefficients and the di are positive integers satisfying the conditions: For any integer d0i > di and any monic polynomial gi (x) of degree i with integral coefficients, gi (θ)/d0i is not integral. Moreover I(γ) = d1 · · · dn−1 [Marcus (1977), Th. 13, Exrcises 38–40, Chap. 2]. Applying this theorem for our case, we conclude that {1, θ, θ2 /2} is an integral basis, I(θ) = 2 and Disc(K) = 22 .33 . To see how the prime 2 splits in A, we compute the minimal polynomial of θ2 /2. It is given by f3 (x) = x3 − 2, hence by (5), 2 - I(θ2 /2) and then 2 is totally ramified in A. (ii) Here f (x) = x3 +4x+8. We have Disc(f ) = −4.43 −27.82 = −26 .31. This shows immediately that 31 - I(θ) and then 31 is ramified in A. Since f (x) ≡ (x + 3)2 (x − 6) (mod 31), then Dedekind’s Theorem

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[Marcus (1977), Th. 27, Chap. 3] implies that the splitting of 31 in A is given by 31 = P12 P2 , where P1 and P2 are prime ideals of A both of residual degree equal to 1. We now look at the prime 2. We have f (x) = x3 + 2(2x + 4). Since x | 2x + 4 in F2 [x], then 2 | I(θ). The element θ/2 is a root of f1 (x) = x3 + x + 1, hence it is an algebraic integer. We conclude that for the prime 2, Mθ (x) = x. It is seen that θ/4 is not integral. Obviously θ2 /4 is integral. From the value of the discriminant of θ and the theorem cited above, we conclude that {1, θ/2, θ2 /4} is an integral basis, I(θ) = 23 and Disc(K) = −31. It is easy to see that the minimal polynomial of θ2 /4 over Q is given by f2 (x) = x3 + 2x2 + x − 1. Since f2 (x) is irreducible over F2 , then the prime 2 is inert in A. Exercise 10.31. Let K be a number field of degree n, A be its ring of integers and p be a prime number. Let θ ∈ A be a primitive element, I(θ) be its index. Show that the following assertions are equivalent (i) p - I(θ). (ii) A = Z[θ] + pA. (iii) A ⊂ Z(p) [θ], where Z(p) is the localized ring of Z in the prime ideal pZ. Solution 10.31. We show the implications (i) ⇒ (ii) ⇒ (iii) ⇒ (i). • (i) ⇒ (ii). Consider the following diagram. A

Z[θ] + pA

Z[θ]

pA

We have I(θ) = (A : Z[θ]) and (A : pA) = pn , hence (A : Z[θ] + pA) divides both I(θ) and (A : pA). It follows that (A : Z[θ] + pA) = 1 and then Z[θ] + pA = A. • (ii) ⇒ (iii). Obviously Z[θ] and pA are contained in Z(p) [θ], hence A ⊂ Z(p) [θ].

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• (iii) ⇒ (i). It is known that A has a basis over Z of the form {1, f1 (θ)/d1 , . . . , fn−1 (θ)/dn−1 }, where fi (x) is a monic polynomial of degree i, with integral coefficients and d1 , . . . , dn−1 are positive integers such that di | di+1 , for i = 1, . . . , n − 2. Moreover I(θ) = d1 · · · dn−1 [Marcus (1977), Th. 13 and Exercise 40 Chap. 1]. Since fi (θ)/di ∈ Z(p) [θ], then p - di , hence p - I(θ). Exercise 10.32. Let K be a number field, A be its ring of integers, θ ∈ A and g(x) = Irr(θ, Q, x). Let p be a prime number such that p is not ramified in K and p - I(θ). Let g1 (x), . . . , gr (x) ∈ Z[x] be distinct monic irreducible polynomials such that g(x) ≡ g1 (x) · · · gr (x) (mod p). (1) Suppose that gi (x) 6≡ x (mod p) for any i ∈ {1, . . . , r} and let di be the order of gi (x) over Fp . (a) Show that the sequence (θn )n≥0 , considered modulo p, is periodic and its period Tp (θ) is equal to lcmri=1 (di ). (b) Show that Tp (θ) divides lcmri=1 (pfi − 1). (c) Show that there exists a primitive element γ ∈ A such that Tp (γ) = lcmri=1 (pfi − 1). (d) Compute T3 (θ) and T7 (θ) when θ is a root of g(x) = x3 + 4x + 1. (2) Suppose that g1 (x) ≡ x (mod p) and let di be the order of gi (x) for i = 2, . . . , r. (a) Show that the sequence (θn )n≥1 , considered modulo p, is periodic and its period Tp (θ) is equal to lcmri=2 (di ). (b) Compute T2 (α) when α is a root of h(x) = x3 − 5x2 + 5x − 2. Solution 10.32. (1)(a) Since p is not ramified, then the splitting of p in A has the form pA = P1 · · · Pr , where P1 , . . . , Pr are prime ideals of A with residual degrees equal to f1 , . . . , fr respectively. Set T = lcmri=1 (di ). For any i = 1, . . . , r, we have θdi ≡ 1 (mod Pi ), hence θT ≡ 1 (mod Pi ). It Qr follows that θT ≡ 1 (mod i=1 Pi ), that is θT ≡ 1 (mod pA). We deduce that for any non negative integer n, θn+T ≡ θn (mod pA). Let S ≥ 1 be an integer such that θS ≡ 1 (mod pA), then θS ≡ 1 (mod Pi ) for any i ∈ {1, . . . , r}. It follows that the order of θ + Pi in A/Pi divides S, that is di | S for any i ∈ {1, . . . , r}. Hence T =

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lcmri=1 (di ) divides S. This implies that the period of the sequence (θn )n≥0 , modulo p, is equal to T = Tp (θ). (b) We have fi = [A/Pi : Fp ], hence the order of θ + Pi in A/Pi divides |A/Pi? | = pfi − 1. We deduce that T = lcmri=1 (di ) divides lcmri=1 (pfi − 1). (c) For any i ∈ {1, . . . , r} let hi (x) ∈ Z[x] be monic of degree fi such that hi (x) has its order over Fp equal to pfi − 1. Suppose that the hi (x) are chosen so that if fi = fj and i 6= j, then hi (x) 6= hj (x). For any i ∈ {1, . . . , r} let γi ∈ A such that γi + Pi is a root of hi (x). Let γ ∈ A such that γ ≡ γi (mod Pi ). Then by Exercise 10.30, γ is a primitive element of K and p - γ. By (a), Tp (γ) = lcmri=1 (pfi − 1). (d) The factorization of g(x) modulo 3 is given by g(x) ≡ (x − 1)(x2 + x − 1), hence 3A = P1 P2 , where P1 and P2 are prime ideals of A with residual degrees equal to 1 and 2 respectively. It follows that d1 = 1 and d2 | 32 − 1. Obviously d2 6= 1. If d2 = 2, then θ2 ≡ 1 (mod 3A), hence θ3 ≡ θ (mod 3A). But θ3 = −4θ − 1 6≡ θ (mod 3A), thus d2 6= 2. The same reasoning shows that d2 6= 4. It follows that T3 (θ) = lcm(d1 , d2 ) = 8. In F7 [x], g(x) is irreducible, hence 7A is prime. It follows that T7 (θ) | 73 −1. Since 73 −1 = 2.32 .19, then the possible values of T7 (θ) are the followings: 1, 2, 3, 6, 9, 18, 19, 2.19, 3.19, 6.19, 32 .19, 73 − 1. Obviously we may exclude the values 1, 2. We have θ3 ≡ 3θ − 1 (mod 7A), θ4 ≡ 3θ2 − θ, θ5 ≡ −θ2 + 2θ − 3 (mod 7A), θ6 ≡ 2θ2 + θ + 1 θ

18

θ

2.19

θ

6.19

2

≡ 3θ + 2 2

(mod 7A), θ9 ≡ θ2 − θ

(mod 7A), θ

≡ 2θ − 3θ + 2 ≡θ−1

19

≡ −3θ − 3

(mod 7A), θ

3.19

(mod 7A),

(mod 7A), 2

≡ 3θ − θ

(mod 7A), θ32 .19 (θ) ≡ −3θ − 3

(mod 7A),

(mod 7A),

hence T7 (θ) = 73 − 1. (2)(a) Since p is not ramified, then the splitting of p in A has the form pA = P1 · · · Pr , where P1 , . . . , Pr are prime ideals of A with residual degrees equal to f1 , . . . , fr respectively, where f1 = 1. Set T = lcmri=2 (di ). For any i = 2, . . . , r, we have θdi ≡ 1 (mod Pi ), hence Qr θT ≡ 1 (mod Pi ). It follows that θT ≡ 1 (mod i=2 Pi ), that is Qr θT − 1 ≡ 0 (mod i=2 Pi ) and then θT +1 ≡ θ (mod p). The proof may be completed by using similar arguments as in (1)(a).

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(b) We have h(x) ≡ x(x2 +x+1) (mod 2) and the polynomial x2 +x+1 is irreducible over F2 . The order of this polynomial is equal 3, hence T2 (α) = 3. This means that θn+3 ≡ θn (mod 2A) for n ≥ 1. Exercise 10.33. Let f (x) = x3 − 2x2 − 1, θ be a root of f (x) in C, K = Q(θ) and A be the ring of integers of K. (1) Show that f (x) is irreducible over Q and that {1, θ, θ2 } is a basis of A over Z. (2) Show that 2A = P1 P2 , where P1 and P2 are prime ideals of A having their residual degrees equals to 1 and 2 respectively. Show that P1 = (θ − 1)A and P2 = (θ2 + θ + 1)A. (3) Show that 3A is prime in A. (4) Let I = 3(θ − 1)2 A + 4(θ2 + θ + 1)A. Determine a basis of I −1 over Z. (5) Show that 36 ∈ I and find α ∈ A such that I = 36A + αA. Solution 10.33. (1) Since the polynomial f (x) has no root in Z, then it is irreducible over Q. According to Exercise 10.5, we have Z[1/θ] = Z[θ], thus it is equivalent to show that {1, 1/θ, 1/θ2 } is a basis of A over Z. This claim is true since the minimal polynomial of 1/θ is given by F (x) = x3 +2x−1 and the discriminant of this polynomial, being equal to −59, is square free. (2) Since f (x) = (x−1)(x2 +x+1)−2x2 , then, by Exercise 10.30, 2 - I(θ) and 2A = P1 P2 , where P1 and P2 are prime ideals of A having their residual degrees equal to 1 and 2 respectively. Since θ − 1 is a root of the polynomial f (x + 1) and since the constant term of this polynomial is equal to f (1), that is to −2, then NK/Q = 2, then P1 = (θ − 1)A. Let γ = θ2 + θ + 1. By the method used in Exercise 5.30(3)(b), we find that the characteristic polynomial (here equal to the minimal polynomial) g(x) of γ is given by g(x) = x3 + 3x2 − 4. Moreover, since g(x) = x2 (x + 1) + 2(x2 − 2), then by Exercise 10.30, 2 | I(γ) and P2 = γA = (θ2 + θ + 1)A. (3) The reduced polynomial of f (x) modulo 3 has no root in F3 , hence it is irreducible over this field and then 3A is a prime ideal of A. (4) Let x ∈ K, then x ∈ I −1 ⇔ xI ⊂ A


⇔ x 3(θ − 1)2 ∈ A


and x 4(θ2 + θ + 1) ∈ A.

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Set x = (x0 + x1 θ + x2 θ2 )/d with x0 , x1 , x2 ∈ Z and d is a positive integer. Using the relations θ3 = 2θ2 + 1, θ4 = 4θ2 + θ + 2, we find the following identities θ2 (x0 + x2 ) + θ(−2x0 + x1 + x2 ) + (x0 + x1 ) and d θ2 (x0 + 3x1 + 7x2 ) + θ(x0 + x1 + x2 ) + (x0 + x1 + 3x2 ) . 4x(θ2 + θ + 1) = d 3x(θ − 1)2 =

We deduce that x ∈ I −1 if and only if d divides the following integers 3(x0 + x2 ),

3(−2x0 + x1 + x2 ),

4(x0 + 3x1 + 7x2 ),

3(x0 + x1 ),

4(x0 + x1 + x2 ),

4(x0 + x1 + 3x2 ).

(a) Suppose that x ∈ I −1 and d is odd, then since d divides the three last integers in the above list, we conclude that d divides each of the integers x0 + 3x1 + 7x2 , x0 + x1 + x2 , x0 + x1 + 3x2 . This implies that d divides x1 , x2 , x3 which means that x ∈ A. (b) Suppose that x ∈ I −1 and d ≡ 2 (mod 4). Let δ = gcd(d, 12) and set d = δd0 , where d0 is odd and δ = 2, or 6. From the conditions on d described above we deduce in particular that d divides each of the following integers x0 + 3x1 + 7x2 , x0 + x1 + x2 , x0 + x1 + 3x2 . It is easy to show successively that d0 | x2 , d0 | x1 and d0 | x0 . Let y0 = x0 /d0 , y1 = x1 /d0 and y2 = x2 /d0 , then δ divides each of the followings integers 3(y0 + y2 ),

3(−2y0 + y1 + y2 ),

4(y0 + 3y1 + 7y2 ),

3(y0 + y1 ),

4(y0 + y1 + y2 ),

4(y0 + y1 + 3y2 ).

• If δ = 2, we get y0 ≡ y1 ≡ y2 (mod 2). Set y1 = y0 + 2z1 and y2 = y0 + 2z2 , then we conclude that x = (x0 + x1 θ + x2 θ2 )/(δd0 )


= y0 + (y0 + 2z1 )θ + (y0 + 2z2 )θ2 /2 = y0 (1 + θ + θ2 )/2 + z1 θ + z2 θ2 .

• If δ = 6, we get 2 | y0 + y2 , y1 + y2 , y0 + y1 , that is y2 ≡ y1 ≡ y0 (mod 2). We also obtain that 3 | y0 + y2 , y0 + y1 + y2 , y0 + y1 , which implies y2 ≡ y1 ≡ y0 ≡ 0 (mod 3). From these conditions modulo 2 and modulo 3, we may set y0 = 3z0 , y1 = 3z0 +6z1 , y2 = 3z0 + 6z2 . As above, we obtain x = z0 (1 + θ + θ2 )/2 + z1 θ + z2 θ2 .

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(c) Suppose that x ∈ I −1 and d ≡ 0 (mod 4). As above, let δ = gcd(d, 12) and set d = δd0 , where d0 is a positive integer and δ = 4, or 12. We know that δd0 divides each of the followings integers 3(x0 + x2 ),

3(−2x0 + x1 + x2 ),

4(x0 + 3x1 + 7x2 ),

3(x0 + x1 ),

4(x0 + x1 + x2 ),

4(x0 + x1 + 3x2 ).

• If δ = 4, then 3 - d0 and conclude that d0 divides each of the integers (x0 + x2 ),

(−2x0 + x1 + x2 ),

(x0 + 3x1 + 7x2 ),

(x0 + x1 ),

(x0 + x1 + x2 ),

(x0 + x1 + 3x2 ).

It follows that d0 divides each of x0 , x1 , x2 . Let y0 = x0 /d0 , y1 = x1 /d0 and y2 = x2 /d0 , then 4 divides each of the followings integers 3(y0 + y2 ),

3(−2y0 + y1 + y2 ),

4(y0 + 3y1 + 7y2 ),

3(y0 + y1 ),

4(y0 + y1 + y2 ),

4(y0 + y1 + 3y2 ),

thus 4 divides each of the integers (y0 + y2 ),

(−2y0 + y1 + y2 ),

(y0 + y1 ).

It follows that y1 ≡ −y0 (mod 4) and y2 ≡ −y0 (mod 4). Set y1 = −y0 + 4z1 and y2 = −y0 + 4z2 , then x = y0 (1 − θ − θ2 )/4 + z1 θ + z2 θ2 . • If δ = 12, similar computations as above show that d0 divides x0 , x1 , x2 . Setting y0 = x0 /d0 , y1 = x1 /d0 and y2 = x2 /d0 , we get y2 ≡ y1 ≡ y0 ≡ 0 (mod 3) and y1 ≡ −y0 (mod 4), y2 ≡ −y0 (mod 4). Set y0 = 3z0 , y1 = −3z0 + 12z1 and y2 = −3z0 + 12z2 , then x = z0 (1 − θ − θ2 )/4 + z1 θ + z2 θ2 . We have proved that x is a linear combination of {1, θ, θ2 }

or {(1 + θ + θ2 )/2, θ, θ2 }

or {(1 − θ − θ2 )/4, θ, θ2 }.

Since (1 + θ + θ2 )/2 = 2(1 − θ − θ2 )/4 + θ + θ2 , then in any case x is a linear combination of the last set of generators. On the other hand since A ⊂ I −1 , then θ and θ2 ∈ I −1 . To obtain the same conclusion for (1 − θ − θ2 )/4, it is sufficient to verify that (1 − θ − θ2 )/4(3θ2 − 6θ + 3) 2

2

and

(1 − θ − θ )/4(4θ + 4θ + 4) ∈ A.

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We have (1 − θ − θ2 )/4(3θ2 − 6θ + 3) = −3(θ2 + θ) ∈ A 2

2

2

and

2

(1 − θ − θ )/4(4θ + 4θ + 4) = (1 − θ − θ )(θ + θ + 1) ∈ A,

hence the result. We conclude that {(1 − θ − θ2 )/4, θ, θ2 } is a basis of I −1 over Z. (5) We have 36θ = 3(4θ2 + 4θ + 4) − 4(3θ2 − 6θ + 3). Since θ is a unit, then 36 ∈ A. Set B = 36I −1 , then we have αA + 36A = I ⇔ αII −1 + 36II −1 = II −1 ⇔ αI −1 + B = I

⇔ αI −1 + P = I, ⇔ αI

−1

6⊂ P,

⇔ α 6∈ IP, 2

for any

for any

for any

2

P|B

P|B

P | B.

The factorizations of 3(θ − 1) and 4(θ + θ + 1) into products of prime ideals of A are given by 3(θ − 1)2 = 3AP12 and 4(θ2 + θ + 1) = P12 P23 , hence I = P12 and then B = (3A)2 P22 . The conditions above on α read α 6∈ IP2 and α 6∈ 3AI. Clearly α = (θ − 1)2 works. Exercise 10.34. Let K be a field of characteristic 0, Ω be an algebraic closure of K and let U be the set of elements β ∈ Ω for which there exists α ∈ Ω∗ such that 0 0 β = α/α , where α is a conjugate of α over K. Call the elements of U, U-numbers. (1) Let F be a finite extension of Q and k be a positive integer. Show that for any prime number p, except finitely of them, xk − p is irreducible over F . (2) Let β ∈ Ω. Suppose that β k ∈ U for some positive integer k. Show that β ∈ U. (3) Let β ∈ Ω, Nβ be the normal closure of K(β) over K, Gβ = Gal(Nβ , K). For any σ ∈ Gβ , let eσ be the smallest positive integer such that Qeσ −1 i σ eσ (β) = β and let P (σ, β) = i=0 σ (β). (a) Show that the following conditions are equivalent.

(i) β ∈ U. (ii) There exists σ ∈ Gβ such that P (σ, β) is a root of unity.

(b) If the equivalent conditions (i) and (ii) hold and P (σ, β) is a k-th 0 root of unity, show that β may be written in the form β = α/α , 0 where α is a conjugate of α over K and αk ∈ Nβ .

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(c) Show that (ii) is equivalent to the following assertion. Qn−1 (iii) There exists σ ∈ Gβ such that i=0 σ i (β) is a root of unity where n is the order of σ.

(4) Let E = K(β) and suppose that E/K is cyclic. Show that β ∈ U if and only if NE/K (β) = 1. Solution 10.34. (1) Let p be a prime number and f (x) = xk − p ∈ F [x]. Then, by [Lang (1965), Th. 16, Chap. 8.9], f (x) is reducible over F if and only if there exists a prime l | k such that p ∈ F l or 4|k and p ∈ −4F 4 . For any prime number l such that l | k, let Pl = {p prime, p ∈ F l }. If 4 | k, let P4 = {p prime, p ∈ −4F 4 }. To get that the set of primes p, for which xk − p is reducible over F , it is sufficient to prove that Pl is finite for any prime factor l of k and that P4 is finite. Suppose that Pl is infinite for some prime factor l of k, then as it is said above, for any p ∈ Pl , there exists α ∈ F such that p = αl , that is F contains the field Q(p1/l ). Since the number of subfields of F is finite, there exist two distinct prime numbers p1 and p2 belonging to Pl , such that p1 6= l, 1/l 1/l p2 6= l and Q(p1 ) = Q(p2 ). Claim. Let p and l be distinct prime numbers and K = Q(p1/l ). Then except possibly for l, the only prime number which is ramified in K is p. Proof. Let α = p1/l and g(x) = Irr(α, Q) = xl − p. We have 0

Disc(g) = (−1)l(l−1)/2 NK/Q (g (α)) = ±(1)ll pl−1 . This shows that the only prime, excluding l, which can be ramified in K is p. We show that effectively p is ramified in K. Using the result of Exercise 10.30, it is seen that p - I(α) and then p is ramified in K. 1/l We deduce from this claim that pi is ramified in Q(pi ) but not in 1/l Q(pj ) for i, j ∈ {1, 2}, i 6= j, contradicting the equality of the fields. We have proved that Pl is finite. The same proof, mutadis mutandis, works for P4 and will be omitted. 0 (2) Let γ ∈ Ω such that β k = γγ0 , where γ is a conjugate of γ over K. Let ρ ∈ Ω such that ρk = γ. Let σ : K(γ) → Ω be the unique K0 embedding such that σ(γ) = γ and let σ ˆ : K(ρ) → Ω be its extension 0 to K(ρ). Since ρk = γ, then σ ˆ (p)k = σ ˆ (γ) = σ(γ) = γ . Therefore ρ ρk k β k = ( σˆ (ρ) = σˆ (ρ) σ (ρ), where k is a k) k . We deduce that β = k ρ/ˆ k-th root of unity in Ω. Let F = K(ρ, k ) and let p be a prime number

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such that xk − p is irreducible over F . Let p1/k be a root of xk − p ρp1/k in Ω, then we may write β in the form β = σˆ (ρ) −1 1/k . Consider the p k

following diagram of fields. F (p1/k )

K(p1/k )

F

K Since F and K(p1/k ) are linearly disjoint over K, there exists a Kembedding τ : F (p1/k ) → Ω such that 1/k τ (p1/k ) = −1 k p

and

τ (ρ) = σ ˆ (ρ). 1/k 1/k We deduce that τ (ρp1/k ) = σ ˆ (ρ)−1 /τ (ρ(p1/k ), k p 0 , hence β = ρp 0 thus β = α/α where α = ρp1/k and α = τ (ρp1/k ) is a conjugate of α. α (3) • (i) ⇒ (ii). Let α ∈ Ω such that β = τ (α) with τ ∈ Gal(Nα , K), then β ∈ Nα , hence Nβ ⊂ Nα . Let σ be the restriction of τ to Nβ , then Qeσ −1 i σ ∈ Gβ . We consider P (σ, β) = i=0 σ (β). Let n be the smallest positive integer such that τ n (α) = α. Since   τ n (α) α α = n+1 = = β, σ n (β) = τ n (β) = τ n τ (α) τ (α) τ (α)

then m | n. Set n = mk. On the one hand we have   Y n n Y α σ i−1 = σ i−1 (β) = (β1 , · · · , βm )k = P (σ, β)k . τ (α) i=1 i=1 On the other hand we have   Y n n n Y Y α i−1 σ = τ i−1 (α)/ τ i (α) = τ 0 (α)/τ n (α) = 1, τ (α) i=1 i=1 i=1 hence P (α, β)k = 1. It follows that P (σ, β) is a root of unity.

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• (ii) ⇒ (i). Let σ ∈ Gβ such that P (σ, β) is a k-th root of unity for some positive integer k and let e be the smallest positive integer such σ e (β) = β. Let N be the Galois closure of K(β) over K, F = Inv(σ) and n = [N : F ], then, | < σ > | = [N : F ] = n. Therefore e | n. We Qn−1 deduce that i=0 σi (β)k = 1. Let γ ∈ Nβ be a primitive element over K and let m be the smallest positive integer such that σ m (γ) = γ, then as σ m fixes γ, then it fixes any element of Nβ , thus m = n. Qn Let βi = σ i−1 (β) and γi = σ i−1 (γ) for i = 1, . . . , n then i=1 β i = 1 and the γi are the distinct conjugate of γ. For any j ∈ {1, 2, . . . , s}, Pn j Qi k let µj = i=1 γi P l=1 βl . Consider the homogeneous system of n j linear equations: i=1 γi xi = 0, j = 1, . . . , n. Its determinant is a Vandermonde determinant, hence non zero. Therefore it has a unique solution, namely the trivial one. It follows that there exists j ∈ {1, . . . , n} such that µj 6= 0. Otherwise x1 = · · · = xn = 0, Qn contradicting the fact that l=1 βlk 6= 0. For this index j, we have β k σ(µj ) = β k σ(γ1j β1k + γ2j β1k β2k + · · · + γnj β1k · · · βnk )

= β k (γ2j β2k + γ3j β2k β3k + · · · + γnj β2j · · · βnk + γ1j β2k · · · βnk β1k ).

Since β1k · · · βnk = 1, then β k σ(µj ) = µj . Therefore β1k = µj /σ(µj ) ∈ U. Using (2), we conclude that β ∈ U. (c) (ii) ⇔ (iii). Since σ n = IdNβ , then σ n (β) = β, hence eσ | n. Set Qn−1 n = eσ · q where q is a positive integer, then i=0 σ i (β) = P (σ, β), Qn−1 hence i=0 σ i (β) is a root of unity if and only if P (σ, β) is. (4) • Sufficiency of the condition. Let σ be a generator of Gal(E, K), n = [E : K], then 1 = NE/K (β) =

n−1 Y

σ i (β),

i=0

hence by (3) (iii), β ∈ U. • Necessity of the condition. α , where τ ∈ Gal(E, K), then Let α ∈ E such that β = τ (α) NE/K (β) = NE/K (α)/ NE/K (τ (α)) = 1. Exercise 10.35. Let K be a number field, A be its ring of integers and p be a prime number. Let P be a prime ideal of A such that NK/Q (P) = pf , where f is a positive integer. Let φ(x) ∈ Z[x] be monic of degree f such that its reduction modulo p is irreducible.

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(1) Show that there exists θ1 ∈ A such that νP (φ(θ1 )) = 1. (2) Let k ≥ 1 be an integer. Show that A/P k ' Fp [x]/φk (x)Fp [x]. Deduce that |(A/P k )∗ | = pf (k−1) (pf − 1). (3) Show that there exists θ ∈ A such that νP (φ(θ)) = 1 and νP 0 (φ(θ)) = 0 for any prime ideal P 0 |p and P 0 6= P. Deduce that P = pA + φ(θ)A = (p, φ(θ)). (4) Let {ω1 , . . . , ωn } be a basis of A over Z and let ~u = (u1 , . . . , un ) be an Pn n-tuple of independent variables over Q. Let η = i=1 ui ωi . Pf −1 (a) Show that η ≡ u)θj (mod P), where Lj (~u) is a linear j=0 Lj (~ form with coefficients in Fp for j = 0, 1, . . . , f − 1. Qf −1 Pf −1 k (b) Let H(~u, x) = k=0 (x − j=0 Lj (~u)θjp ). Show that H(~u, x) ∈ Fp [~u, x] and H(~u, x) is irreducible over Fp . ˆ u, x) ∈ Z[~u, x] be a monic lift (in x) of H(u, x) and let (c) Let H(~ α1 , . . . , αd ∈ A be all the non zero coefficients of the monomials ui11 · · · uinn in H(~u, η). Show that P = (p, α1 , . . . , αd ). Solution 10.35. (1) Let θ1 ∈ A such that θ1 + P is a primitive element of A/P over FP and φ(x) is its minimal polynomial over Fp , then φ(θ1 ) ≡ 0 (mod P). If φ(θ1 ) 6≡ 0 (mod P 2 ), the proof is complete. Otherwise, let τ ∈ P \ P 2 and µ = θ1 + τ, then φ(µ) ≡ 0 (mod P) and φ(µ) 6≡ 0 (mod P 2 ). Therefore replacing θ1 by µ if necessary, we may suppose that νP (φ(θ1 )) = 1. (2) Consider the map Ψ : Fp [x] → A/P k such that Ψ(g(x)) = g(θ1 ) + P k . Obviously Ψ is a morphism of rings. Let g(x) ∈ Fp [x]. If φk (x) | g(x), then obviously g(θ1 ) ≡ 0 (mod P k ). We prove the converse by induction on k. The claim is true for k = 1. Since φ(x) is the minimal polynomial of θ1 + P in A/P. Suppose the claim is true for k − 1 and suppose that g(θ1 ) ≡ 0 (mod P k ). Let g(x) = φ(x)q(x) + r(x), where q(x), r(x) ∈ Fp [x] and deg r < deg φ. Since r(θ1 ) ≡ 0 (mod P), then r(x) = 0, that is g(x) = φ(x)q(x). Since νP (φ(θ1 )) = 1, then q(θ1 ) ≡ 0 (mod P k−1 ). Therefore φk−1 (x)|q(x) by the inductive hypothesis. It follows that φk (x) | g(x). We deduce that Ker Ψ = φk (x)Fp [x] and then Fp [x]/φk (x)Fp [x] ' Im Ψ. Let f = deg φ = [A/P : Fp ], then |Fp [x]/φf (x)Fp [x]| = pkf = |A/P k |,

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hence Im Ψ = A/P k and the isomorphism is established. The elements of the form a0 (x) + a1 (x)φ(x) + · · · + ak−1 (x)φk−1 (x),

where ai (x) ∈ Fp [x], deg ai < deg φ = f , for i = 0, . . . , k − 1, constitute a complete set of representatives of the cosets of the elements of Fp [x] modulo φk (x)Fp [x]. Any element of this form is a unit in Fp [x]/φk (x)Fq if and only if a0 (x) 6= 0. Therefore the number of units is equal to (pf − 1)(pf )k−1 = (pf − 1)pf (k−1) . (3) Let I be the ideal of A such that pA = P e I and P - I and let γ ∈ I f f then γ 6∈ P, and γ is a unit in A/P 2 , hence γ (p −1)p ≡ 1 (mod P 2 ). f f Let θ = θ1 γ (p −1)p , then θ ≡ θ1 (mod P 2 ), hence νP (φ(θ)) = 1. On the other hand let P 0 be a prime ideal of A such that P 0 | I. We have φ(θ) = θf + af −1 θf −1 + · · · + a0 , where a0 , . . . , af −1 ∈ Fp and a0 6= 0 hence φ(θ) ≡ a0 (mod P 0 ) 6≡ 0 (mod P 0 ). It is clear that the only ideal of A which divides both pA and φ(θ)A is equal to P, hence P = pA + φ(θ)A. Pf −1 (4)(a) For i = 1, ..., n, we have ωi ≡ j=0 aji θj (mod P), where aji ∈ Z, Pn Pf −1 j j hence η ≡ i=1 ui j=0 ai θ (mod P), thus η≡

f −1 X n X

(ui aji )θj

j=0 i=1

(mod P).

Pn Pf −1 Set Lj (~u) = i=1 ui aji , then η ≡ j=0 (~u)θj (mod P). (b) Consider the following diagram of fields: Fp (~u, θ)

Fp (θ)

Fp (~u)

Fp Since Fp (θ) is algebraic over Fp and Fp (~u) is purely transcendental over Fp , then [Fp (~u, θ) : Fp (~u)] = [Fp (θ) : Fp ] = f.

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Clearly the characteristic polynomial over Fp (~u) of η is given by   f −1 fY −1 X k x − Lj (~u)θjp  . H(~u, x) = k=0

j=0

Pn ∗ Let ~u = ∈ Z such that θ = i=1 ui wi , that is L1 (~u∗ ) = 1 = 0 for j 6= 1, then H(u∗ , x) = φ(x). Since this last polynomial is irreducible over Fp , then so is H(~u, x). Pn ˆ ∗ , x) ≡ φ(x) (c) Let ~u∗ = (u∗1 , ..., u∗n ) such that θ = i=1 u∗i wi , then H(u (mod p), hence ?

(u∗1 , ..., u∗n ) and Lj (~u∗ )

n

ˆ ∗ , θ)). P = (p, φ(θ)) = (p, H(u

It follows that P = (p, α1 , . . . , αd ) because if a prime ideal Q lyˆ ∗ , θ) which is a ing over pZ, divides α1 , . . . , αd then it divides H(u contradiction. Exercise 10.36. Let f (x) ∈ Z[x] be monic irreducible of degree n, θ be a root of f , K = Q(θ) and A be the ring of integers of K. Let P be the set of the prime numbers p such that p | I(θ). (1) Let ω0 , ω1 , . . . , ωn−1 be an integral basis of the form ωi = fi (θ)/di , where fi (x) is a monic polynomial with integral coefficients of degree i and d0 , . . . , dn−1 are positive integers such that 1 = d0 | d1 | . . . | dn−1 Qn−1 and I(θ) = i=0 di . This basis exists by [Marcus (1977), Th. 13 and Exercise 40, Chap. 1]. Let B = Z[θ] and for any p ∈ P let Bp = {γ ∈ A, there exists h ∈ N, ph γ ∈ B}.

Show that {fi (θ)/pνp (di ) , i = 0, . . . , n − 1} is a Z-basis of Bp . (p) (p) (2) For any p ∈ P, let {1, f1 /ph1 (p) , . . . , fn−1 /phn−1 (p) } be a Z-basis of Bp , (p) where fi (x) is a monic polynomial of degree i with integral coefficients and h1 (p), . . . , hn−1 (p) are non negative integers such that h1 (p) ≤ · · · ≤ hn−1 (p). Show that for any i ∈ {1, . . . , n−1}, there exists a monic polynomial fi (x) ∈ Z[x], of degree i such that for any p ∈ P, fi (x) ≡ (p) } is , . . . , Q fn−1h(θ) fi (x) (mod phi (p) ). Show that {1, Q f1 (θ) n−1 (p) ph1 (p) p∈P

p∈P

p

an integral basis of K. (3) Let µ be a root of g(x) = x3 − 5x2 − 12x − 64, E = Q(µ), A be the ring of integers of E and B = Z[µ]. (a) Show that g(x) is irreducible over Q and Disc(g) = −42 · 52 · 503. (b) Show that I(µ) = 20.

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(c) Show that {1, µ, (µ2 − µ)/4} (resp. {1, µ, (µ2 + 2µ + 2)/5}) is a Z-basis of B2 (resp. B5 ) and deduce an integral basis of E. Solution 10.36. Pk (1) Let γ ∈ Bp , γ = i=0 λi fi (θ)/di , where 1 ≤ k ≤ n − 1, λi ∈ Z and λk 6= 0. We prove by induction on k that λ is a linear combination of 1, f1 (θ)/pνp (d1 ) , . . . , fn−1 (θ)/pνp (dn−1 ) with integral coefficients. The result is trivial for k = 0. Suppose that the result is true for any l < k. Let h be a non negative integer such that ph γ ∈ B, then there exists b0 , . . . , bk ∈ Z such that k X

λi ph fi (θ)/di =

i=0

Setting dk =

pνp (dk ) d0k ,

k X

bi θi .

i=0

we deduce in particular that λk ph = bk pνp (dk ) d0k .

This shows that d0k | λk . We have hence

Pk−1 i=0

λk fk (θ)/dk = (λk /d0k )fk (θ)/pνp (dk ) ∈ Bp ,

λi fi (θ)/di ∈ B. From our assumptions, it follows that k−1 X

λi fi (θ)/di =

i=0

k−1 X

µi fi (θ)/pνp (di ) ,

i=0

where µi ∈ Z for i = 1, . . . , k − 1. We conclude that γ=

k X

µi fi (θ)/pνp (di )

i=0

with µk =

λk /d0k .

Therefore n o 1, f1 (θ)/pνp (d1 ) , . . . , fn−1 (θ)/pνp (dn−1 )

is a generating set of Bp . Obviously, this set is free over Z. Therefore it is a Z-basis of Bp . (2) The existence of the polynomials fi (x) is established in Exercise 1.30. We know that A has a Z-basis of the form 1, g1 (θ)/d1 , . . . , gn−1 (θ)/dn−1 , where gi (x) is a monic polynomial of degree i with integral coefficients and d1 , . . . , dn−1 are positive integers such that d1 | d2 | dn−1 and I(θ) = d1 d2 · · · dn−1 [Marcus (1977), Th. 13 and Exercise 40 Chap. 2]. Moreover any of the polynomials gi (x) may be replaced by any monic polynomial hi (x) of degree i with

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integral coefficients satisfying hi (θ) ≡ 0 (mod di ) [Marcus (1977), Exercise 39, Chap. 2]. For our purpose we take hi (x) = fi (x). The proof Q will be complete if it is shown that di = p∈P phi (p) and fi (θ) ≡ 0 Q (mod p∈P phi (p) ) for i = 1, . . . , n − 1. These congruences are obvious and we omit their proof. Let p ∈ P. Suppose that pli (p) || di , then (di /pli (p) )gi (θ) = gi (θ)/pli (p) ∈ Bp ,

hence li (p) ≤ hi (p). On the other hand, fi (θ)/phi (p) ∈ A, hence hi (p) ≤ Q li (p). It follows that di = p∈P phi (p) for i = 1, . . . , n − 1. (3)(a) The reduction modulo 3 of g(x) is irreducible over F3 , hence g(x) is irreducible over Q. We omit the details of the computations for the discriminant of g(x). One may apply the general formula for the discriminant of a polynomial of degree 3. Alternatively one may proceed as follows. We have Disc(g) = − NE/Q (g 0 (µ). We compute the characteristic polynomial, say u(x), of g 0 (µ) by the method used in Exercise 2.7(4), then Disc(g) = b0 , where b0 is the constant coefficient of u(x). (b) Since Disc(g) = −42 · 52 · 503, then 503 - I(µ), 0 ≤ ν2 (I(µ)) ≤ 2 and 0 ≤ ν5 (I(µ)) ≤ 1. We have g(x) = x2 (x − 1) − 4(x2 − 3x − 16) and x | 2(x2 − 3x − 16) in F2 [x], hence, by Exercise 10.30, 2 | I(µ). We compute the minimal polynomial h(x) of α = (µ2 − µ)/4 by using the method of Exercise 2.7(4), we find h(x) = x3 − 5x2 − 34x − 72. Therefore α ∈ A and ν2 (I(µ)) = 2. We also have g(x) = (x − 1)(x − 2)2 − 5(x2 + 2x + 12) and x − 2 | (x2 + 2x + 12) in F5 [x], hence 5 | I(µ). Therefore ν5 (I(µ)) = 1. We deduce that I(µ) = 20. (c) It is clear that B2 = Z + Zµ + Zµ(µ − 1)/2, hence {1, µ, µ(µ − 1)/4} is a Z-basis of B2 . Similarly B5 = Z + Zµ + Z(µ2 + 2µ + 2)/5, hence {1, µ, (µ2 + 2µ + 2)/5} is a Z-basis of B5 . To compute an integral basis, we use the method of (2). We compute an integral basis of the form {1, µ, g2 (µ)/20}, where g2 (x) is a monic polynomial of degree 2, with integral coefficients and satisfying the following conditions: g2 (x) ≡ x2 − x (mod 4Z[x]) and g2 (x) ≡ x2 + 2x + 2 (mod 5Z[x]). We easily find a solution, namely g2 (x) = x2 + 7x − 8. Therefore {1, µ, (µ2 + 7µ − 8)/20} is an integral basis. Exercise 10.37. Let A be an integral domain, K be its fraction field and f (x) = a0 xn + · · · + an ∈ A[x] \ A. Let θ be a root of f in an algebraic closure of K. For i = 1, . . . , n − 1, let ui (θ) = a0 θi + a1 θi−1 + · · · + ai .

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(1) Show that γk := uk (θ) is integral over A. (2) Show that θγk is integral over A. (3) Let B = A + Aγ1 + · · · + Aγn−1 . Show that B is a ring. Solution 10.37. (1) • First proof. Consider the polynomial −a0 −a1 0 −a0 ... ... 0 0 Rk (y) = 1 a1 /a0 0 1 ... . .. 0 0


Rk (y) = Resx y − uk (x), a10 f (x) . Then ... y − ak 0 ... 0 ... −ak−1 y − ak . . . 0 ... ... ... ... . . . ... 0 −a0 . . . y − ak . ... an /a0 0 ... 0 . . . an−1 /a0 an /a0 . . . 0 ... ... ... ... . . . ... 0 1 . . . an /a0

Denote by bji the coefficient appearing in the above determinant at the intersection of the i-th row with the j-th column. Rk (y) is a sum of n + k terms. One of them is ±bk+1 bk+2 · · · bk+n b1n+1 b2n+2 · · · bkn+k = ±(y − ak )n (1)k . n 1 2 From this it appears that Rk (y) is a polynomial in y of degree n and its leading coefficient is equal to ±1. Moreover uk (θ) is a root of this polynomial. Hence to get the result it is sufficient to prove that Rk (y) ∈ A[y]. We may write f (x) in the form f (x) = xn−k uk (x) + fˆ(x), where fˆ(x) is a polynomial of degree at most n − (k + 1) with coefficients in A. We have   1 Rk (y) = Resx y − uk (x), f (x) a0   1 n−k ˆ uk (x) + f (x)) = Resx y − uk (x), (x a0 Y 1 = (−a0 )n (ξ n−k uk (ξ) + fˆ(ξ)) a0 ξ Y = (−1)n an−k (ξ n−k uk (ξ) + fˆ(ξ)) 0 ξ

i

n−k

= (−1) (−a0 )

Y (ξ n−k y + fˆ(ξ)) ξ


= (−1) Resx y − uk (x), yxn−k + fˆ(x)) . i

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Here ξ denotes any root of uk (x) − y in an algebraic closure of K(y). The polynomials in x, for which we compute the resultant, in the last equality above, have their coefficients in A[y], hence Rk (y) ∈ A[y]. • Second proof. We have γk = a0 θk + a1 θk−1 + · · · + ak−1 θ + ak

θγk = a0 θk+1 + a1 θk + · · · + ak−1 θ2 + ak θ

................................................................ θ

n−k−1

γk = a0 θn−1 + a1 θn−2 + · · · + ak−1 θn−k + ak θn−k−1

θn−k γk = −ak+1 θn−k−1 − ak+2 θn−k−2 − · · · + an−1 θ − an .......................................................................................

θn−1 γk = −ak+1 θn−2 − ak+2 θn−3 − · · · − an−1 θk − an θk−1 . This shows that 1, θ, . . . , θn−1 is a non zero solution of the following homogeneous system of equations: (ak − γk )x0 + ak−1 x1 + · · · + a0 xk = 0

(ak − γk )x1 + ak−1 x2 + · · · + a0 xk+1 = 0 ..........................................................

(ak − γk )xn−k−1 + ak−1 xn−k + · · · + a0 xn−1 = 0

γk xn−k + ak+1 xn−k−1 + · · · + an x0 = 0 .........................................................

γk xn−1 + ak+1 xn−2 + · · · + an xk−1 = 0. The determinant of this system must be zero. Therefore γk is a root of a polynomial with coefficients in A whose leading coefficient is (−1)k , hence γk is integral over A. (2) It is easy to verify that γk+1 = θγk + ak+1 , hence θγk = γk+1 − ak+1 . It follows, by (1), that θγk is integral over A. (3) Clearly B is an additive group. For completing the proof it is sufficient to show that γi γj ∈ B for all i, j ∈ {1, . . . , n − 1}. We use a formula relating γi with γi−1 proved in (2). We obtain γi γj = (θγi−1 + ai )γj = γi−1 (γj+1 − aj+1 ) + ai γj and the conclusion follows by induction on i.

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Exercise 10.38. Let A be an integral domain, K be its fraction field and E be an extension of K. Let f (x) ∈ K[x] be monic and non constant such that if the characteristic p is positive, then p - deg f . Suppose that f (x) = g(h(x)), where g(x) and h(x) are monic non constant polynomials with coefficients in E such that h(0) = 0. (1) Show that g(x) and h(x) ∈ K[x]. (2) If f (x) ∈ A[x], show that the coefficients of g(x) and h(x) belong to the integral closure of A in K. (3) Let f1 (x) ∈ K[x]. Suppose that f1 (x) = g1 (h1 (x)), where g1 (x) and h1 (x) are non constant polynomials with coefficients in K. Show that there exist g(x), h(x) ∈ K[x], non constant such that, h is monic, h(0) = 0, deg g1 = deg g, deg h1 = deg h and f1 (x) = g(h(x)). (4) Show that the conclusion in (1) does not (necessarily) hold if we omit the condition on the characteristic of K. Solution 10.38. (1) Let m = deg g, n = deg h, g(x) = xm + b1 xm−1 + · · · + bm ,

h(x) = xn + c1 xn−1 + · · · + cn−1 x

f (x) = x

mn

+ a1 x

mn−1

and

+ · · · + amn ,

where a1 , . . . , amn ∈ K and b1 , . . . , bm , c1 , . . . , cn−1 ∈ E. Identifying the coefficients of xmn−1 , . . . , xmn−(n−1) in the identity f (x) = g(h(x)), we obtain the following equations: a1 = mc1

and

ai = mci + Fi (c1 , . . . , ci−1 )

for i = 2, . . . , n − 1,

where Fi ∈ Z[x1 , ..., xi−1 ] is homogeneous of degree i. Since p - degf , then p - m and c1 = a1 /m ∈ K. By induction, we conclude that c1 , c2 , . . . , cn−1 ∈ K. We next identify in the same identity the coefficients of xmn−n , . . . , xmn−nm and we obtain the following equations: b1 + G1 (c1 , . . . , cn−1 ) = an bi + Gi (b1 , . . . , bi−1 , c1 , . . . , cn−1 ) = ani

for i = 2, . . . , m,

where Gi ∈ Z[y1 , . . . , yi−1 , x1 , . . . , xn−1 ] is homogeneous of degree m−i. By induction we see that bi ∈ K for i = 1, . . . , m. Therefore g(x) and h(x) ∈ K[x].

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(2) Let α1 , . . . , αm be the roots of g(x) is an algebraic closure of K, then f (x) = (h(x) − α1 ) · · · (h(x) − αm ). Let i ∈ {1, . . . , m} and β be any root of hi (x) = h(x) − αi , then β is a root of f (x). Since f (x) ∈ A[x], we conclude that β is integral over A. Therefore the coefficients of hi (x) are integral over A. By (1), h(x) ∈ K[x], hence the coefficients of h(x) belong to the integral closure of A in K. Now α1 , . . . , αm are integral over A, hence all the coefficients of g(x) are integral over A. By (1) these coefficients are in K, hence belong to the integral closure of A in K. (3) Let a, b ∈ E such that the polynomial h(x) = ah(x)+b is monic and has a zero constant coefficient. Let g(x) = g1 ( x−b a ), then these polynomials g and h satisfy the required conditions. (4) Let A = K = F2 and let α be a root of u(x) = x2 + x + 1 in an algebraic closure of F2 . Let f (x) = x4 + x ∈ F2 [x], then f (x) = g(h(x)), with g(x) = x2 + (α + 1)x

and h(x) = x2 + αx.

These polynomials have their coefficients in F2 (α) but not in F2 . Notice that f (x) = g1 (h1 (x)), where g1 (x) = h1 (x) = x2 + x. Exercise 10.39. (1) Let L be a field, Ω be an algebraic closure of L, (aij )(i,j) , (bij )(i,j) be two sets of indeterminates with 0 ≤ i < r, 0 ≤ j < r and i + P j > r. Let A = L[(aij ), (bij )]. Set f ((aij ), x, y) = (i,j) aij xi y j and P g((bij ), x, y) = bij xi y j . Let φ(x, y) ∈ L[x, y] such that deg φ = r. Let I be the ideal of A generated by the finite set B formed by the coefficients of the polynomial f (x, y)g(x, y) − φ(x, y) ∈ A[x, y]. Show that the following conditions are equivalent. (i) φ(x, y) is irreducible over Ω. (ii) I has no zero in Ω. (iii) I = A. (2) Let L be a number field, O be its ring of integers, φ(x, y) ∈ L[x, y] absolutely irreducible. Show that for almost all prime ideals P of O, φ(x, y) (mod P) is irreducible over an algebraic closure of O/P. Solution 10.39. (1) φ(x, y) is irreducible over Ω if and only if for any families (a∗ij )(i,j) and (b∗ij )(i,j) of elements of Ω, φ(x, y) − f ((a∗ij ), x, y)g((a∗ij ), x, y) 6= 0 if and only if (ii).

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(ii) and (iii) are equivalent by Hilbert’s Nullstellensatz [Lang (1965), Chap. 10.2]. (2) Let r = deg φ(x, y) and let A, B, I be the sets defined in (1). Since φ is absolutely irreducible then I = A (by iii), hence there exist a positive Pk integer k, λ1 , . . . , λk ∈ A and µ1 , . . . , µk ∈ B such that 1 = i=1 λi µi . u((aij ),(bij )) , where d Any of these λi and any of these µi has the form d is a positive integer and u((aij ), (bij )) is a polynomial in the variables aij , bij with coefficients in O. We first exclude the prime ideals of A which divides some denominator of a coefficient of φ. Exclude also the primes P for which deg φ¯ < r, where φ¯ is the reduced polynomial of φ modulo P. Finally exclude the primes P which divide any denominator d of λi or µi in the preceding representation. Clearly the excluded ¯ B, ¯ I¯ primes are finite in number. For any other prime P, we define A, similarly as A, B, I but now for the field O/P and for the polynomial ¯ y) of degree r. If we reduce modulo P the above Bezout’s identity, φ(x, Pk ¯ ¯ Therefore φ¯ is absolutely we obtain ¯ 1 = ¯i , hence I¯ = A. i=1 λi µ irreducible accordingly to (1). Exercise 10.40. Let f (x) = xn + · · · + a1 x + a0 ∈ Z[x] be irreducible, θ be a root of f and K = Q(θ). Let k be a positive integer with k < n and p be a prime number such that pk || a0 and pk+1−i | ai for i = 1, . . . , k. Show that p ramifies in K. Solution 10.40. Suppose that p does not ramify in K, then pA = P1 · · · Pr , where A is the ring of integers of K and P1 , . . . , Pr are distinct prime ideals of A. We have a0 A = P1k · · · Prk I, where I is an ideal of A whose norm is coprime with p. Since NK/Q (θ) = ±a0 , then the ideal θA is divisible by at least one Pi say P. As P k+1−i | ai for i = 1, . . . , k, we have a0 = a0 − f (θ)

= (−a1 θ − . . . − ak θk ) − (ak+1 θk+1 + · · · + an θn ) ≡0

(mod P k+1 ),

contradicting pk || a0 . Thus p ramifies in K. Exercise 10.41. ⊂ = If A and B are sets we write A ∼ B (resp. A ∼ B) to mean that all the elements of A, but finitely of them, are elements of B (resp. A and B have the same elements except a finite number of elements). For any non

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constant F (x) ∈ Z[x] of degree m define D(F ) and Di (F ) for i = 1, . . . , n by D(F ) = {p prime, there exists x ∈ Z,

F (x) ≡ 0

(mod p)}

Di (F ) = {p prime, the equation F (x) ≡ 0 (mod p) has exactly i solutions}. (1) Show that D(F ) is infinite. (2) Recall the following result from [Marcus (1977), Th. 13, Chap. 1 and Exercise 40, Chap. 1]. Let K be a number field of degree m, A be its ring of integers and γ ∈ A be a primitive element over Q. Then A has a basis over Z of the form {ω0 = 1, ω1 = u1 (γ)/d1 , . . . , ωm−1 = um−1 (γ)/dm−1 },

where ui (x) ∈ Z[x] and di is a positive integer for i = 1, . . . , m − 1. Moreover d1 | d2 | · · · | dm−1 and I(γ) = d1 d2 · · · dm−1 . This shows that any element φ ∈ A may be expressed in the form φ = A(γ)/d, with A[x] ∈ Z[x] and d is a positive divisor of I(γ). Let f (x) and g(x) be non constant polynomials with integral coefficients. Suppose that f and g are monic and irreducible over Q. Let α and β be roots of f (x) and g(x) respectively in C. If Q(α) ⊂ Q(β), show that if p ∈ D(g) and p - I(β), then p ∈ D(f ). Deduce that ⊂ D(g) ∼ D(f ). Show that the inclusion may be strict. (3) Let f (x) and g(x) be non constant polynomials with integral coefficients. Suppose that f and g are monic and irreducible over Q. Let α and β be roots of f (x) and g(x) respectively in C. If Q(α) = Q(β), show that if p - I(α) I(β), then p ∈ D(g) ⇔ p ∈ D(f )

p ∈ Di (g) ⇔ p ∈ Di (f ), =

and =

for i = 1, . . . , n. Deduce that D(f ) ∼ D(g) and Di (f ) ∼ Di (g) for i = 1, . . . , n. (4) Suppose that f is monic irreducible of degree n and let α be one of its roots. Suppose that Q(α)/Q is normal. If p is a prime number such that p - Disc(f ), show that p ∈ D(f ) ⇔ p ∈ Dn (f ). Deduce that = D(f ) ∼ Dn (f ). Show that if γ is a root of F (x) = x4 + x2 + x − 2, then Q(γ)/Q is not normal. (5) Let f (x) and g(x) be non constant polynomials with integral coefficients. Suppose that f and g are monic and irreducible over Q. Let α and β be roots of f (x) and g(x) respectively in C. Suppose that Q(β)/Q

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is normal and that Q(α) ⊂ Q(β). Let n = deg f . Let p be a prime number such that p - I(β) Disc(f ). Show that p ∈ D(g) ⇒ p ∈ Dn (f ) ⊂ and that D(g) ∼ Dn (f ). Deduce that if h(x) is a monic irreducible polynomial over Q of degree n, then Dn (h) is infinite. (6) Let K/Q be a number field of degree n, L be a normal closure of K. Let α and β be primitive elements of K and L respectively which are supposed to be algebraic integers. Let f (x) = Irr(α, Q) and g(x) = Irr(β, Q). Let p be a prime number such that p - I(β)Disc(f ). Show = p ∈ D(g) ⇔ p ∈ Dn (f ) and that D(g) ∼ Dn (f ). 3 3 (7) Let f1 (x) = x − 18x − 6, f2 (x) = x − 36x − 78, f3 (x) = x3 − 54x − 150 and let α1 , α2 , α3 be a root of f1 , f2 , f3 respectively. Show that these polynomials are irreducible over Q and Disc(f1 ) = Disc(f2 ) = Disc(f3 ) = 22 .35 .23. By looking at the primes 5 and 11, show that the fields Q(α1 ), Q(α2 ) and Q(α3 ) are distinct. Solution 10.41. (1) Let m = deg F and set F (x) = bm xm + bm−1 xm−1 + · · · + b0 , where bi ∈ Z for any i ∈ {0, . . . , m}. If b0 = 0, then the result is clear. Suppose that b0 6= 0. There exists x0 ∈ Z such that f (x0 ) 6= 1 and f (x0 ) 6= −1, hence D(F ) 6= ∅ since it contains a prime divisor of F (x0 ). Suppose that this set is finite and let p1 , . . . , pr be its elements and set a = p1 · · · pr . We have F (acx) = bm (ab0 x)m + bm−1 (ab0 x)m−1 + · · · + b1 (ab0 x) + b0
= b0 am bm−1 bm xm + · · · + b1 ax + 1 0 := b0 G(x),

hence D(G) ⊂ D(F ). For the same reason as for F , we have D(G) 6= ∅. Let p ∈ D(G), then p | a and there x ∈ Z such that p | am bm−1 bm xm + · · · + b1 ax + 1, 0 hence p | 1 which is a contradiction. Therefore D(F ) is infinite. (2) Let n = deg f . The element α may be written in the form α = u(β)/d, where u(x) ∈ Z[x] and d is a positive integer dividing I(β) and such that d and the coefficients of u(x) are coprime. Since g(x) | f (u(x)/d) in Q[x], there exists q(x) ∈ Z[x] such that f (u(x)/d) = g(x)q(x)/dn .

(Eq 1)

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From this identity it is seen that for any prime p - d, we have p ∈ ⊂ D(g) ⇒ p ∈ D(f ), hence D(g) ∼ D(f ). 2 Suppose that f (x) = x + 9 and g(x) = x2 + 1. Let α = 3i and β = i, then obviously Q(α) ⊂ Q(α). We have f (0) ≡ 0 (mod 3), so that 3 ∈ D(f ). On the other hand, we have g(0) ≡ 1 (mod 3) and g(±1) ≡ −1 (mod 3), hence 3 6∈ D(g). this proves that the inclusion proved above may be strict. (3) Our assumptions allow us to adjoin to the preceding equation the following one   f (x)h(x) v(x) , (Eq 2) = g 0 d (d0 )n where v(x) and h(x) ∈ Z[x]. We have already proved that if p - I(β) ⊂ then p ∈ D(g) ⇒ p ∈ D(f ), thus D(g) ∼ D(f ). Using (2), we obtain = similar conclusions when permuting α and β, hence D(g) ∼ D(f ). Let p ∈ Di (g) such that p - d and let x1 , . . . , xi be the distinct zeros of g(x) modulo p. By (Eq 1), it is seen that u(xi )/d is a root of f (x) 0 modulo p. We show that if moreover p - d , these roots are distinct modulo p. We have d10 v(u(β)/d) = β, hence 1 v(u(x)/d) = x + g(x)a(x)/(d0 de ), (Eq 3) d0 where a(x) ∈ Z[x] and e is a non negative integer. Suppose that u(xi )/d ≡ u(xj )/d (mod p), then (Eq 3) shows that xi ≡ xj (mod p). Therefore the number of distinct roots of f (x) modulo p is at least equal = to i. Inverting the roles of f (x) and g(x), we obtain Di (f ) ∼ Di (g) for i = 1, . . . , n. ⊂ (4) The inclusion Dn (f ) ∼ D(f ) is trivial. We show the reverse inclusion. Set f (x) = xn + an−1 xn−1 + · · · + a0 , with a0 , . . . , an−1 ∈ Z and let α1 = α, α2 = u2 (α)/d, . . . , αn = un (α)/d, where d is a positive integer dividing I(α) and u2 (x), . . . , un (x) ∈ Z[x]. Let     u2 (x) un−1 (x) F (x, y) = (y − x) y − ··· y − d d An−1 (x) n−1 A0 (x) = yn + y + · · · + n−1 , (Eq 4) d d then Ai (x) ∈ Z[x] for i = 0, . . . , n − 1 and An−1 (α) n−1 A0 (α) F (α, y) = f (y) = y n + y + · · · + n−1 , d d n−j hence Aj (α)/d = aj for j = 0, . . . , n − 1. Therefore f (x) divides Aj (x) − dn−j aj in Q[x] for j = 0, . . . , n − 1. Let qj (x) ∈ Q[x] such that

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Aj (x) − dn−j aj = f (x)qj (x). Since f (x) is monic then qj (x) ∈ Z[x] for j = 0, . . . , n − 1. We have     q0 (x) qn−1 (x) n−1 n y + · · · + a0 + f (x) n−1 F (x, y) = y + an−1 + f (x) d d G(x, y) = f (y) + f (x) n−1 , (Eq 5) d with G(x, y) ∈ Z[x, y]. Let p ∈ D(f ), then there exists a ∈ Z such that f (a) ≡ 0 (mod p). If p - d, then by By (Eq 4) and (Eq 5), we have F (a, y) = (y − a)(y − u2 (a)/d) · · · (y − un−1 (a))/d ≡ f (y)

(mod p).

Since p - Disc(f ), then a, u2 (a)/d, . . . , un−1 (a)/d are distinct modulo p. Therefore p ∈ Dn (f ). We apply the result for F (x) = x4 + x2 + x − 2. We omit the proof that F (x) is irreducible over Q. We have F (x) ≡ x(x3 + x + 1) (mod 2) and the factor x3 + x + 1 is irreducible modulo 2. This shows that F (x) is separable modulo 2, thus 2 - Disc(F ). Obviously 2 ∈ D(F ) but 2 6∈ D4 (F ). We deduce that Q(γ)/Q is not normal. (5) Let p ∈ D(g) such that p - Disc(f ) I(β). Set f (x) = xn + an−1 xn−1 + · · · + a0 .

Let α1 , . . . αn be the roots of f which are in Q(β). Let ui (x) ∈ Z[x] for i = 1, . . . , n such that αi = ui (β)/d where d is a divisor of I(β). Let F (x, y) =

n Y

(y − ui (x)/d) = y n + (Bn−1 (x)/d)y n−1 + · · · + B0 (x)/dn .

i=1

Then F (β, y) =

n Y

(y −αi ) = f (y) = y n +(Bn−1 (β)/d)y n−1 +· · ·+B0 (β)/dn .

i=1

It follows that Bj (β)/dn−j − aj = 0 for j = 0, . . . , n − 1. Therefore g(x) | Bj (x) − aj dn−j in Z[x]. Set Bj (x) − aj dn−j = g(x)qj (x) with qj (x) ∈ Z[x]. Then n Y

(y − ui (x)/d) = f (y) + g(x)H(x, y)/dn ,

i=1

where H(x, y) ∈ Z[x, y]. Let x0 ∈ Z such that g(x0 ) ≡ 0 (mod p). Since p - d and p - Disc(f ) then f (y) factorizes into distinct linear factors modulo p, which implies that p ∈ Dn (f ). We deduce that ⊂ D(g) ∼ Dn (f ).

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Let α be a root of h, L be a Galois closure of Q(α) over Q, β be a primitive element of L and g(x) be the minimal polynomial of β. By ⊂ the first part of (5), we have D(g) ∼ Dn (h). Since D(g) is infinite by (1), then so is Dn (h). ⊂ (6) By (5), we have D(g) ∼ Dn (f ). We prove the reverse inclusion. Let p ∈ Dn (f ) and let α1 , . . . αn be the roots of f , then n n Y Y f (x) = (x − αi ) ≡ (x − ai ) (mod p), i=1

i=1

where the ai are distinct integers. We deduce that σj (α1 , . . . , αn ) ≡ σj (a1 , . . . , an ) (mod p), where σj is the elementary symmetric polynomial of degree j. The primitive element theorem asserts that there Pn exists a primitive element of L/Q of the form i=1 ci αi , where ci ∈ Z Pn for i = 1, . . . , n. By (3), we may suppose that β = i=1 ci αi . Consider the polynomials ! n Y X X φ(x) = x− ci ατ (i) = dk xk and i=1

ψ(x) =

Y

x−

n X i=1

ci aτ (i)

!

=

X

δ k xk ,

where the products run over the permutations τ of the set {1, . . . , n}. Since dk is a symmetric polynomial of α1 , . . . , αn and δk is a symmetric polynomial of a1 , . . . , an , then δk ≡ dk (mod p). Therefore φ(x) ≡ ψ(x) (mod p). Since φ(β) = 0, then g(x) | φ(x) in Z[x], hence g(x) | φ(x), where g(x) (resp. φ(x)) denotes the reduced polynomial modulo p of g (resp. φ). Therefore g(x) | ψ(x). Since ψ(x) is a product of linear factors (distinct or not), so is g(x), which implies p ∈ D(g). (7) Eisenstein’s irreducibility criterion for the prime p = 2 applies for the three polynomials, hence they are irreducible. It is known that the discriminant of a polynomial P (x) = x3 + ax + b is given by Disc(P ) = −4a3 − 27b2 . Using this formula, we obtain Disc(f1 ) = Disc(f2 ) = Disc(f3 ) = 22 .35 .23.

We have f1 (x) ≡ (x + 1)(x + 2)(x − 3) (mod 11) and both f2 , f3 are irreducible modulo 11, hence 11 ∈ D(f1 ), 11 6∈ D(f2 ), 11 6∈ D(f3 ). Therefore D(f1 ) 6⊂ D(f2 ) and D(f1 ) 6⊂ D(f3 ). On the other hand, we have f3 (x) ≡ x(x − 2)(x + 2) (mod 5) and f2 (x) is irreducible modulo 5, hence 5 ∈ D(f3 ) and 5 6∈ D(f2 ). Therefore D(f3 ) 6⊂ D(f2 ) and the proof is complete by (3).

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Exercise 10.42. Let θ1 , θ2 , θ3 be the roots in R of f (x) = x3 − 3x + 1 such that θ1 < θ2 < θ3 . (1) Show that θ1 = −1.879..., θ2 = 0.347... and θ3 = 1.529.... (2) Show that θ2 = θ12 − 2 and θ3 = −θ12 − θ1 + 2. (3) Let αi = θi2 for i = 1, 2, 3, a = θ1 θ2 α3 + θ2 θ3 α1 + θ3 θ1 α2 and b = θ1 θ2 α3 + θ2 θ3 α2 + θ3 θ1 α1 . Show that among a and b one and only one of them is a rational integer. Compute this integer. Solution 10.42. (1) Using a simple calculator, we find f (−1.879) = 0.002925561 > 0, f (0.347) = 0.000781923 > 0, f (1.529) = −0.012441111 < 0

f (−1.88 = −0.004672) < 0,

f (0.348) = −0.001855808 < 0, and f (1.53) = 0.08423 > 0,

hence θ1 = −1.879..., θ2 = 0.347... and θ3 = 1.529.... (2) Since f (x) is irreducible over F2 , then it is irreducible over Q. Set θ1 = θ. Using the identities θ3 = 3θ − 1,

θ4 = 3θ2 − θ,

θ5 = −θ2 + 9θ − 3

and θ6 = 9θ2 − 6θ + 1,

one verifies easily that f (θ2 − 2) = f (−θ2 − θ + 2) = 0. Obviously θ1 , θ2 and θ3 are distinct. Therefore they constitute the complete set of roots of f (x). An alternative way to prove the same assertions is to show that f (x) | f (x2 − 2). In that case it is obvious that θ2 − 2 is a root of f (x). Moreover the divisibility property of polynomials shows also that f (θ22 − 2) = 0. Straightforward computations shows that θ22 − 2 = −θ2 − θ + 2. (3) • First proof. We have θ12 = θ2

θ22 = (θ2 − 2)2 = −θ2 − θ + 4 θ32 = (−θ2 − θ + 2)2 = θ + 2.

Using these relations and also the expressions of θ3 , θ4 , θ5 and θ6 as polynomials in θ of degree at most 2, we find a = 0 and b = 3(θ2 −4). Thus a is a rational integer and b is not. • Second proof. Using the approximate values of θ1 , θ2 and θ3 respectively, we get a = 0.0029... and b = −11.6039.... Therefore we may conclude that b is not an integer. The approximate value of

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a does not allow us to conclude that a is an integer. From (2), we see that Q(θ) is a cyclic extension of Q of degree 3, whose Galois group is generated by the automorphism σ such that σ(θ) = θ2 = θ2 − 2. It is clear that a is an algebraic integer. Since σ(a) = σ(θ1 )σ(θ2 )σ(α3 ) + σ(θ2 )σ(θ3 )σ(α1 ) + σ(θ3 )σ(θ1 )σ(α2 ) = a, then a ∈ Q. It follows that a is a rational integer and its approximate value shows that a = 0. Exercise 10.43. Let f (x) = x3 − 3x + 1, θ be a root of f (x) in C, K = Q(θ) and A be the ring of integers of K. (1) Show that f (x) is irreducible over Q. Show that the roots of f (x) are given by θ1 = θ, θ2 = θ2 − 2 and θ3 = −θ2 − θ + 2. (2) Show that Disc(f ) = 34 and that 3 - I(θ). Deduce that A = Z[θ] and Disc(K) = 34 . (3) Show that 2A is prime, 3A = P 3 , where P is a prime ideal of A with residual degree equal to 1. Show that P = (θ + 1)A. Show that DK = P 4 = (3θ + 3)A. (4) Show that 17A = P1 P2 P3 , where P1 , P2 and P3 are prime ideals of A of residual degree equal to 1. Show that P1 = (θ + 3)A, P2 = (θ2 + 1)A and P3 = (−θ2 − θ + 5)A. (5) Compute a basis over Z of DK . Solution 10.43. (1) Since f (x) is irreducible over F2 , then it is irreducible over Q. Using the identities θ3 = 3θ − 1, θ4 = 3θ2 − θ, θ5 = −θ2 + 9θ − 3 and θ6 = 9θ2 − 6θ + 1,

one verifies easily that f (θ2 − 2) = f (−θ2 − θ + 2) = 0. Obviously θ1 , θ2 and θ3 are distinct. Therefore they constitute the complete set of roots of f (x). An alternative way to prove the same assertions is to show that f (x) | f (x2 − 2). In that case it is obvious that θ2 − 2 is a root of f (x). Moreover the divisibility property of polynomials shows also that f (θ22 − 2) = 0. Straightforward computations shows that θ22 − 2 = −θ2 − θ + 2. (2) We have Disc(f ) = −4(−3)3 − 27 = 34 . We find that f (x) ≡ (x + 1)3 (mod 3) and f (x) = (x + 1)3 − 3x(x + 2). Since x + 1 - x(x + 2) in F3 [x], then, by Exercise 10.30 3 - I(θ). It follows that A = Z[θ] and Disc(K) = 34 .

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(3) Since f (x) ≡ x3 + x + 1 (mod 2) and since this last polynomial is irreducible over F2 , then by Dedekind’s Theorem [Marcus (1977), Th. 27, Chap. 3], 2A is prime, that is the prime 2 is inert in A. Here also, for the prime 3, Dedekind’s Theorem applies since 3 - I(θ) and we conclude that 3A = P 3 , where P is a prime ideal of A with residual degree equal to 1. Moreover P = (3, θ + 1). We compute the norm of θ + 1. Let β = θ + 1, then β is a root of g(x) = (x − 1)3 − 3(x − 1) + 1 = x3 − 3x2 + 3. We conclude that NK/Q (θ + 1) = −3. It follows that P = (θ + 1)A. Since A = Z[θ] and since θ − 1 is a unit, then 0

DK = f (θ)A = (3θ2 − 3)A = 3(θ − 1)(θ + 1)A = 3(θ + 1)A = P 4 . (4) We have f (x) ≡ (x + 3)(x + 4)(x − 7) (mod 17), hence 17A = P1 P2 P3 , where P1 , P2 and P3 are prime ideals of A of residual degree equal to 1. Moreover P1 = (17, θ + 3), P2 = (17, θ + 4)

and

P3 = (17, θ − 7). By the method used in (3), we compute the norm of θ + 3 and we find NK/Q (θ + 3) = 17. Therefore P1 = (θ + 3)A. Since Disc(K) is a square, then K is a Galois extension of Q. It follows that the prime ideals of A lying over 17Z are conjugate. Thus P2 = (θ2 + 3)A = (θ2 + 1)A and P3 = (θ3 + 3)A = (−θ2 − θ + 5)A. (5) Let Z[α]v = {x ∈ K, TrK/Q (xZ[α]) ⊂ Z}, then D = (Z[α]v )−1 . We begin with the determination of a Z-basis of Z[α]v . Since f (x) = x3 − 3x + 1 = (x − θ)(x2 + θx + θ2 − 3), then according to [Lang (1965), Prop. 1, Chap. 8.6], Z[α]v is generated 0 0 0 0 over Z by 1/f (θ), θ/f (θ) and (θ2 − 3)/f (θ). We have 1/f (θ) = 1/(3θ2 − 3) = (2θ2 + θ − 4)/9. It follows that θ(2θ2 + θ − 4) (θ2 − 3)(2θ2 + θ − 4) (2θ2 + θ − 4) +Z +Z 9 9 9 = Z(2θ2 + θ − 4)/9 + Z(θ2 + 2θ − 2)/9 + Z(−4θ2 − 2θ + 11)/9.

Z[α]v = Z

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Let x = a + bθ + cθ2 ∈ Z[θ], then x ∈ D if and only if (2θ2 + θ − 4) ∈ Z[θ], 9 (θ2 + 2θ − 2) x ∈ Z[θ] and 9 (−4θ2 − 2θ + 11) ∈ Z[θ]. x 9 x

Easy calculations show that (2θ2 + θ − 4) u(θ) = , 9 9 (θ2 + 2θ − 2) v(θ) = and x 9 9 2 w(θ) (−4θ − 2θ + 11) = x 9 9 x

with (2a + b + 2c)θ2 + (a + 2b + c)θ − 4a − 2b − c , 9 2 (a + 2b + c)θ + (2a + b + 5c)θ − 2a − b − 2c v(θ) = and 9 (−4a − 2b − 11c)θ2 + (−2a − 11b − 2c)θ + a + 4b + 2c w(θ) = . 9 u(θ) =

Therefore, the conditions for x to belong to D are given by 2a + b + 2c ≡ 0

(mod 9),

−4a − 2b − c ≡ 0

(mod 9),

2a + b + 5c ≡ 0

(mod 9),

−4a − 2b − 11c ≡ 0

(mod 9),

a + 4b + 2c ≡ 0

(mod 9).

a + 2b + c ≡ 0

(mod 9),

a + 2b + c ≡ 0

(mod 9),

−2a − b − 2c ≡ 0

(mod 9),

−2a − 11b − 2c ≡ 0

(mod 9)

and

The second and the third equations imply that 3a ≡ 0 (mod 9), that is a ≡ 0 (mod 3). Set a = 3a1 . The first two equations then become b + 2c ≡ 3a1 (mod 9) and 2b + c ≡ −3a1 (mod 9). We deduce that

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c ≡ 0 (mod 3). Set c = 3c1 . Then b ≡ 3(a1 + c1 ) (mod 9). Set b = 3(a1 + c1 ) + 9b1 , then x = 3a1 + (3(a1 + c1 ) + 9b1 )θ + 3c1 θ2 = a1 (3 + 3θ) + b1 (9θ) + c1 (θ + 3θ2 ). This shows that D is generated over Z by {3 + 3θ, 9θ, θ + 3θ2 }. This set is obviously free over Z, so it is a basis over Z. Exercise 10.44. Let p ≡ 1 (mod 8) be a prime number, f (x) = x4 − p. Let θ ∈ C be a root of f (x), K = Q(θ) and A be the ring of integers of K. (1) Show that α = (θ2 + 1)/2 and β = (θ3 + θ2 + θ + 1)/4 are algebraic integers. (2) Show that Disc(θ) = −28 p3 . (3) Show that p - I(θ). (4) Show that 23 | I(θ). (5) Deduce that I(θ) = 23 , Disc(K) = −22 p3 and {1, θ, α, β} is an integral basis of K. (6) If p ≡ 1 (mod 16), show that the splitting of the prime 2 in A is given 0 00 by (2) = P12 P1 P1 , where the prime ideals appearing there have their residual degree equal to 1. Solution 10.44. √ √ (1) Since α = (1 ± p)/2, then α is an integer of the field Q( p). The element β is a root of the polynomial 4 4 3 g(x) = x4 − x3 − (p − 1)x2 − (p − 1)2 x − (p − 1)3 , 8 64 64 hence it is an algebraic integer. (2) We have
3 Disc(θ) = NK/Q (f 0 (θ)) = NK/Q (4θ3 ) = 44 NK/Q (θ) = −28 p3 . (3) We have f (x) ≡ x4 (mod p), f (x) = x4 − p(1) and p - 1, hence p - I(θ). (4) We know that the ring of integers of K has a basis of the form {1, f1 (θ)/d1 , f2 (θ)/d2 , f3 (θ)/d3 }, where for every i, fi (x) is a monic polynomial of degree i, with integral coefficients and di is a positive integer. Moreover I(θ) = d1 d2 d3 and the coefficients of fi (x) are determined modulo di [Marcus (1977), Th. 13, Exercises 39 and 40, Chap. 2]. Since f (x) ≡ (x − 1)4 (mod 2), then for any prime ideal P lying over 2Z, we have θ ≡ 1 (mod P). We deduce

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that if 2 | d1 , then f1 (θ) ≡ 0 (mod 2), hence f1 (θ) ≡ 0 (mod P), thus f1 (x) = x + 1. Let µ = θ + 1, then θ = µ − 1 and (µ − 1)4 − p = 0. It follows that µ4 − 4µ3 + 6µ2 − 4µ + 1 − p = 0. From this equation it is seen that µ/2 is not integral. We conclude that 2 - d1 . We know that α = (θ2 + 1)/2 is integral, so that 2 | d2 . Does 22 | d2 ? We must look at f2 (x) = x2 + 1, x2 − 1, x2 + 2x + 1, x2 + 2x − 1.

The minimal polynomial of α = (θ2 + 1)/2 is given by g(x) = x2 − x + (1 − p)/4, thus it is seen that α/2 = (θ2 + 1)/4 is not integral. Let γ = (θ2 − 1)/2 = α − 1, then γ 2 + γ + (1 − p)/4 = 0. This implies that (θ2 − 1)/4 is not integral. Let ρ = (θ2 + 2θ + 1)/2 = µ2 /2. We find that ρ satisfies the following equation: (1 − p)2 (35 − p) 2 7 − 3p ρ + ρ+ = 0. 2 2 16 We deduce that the characteristic polynomial of ρ/2 is given by ρ4 − 2ρ3 +

(35 − p) 2 7 − 3p (1 − p)2 x + x+ . 8 16 162 Since p ≡ 1 (mod 8), then p 6≡ 35 (mod 8), so that ρ/2 is not integral. Let ν = (θ2 + 2θ − 1)/2, then ν = ρ − 1. Using the equation of ρ, we find that of ν: (35 − p) 2 ν 4 +2ν 3 + ν +(40−4p)ν +6−3p+(1−p)2 /16+(35−p)/2 = 0. 2 It follows that ν/2 is a root of the polynomial h(x) = x4 − x3 +

(35 − p) 2 (10 − p) x + x + c, 8 2 where c is an integer. Since p ≡ 1 (mod 8), then p ≡ 1 (mod 2), hence p 6≡ 10 (mod 2), thus ν/2 is not integral. u(x) = x4 + x3 +

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We conclude that 2 | d2 , but 22 - d2 . We now discuss the 2-adic valuation of d3 . Since β is integral, then 22 | d3 . Does 23 | d3 ? Since the coefficients of f3 (x) are determined modulo 8, we must consider the following possibilities for f3 (x): f31 (x) = x3 + x2 + x + 1,

f32 = x3 + x2 + x + 5,

f33 (x) = x3 + 5x2 + x + 1,

f34 (x) = x3 + 5x2 + x + 5,

f35 (x) = x3 + x2 + 5x + 1,

f36 (x) = x3 + x2 + 5x + 5,

f37 (x) = x3 + 5x2 + 5x + 1,

f38 (x) = x3 + 5x2 + 5x + 5.

The following table gives the characteristic polynomial of f3i (θ). Table 1 f3i (θ) f31 (θ)

Char(f3i (θ), Q) x4 − 4x3 + (6 − 6p)x2 + (−4 + 8p − 4p2 )x + 1 − 3p + 3p2 −p3

f32 (θ)

x4 − 20x3 + (150 − 6p)x2 + (−500 + 56p − 4p2 )x + 625 −131p + 19p2 − p3

f33 (θ)

x4 − 4x3 + (6 − 54p)x2 + (−4 + 88p − 20p2 )x + 1 − 35p +547p2 − p3

f34 (θ)

x4 − 20x3 + (150 − 54p)x2 + (−500 + 520p − 20p2 )x + 625 −1251p + 627p2 − p3

f35 (θ)

x4 − 4x3 + (6 − 22p)x2 + (−4 − 56p − 4p2 )x + 1 − 547p +35p2 − p3

f36 (θ)

x4 − 20x3 + (150 − 22p)x2 + (−500 + 120p − 4p2 )x + 625 −675p + 51p2 − p3

f37 (θ)

x4 − 4x3 + (6 − 70p)x2 + (−46360p − 20p2 )x + 1 − 195p +195p2 − p3

f38 (θ)

x4 − 20x3 + (150 − 70p)x2 + (−500 + 200p − 20p2 )x + 625 +125p + 275p2 − p3

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This table shows that the characteristic polynomial of f3i (θ) is given by F (x) = x4 − 4x3 + lower degree terms or

F (x) = x4 − 20x3 + lower degree terms. In the first case the characteristic polynomial of f3i (θ)/8 is given by G(x) = x4 − (1/2)x3 + lower degree terms, hence f3i (θ)/8 is not an algebraic integer, thus 23 - d3 in this case. Similarly we get the same conclusion in the second case. It follows that I(θ) = d1 d2 d3 = 23 . Remark. The polynomials appearing in this table suggest that we may avoid computing them explicitly. In fact, finding TrQ(θ)/Q (f3i (θ)) is sufficient for our needs. (5) From (2) and (3), we conclude that p3 || Disc(K). From (2) and (4), we get 22 || Disc(K). It follows from (2) that I(θ) = 23 and Disc(K) = −22 p3 . Consider the inclusions Z[θ] ⊂ Z[α, β] ⊂ A. It is easy to compute the matrix which expresses the coordinates of θi , i = 0, . . . , 3, in the basis {1, θ, α, β} of K over Q. Its determinant is equal to 8. Therefore (Z[α, β] : Z[θ]) = 8 = I(θ). We deduce that A = Z[α, β] and then {1, θ, α, β} is a basis of A over Z. √ (6) Obviously, K contains the quadratic field E = Q( p) = Q(θ2 ). Let B be the ring of integers of E. Since p ≡ 1 (mod 8), then 2B = Q1 Q2 , where Q1 and Q2 are prime ideals of B having their residual degree equal to 1. From (5), we know that 2 is ramified in A. It follows that the splitting of 2 in A takes one of the following forms. 0

(i) 2A = P12 P12 . (ii) 2A = P12 P2 . 0 00 (iii) 2A = P12 P1 P1 .

Here all the ideals appearing in the splitting of 2, except P2 , have their residual degree equal to 1. The residual degree of P2 is equal to 2. In order to decide which of the splittings (i), (ii) and (iii) is the right one, we show that for any γ ∈ A, we have γ 2 ≡ γ (mod 2) or γ 2 ≡ γ + 1 − θ (mod 2). It is easy to prove the following formulas: θ2 = 2α − 1 and θ3 = 4β − 2α − θ. Using these relations, we get p−1 p−1 p−1 θ3 + θ2 + θ + 1 2 ) =β+ α+ θ+ and β2 = ( 4 8 8 8 2 θ +1 2 p−1 α2 = ( ) =α+ . 2 4

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Since p ≡ 1 (mod 16), using the preceding identities, we get θ2 ≡ 1 (mod 2), α2 ≡ α (mod 2) and β 2 ≡ β (mod 2). We deduce that for any γ = a + bθ + cα + dβ, we have γ 2 ≡ a + bθ2 + cα2 + dβ 2 ≡ a + b + cα + dβ ≡ γ + b(1 − θ) (mod 2). Hence γ 2 ≡ γ (mod 2) if b ≡ 0 (mod 2) and γ 2 ≡ γ + 1 − θ (mod 2) if b ≡ 1 (mod 2). The element δ = θ − 1 will be of some help in the proof. Its minimal polynomial is given by u(x) = x4 + 4x3 + 6x2 + 4x + 1 − p. From this, it is seen that NK/Q (δ) = 1 − p ≡ 0 (mod 16) and (δ − 1)/2 is not an algebraic integer. • Suppose that the splitting (ii) holds, that is 2A = P12 P2 . From u(x), we see that θ ≡ 1 (mod P1 ) and θ ≡ 1 (mod P2 ). Since θ 6≡ 1 (mod 2), then P1 2 - θ − 1. Since NK/Q (θ − 1) = NK/Q (δ) ≡ 0 (mod 16), 2 then P2 | θ − 1. Choose elements γ1 and γ2 of A such that γ1 ∈ P1 2 and γ2 + P2 is a primitive element over F2 of the field A/P2 , which is isomorphic to F4 . By the Chinese remainder theorem, there exists γ ∈ A such that γ ≡ γ1 (mod P12 ) and γ ≡ γ2 (mod P2 ). Hence γ ≡ 0 (mod P12 ) and γ 2 + γ + 1 ≡ 0 (mod P2 ). Therefore γ 3 + γ 2 + γ ≡ 0 (mod 2). We have seen that γ 2 ≡ γ (mod 2) or γ 2 ≡ γ + 1 − θ (mod 2). If γ 2 ≡ γ (mod 2), then γ 3 ≡ 0 (mod 2), hence γ 3 ≡ 0 (mod P2 ), which is a contradiction. If γ 2 ≡ γ + 1 − θ (mod 2), then γ 3 + 1 − θ ≡ 0 (mod 2). Since P2 | 1 − θ, then P2 | γ 3 . Therefore P2 | γ, which again is a contradiction. We have shown that the splitting (ii) does not hold. 0 • Suppose that the splitting (i) holds, that is 2A = P12 P12 . We have 0 θ ≡ 1 (mod P1 ) and θ ≡ 1 (mod P1 ). Since θ 6≡ 1 (mod 2), then 0 P12 - θ − 1 or P12 - θ − 1. Without loss of generality, we may suppose 0 that P12 - θ − 1. Since NK/Q (θ − 1) ≡ 0 (mod 16), then P13 | θ − 1. Choose γ ∈ A such that P1 || γ − 1. Recall that we proved that any element of A satisfies a congruence equation modulo 2 of two possible forms. Since γ 2 − γ = γ(γ − 1) 6≡ 0 (mod 2), then the first form of these equations does not hold. Suppose that the second form is satisfied, that is γ 2 − γ ≡ 1 − θ (mod 2), then γ 2 − γ ≡ 1 − θ (mod P12 ). The P1 -adic valuation of the left side of this congruence equation is equal to 1, while the valuation of the right side is at least equal to 3, hence a contradiction. 0

00

We conclude that 2A = P12 P1 P1 .

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Exercise 10.45. Let K be a number field with class number h = 2, A be its ring of integers. (1) Let ρ ∈ A be an irreducible element. Show that ρ is prime or ρA = P1 P2 , where P1 and P2 are prime ideals of A, distinct or not. (2) Let α ∈ A? . Let α = π1 · · · πr and α = χ1 · · · χs be two factorizations of α into products of (distinct or not) irreducible elements of A. Let r1 (resp. s1 ) be the number of (distinct or not) prime elements appearing in the first (resp. second) factorization and suppose that these primes are π1 , . . . , πr1 and χ1 , . . . , χs1 respectively. Show that r1 = s1 and reordering if necessary χ1 , . . . , χr1 , we have χi and πi are associate for i = 1, . . . , r1 . Deduce that r = s. Solution 10.45. (1) Suppose that ρ is not prime and let ρA = P1 · · · Pr be the factorization of ρA as a product of (distinct or not) prime ideals of A, then r ≥ 2. Suppose that r is odd, say r = 2k + 1. Since h = 2, then Pi Pi+1 = αi A, where αi ∈ A \ A? for i = 1, 3 . . . , 2k − 1. It follows that P2k+1 is principal say P2k+1 = α2k+1 A, thus ρ = α1 α3 · · · α2k+1 with  ∈ A? , contradicting the assumption that ρ is irreducible. We conclude that r is even, say r = 2k. As above, mutadis mutandis, we obtain ρ = α1 α3 · · · α2k−1 , where  ∈ A? . This implies that k = 1; otherwise ρ would be reducible. We conclude that r = 2. (2) We may assume that r1 ≤ s1 . Since χ1 is prime and divides π1 · · · πr , then it divides some πi , thus it is an associate of this πi . Since πj is not prime if j > r1 , then i ≤ r1 . Therefore we may assume that i = 1, that is χ1 and π1 are associated. Set π1 = 1 π1 , then π2 · · · πr = 1 χ2 · · · χs . Iterating the same reasoning as above, we obtain that χj is associate to πj , for j = 1, . . . , r1 . Set πj = j χj for j = 1, . . . , r1 , then πr1 +1 · · · πr = 1 · · · r1 χr1 +1 · · · χs . If r1 < s1 , then the prime element χr1 +1 will divide some irreducible (but not prime) element appearing on the left side of this identity, which is a contradiction. We conclude that r1 = s1 and χj is associate to πj for j = 1, . . . , r1 . We also have got the following relation πr1 +1 · · · πr = 1 · · · r1 χr1 +1 · · · χs , where 1 , . . . , r1 are units of A and all the factors πi , χj are irreducible elements but not prime. It follows that πr1 +1 A · · · πr A = χr1 +1 A · · · χs A. According to (1), any ideal appearing in this identity is a product of exactly to prime ideals in A. Thanks to the uniqueness of the factorization into prime ideals in A, we get 2(r − r1 ) = 2(s − r1 ), hence r = s.

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Exercise 10.46. √ Let K = Q( p) where p = 2 or p is a prime number congruent to 1 modulo 4. (1) Let I be an ideal of A such that I = I 0 , where I 0 is the ideal of A obtained from I by applying the automorphism σ of K such that √ √ σ( p) = − p. Show that there exists c ∈ Z such that I = cA or √ I = c pA. (2) Let  be a fundamental unit of K. Show that NK/Q () = −1. (3) Show that the class number h of K is odd. Solution 10.46. 0

(1) If a prime ideal P of A divides I and e = νP (I), then e = νP (I ). Notice that in A, there exists one and only one prime ideal ramified in √ A, namely pA. Write the factorization of I into a product of prime √ ideals in the form I = Π1 Π2 ( pA)k , where Y Y 0 Π1 = P e and Π2 = P hP h, deg P=2

deg P=1

then clearly Π1 = aA and Π2 = bA, where a and b are positive integers. √ If k is even, say k = 2l, then ( pA)k = pl A, hence I = abpl A, thus I = cA with c ∈ Z. √ √ √ If k is odd, say k = 2l + 1, then pk A = pl pA, hence I = abpl pA, √ thus I = c pA with c ∈ Z. (2) Suppose that NK/Q () = 1, then by Hilbert’s Theorem 90, [Lang (1965), 0 0 Chap. 8.6], there exists γ ∈ A such that  = γ/γ , where γ is the 0 conjugate of γ distinct from γ. We deduce that γA = γ A. According √ to (1), γ = ηc or γ = ηc p with c ∈ Z and η is a unit of A. In the 0 0 first case, we have  = γ/γ = η/η = ±η 2 . In the second case, we get 0 0 2  = γ/γ = −η/η = ±η . In any case we reached a contradiction since  is a fundamental unit. (3) Suppose that h is even, then there exists a class [A] of order 2 in the group of ideal classes, that is [A] 6= [A] and [A]2 = [A]. This implies 0 0 that [A]−1 = [A]. Since [A][A ] = [AA ] = [aA], with a ∈ Z, then 0 [A ] = [A−1 ] = [A]. It follows that there exist α and β ∈ A such that 0 αA = βA . We deduce that NK/Q (α) = ± NK/Q (β). If the sign in this identity is +, then NK/Q (α/β) = 1, which implies, by 0 Hilbert’s Theorem 90 [Lang (1965), Chap. 8.6], α/β = γ/γ , with γ ∈ 0 0 0 0 0 0 A. We deduce that γ αA = γβA = γ βA , thus γA = γ A = (γA) . 0 We have found in this case an ideal I equivalent to A such that I = I .

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If the sign in this identity is −, then NK/Q (α/β) = 1, hence α/β = 0 0 0 0 γ/γ , with γ ∈ A. We deduce that γ αA = γβA = γ βA , thus 0 0 0 γA = γA = γ A = (γA) . In this case also we have found an ideal 0 I equivalent to A such that I = I . By (1), we conclude that I is principal ant then [A] is of order 1, which is a contradiction.

Exercise 10.47. Let f (x) = x3 − 7x + 10. (1) Show that f (x) is irreducible over Q. Show that this polynomial has exactly one real root. Denote it by θ. (2) Let K = Q(θ) and A be the ring of integers of K. Show that {1, θ, ω} 2 . Deduce that any α ∈ A may is a basis of A over Z where ω = θ −θ+2 4 a+bθ+cθ 2 be written in the form α = , where a, b, c ∈ Z, a ≡ 0 (mod 2), 4 a + 2b ≡ a − 2c ≡ b + c ≡ 0 (mod 4) and conversely any element α of this form with these conditions on a, b, c, belongs to A. 2 0 00 (3) Let  = 2+θ−θ = 1 − ω. For any α ∈ K denote by α and α be the 4 images of α by the two non real embeddings of K into C. 0 00 0 0 Show that |θ| < 3.2, |θ | < 1.8, |θ − θ | < 1.6, |θ − θ| < 4.9, || < 2.9 0 and | | < 0.6. Deduce that  is a fundamental unit of A. (4) Show that h(K) = 1. (5) Show that A = Z[ω].

Solution 10.47. (1) Since f (x) = x3 − x + 1 (mod 3) and since x3 − x + 1 is irreducible over F3 , then f (x) is irreducible over Q. We omit the proof that f (x) has a unique real root. (2) It is easy to show that the characteristic polynomial of ω is given by g(x) = x3 − 5x2 + 5x − 2, hence ω is an algebraic integer. The index of Z[θ] in Z + Zθ + Zω is equal to the determinant 1 0 2 ∆ = 0 1 −1 , hence Disc(1, θ, θ2 ) = 42 Disc(1, θ, ω). 0 0 4 Since Disc(1, θ, θ2 ) = Disc(f ) = −42 · 83, then Disc(1, θ, ω) = −83. Therefore Disc(K) = −83 and {1, θ, ω} is an integral basis.

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Let α ∈ A, then α = x + yθ + zω   2 − θ + θ2 = x + yθ + z 4 4x + 2z + (4y − z)θ + zθ2 = , 4 2

hence α has the form α = a+bθ+cθ , where a = 4x+2z, b = 4y−z, c = z. 4 Moreover these identities show that a ≡ 0 (mod 2), a + 2b ≡ 0 (mod 4) and b + c ≡ 0 (mod 4). Conversely, given a, b, c ∈ Z satisfying these congruences, we may compute x, y, z ∈ Z. (3) Let z = x + iy with x, y ∈ R be a root of f (x), then f (z) = 0 is equivalent to the following system of equations x3 − 3xy 2 − 7x + 10 = 0

y(3x2 − y 2 − 7) = 0.

If y = 0, then z = x = θ. If y 6= 0, then 3

3x2 − y 2 − 7 = 0

and

2

x − 3xy − 7x + 10 = 0,

hence y 2 = 3x2 − 7 and x is the unique real root of the equation 4x3 − 7x − 5 = 0. It is easy to show that the characteristic polynomial of  is equal to h(x) = x3 +2x2 −2x+1, hence  is a unit in A. If  is not a fundamental unit then there exist an integer n ≥ 2 and integers a, b, c 2 )n with a ≡ 0 (mod 2), a + 2b ≡ a − 2c ≡ such that ± = ( a+bθ+cθ 4 b + c ≡ 0 (mod 4). Moreover, we may suppose that c ≤ 0. We deduce that a + bθ + cθ2 = 4(±)1/n and conjugating this equation, we have 0

02

0

a + bθ + cθ = 4(±) 1/n 00

a + bθ + cθ

00

2

= 4(±00 )1/n .

The  determinant  of this system of linear equations in a, b, c is given by 1 θ θ2 p √ δ = 1 θ0 θ02  = Disc(θ) = ±4i 83. Solving these equations, we 1 θ00 θ002 obtain: 002 001/n 02 4 2 2 b = (±)1/n (θ0 − θ00 ) + (±0 )1/n (θ − θ2 ) + (± )(θ2 − θ )) δ 00 01/n 001/n 0 4 c = (±)1/n (θ − θ0 ) + (±) (θ − θ00 ) + (±) (θ − θ)). δ

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00

00

00

361 0

00

0

Since  = 0 , θ = θ0 , θ − θ = θ − θ0 , then | | = | |, |θ | = |θ | and 00 |θ − θ | = |θ − θ0 |. We deduce that √ 0 00 0 0 0 |b| ≤ 1/ 83(|1/n ||θ||θ − θ | + 2| |1/n |θ ||θ − θ|) √ 0 00 0 0 ≤ 1/ 83(|θ|||1/2 |θ − θ | + 2|θ ||θ − θ|)  26.4 √  √ ≤ 1/ 83 (3.2) · 2.9 · (1.6) + 2 · (1.8) · (4.9) < < 2.89 9.11 and √ 0 00 0 |c| ≤ 1/ 83(||1/2 |θ − θ | + 2|θ − θ|)   √ √ < 1/ 83 2.9 · (1.6) + 2 · (4.9)

12.536 < 1.4. 9.11 We conclude that b = 0 or b = ±1 or b = ±2 and since c < 0, then c = 0 or c = −1. The condition b + c ≡ 0 (mod 4) implies that (b, c) = (0, 0) or (b, c) = (1, −1), since  6∈ Q, we immediately reject the case (b, c) = (0, 0). In the second case, we have 4n (±) = (a + θ − θ2 )n , hence NK/Q (a + θ − θ2 ) = ±64. But we have a + θ − θ2 = a − 2 + 4. Let µ = a + θ − θ2 , then  = µ−(a−2) . Using the characteristic polynomial 4 of , we obtain  3  2   µ − (a − 2) µ − (a − 2) µ − (a − 2) +2 −2 + 1 = 0, 4 4 4
1. (1) Show that | Disc(K)|/4 < 3 + 7. (2) Deduce that the real root θ of f (x) = x3 − 2x2 − 1 is a fundamental unit of the ring of integers of the field Q(θ). Solution 10.49. (1) Let σ : K → C be one of the complex embeddings, then NK/Q () = σ()σ() = |σ()|2 ,

hence NK/Q () = 1. Let ρ = |σ()| = √1 . Set σ() = ρeiθ , σ() = ρe−iθ and  = ρ12 . We have Disc(1, , 2 ) = I()2 Disc(K), hence | Disc(K)| ≤

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| Disc(1, , 2 )|. We have 1  2 Disc(1, , 2 ) = 1 σ() σ()2 1 σ() σ()2

= [(σ() − σ())( − σ())( − σ())]2   1 1 = 2iρ(sin θ) − ρ(cos θ + i sin θ) − ρ(cos θ − i sin θ) ρ2 ρ2 !!2 2  1 2 2 2 2 − ρ(cos θ) + ρ (sin θ) = −4ρ (sin θ) ρ2  2 1 2 cos θ 2 = −4ρ2 (sin2 θ)ρ2 − + ρ ρ4 ρ  2 1 2 3 = −4(sin θ) + ρ − 2(cos θ) . ρ3

It follows that

| Disc(K)|/4 ≤ | Disc(1, , 2 )/4 = (sin2 θ)



1 + ρ3 − 2(cos θ) ρ3

Let c = cos θ and z = ρ3 + ρ13 , then z − 2 = hence z > 2. We deduce that

ρ6 −2ρ+1 ρ3

=

2

(ρ3 −1)2 ρ3

.

> 0,

| Disc(K)|/4 ≤ (1 − c2 )(z − 2c)2

= (z 2 − 4cz + 4c2 )(1 − c2 )

= z 2 − c2 z 2 − 4cz(1 − c2 ) + 4c2 (1 − c2 )

= z 2 − (cz + 2(1 − c2 ))2 + 4(1 − c2 )2 + 4c2 (1 − c2 )  2 1 = z 2 + 4 − cz + 2(1 − c2 − 4c2 < z 2 + 4 = ρ6 + 6 + 6 ρ < 3 + 7.

(2) We have Disc(f ) = (−1)3 NK/Q f 0 (θ) = − NK/Q (3θ2 − 4θ)

= − NK/Q (θ)NK/Q (3θ − 4) = − NK/Q (3θ − 4).

β+4 3 β+4 2 Let β = 3θ − 4, thus θ = β+4 3 , then ( 3 ) − 2( 3 ) − 1 = 0, hence (β + 4)3 − 6(β + 4)2 − 27 = 0. Therefore β 3 + 6β 2 − 59 = 0. We conclude

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that Disc(f ) = −59. Since 59 is a prime number, then Disc(K) = −59 and {1, θ, θ2 } is a basis of the ring of integer of K = Q(θ) over Z. Let  be the fundamental unit of K such that  > 1. By (1), 3 + 7 > | Disc(K)|/4 > 14.75, hence  > 1.95 and then 2 > 3.8. We have f (2) = −1 < 0, f (3) = 8 > 0, hence θ ∈]2, 3[. Since g(3.8) = 3.82 · (1.8) − 1 > θ > 0, then θ < 3.8 < 2 . Since θ is a unit greater than 1, then it is the unique fundamental unit of K which is greater than 1. Exercise 10.50. Let f (x) = x3 + 4x + 1. (1) Show that f (x) is irreducible over Q. (2) Show that f (x) has exactly one real root. Let θ be this root and let K = Q(θ) and A be the ring of integers of K. Show that Disc(K) = −283 and {1, θ, θ2 } is a basis of A over Z. 0 0 (3) Show that 283A = R2 R , DK = R, where R and R are prime ideals of A with residual degree equal to 1. Show that DK is principal and determine for it a generator. (4) Show that −1 θ is the fundamental unit of A, greater than 1. Hint. One may use the preceding exercise. (5) Show that h(K) = 2. Solution 10.50. (1) Obvious. (2) We have f 0 (x) = 3x2 + 4 > 0, hence f (x) is strictly increasing from −∞ to +∞. Therefore the equation f (x) = 0 has one and only one real solution. We have Disc(1, θ, θ2 ) = Disc(f ) = −4(43 ) − 27 = −283. Since 283 is a prime number then Disc(K) = −283 and {1, θ, θ2 } is a basis of A over Z. 0 (3) Since 283 is ramified in K, then 283A = R2 R or 283A = R3 . In the second case R3 | DK and then 2832 | Disc K, which is a contradiction. 0 We conclude that 283A = R2 R . We deduce that DK = R. Since 0 A = Z[θ], then DK is principal and DK = f (θ)A = (3θ2 + 4)A. (4) It is easy to show that the minimal polynomial of −1/θ is given by g(x) = x3 − 4x2 − 1, hence −1/θ is a unit of A. Moreover, we have g 0 (x) = 3x2 − 8x and we give the table of variations of the function g.

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x

−∞

f 0 (x)

8 3

0

+∞

−

+

+ +∞

−1

f (x)

8 g( ) < 0 3

−∞

We have g(4) = −1 < 0 and g(5) = 24 > 0, hence −1/θ ∈]4, 5[. Let  be the fundamental unit of A such that  > 1 then by Exercise 10.49, 283 = 70.75, 4 hence 3 > 63.75. We deduce that 2 > 15.21 > −1/θ. Since −1/θ is a unit > 1 then −1/θ = n for some positive integer n. Suppose that n ≥ 2, then −1/θ = n ≥ 2 contradicting the inequality already proved −1/θ < 2 . It follows that n = 1 and then −1/θ is a fundamental unit of A. (5) In order to compute h(K), we use the result [Marcus (1977), Cor. 2, Chap. 5] stating that every class of ideals contains an integral ideal I such that 4 3! p | Disc K| < 5, NK/Q (I) < π 33 hence we must look at the ideals of norm 2, 3 and 4 respectively. We have 3 + 7 > | Disc(K)|/4 =

f (x) ≡ x3 + 1 ≡ (x + 1)(x2 + x + 1)

(mod 2),

hence 2A = P1 P2 , where P1 and P2 are prime ideals of A of residual degree equal to 1 and 2 respectively. Moreover P1 = (2, θ + 1) and P2 = (2, θ2 + θ + 1). We write I ∼ J when I and J have the same class. The splitting of 2 in A shows that P1 ∼ P2 . We have f (x) ≡ x3 + x + 1 ≡ (x − 1)(x2 + x − 1)

(mod 3),

hence 3A = Q1 Q2 and Q1 ∼ Q2 , where Q1 = (3, θ − 1)

and Q2 = (3, θ2 + θ − 1).

It is easy to show that the characteristic polynomial of θ2 + θ + 1 is given by x3 + 5x + 10x − 12, hence (θ2 + θ + 1)A = P2 Q1 . It follows that Q2 ∼ Q1 ∼ P2 ∼ P1 . Therefore h(K) = 1 or h(K) = 2 according to

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P1 is principal or not. It is easy to show that the minimal polynomial (and characteristic polynomial) of θ + 1 is given by x3 − 3x2 + 7x − 4, hence NK/Q (θ + 1) = 4. It follows that (θ + 1)A = P12 because θ + 1 ≡ 0 (mod P1 ) but θ + 1 6≡ 0 (mod P2 ). Suppose by contradiction that P1 is principal and let γ ∈ A such that P1 = γA, then (θ + 1)A = (γ 2 A), hence there exists an integer n such that θ + 1 = ±γ 2 .n , where  is the fundamental unit of A computed above, that is  = −1/θ. If n is even, say n = 2k, then θ + 1 = ±γ 2 2k = ±(γk )2 := ±(a + bθ + cθ2 )2 , where a, b, c ∈ Z. If n is odd, say n = 2k + 1, then θ + 1 = ±γ 2 2k+1 , hence θ+1 = ±(γk )2 = ±(a + bθ + cθ2 )2 .  Straightforward computations show that (a + bθ + cθ2 )2 = a2 − 2bc + (−c2 + 2ab − 8bc)θ + (b2 − 4c2 + 2ac)θ2 . • case θ + 1 = (a + bθ + cθ2 )2 . In this case, we obtain the following equations a2 − 2bc = 1, −c2 + 2ab − 8bc = 1, b2 − 4c2 + 2ac = 0. The third equation shows that 2 | b, the same equation then implies 2 | ac. The first equation shows that 2 - a, hence 2 | c. The second equation then implies c2 ≡ −1 (mod 4) which is a contradiction. • case θ + 1 = −(a + bθ + cθ2 )2 . Here, we obtain the following equations a2 − 2bc = −1, −c2 + 2ab − 8bc = −1, b2 − 4c2 + 2ac = 0. The last equation shows that 2 | b. The first equation then implies a2 ≡ −1 (mod 4) which is a contradiction. 2 2 2 • case θ+1  = −θ − θ = (a + bθ + cθ ) . We get the following equations a2 − 2bc = 0, −c2 + 2ab − 8bc = −1, b2 − 4c2 + 2ac = −1. The first equation shows that 2 | a. The last equation then implies b2 ≡ −1 (mod 4) which is a contradiction. 2 2 2 • case θ+1  = −θ − θ = −(a + bθ + cθ ) . Here we get the following equations a2 − 2bc = 0, −c2 + 2ab − 8bc = 1, b2 − 4c2 + 2ac = 1. The first equation shows that 2 | a and then 2 | bc. The second equation implies gcd(b, c) = gcd(a, c) = 1. We deduce that 2 - c, 2 | a and 2 | b. These conditions contradict the third equation. We conclude that P1 is not principal and h(K) = 2.

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Exercise 10.51. Let K be a number field of degree n ≥ 2 and A be its ring of integers. Suppose that the following property is satisfied. (P) There exist a prime number p and θ ∈ A such that p | θ2 , p2 - θ2 and p2 | θ3 . (1) Show that p is ramified in K. (2) Show that n 6= 2. (3) Show that (P) holds for the prime 2 in the field Q(θ), where θ is a root of f (x) = x3 − 12x + 12. Deduce that 2 is totally ramified in Q(θ). Solution 10.51. (1) Suppose that p is not ramified in K and let P be a prime ideal of A lying of over pZ. Since P | θ2 , then P | θ, hence p | θ, which contradicts the hypothesis p2 - θ2 . We conclude that p is ramified in K. (2) Suppose that n = 2 and (P) holds in K for some prime p, then p is ramified, say pA = P 2 . Since P 2 |θ2 , then P | θ. Let k = νP (θ). Since p2 - θ2 , then P 4 - θ2 , hence k = 1. We deduce that νP (θ3 ) = 3, contradicting the condition p2 | θ3 . (3) Let µ = θ2 , then it is easy to show that the minimal polynomial of µ over Q is given by g(x) = x3 − 8 · 3x2 + 42 · 32 x − 42 · 32 . Therefore it is seen that µ/2 is an algebraic integer but not µ/4. Since θ3 = 4(3θ − 3), then obviously 4 | θ3 . Since the property (P) holds for the prime 2 in K, then 2 is ramified in K. 0 0 Suppose that 2A = P12 P1 , where P1 and P1 are prime ideals of A of 0 0 residual degree equal to 1. Since P12 P1 | θ2 , then P1 | θ and P1 | θ. 0 Since P14 P12 - θ2 , then P14 - θ2 . Therefore νP1 (θ) = 1. It follows that 0 νP1 (θ3 ) = 3, contradicting the fact that P14 P12 | θ3 . We conclude that 2 is totally ramified in A. Exercise 10.52. Let f (x) = x5 − 5x − 5, θ be a root of f (x) in C, K = Q(θ) and A be its ring of integers. (1) Show that f (x) is irreducible over Q. Show that Disc(f ) = 55 · 32 · 41, Disc(K) = 55 · 41 and A = Z + Zθ + Zθ2 + Zθ3 + Zω, where ω = (θ4 + θ3 + θ2 + θ − 1)/3.

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(2) Show that the characteristic polynomial over Q of ω is given by h(x) = x5 − 5x4 − 10x2 − 10x − 3. Using the factorization of f (x) in F3 [x] and looking at h(x), show that the splitting of the prime 3 in A has 0 0 the form 3A = P1 P1 P3 , where P1 , P1 are prime ideals of A of residual degree equal to 1 and P3 is a prime ideal of A of residual degree equal to 3. Show that 0

P1 = (3, ω), P1 = (3, ω + 1),

P3 = (3, ω − 1) = (3, θ3 − θ2 + 1)

and

0

P1 P1 = (3, θ − 1). 0

(3) Show that P1 , P1 and P3 are principal and find for each of these ideals a generator. (4) Show 5A = P 5 where P is a prime ideal of A of residual degree equal to 1. Show that P = θA. (5) Show that 2A = Q2 Q3 , where Q2 , Q3 are prime ideals of A having their residual degrees equal to 2 and 3 respectively. Show that Q2 = (2, θ2 + θ + 1)

and

Q3 = (2, θ3 + θ2 + 1).

(6) Show that the ideals Q2 and Q3 are principal and determine for each of them a generator. (7) Determine bases over Z of Q2 and Q3 . Solution 10.52. (1) Eisenstein’s irreducibility theorem shows that f (x) is irreducible over Q. We have Disc(θ) = NK/Q (f 0 (θ)) = NK/Q (5(θ4 −1)) = 55 NK/Q (θ2 −1) NK/Q (θ2 +1).

It is easy to see that the characteristic polynomial of θ2 over Q is given by g(x) = x5 − 10x3 + 25x − 25. Therefore θ2 + 1 (resp. θ2 − 1) is a root of g(x − 1) (resp. g(x + 1)). We deduce that NK/Q (θ2 + 1) = 41 and NK/Q (θ2 − 1) = 32 . Therefore Disc(θ) = 55 · 32 · 41. We have f (x) = x5 − 5(x + 1) and since x - x + 1 in F5 [x], then by Exercise 10.30, 5 - I(θ), thus ν5 (Disc(K)) = 5. We have f (x) = (x − 1)2 (x3 − x2 + 1) + 3(x4 − x3 − x − 2)

and since x − 1|x4 − x3 − x − 2 in F3 [x], then, by Exercise 10.30, 3 | I(θ). From the identity Disc(θ) = I(θ)2 Disc(K), we conclude that 3 - Disc(K). From the same identity we conclude that 41 | Disc(K). It follows that Disc(K) = 55 · 41. Since f (x) ≡ (x − 1)2 (x3 − x2 + 1)

(mod 3),

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and since x3 − x2 + 1 is irreducible over F3 and 3 | I(θ), then (θ − 3 2 1)(θ3 − θ2 + 1) ≡ 0 (mod 3), that is (θ−1)(θ3−θ +1) ∈ A. We have

(θ − 1)(θ3 − θ2 + 1) θ4 + θ3 + θ2 + θ − 1 = + θ3 , 3 3 hence ω ∈ A. Consider the matrix M expressing the coordinates of 1, θ, θ2 , θ3 , θ4 in the basis {1, θ, θ2 , θ3 , ω}. It is given by:   1 0 0 0 1 0 1 0 0 −1    M = 0 0 1 0 −1 . 0 0 0 1 −1 0 0 0 0 3 Its determinant is equal to 3 and we have ω=

Disc(1, θ, θ2 , θ3 , θ4 ) = Det M 2 Disc(1, θ, θ2 , θ3 , ω), hence Disc(1, θ, θ2 , θ3 , ω) = 55 · 41 = Disc(K).

It follows that {1, θ, θ2 , θ3 , ω} is an integral basis. (2) Let γ = 3ω = θ4 + θ3 + θ2 + θ − 1. Using the relation θ5 = 5θ + 5, we obtain θγ = θ4 + θ3 + θ2 + 4θ + 5 θ2 γ = θ4 + θ3 + 4θ2 + 10θ + 5 θ3 γ = θ4 + 4θ3 + 10θ2 + 10θ + 5 θ4 γ = 4θ4 + 10θ3 + 10θ2 + 10θ + 5. This shows that 1, θ, θ2 , θ3 , θ4 is a non trivial solution of the following homogeneous system of linear equations: (−1 − γ)x0 + x1 + x2 + x3 + x4 = 0

5x0 + (4 − γ)x1 + x2 + x3 + x4 = 0

5x0 + 10x1 + (4 − γ)x2 + x3 + x4 = 0

5x0 + 10x1 + 10x2 + (4 − γ)x3 + x4 = 0

5x0 + 10x1 + 10x2 + 10x3 + (4 − γ)x4 = 0.

This implies that −1 − γ 5 5 5 5

1 4−γ 10 10 10

1 1 4−γ 10 10

1 1 1 4−γ 10

1 1 1 = 0. 1 4 − γ

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Standard computations on determinants show that the characteristic polynomial of γ is given by h1 (x) = x5 − 5 · 3x4 − 10 · 33 x2 − 10 · 34 x − 36 . It follows that the characteristic polynomial of ω is equal to h1 (3x) = x5 − 5x4 − 10x2 − 10x − 3. 35 We have seen in (1) that 3 | I(θ) and h(x) =

f (x) ≡ (x − 1)2 (x3 − x2 + 1)

(mod 3).

This shows that there exists one prime ideal of A lying over 3Z, having its residual degree equal to 3. This also shows that there exist two prime ideals lying over 3Z of residual degree equal to 1 or one prime ideal of residual degree equal to 2. But looking at h(x), we conclude that NK/Q (ω) = 3. Therefore ωA is a prime ideal lying over 3Z with residual degree equal to 1. We conclude that there exist two prime ideals lying over 3Z having their residual degree equal to 1. Therefore 0 3A = P1 P1 P3 . The congruence modulo 3, satisfied by f (x) shows that 0

P1 P1 = (3, θ − 1)

and P3 = (3, θ3 − θ2 + 1).

Since h(x) ≡ x(x + 1)(x − 1)3 (mod 3), we conclude that P3 = (3, ω − 1), P1 = (3, ω)

0

and P1 = (3, ω + 1).

(3) We have seen above that NK/Q (ω) = 3, hence P1 = ωA, thus P1 is principal. Since the characteristic polynomial of ω − 1 over Q is equal to h(x + 1), then NK/Q (ω − 1) = 33 . 0 Therefore P3 = (ω−1)A, and then P3 is principal. Since 3A = P1 P3 P1 , 0 0 then P1 is also principal. Let µ be a generator of P1 , then 3A = ω(ω − 1)µA. It follows that 3 = ω(ω − 1)µ, where  is a unit of A, 3 . Since µ and µ generates the same ideal, we may hence µ = ω(ω−1) 3 suppose that  = 1 and then µ = ω(ω−1) . Using the relation θ5 = 5θ+5, we find 2θ4 + 5θ3 + 8θ2 + 11θ + 7 , ω2 = 3 hence θ4 + 4θ3 + 7θ2 + 10θ + 8 ω2 − ω = . 3 We deduce that 3 9 = 4 . 2 3 ω −ω θ + 4θ + 7θ2 + 10θ + 8

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By performing three Euclidean divisions, we may compute the gcd of the polynomials x5 − 5x − 5 and x4 + 4x3 + 7x2 + 10x + 8 and then obtain a Bezout’s identity relating these polynomials. Substituting in this identity θ for x, we obtain 27 = θ4 −2θ3 +4θ2 −5θ−1 = 3ω −3θ3 +3θ2 −6θ. θ4 + 4θ3 + 7θ2 + 10θ + 8 It follows that µ=

θ4

+

4θ3

9 = ω − θ3 + θ2 − 2θ, + 7θ2 + 10θ + 8

0

thus P1 = (ω − θ3 + θ2 − 2θ)A. (4) Since f (x) = x5 − 5(x + 1), then 5 - I(θ) and 5 is totally ramified in A, that is 5A = P 5 , where P is a prime ideal of A of residual degree equal to 1. Moreover since NK/Q (θ) = 5, then P = θA, thus P is principal. (5) The factorization of f (x) over F2 into a product of irreducible factors is given by f (x) ≡ (x2 + x + 1)(x3 + x2 + 1)

(mod 2),

hence 2A = Q2 Q3 , where the residual degrees of Q2 and Q3 are equal to 2 and 3 respectively. Moreover, we have Q2 = (2, θ2 + θ + 1)

and

Q3 = (2, θ3 + θ2 + 1).

(6) We compute NK/Q (θ2 + θ + 1). Set β = θ2 + θ + 1. Then θβ = θ3 + θ2 + θ θ2 β = θ4 + θ3 + θ2 θ3 β = θ4 + θ3 + 5θ + 5 θ4 β = θ4 + 5θ2 + 10θ + 5, hence 1, θ, θ2 , θ3 , θ4 is a non trivial solution of the following homogeneous system of linear equations (1 − β)x0 + x1 + x2 = 0

(1 − β)x1 + x2 + x3 = 0

(1 − β)x2 + x3 + x4 = 0

5x0 + 5x1 + (1 − β)x3 + x4 = 0

5x0 + 10x1 + 5x2 + (1 − β)x4 = 0.

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It follows that 1 − β 0 0 5 5

1 1−β 0 5 10

1 1 1−β 0 5

373

0 0 1 = 0. 1 1 − β

0 1 1 1−β 0

Straightforward computations show that the characteristic polynomial of β over Q is given by q(x) = x5 − 3x4 + 24x3 − 11x2 − 36x + 36, hence NK/Q (β) = −32 · 22 . It follows that (θ2 + θ + 1)A = Q2 P12 2

(θ + θ + 1)A =

or

0

Q2 P12

or 0

(θ2 + θ + 1)A = Q2 P1 P1 . 0

0

0

Since by (3), P1 and P1 are principal then so are P12 , P12 and P1 P1 . It follows that Q2 is principal and then Q3 is also principal. Moreover Q2 is generated by one and only one of the following elements: (θ2 + θ + 1)/ω 2 , (θ2 + θ + 1)/(ω − θ3 + θ2 − 2θ)2

and

(θ2 + θ + 1)/ω(ω − θ3 + θ2 − 2θ),

according to the above possible factorizations of (θ2 + θ + 1)A. Indeed, we must find among these three elements the only one which is an 0 0 algebraic integer. We have seen that P1 P1 = (3, θ − 1), hence P1 P1 | 0 θ − 1. It follows that P12 P12 | (θ − 1)2 . From the splitting of the prime 3 it is seen that νP1 (3θ) = νP 0 (3θ) = 1. Now the identity 1

2

θ + θ + 1 = (θ − 1)2 + 3θ 0

shows that P12 - θ2 + θ + 1 and P12 - θ2 + θ + 1. It follows that we have the third possibility for the generator, namely Q2 =

(θ2 + θ + 1) A = (1 + θ + ω)A. ω(ω − θ3 + θ2 − 2θ)

Since 2A = Q2 Q3 , then 2/(1 + θ + ω) is a generator of Q3 , thus Q3 = (−3 − θ2 − θ3 + 2ω)A.

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(7) We know that A/Q2 is isomorphic to F22 , so there exists a surjective morphism of rings φ : A → F22 such that Ker φ = Q2 . To define φ it is sufficient to define φ on Z and also the value of φ(θ). Obviously for any a ∈ Z, φ(a) = a ¯, where a ¯ denotes the class of a modulo 2. 2 Since θ + θ + 1 = 0 in A/Q2 , then φ(θ2 + θ + 1) = 0 in F22 , hence φ(θ)2 + φ(θ) + 1 = 0 in F22 , i.e. φ(θ) is a root x2 + x + 1. Set ρ = φ(θ). Let α = a0 + a1 θ + a2 θ2 + a3 θ3 + a4 ω be an element of A, then α ∈ Q2 ⇔ α ∈ Ker(φ)

⇔ a0 + a1 ρ + a2 ρ2 + a3 ρ3 + a4 (ρ4 + ρ3 + ρ2 + ρ + 1) = 0

⇔ a0 + a2 + a3 + a4 + (a1 + a2 + a4 )ρ = 0  a0 + a2 + a3 + a4 = 0 ⇔ a1 + a2 + a4 = 0  a1 = a2 + a4 + 2λ ⇔ a0 = a2 + a4 + a3 + 2µ where λ and µ in Z. We deduce that

α ∈ Q2 ⇔ α = (a2 + a4 + a3 + 2µ) + (a2 + a4 + 2λ)θ + a2 θ2 + a3 θ3 + a4 ω ⇔ α = a2 (1 + θ + θ2 ) + a3 (1 + θ3 ) + a4 (1 + θ + ω) + 2µ + λ(2θ).

We have

θ4 + θ3 + θ2 + θ − 1 +θ+1 3 θ4 + θ3 + θ2 + 4θ + 2 = 3 4 3 ≡ θ + θ + θ2 (mod 2)

1+θ+ω =

≡0

(mod P2 ).

Clearly all the elements 1 + θ + θ2 , 1 + θ3 , 2, 2θ and 1 + θ + ω belong to P2 . Therefore they constitute a set of generators of P2 as a Z module. Obviously this set is free over Z. Therefore it is a basis of P2 over Z. We use the same method for the determination of a Z- basis of P3 . Let γ = a0 + a1 θ + a2 θ2 + a3 θ3 + a4 ω with ai ∈ Z for i = 0, 1, . . . , 4. Then γ ∈ P3 if and only if a0 + a1 τ + a2 τ 2 + a3 τ 3 + a4 (τ 4 + τ 3 + τ 2 + τ + 1) = 0,

where τ is a root of x3 + x2 + 1 in F23 . We deduce that γ ∈ P3 ⇔ a0 + a1 τ + a2 τ 2 + a3 (τ 2 + 1) + a4 (τ 2 + 1) = 0 ⇔ a0 + a3 + a4 = 0, a1 = 0, a2 + a3 + a4 = 0

⇔ a1 = 2λ, a2 = a3 + a4 + 2µ, a0 = a3 + a4 + 2ν,

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where λ, µ and ν ∈ Z. We conclude (and we omit the proof) that {2, 2θ, 2θ2 , 1 + θ2 + θ3 , 1 + θ2 + ω} is a Z-basis of P3 . Exercise 10.53. Let K be a number field, A be its ring of integers, f (x) = an xn + · · · + a0 , g(x) = bm xm + · · · + b0 and h(x) = ck xk + · · · = c0 be polynomials with coefficients in K such that f (x) = g(x)h(x). Consider the fractional ideals of K, A = (a0 , . . . , an ),

B = (b0 , . . . , bm )

and C = (c0 , . . . , ck ).

(1) Show that A = BC. (2) Deduce that if a0 , . . . , an ∈ A, then bi cj ∈ A for any i ∈ {1, . . . , m} and any j ∈ {1, . . . , k}. (3) Suppose that all the coefficients ah belong to A ant that there exists α ∈ A such that α | ah for h = 0, . . . , n. Show that bi cj ∈ A and α | bi cj for any (i, j). (4) Let F (x) be a non constant polynomial with coefficients in A and let a ∈ A be its leading coefficient. Show that if F (x) is reducible in K[x] then aF (x) is a product of two non constant polynomials in A[x]. Solution 10.53. (1) Let P be a prime ideal of A, e1 = νP (B) and e2 = νP (C). To get the result, it is sufficient to prove that νP (A) = e1 + e2 . By definition of e1 and e2 , for any i and j, νP (bi ) ≥ e1 , νP (cj ) ≥ e2 and there exist i0 and j0 such that νP (bi0 ) = e1 , νP (cj0 ) = e2 . Moreover we may suppose that P i0 and j0 are minimal. Since ah = i+j=h bi cj , then νP (ah ) ≥ e1 + e2 and νP (ai0 +j0 ) = νP (· · · + bi0 −1 cj0 +1 + bi0 cj0 + bi0 +1 cj0 −1 + · · · ) = e1 + e2 . (2) Since bi ∈ B, then bi A = BI, where I is an ideal of A. Similarly, cj A = CJ, where J is an ideal of A. From (1), we deduce that bi cj A = BCIJ = AIJ, hence bi cj ∈ A. (3) Apply the result of (2), for the polynomials f (x)/α, g(x)/α and h(x).

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(4) Since F (x) is reducible in K[x], then it is so for the monic polynomial F (x)/a. Let G(x) and H(x) ∈ K[x] be monic such
that F (x)/a =
G(x)H(x). Multiplying by a2 , we get aF (x) = aG(x) aH(x) . Set G(x) = bm xm + bm−1 xm−1 + · · · + b0 k

H(x) = ck x + ck−1 x

k−1

and

+ · · · + c0

with bm = ck = 1, then by (3) abi acj ∈ A and a | abi acj for any (i, j). In particular a | a2 bi and a | a2 cj for any i, j. We deduce that aci ∈ A and acj ∈ A for any i, j. This implies that aG(x) and aH(x) ∈ A[x], showing that aF (x) is a product of two non constant polynomials with coefficients in A. Remarks. (1) generalizes the classical result on the content of a product of polynomials with coefficients in the fraction field of a factorial domain. (2) is a generalization of Exercise 1.10. Exercise 10.54. Let K be a number field and A be its ring of integers. Let f (x) be a polynomial of degree n with integral coefficients, an and a0 be its leading and constant coefficients respectively. Suppose that f (x) is irreducible in Z[x] and an or a0 is irreducible in A. Show that f (x) is irreducible in A[x]. Solution 10.54. • First case: an irreducible. By contradiction, suppose that f (x) is reducible in A[x]. If f (x) = λh(x) with h(x) ∈ A[x] and λ a non zero constant and a non unit. Then λ divides each coefficient ai of f (x). But since f (x is irreducible over Z, then f (x) is primitive, which implies, the P existence of coefficients ui such that ui ai = 1. Therefore λ | 1, that is λ = ±1, hence a contradiction. Suppose next that f (x) = g(x)h(x), where g(x) and h(x) are non constant polynomials with coefficients in A. Let b and c be the leading coefficients of g(x) and h(x) respectively, then an = bc. Since an is irreducible in A, then b =  or c =  with  = ±1. We may suppose that b = . We have f (x) = −1 g(x)h(x) = g1 (x)h1 (x), where now g1 (x) is monic. Let α be a root of g1 (x), then α is integral over A, hence integral over Z. Since α is a root of f (x) and since this polynomial is irreducible over Z, then a = ±1 contradicting our assumptions.

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• Second case: a0 irreducible. Let g(x) = xn f (1/x). The leading coefficient of this polynomial is equal to a0 , thus irreducible in A. We apply for g(x), what was proved in the first case and we conclude that g(x) is irreducible in A[x], which in turn implies that f (x) is irreducible in A[x]. Exercise 10.55. Let K be a number field and A be its ring of integers. (1) Let h be the class number of K and {I1 , . . . , Ih } be a complete set of representatives of the ideal classes. Let m0 be a positive integer divisible by I1 , . . . , Ih . For any g(x) ∈ A[x], let c(g) be the ideal of A generated by the coefficients of g(x). Let I(g) ∈ {I1 , . . . , Ih } such that c(g)I(g) is principal and let α(g) ∈ A such that c(g)I(g) = α(g)A.

(Eq 1)

Let J(g) be the ideal of A such that m0 A = I(g)J(g).

(Eq 2)

(a) Let gˆ(x) = m0 g(x)/α(g). Show that gˆ(x) ∈ A[x] and c(ˆ g ) = J(g).

(Eq 3)

(b) Let f (x) ∈ A[x] such that f (x) is reducible over K. Show that m30 f (x) may be decomposed in A[x] as a product of two non constant polynomials. √ (2) Let f (x) = 3x2 + 4x + 3. Show that f is reducible over Q( −5), √ but irreducible over Z[ −5]. Show 5f (x) cannot be decomposed as a √ product of two non constant polynomials in Z[ −5][x]. √ (3) Factorize 2f (x), 3f (x) and 7f (x) in Z[ −5][x]. (4) For any g(x) ∈ A[x] such that g(x) is reducible over K, let D(g) = {0} ∪ {d ∈ Z \ {0},

dg(x) = g1 (x)g2 (x)

with

g1 (x)

and g2 (x) ∈ A[x] \ A}.

Show that, in general, D(g) is not an ideal of Z. (5) Let g(x) be a non constant polynomial with coefficients in A and let a be its leading coefficient. Show that if g(x) is reducible over K, then a ∈ D(g).

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Solution 10.55. (1)(a) Set g(x) = bk xk + bk−1 xk−1 + · · · + b0 , where bj ∈ A for j = 0, . . . , k. Let P be a prime ideal of A such that P e || α(g). From (Eq 1), P e || c(g)I(g). It follows that P e1 || c(g) and P e2 || I(g), with e1 + e2 = e. Since I(g) | m0 A, then P e2 | m0 A, hence P e | m0 bj for j = 0, . . . , k. Thus gˆ(x) ∈ A[x]. We have c(ˆ g ) = (m0 A)c(g)(α(g)A)−1 , hence by (Eq 1) and (Eq 2), c(ˆ g ) = I(g)J(g)(α(g)A)(I(g))−1 (α(g)A)−1 = J(g). (b) Let I(f ), c(f ), α(f ) and fˆ(x) as defined in the statement of the exercise. We have m0 f (x) = α(f )fˆ(x). Since f (x) is reducible over K, then so is fˆ(x). Let fˆ(x) = g(x)h(x) be a non trivial factorization of fˆ(x) in K[x], then g(x) = g1 (x)/d

and h(x) = h1 (x)/δ,

where d and δ are positive integers and g1 (x), h1 (x) ∈ A[x]. By (a), we have g1 (x) = α(g1 )ˆ g1 /m0

ˆ 1 /m0 , and h1 (x) = α(h1 )h

hence ˆ 1 (x)/m2 dδ = λˆ ˆ 1 (x), fˆ = g1 (x)h1 x)/dδ = α(g1 )α(h1 )ˆ g1 (x)h g1 (x)h 0 ˆ 1 ). where λ = α(g1 )α(h1 )/m20 dδ. Therefore, c(fˆ) = (λA)c(ˆ g1 )c(h Since m0 A = I(g1 )J(g1 ) = I(h1 )J(h1 ), then J(g1 ) and J(h1 ) divide m0 A. Since (m0 A)2 c(fˆ)(J(g1 ))−1 (J(h1 ))−1 = (m0 A)2 λA, then m20 λ ∈ A. We now have,

ˆ 1 (x), m30 f (x) = m20 α(f )fˆ(x) = α(f )(m20 λ)ˆ g1 (x)h

hence m30 f (x) is reducible in A[x]. √

√

(2) We have f (x) = 3(x + 2+ 3 −5 )(x + 2− 3 −5 ) hence f (x) is reducible √ over Q( −5). By contradiction, suppose that f (x) is reducible over √ Z[ −5]. Let f (x) = (αx + β)(γx + δ) be a factorization of f (x) in A[x], then αγ = 3, βδ = 3 and αδ + βγ = 4. The first two equations imply that

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(i) (ii) (iiii) (a) (b) (c)

NK/Q (α) = 1, NK/Q (α) = 3, NK/Q (α) = 9, NK/Q (β) = 1, NK/Q (β) = 3, NK/Q (β) = 9,

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NK/Q (γ) = 9 or NK/Q (γ) = 3 or NK/Q (γ) = 1 and NK/Q (δ) = 9 or NK/Q (δ) = 3 or NK/Q (δ) = 1.

√ Clearly 3 is not the norm of an integer of Q( −5). Therefore we may exclude (ii) and (b). By symmetry we must consider only the combinations (i)-(a) and (i)-(c). Suppose that NK/Q (α) = 1,

NK/Q (γ) = 9

NK/Q (β) = 1,

NK/Q (δ) = 9.

and

√ Then α = 1 , β = 2 , γ = 33 or γ = 3 (2 + 4 −5) and δ = 35 or √ δ = 5 (2 + 6 −5) where i = ±1. In any case αx + β = 1 x + 2 , so that ±1 is a root of f (x), which clearly is not true. Therefore f (x) is √ irreducible in Z( −5). Suppose that 5f (x) = (αx + β)(γx + δ) is a factorization of f (x) in A[x] then αγ = 15, βδ = 15 and αδ + βγ = 20. We omit the proof that 0 the splittings of 5A and 3A in A are given by 5A = P 2 and 3A = QQ , 0 where P, Q and Q are prime ideals of A of the first degree. The 0 first two equations may be written in the form αAγA = P 2 QQ and 0 βδ = P 2 QQ . Suppose that α =  and γ = 15 with  = ±1. The third equation gives δ = 20 − 15β. From the second equation, we obtain the √ following: 3β 2 − 4β + 3 = 0. It follows that β = (2 ± −5)/3, which is not an element of A. Thus, this possibility is excluded. Clearly by the same method, we may exclude the possibility β =  (resp. γ = , √ δ = ). Since the equation NK/Q (a + b −5) = a2 + 5b2 = 3 has no integral solutions, then αA, βA, γA and δA cannot be equal to Q nor 0 Q . It remains to consider the following cases. √ 0 (a) αA = QQ and γA = −5A or √ 0 (b) αA = −5A and γA = QQ √ 0 (c) βA = QQ and δA = −5A or √ 0 (d) βA = −5A and δA = QQ . If (a) and (b) hold, then Q divides the left hand side of the third equation while this ideal does not divide the right hand side. We thus get a contradiction. Similarly, we may reject the combination (c)-(d). √ If we have the combination (a)-(d), then −5A divides the right hand side of the third equation and also βγ but not αδ, which is a

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contradiction. Similarly, we may reject the combination (c)-(b). Therefore 5f (x) is not a product of two non constant polynomials in A[x]. (3) Solving equations, we find the following factorizations of 2f (x), 3f (x) √ and 7f (x) in Z[ −5][x]: √ √ √ √

2f (x) = ( −5 + 1)x + ( −5 − 1) (− −5 + 1)x − ( −5 + 1) , √
√
3f (x) = 3x + 2 + −5 3x + 2 − −5 , √ √ √ √

7f (x) = (2 −5 + 1)x + ( −5 + 4) (−2 −5 + 1)x − (− −5 + 4) . (4) In (3), we have seen that 2f (x) and 3f (x) are products of non constant polynomials in A[x], so that 2 and 3 ∈ D(f ). If this set where an ideal, then f (x) would be also an element of this set, contradicting (2). Thus, D(f ) is not an ideal of Z. (5) See the proof in Exercise 10.54, (4)

Questions. (1) Let K be a number field such that h(K) ≥ 2, A be its ring of integers and n ≥ 2 be an integer. Does there exist a polynomial g(x) ∈ A[x] of degree n, reducible in K[x] such that g(x) is not a product of two non constant polynomials in A[x]? (2) Let f (x) and g(x) be polynomials with coefficients in A, reducible in K[x]. Suppose that they cannot be written as products of non constant polynomials in A[x]. What could be said about the relation between D(f ) and D(g)? (3) Let g(x) ∈ A[x] be irreducible but reducible in K[x]. Let p be a prime number such that any of its prime ideal factors is principal. Is pf (x) a product of non constant polynomials in A[x]? Exercise 10.56. √ Let p be an odd prime number and K = Q( −p). Suppose that h(K) = 1. (1) (2) (3) (4)

Show that p ≡ 3 (mod 8). Let q be a prime number such that q < (p + 1)/4. Show that ( pq ) = −1. Show that (p + 1)/4 is prime. Let f (X) = X 2 + X + (p + 1)/4. Show that f (x) is a prime number for any x ∈ {0, 1, . . . , p−3 4 − 1}.

Solution 10.56. (1) Let A be the ring of integers of K. Suppose that p ≡ 1 (mod 4), then −p ≡ 3 (mod 4); therefore p is ramified in A. Since h(K) = 1, then
2 √ there exist a, b ∈ Z such that 2A = (a + b −p)A . It follows that

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2 = a2 + pb2 , which is impossible. We conclude that p ≡ 3 (mod 4). Suppose that p ≡ 7 (mod 8), then −p ≡ 1 (mod 8). Therefore 2A = PP 0 , where P and P 0 are conjugate prime ideals of A having there residual degree equal to√1.
Since h(K) = 1, then there exist a, b ∈ Z such that P = a + b 1+ 2 −p A. We deduce that  √  √  1 + −p 1 − −p 2= a+b a+b , 2 2

hence 8 = (2a + b)2 + pb2 . Since p > 7, this equation is impossible. Therefore p ≡ 3 (mod 8). (2) Notice that the condition q < (p + 1)/4 implies that q 6= p. Suppose −1 p that ( pq ) = 1. On the one hand, we have ( −p q ) = ( q )( q ). On the other p−1 q−1

hand, by the quadratic reciprocity law, we have ( pq )( pq ) = (−1) 2 2 . q Hence ( −p q ) = ( p ) = 1. It follows that qA is a product of two conjugate prime ideals of A. Therefore there exist a, b ∈ Z such that  √  1 + −p NK/Q a + b = q. 2

We deduce that 4q = (2a + b)2 + pb2 ≥ p. Since b 6= 0, this inequality contradicts our assumption. We conclude that ( pq ) = −1. (3) Suppose that the integer (p + 1)/4 is not prime and let q be one of its prime factors. By (2), we have ( pq ) = −1. Using, as previously the quadratic reciprocity law, we obtain ( −p q ) = −1. Therefore q is √

√

inert in A. Since q | (p + 1)/4 and (p + 1)/4 = 1+ 2 −p 1− 2 −p , then √ √ q | 1+ 2 −p or q | 1− 2 −p . We deduce that there exist a, b ∈ Z such √ √ that 1± 2 −p = q(a + b 1+ 2 −p ). From this, we obtain qb = ±1, which is impossible. This proves that (p + 1)/4 is prime. (4) Suppose that there exists x0 ∈ {0, 1, . . . , p−3 4 − 1} such that f (x0 ) is not prime and let q be a prime factor of f (x0 ). Clearly f (x0 ) is odd so q is odd. Let a ∈ Z, a ≥ 2 such that f (x0 ) = aq. We may suppose that q 2 ≤ f (x0 ) = x20 + x0 + (p + 1)/4.

We have 4q 2 ≤ 4x20 + 4x0 + 1 + p 2  p−3 2 = (2x0 + 1) + p < +1 +p 2  2 p+1 = (p2 + 1)/4 < . 2

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Therefore q < (p + 1)/4 and then, by (2), ( pq ) = −1. We deduce that      −p −1 p = q q q    p−1 q−1 −1 q = (−1) 2 2 q p   p−1 q−1 q = (−1)( 2 +1) 2 = −1. p On the other hand, we have 4aq = (2x0 + 1)2 + p, hence ( −p q ) = 1 which is a contradiction. Exercise 10.57. Let p be a prime number such that p ≡ −1 (mod 4), θ = e2iπ/p and K = Q(θ). (1) Show that K contains one and only one quadratic subfield, namely √ F = Q(i p) and that Gal(K, F ) is formed by the automorphisms σ of K such that σ(θ) = θr with 1 ≤ r ≤ p − 1 and ( pr ) = 1. (2) Show that −1 is not a square in K. (3) Suppose that p ≡ −1 (mod 8). Show that −1 is not a sum of two squares in K. Solution 10.57. (1) It is known that K is a cyclic extension of Q of degree p − 1. The group G = Gal(K, Q) is formed by the automorphisms σ of K such that σ(θ) = θr for some r ∈ {1, . . . , p − 1}. Let     r H = σ ∈ G, σ(θ) = θr , =1 . p It is clear that H is the unique subgroup of G of index 2. Therefore K contains a unique quadratic subfield F , namely the invariant field of √ H. We show using two different methods that F = Q(i p). • First method. Let φp (x) be the minimal polynomial of θ over Q, then φp (x) = xp−1 + · · · + x + 1. The discriminant of this polynomial is given by Y p−1 Disc(φp ) = (θi − θj )2 = (−1) 2 pp−2 . 1≤i pl−1 , which is impossible. 4pb2 leads to b12 + 4p Exercise 10.60. Let a ∈ Z such that a ≥ 2, d = (27a4 + 1)/4 and f (x) = x3 − x2 −

9a4 − 1 x + a4 . 4

(1) Show that f (x) is irreducible over Q. Hint. One may use the identity: d f (x) = (x − 1/3)3 − (x − 4/9). 3 (2) Show that Disc(f ) = a4 d2 . Deduce that Gal(f, Q) ' Z/3Z. [Recall that the discriminant of the polynomial g(x) = a0 x3 + a1 x2 + a2 x + a3 is given by: Disc(g) = a21 a22 − 4a0 a32 − 4a31 a3 − 27a20 a23 + 18a0 a1 a2 a3 ]. (3) Let α be a root of f and K = Q(α). Show that
f (u(x)) ≡ 0 2 4 2 (mod f (x)Q[x]), where u(x) = x2 − a 2+1 x − 3a 2−a /a2 . Deduce that the roots of f are α, β = u(α) and γ = 1 − α − β. Let g(x) = f (u(x))/f (x). Show that this polynomial is irreducible over Q and compute its roots. (4) Show that {1, α, β} is a basis of K over Q and that Disc(1, α, β) = d2 . (5) Suppose that d is square free. Show that no prime divisor of d is a factor of the index of α. Deduce that {1, α, β} is an integral basis of K and that Disc(K) = d2 .

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Solution 10.60. (1) Suppose that the polynomial f (x) is reducible over Q, then it has a root x0 ∈ Z an this root satisfies x0 = 0 or x0 | a4 . We have f (0) = a4 6= 0, hence x0 6= 0 and x0 | a4 . We use the identity (3x0 − 1)3 = d(9x0 − 4). It is clear that 9x0 − 4 6∈ {0, −1, 1}. Let p be a prime divisor of 9x0 − 4, then p 6= 3 and p | 3x0 − 1. It follows that x0 ≡ 4/9 (mod p) and x0 ≡ 3/9 (mod p). Therefore 4 ≡ 3 (mod p) which is a contradiction. We conclude that f (x) is irreducible over Q. (2) Straightforward computations. (3) We omit the verification that f (x) | f (u(x)) in Q[x]. Let g(x) ∈ Q[x] such that f (u(x)) = f (x)g(x), then f (u(α)) = f (α)g(α) = 0, hence β = u(α) is a root of f (x). Since α is not a root of a quadratic equation, then β 6= α. Let γ be the third root of f (x). Since α + β + γ = 1, then γ = 1 − α − β. We have f (u(x)) = (u(x) − α)(u(x) − β)(u(x) − γ) = f (x)g(x).

Suppose that g(x) is reducible over Q, then it has a rational root, say r. This implies that u(r) = α or u(r) = β or u(r) = γ, which contradicts the irreducibility of f (x). Thus g(x) is irreducible over Q. Since u(α) = β, u(β) = γ and u(γ) = α, then the roots of g(x) are given by (a2 + 1)/2 − α,

(a2 + 1)/2 − β and (a2 + 1)/2 − γ. 4 2
2 (4) Since β = α2 − a 2+1 α − 3a 2−a /a2 , then it is clear that we may express 1, α, α2 in terms of 1, α, β, hence {1, α, β} is a basis of K over Q. We have [Marcus (1977), Th. 6, Chap. 2] Tr(1) Tr(α) Tr(β) Disc(1, α, β) = Tr(α) Tr(α2 ) Tr(βα) . Tr(β) Tr(αβ) Tr(β 2 ) Here, Tr denotes the trace of elements of K over Q. From the minimal polynomial of α, it is seen that Tr(α) = Tr(β) = 1. Let µ = α2 . Since α3 − α2 − α(9a4 − 1)/4 + a4 = 0, then α(µ − (9a4 − 1)/4) = µ − a4 ,

hence µ(µ − (9a4 − 1)/4)2 = (µ − a4 )2 . Therefore,
2 µ µ − (9a4 − 1)/4 − (µ − a4 )2 = 0.

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From this equation it is seen that Tr(β 2 ) = Tr(α2 ) = Tr(µ) = (1 + 9a4 )/2. We have αβ = hence



 1 − a2 2 3a4 + 2a2 − 1 α + α − a4 /a2 , 2 4

  3a4 + 2a2 − 1 1 1 − a2 2 4 Tr(α ) + Tr(α) − a = (1−9a4 )/4. Tr(αβ) = 2 a 2 4 It follows that 3 2 1 1  1 + 27a4 4 4 = d2 . Disc(1, α, β) = 1 (1 + 9a )/2 (1 − 9a )/4 = 4 1 (1 − 9a4 )/4 (1 + 9a4 )/2

(5) Let p be a prime factor of d. Obviously, p 6= 3. From the identity given d in (1), it is seen that f (x) ≡ (x−1/3)3 (mod p) and x−1/3 - 3p (x−4/9) in Fp [x], thus p - I(α) by Exercise 10.30. We have Z[α] ⊂ Z[α, β] ⊂ A. Using the values of the discriminants computed above, we obtain Disc(1, α, α2 ) = Disc(f ) = a4 d2 = I(α)2 Disc(K) and  2 I(α) 2 2 Disc(1, α, β) = d = (A : Z[α, β]) Disc(K) = Disc(K). (A : Z[α, β])

I(α) Since gcd(d, I(α)) = 1, then gcd(d, (A:Z[α,β]) ) = 1, hence Disc(K) = d2 and {1, α, β} is a basis of A over Z.

Exercise 10.61. Let d be a cube free integer such that d ≥ 2. Write d in the√form d = ab2 , √ 3 3 where a, b ∈ N, gcd(a, b) = 1 and ab square free. Let α = d, β = a2 b, K = Q(α), and A be its ring of integers. (1) Show Zβ. (2) Show (3) Let p (4) Show

that {1, α, β} is a basis of K over Q and that Z[α, β] = Z + Zα + that Disc(1, α, β) = −27a2 b2 and (A : Z[α, β]) | 3ab. be a prime divisor of ab. Show that p is totally ramified in A. that ( P3 if a2 6≡ b2 (mod 9) 3A = , 2 P1 P2 if a2 ≡ b2 (mod 9)

where P, P1 and P2 are prime ideals of A having their residual degree equal to 1.

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(5) Let p be a prime divisor of ab. Show that p - (A : Z[α, β]). (6) If a2 6≡ b2 (mod 9), show that A = Z + Zα + Zβ and Disc(K) = −27a2 b2 . If a2 ≡ b2 (mod 9), show that A = Zα + Zβ + Zγ and Disc(K) = −3a2 b2 , where γ = (1 + aα + bβ)/3. Solution 10.61. (1) We have β = α2 /b. Since {1, α, α2 } is a basis of K over Q, then so is {1, α, β}. It is clear that Z + Zα + Zβ ⊂ Z[α, β]. To get the other inclusion it is sufficient to show that α2 , β 2 and αβ ∈ Z + Zα + Zβ. This claim is verified since α2 = bβ, β 2 = aα and αβ = ab. (2) Let j be primitive cube root of unity in C. The embeddings σ1 , σ2 , σ3 of K into C are given by σ1 = IdK , σ2 (α) = jα and σ3 (α) = j 2 α. Since β = α2 /b, then σ2 (β) = j 2 α2 /b = j 2 β and σ3 (β) = jα2 /b = jβ. It follows that [Marcus (1977), Chap. 2] 2 2 1 1 α β 1 1 Disc(1, α, β) = 1 jα j 2 β = α2 β 2 1 j j 2 2 1 j2α jβ 1 j j  2 = a2 b2 (j − 1)(j 2 − 1)(j 2 − j) = −27a2 b2 .

We use the formula relating the absolute discriminant of a number field, the index of an order of this field and the discriminant of a Z-basis of this order. We obtain −27a2 b2 = Disc(1, α, β) = (A : Z[α, β])2 Disc(K), hence 34 a2 b2 = (A : Z[α, β])2 (−3 Disc(K)).

It follows that −3 Disc(K) is a square and (A : Z[α, β])2 | 32 a2 b2 . Therefore (A : Z[α, β]) | 3ab. (3) Suppose that p | a and let P be a prime ideal of A lying over pZ. Since α3 = ab2 , then P | α, hence P 3 | ab2 . Since gcd(a, b) = 1, then ab2 = pc, where c ∈ Z and p - c. We have P 3 | pc, hence P 3 | p. We conclude that pA = P 3 . If p | b, we use a similar proof on replacing α by β. (4) Since −3 Disc(K) is a squares, then 3 | Disc(K), therefore 3 is ramified in A and the splitting of 3A in A must have one of the following forms 3A = P 3 or 3A = P12 P2 , where the ideals P1 , P2 and P are of the first degree and P1 6= P2 . If 3 | ab, then since ab is square free, we have

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a2 6≡ b2 (mod 9). Therefore by (3), 3 is totally ramified in A and the proof is complete in this case. Suppose now that 3 - ab, then a2 ≡ b2 (mod 3). Let θ = α − a, then θ satisfies the following equation θ3 + 3aθα + a(a2 − b2 ) = 0.

(Eq 1)

Suppose first that 3A = P 3 and let s = νP (θ). We show that 1 ≤ s ≤ 2. From this equation it is seen that 3 | θ3 , hence P | θ. Therefore s ≥ 1. Suppose that 3 | θ, then θ/3 is an algebraic integer whose minimal polynomial over Q is given by 3a2 a(a2 − b2 ) a x+ = 0. g(x) = x3 + x2 + 3 9 27 It follows that 3 | a, which is a contradiction. We conclude that s ≤ 2. Suppose that s = 2. From the identity α3 = ab2 , we deduce that P - α. We have νP (θ3 ) = 6, νP (3aθα) = 5, which implies, by using (Eq 1), νP (a2 − b2 ) ≥ 5. Therefore 32 | a2 − b2 . (Eq 1) shows the incompatibility of of these P-adic valuations. We conclude that s = 1 and again, by (Eq 1), a2 6≡ b2 (mod 9). Suppose now that 3A = P12 P2 , where the ideals P1 , P2 are of the first degree and P1 6= P2 . From Equation (4) we conclude that P2 | θ, hence νP2 (θ3 + 3aθα) ≥ 2. Using (Eq 1) (4), we obtain νP2 (a2 − b2 ) ≥ 2. We conclude that a2 ≡ b2 (mod 9). (5) By contradiction, suppose that p | (A : Z[α, β]). Then, by Lagrange’s Theorem, there exists ω ∈ A \ Z[α, β] such that pω ∈ Z[α, β]. Set pω = x + yα + zβ

(Eq 2)

with x, y, z ∈ Z. Since p | ab, then by (3), p is totally ramified in A. Let P be the unique prime ideal of A lying over pZ. Suppose that p | a. Since α3 = ab2 , then νP (α) = 1. Similarly, we have νP (β) = 2. (Eq 2) implies p | x and then p | y and then p | z. It follows that ω = (x/p) + (y/p)α + (z/p)β ∈ Z[α, β], which is a contradiction. Using similar arguments, one reach a contradiction in the case where p | b. We conclude that p - (A : Z[α, β]).

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(6) Recall that −27a2 b2 = Disc(1, α, β) = (A : Z[α, β])2 Disc(K). Using (5), we conclude that a2 b2 | Disc(K). (4) implies that 3 | Disc(K). We deduce that (A : Z[α, β]) = 1, Disc(K) = −27a2 b2 and then A = Z + Zα + Zβ or (A : Z[α, β]) = 3, Disc(K) = −3a2 b2 and A strictly contains Z + Zα + Zβ. Suppose that a2 6≡ b2 (mod 9), then by (4), 3 is totally ramified in A. If moreover 3 | ab, then by (5), 3 - (A : Z[α, β]). Therefore A = Z[α, β] and the proof is complete in this case. Now, if moreover 3 - ab, then a2 ≡ b2 (mod 3). Equation (Eq 1) shows that P | θ but P 2 - θ. Let ω ∈ A. We want to show that ω ∈ Z + Zα + Zβ. Since 3ω ∈ Z + Zα + Zβ, set 3ω = x + yα + zβ with x, y, z ∈ Z. We have 3bω = bx + by(θ + a) + z(θ + a)2 = bx + bya + za2 + (by + 2za)θ + zθ2 := A + Bθ + Cθ2 , hence 3 | A and then 3 | B. We deduce that 3 | C. It follows that x ≡ y ≡ z ≡ 0 (mod 3). Therefore ω ∈ Z[α, β] and then A = Z[α, β] and Disc(K) = −27a2 b2 . Suppose now that a2 ≡ b2 (mod 9) (which implies 3 - ab). By (4), the splitting of 3A is given by 3A = P12 P2 , where the prime ideals P1 and P2 are distinct. (Eq 1) shows that P1 P2 | θ and then P12 P22 | θ2 , hence 3 | θ2 . Since θ2 = (α − a)2

= α2 − 2aα + a2 = bβ − 2aα + a2

= (1 + aα + bβ) + (a2 − 1 − 3aα)

and since 3 | a2 − 1, then 1 + aα + bβ ≡ 0 (mod 3). It follows that the number γ = (1+aα+bβ)/3 is an algebraic integer and γ 6∈ Z+Zα+Zβ. We have Z[α, β](Z[α, β, γ] ⊂ A and since (A : Z[α, β]) ≤ 3, it follows that this index is equal to 3, A = Z[α, β, γ] and Disc(K) = −3a2 b2 . It remains to show that Z[α, β, γ] = Zα + Zβ + Zγ. Clearly the set appearing on the right side of this equality is contained in the other. To get the reverse inclusion it is sufficient to prove that the elements 1, α2 , β 2 , γ 2 , αβ, αγ, βγ belong to the set on the right side. It is easy to

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show the following relations: 1 = 3γ − aα − bβ

α2 = bβ

β 2 = aα αβ = ab(3γ − aα − bβ)

ab(1 − b2 1 − a 2 b2 α+ β + ab2 γ 3 3 1 − a2 b2 ab(1 − a2 βγ = β+ β + a2 bγ 3 3 b(1 + a2 − 2a2 b2 ) 1 + 2a2 b2 a(1 + b2 − 2a2 b2 ) α+ β+ γ. γ2 = 9 9 3

αγ =

Clearly 1 − a2 b2 ≡ 0 (mod 3). It remains to prove that 1 + b2 − 2a2 b2 ≡ 0 2

2 2

1 + a − 2a b ≡ 0

(mod 9)

and

(mod 9).

By symmetry it is sufficient to prove the first congruence. This follows easily from the following: a2 ≡ b2 (mod 9) ≡ 1 or 4 or −2 (mod 9). Exercise 10.62. (1) Let K be a number field of degree n, A be its ring of integers, {ω1 , . . . ωn } be an integral basis and let p be a prime number which is not ramified in A. Show that the following assertions are equivalent. (i) For any i ∈ {1, . . . , n}, ωip ≡ ω (mod pA). (ii) For any θ ∈ A, θp ≡ θ (mod pA). (iii) The prime p completely (totally) splits in A. (2) Suppose that the preceding equivalent conditions hold and that p < n. Show that, for any θ ∈ A, p | I(θ). (3) Let f (x) = x3 − x2 − 2x − 8, θ ∈ C be a root of f and E = Q(θ). Show that (a) f (x) is irreducible over Q. (b) Disc(f ) = −22 · 503. (c) 2 | I(θ) and µ := θ(θ − 1)/2 is an algebraic integer. Hint. One may use Exercise 10.30. (d) {1, θ, µ} is an integral basis. (e) The prime 2 completely splits in E and 2 | I(θ) for any algebraic integer θ ∈ E.

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(f) Let P1 , P2 , P3 be the prime ideals of E lying over 2Z. Show that we may suppose that they are numbered in order to satisfy the following conditions:     θ ≡ 0 (mod P )   1  µ ≡ 0 (mod P1 ) θ ≡ 0 (mod P2 ) and µ ≡ 1 (mod P2 ) .     θ ≡ 1 (mod P ) µ ≡ 1 (mod P ) 3

3

For each of these three prime ideals of E, determine a basis over Z. (g) Let I = θA + µA. Show that I = P1 . Solution 10.62. (1) • (i) ⇒ (ii). Let θ =

Pn

i=1

θp ≡

ai ωi be an element of A, then n X i=1

ai ωip ≡ θ

(mod p).

• (ii) ⇒ (iii). Since the prime p is not ramified in A, its splitting in A has the form pA = P1 · · · Pr , where r is a positive integer and P1 , . . . Pr are distinct prime ideals of A. Let j ∈ {1, . . . , r}. For any θ ∈ A, θp ≡ θ (mod Pj ). This may be written in the form: for any ¯ This implies that the field A/Pj has cardinality θ¯ ∈ A/Pj , θp = θ. equal to p. Therefore A/Pj ' Fp and the residual degree of Pj is equal to 1. • (iii) ⇒ (i). Let pA = P1 · · · Pr be the splitting of p in A, where r is a positive integer and P1 , . . . Pr are distinct prime ideals of A all of residual degree equal to 1. Fix i ∈ {1, . . . , n}. For any j ∈ {1, . . . , r}, we have ωip ≡ ω (mod P)| , hence ωip ≡ ω (mod pA). (2) By (ii), we have for any θ ∈ A, θp ≡ θ (mod pA), hence modulo p, θ satisfies an algebraic equation of degree less than n. Therefore p | I(θ) by Exercise 10.29. (3)(a) If f (x) where reducible over Q, then it will have a root n = 0 or n ∈ Z with n | −8. The possible values of n are n = ±1, ±2, ±4, ±8. It is straightforward to verify that f (x) does not vanish for any of these values of n nor for 0. Therefore f (x) is irreducible over Q. 0 00 (b) Let θ, θ, θ be the roots of f (x). We have 0

Disc(f ) = − NE/Q (f (θ))

= − NE/Q (3θ2 − 2θ − 2)

= −(3θ2 − 2θ − 2)(3θ02 − 2θ0 − 2)(3θ002 − 2θ00 − 2).

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It is possible to expand this product and to express it as a polynomial in the elementary symmetric functions of the roots of f (x) and then find the value of Disc(f ). We will not go further in this way. Instead of that, set α = 3θ2 − 2θ − 2 and multiply this equation successively by θ and θ2 . Using the equation satisfied by θ, we get the following equations: 3θ2 − 2θ + (−2 − α) = 0 θ2 + (4 − α)θ + 24 = 0

(5 − α)θ2 + 26θ + 8 = 0. This implies that x1 = 1, x2 = θ, x3 = θ3 is a non zero solution of the homogeneous system of equations 3x3 − 2x2 + (−2 − α)x1 = 0

x3 + (4 − α)x2 + 24x1 = 0

(5 − α)x3 + 26x2 + 8x1 = 0. It follows that the determinant of this system is 0, that is 3 −2 (−2 − α) 1 (4 − α) 24 = 0. (5 − α) 26 8 We get the equation α3 − 7α2 − 22 · 503 = 0. Therefore 0

NE/Q (f (θ)) = 22 · 503 and Disc(f ) = −22 · 503. (c) We may write f (x) in the form f (x) = x2 (x − 1) − 2(x + 4) and we see that x | x+4 in F2 [x]. Therefore, by Exercise 10.30, 2 | I(θ). Using the same method as in (2),(b), we compute the minimal polynomial g(x) over Q, of µ and we find g(x) = x3 − 2x2 + 3x − 10. This polynomial is monic and has integral coefficients, hence µ is an algebraic integer. (d) Consider the inclusions Z[θ] ⊂ Z[θ, µ] ⊂ A. By (b) and (c), the index of Z[θ] in A is equal to 2. The index of Z[θ] in Z[θ, α] is equal to the determinant of the matrix expressing 1, θ, θ2 in terms of 1, θ, α. This determinant is equal to 2, hence A = Z[θ, α]. Therefore {1, θ, α} is an integral basis.

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(e) Since the rational prime 2 is not ramified, we may use (1). We have θ2 = θ + 2µ, hence θ2 ≡ θ (mod 2). Straightforward computations, lead to µ2 = −2 + 2θ + µ, hence µ2 ≡ µ (mod 2). According to (1), the prime 2 completely splits in E. (f) From (e), we have 2A = P1 P2 P3 , where the prime ideals Pi , i = 1, 2, 3 are distinct and have their residual degree equal to 1. The minimal polynomial f (x) of θ over Q may be written in the form: f (x) = x2 (x − 1) − 2(x − 4). According to Exercise 10.30, 2 | I(θ) and we may suppose that    θ ≡ 0 (mod P1 ) (Eq 1) θ ≡ 0 (mod P2 ) .   θ ≡ 1 (mod P ) 3

Similarly, the minimal polynomial g(x) = x3 − 2x2 + 3x − 10 of µ over Q, may be written in the form g(x) = x(x + 1)2 + 2(−2x2 + x − 5),

   µ ≡ 1 (mod Pi1 )

µ ≡ 1 (mod Pi2 ) , where the indices i1 , i2 , i3 are distinct   µ ≡ 0 (mod P ) i3 and belong to {1, 2, 3}. Since

hence

θµ = θ2 (θ − 1)/2 = (θ3 − θ2 )/2 = θ + 4

and since θ 6≡ 0 (mod P3 ), then θµ 6≡ 0 (mod P3 ). We deduce that µ 6≡ 0 (mod P3 ), thus µ ≡ 1 (mod P3 ). Therefore, we may suppose that    µ ≡ 0 (mod P1 ) (Eq 2) µ ≡ 1 (mod P2 ) .   µ ≡ 1 (mod P ) 3

We begin with the computation of a Z-basis of P1 . Since A/P1 is isomorphic to Z/2Z, then there exists a morphism of rings φ : A → Z/2Z whose kernel is equal to P1 . Since A = Z + Zθ + Zµ, then in order to define φ, it is sufficient to define φ on Z and to find explicitly the respective values of φ(θ) and φ(µ). Obviously, for any a ∈ Z, we must have φ(a) = a ¯, where a ¯ = 0 or a ¯ = 1, according to a is even or a is odd. From the congruences satisfied by θ and µ described

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above, we must have φ(θ) = φ(µ) = 0. Let γ = a + bθ + cµ be an element of A, then γ ∈ P1 ⇔ φ(γ = 0) ⇔a ¯=0 ⇔a≡0

⇔ a = 2λ

(mod 2) with

λ∈Z

⇔ γ = 2λ + bθ + cµ with

λ ∈ Z.

Thus P1 is generated over Z by {2, θ, µ}. Obviously this set is free over Z, hence {2, θ, µ} is a basis of P1 . We use a similar reasoning for P2 and keep the same notations as above. We have γ ∈ P2 ⇔ φ(γ = 0)

⇔a ¯ + c¯ = 0

⇔a+c≡0

⇔ a = c + 2λ

(mod 2) with

λ∈Z

⇔ γ = 2λ + bθ + c(1 + µ)

with λ ∈ Z.

Hence P2 is generated over Z by {2, θ, µ + 1}. Obviously this set is free over Z, hence {2, θ, µ + 1} is a basis of P2 . We omit the computations for P3 . It is seen that {2, θ + 1, µ + 1} is a basis of P3 . (g) From (Eq 1), and since NE/Q (θ) = 8, we have θA = P1 e1 P2 e2 with e1 , e2 ≥ 1 and e1 + e2 = 3. From (Eq 2), and since NE/Q (µ) = 10, we have µA = P1 Q, where Q is a prime ideal lying over 5Z. We use the following identity: θµ = θ + 4.

(Eq 3)

It is easy to show that NE/Q (θ + 4) = 5.24 . Suppose that e1 = 1. Then νP1 (θµ) = 2 and νP1 (θ + 4) = 1, which is a contradiction to (Eq 3). It follows that e1 = 2, e2 = 1, θA = P1 2 P2 and µA = P1 Q, thus I = P1 . Remark. We may prove the same result as follows. Let γ = a + bθ + cµ ∈ A,

then γ ≡ a (mod I). We have

2 = µ(10/µ) − θ(8/θ) = µ(µ2 − 2µ + 3) − θ(θ2 − θ − 2),

hence 2 ∈ I. We deduce that 2Z ⊂ I. Let n = 2k + 1 be an odd integer, then n ≡ 1 (mod I). The question now reduces to does 1

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belong to I? Or do we have the equality I = A? Since θA ⊂ P1 and µA ⊂ P1 (see (Eq 1), (Eq 2)), then the answer is negative and then A/I ' Z/2Z. It follows that I is a prime ideal lying over 2Z. By (Eq 1), (Eq 2) again we conclude that I = P1 . Exercise 10.63. Let K be a number field of degree n, A be its ring of integers, p be a prime number and P be a prime ideal of A lying over pZ. For any rational integer m ≥ 1 and any α ∈ K, let   α α(α − 1) · · · (α − (m − 1)) . = m! m Let KP = {α ∈ K, νP (α) ≥ 0}. (1) Show that the following propositions are equivalent.
α (i) For all integers m ≥ 1 and all α ∈ KP , m ∈ KP . (ii) The residual degree of P is equal to 1 and P 2 - p.

(2) For any α ∈ K, denote by Den(α) the denominator of α, that is the smallest positive integer d such that dα is integral. Let Kp = {α ∈ K, p - Den(α)}.

Show that Kp = ∩P|p KP . Deduce that the following assertions are equivalent.
α (iii) For all integers m ≥ 1 and all α ∈ Kp , m ∈ Kp . (iv) The splitting of pA as a product of prime ideals of A is given by pA = P1 P2 · · · Pn , where the ideals Pi , i = 1, . . . , r are distinct and of the first degree over pZ. Solution 10.63. (1) • (i) ⇒ (ii). Suppose that P is of degree f ≥ 2, then there exists α ∈ A ⊂ K
P such that α 6≡ k (mod P) for k = 0, 1 . . . , p − 1. Therefore αp 6∈ KP . Suppose that P is of the first degree and
P 2 | p. Let α ∈ A such that αA + pA = P. Consider αp . The 2 numerator of this fraction is divisible by P but
not by P , while the α 2 denominator is divisible by P . Therefore p 6∈ KP . • (ii) ⇒ (i). We first show that for any positive integer s ≥ 1, the numbers 0, 1, . . . , ps − 1 form a complete residue system modulo P s . We have |A/P s | = NK/Q (P s ) = ps ,

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hence it is sufficient to prove that the elements in the preceding list are distinct modulo P s . If not, then ps | m for some m, 1 ≤ m < ps , which is a contradiction. Let α ∈ KP and let m ≥ 1 be an integer. It follows from the preceding that for any s ≥ 1, the numbers α, α − 1, . . . , α − (ps − 1) form a
complete system of α residues modulo P s . Consider the fraction m . In the sequence α, α − 1, . . . , α − (m − 1), there are bm/pc multiples of P, bm/p2 c multiples of P 2 and so on. Therefore the numerator of the fraction is divisible by P k , where k = bm/pc + bm/p2 c + · · · . k It is known that m! is divisible by exactly
p [Ribenboim (2001), α Lemma 3, Chap. 17.2]. It follows that m ∈ KP . (2) Let α ∈ Kp , then α = θ/d, where θ is an integer of K, d is the denominator of α and p - d. It follows that for any prime ideal P lying over pZ, we have P - d, hence α ∈ ∩P|p KP . We prove the reverse inclusion. Let α be an element of this intersection, then νP (α) ≥ 0 for any P | p, hence p - Den(α), thus α ∈ Kp . Now the equivalence of (iii) and (iv) follows from (1).

Exercise 10.64. Let K be a field, f (x) = xn + an−1 xn−1 + · · · + +a0 be a polynomial with coefficients in K. The matrix   0 0 ... 0 −a0 1 0 ... 0 −a1    M = 0 1 . . . 0 −a2    . . . . . . . . . . . . ...  0

0

...

1

−an−1

is called the companion matrix of f (x). For any matrix N ∈ Mn (K), we denote by χ(N ) the characteristic polynomial of N .

(1) Show that χ(M ) = f (x). (2) Suppose that f (x) is irreducible over K and let α be a root of f (x) in an algebraic closure of K. Let N ∈ Mn (K) such that f (N ) = 0. Show that K[α] ' K[N ]. (3) Show that the finite field F8 may be represented as a field of matrices belonging to M3 (F2 ). Write explicitly these matrices. (4) Suppose that K = Q and let f (x) = xn + an−1 xn−1 + · · · + a0 ∈ Z[x], α be a root of f (x) in C and N ∈ Mn (Z) such that f (N ) = 0. Suppose

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that f (x) is irreducible over Q. Let u(α) ∈ Q[α], where u(x) ∈ Q[x] and deg u ≤ n − 1. Show that u(α) is an algebraic integer if and only if χ(u(N )) ∈ Z[x]. (5) Let g(x) ∈ Z[x] be monic, irreducible of degree n and θ be a root of g(x). Let F = Q(θ) and A be the ring of integers of F . Show that the following conditions are equivalent. (i) A 6= Z[θ]. (ii) There exist a prime number p, a positive integer k, matrices M, N ∈ Mn (Z) and b0 , . . . , bk ∈ Z such that bk 6= 0, g(M ) = 0 and b0 I + b1 M + · · · + bk M k = pN.

Solution 10.64. (1) • First method. We have



x −1  χ(M ) = Det(xI − M ) = Det  0 . . . 0

0 x −1 ... 0

... ... ... ... ...

0 0 0 ... −1

 a0  a1  . a2  ...  x + an−1

Expanding this determinant through its first row, we obtain χ(M ) = Det(xI − M )  x 0 −1 x  = x Det  −1 0 . . . . . . 0 0

... ... ... ... ... 

−1 0  + (−1)n+1 a0 Det  0 . . . 0  x 0 −1 x  = a0 + x Det  −1 0 . . . . . . 0 0

0 0 0 ... −1

 a1  a2   a3  ...  x + an−1

 0 0  0  . . . −1 

x −1 0 ... 0

... ... ... ... ...

0 0 0 ... 0

... ... ... ... ...

0 0 0 ... −1

a1  a2   a3  ...  x + an−1

and the proof may be completed by induction.

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• Second method. Let a ˜0 , . . . , a ˜n−1 be indeterminates algebraically independent over K, E = K(˜ a0 , . . . , a ˜n−1 ), f˜(x) = xn + a ˜n−1 xn−1 + · · · + a ˜0

˜ be the matrix obtained from M by replacing ai by a and M ˜i for i = 0, . . . , n − 1. Let B = {e1 , . . . , en } be the canonical basis of E n over E and let T : E n → E n be the unique E-linear map such that T (ei ) = ei+1 for i = 1, . . . , n − 1 and T (en ) = −˜ a0 e1 − · · · − a ˜n−1 en , ˜. then clearly the matrix of T relative to the basis B is equal to M We first show that f˜(T ) = 0. We proceed by induction. We have f˜(T )(e1 ) = (−˜ a0 IdE n − a ˜1 T − · · · − a ˜n−1 T n−1 − T n )(e1 ) = 0.

Suppose that f˜(T )(ej ) = 0 for j = 1, . . . , k. Then

f˜(T )(ek+1 ) = f˜(T )(T (ek )) = T (f˜(T )(ek )) = T (0) = 0. ˜ ) = 0. Let P˜ (x) be the minimal It follows that f˜(T ) = 0, hence f˜(M ˜ polynomial of M over E. On the one hand P˜ (x) | f˜(x) in E[x]. Since f˜(x) is irreducible in K[˜ a0 , . . . , a ˜n−1 , x], then it is irreducible in E[x]. ˜) It follows that f˜(x) = P˜ (x). On the other hand, P˜ (x) | Det(xI − M in E[x]. Since these polynomials are both monic of the same degree n, then ˜ ). f˜(x) = P˜ (x) = Det(xI − M Consider the unique morphism of rings φ : K[˜ a0 , . . . , a ˜n−1 , x] → K such that φ(λ) = λ for any λ ∈ K, φ(x) = x and φ(˜ ai ) = ai for i = 0, . . . , n − 1. Applying φ to the preceding identity, we obtain f (x) = Det(xI − M ) = χ(M ).

¯ and τ : K[x] → Mn (K), defined (2) Consider the maps σ : K[x] → K by σ(u(x)) = u(α) and τ (u(x)) = u(M ), then clearly σ and τ are morphisms of rings, Ker σ = Ker τ = f (x)K[x],

Im σ = K[α]

and

Im τ = K[M ].

The first isomorphism theorem implies that K[M ] ' K[x]/f (x)K[x] ' K[θ]. (3) The polynomial f (x) = x3 − x + 1 is irreducible over F2 . Let θ be a root of f (x) in an algebraic closure of F2 , then F8 ' F2 [θ] ' F2 [M ],

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where M is any matrix belonging to M3 (F2 ) such that f (M ) = 0. We may take M to be the companion matrix of f (x), that is M =   0 0 −1 1 0 1  . Then 0 1 0 F8 ' {0, I, M, I + M, M 2 , I + M 2 , M + M 2 , I + M + M 2 }.

(4) We use the isomorphism of fields ψ : Q[θ] → Q[M ] such that ψ(θ) = M . Suppose that β := u(θ) is integral over Z, then g(β) = 0, where g(x) ∈ Z[x] is the minimal polynomial of β over Q. Applying ψ, we obtain g(u(N )) = 0. Therefore g(x) is the minimal polynomial of u(N ). Since g(x) is irreducible, then χ(u(N )) is a power of g(x), hence χ(u(N )) ∈ Z[x]. We prove the converse. Suppose that χ(u(N )) ∈ Z[x], then χ(u(N ))(N ) = 0 and applying ψ −1 , we obtain χ(u(N ))(β) = 0, which implies that β is integral over Z. (5) • (i) ⇒ (ii). Let B = {ω1 , . . . , ωn } be an integral basis of F . Since A 6= Z[θ], there exists β ∈ A \ Z[θ] of the form β = (b0 + b1 θ + bk θk )/p, where p is a prime number and b0 , . . . , bk are integers such that bk 6≡ 0 (mod p). Let M be the matrix in base B of the Q-endomorphism mθ of F such that mθ (x) = θx. Exercise 2.5 shows that N := (b0 + b1 M + · · · + bk M k )/p is the matrix representing the endomorphism multiplication by β in the basis B. So it has integral entries and b0 + b1 M + · · · + bk M k = pN. We show that g(M ) = 0. We have χ(M )(x) = Det(xI − M ) = Det(xIdF − mθ ), hence χ(M )(θ) = Det(θIdF − mθ ). Since (θIdF − mθ )(1) = 0, then χ(M )(θ) = 0. Therefore g(x) | χ(M )(x). Since these polynomials are both monic of degree n, then g(x) = χ(M )(x), which implies that g(M ) = 0.

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• (ii) ⇒ (i). Let a prime number p, matrices M, N ∈ Mn (Z) and b0 , . . . , bk ∈ Z such that bk 6≡ 0 (mod p), g(M ) = 0 and b0 I + b1 M + · · · + bk M k = pN. Let u(x) = (b0 + b1 x + · · · + bk xk )/p, then χ(u(M )(x) = χ(N )(x) ∈ Z[x], hence u(θ) is an algebraic integer by (4). Moreover u(θ) 6∈ Z[θ], hence A 6= Z[θ]. Exercise 10.65. (1) Let A be a Dedekind domain, B1 , . . . , Bk , be ideals of A. Show that k−1 ∩k−1 i=1 (Bi + Bk ) = ∩i=1 Bi + Bk .

(2) Let a1 , . . . , ak ∈ A. Show that the following system of congruence equations x ≡ ai (mod Bi ) for i = 1, . . . , k is solvable in A if and only if for any i, j, i 6= j, ai ≡ aj (mod gcd(Bi , Bj )). (3) Let (ai , bi ) for i = 1, . . . , k be distinct couples of integers such that 0 ≤ ai < bi . We say that this set of couples is a disjoint covering system if for any non negative integer n, there exists a unique i ∈ {1, . . . , k} such that n ≡ ai (mod bi ). Show that (ai , bi ), for i = 1, . . . , k, is a disjoint covering system if and only if the following conditions hold. P (i) 1/bi = 1. (ii) For any i, j ∈ {1, . . . , k}, i 6= j, we have ai ≡ aj (mod gcd(bi , bj )). (4) Let (ai , bi ) for i = 1, . . . , k be distinct couples of integers such that 0 0 ≤ ai < bi . For i = 1, . . . , m, let ai = bi − ai − 1. Show that (ai , bi ) 0 is a disjoint covering system if and only if (ai , bi ) satisfies the same property. (5) Let (ai , bi ) for i = 1, . . . , k be distinct couples of integers such that 0 ≤ ai < bi and let N = lcm(b1 , . . . , bk ). (a) Show that (ai , bi ), for i = 1, . . . , k, is a disjoint covering system if and only if for any integer n, 0 ≤ n ≤ N − 1, there exists a unique i ∈ {1, . . . , k} such that n ≡ ai (mod bi ).

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(b) Let α ∈ C be transcendental or algebraic over Q of degree greater than N . Prove the following identities. N −1 X αt = (αN − 1)/(α − 1), k X j=1



t=0

N/bj −1



X s=0



αaj +sbj  =

k X j=1

αaj (αN − 1)/(αbj − 1).

Deduce that (ai , bi ), for i = 1, . . . , k, is a disjoint covering system if and only if αa1 /(αb1 − 1) + · · · + αak /(αbk − 1) = 1/(α − 1). Solution 10.65. (1) For any i = 1, . . . , k − 1, Bi ⊂ Bi + Bk , hence

k−1 ∩k−1 i=1 Bi ⊂ ∩i=1 (Bi + Bk ).

Moreover since Bk ⊂ Bi + Bk for i = 1, . . . , k − 1, then We deduce that

Bk ⊂ ∩k−1 i=1 (Bi + Bk ).

k−1 ∩k−1 i=1 Bi + Bk ⊂ ∩i=1 (Bi + Bk ).

We prove the reverse inclusion. Let P be a prime ideal dividing ∩k−1 i=1 (Bi + Bk ), with a positive exponent equal to e. Then there exists i0 ∈ {1, . . . , k − 1} such that P e | Bi0 + Bk . It follows that P e | Bi0 e and P e | Bk . We deduce that P e | ∩k−1 i=1 Bi and P | Bk and then the desired inclusion. (2) • Necessity of the conditions. Let x ∈ A be a solution of the system of congruence equations and let i and j ∈ {1, . . . , k} such that i 6= j, then ai ≡ x ≡ aj (mod gcd(Bi , Bj )). • Sufficiency of the conditions. Notice that gcd(Bi , Bj ) = Bi + Bj . We prove the result by induction on k. Suppose first that k = 2. Since a1 ≡ a2 (mod Bi + Bj ), then a1 − a2 = u + v, where u ∈ B1 and v ∈ B2 . Let x = a1 − u = a2 + v, then x satisfies the congruence equations. Suppose that the result holds for any number of ideals of A smaller than k. Then, there exists c ∈ A such that c ≡ ai (mod Bi ) for i = 1, . . . , k − 1. Consider the following system consisting of two congruence equations x≡y

x ≡ an

(mod ∩k−1 i=1 Bi ) (mod Bk ).

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To use the result proved for two ideals, we must show that y ≡ an

(mod ∩k−1 i=1 Bi + Bk ).

For any i = 1, . . . , k − 1, we have y − ak = (y − ai ) + (ai − ak ) ∈ Bi + (Bi + Bk ) = Bi + Bk . Hence by (1), we have k−1 y − ak ∈ ∩k−1 i=1 (Bi + Bk ) = ∩i=1 Bi + Bk

and this proves the verification to be done. (3) We begin with a remark. Let b = lcm(b1 , . . . , bk ), Pi = ai + bi Z for i = 1, . . . , k and A = {0, 1, . . . , b − 1}, then |Pi ∩ A|/|A| = |{ai , ai + b, . . . , ai + (qi − 1)b}|/b, where qi is defined by the relation b = bi qi . Hence |Pi ∩ A|/|A| = 1/bi . • Necessity of the conditions. The conditions (ii) follows from (1). Since the covering system is disjoint, then A = ∪ki=1 (Pi ∩ A) and this union is a disjoint one. It follows that ! k k X X k 1 = | ∪i=1 (Pi ∩ A)|/|A| = |(Pi ∩ A)| /|A| 1/bi , i=1

i=1

hence (i). • Sufficiency of the conditions. (ii) shows that (ai , bi ) for i = 1, . . . , k is a covering system of Z. (i) implies that the covering system is disjoint on A, hence disjoint on Z. (4) We use the result in (3). It is clear that the condition (i) is satisfied by both systems or none. The second condition (ii), follows from the 0 0 identities ai − aj = bi − bj − (ai − aj ) for any i 6= j. (5)(a) The necessity of the condition is obvious. We prove the sufficiency. Let n ∈ Z, 0 ≤ r < N such that n ≡ r (mod N ) and let i be the unique integer in {1, . . . , k} such that r ≡ ai (mod bi ). Then n≡r

≡r

≡ ai

(mod N ) (mod bi ) (mod bi ).

Suppose that n ≡ ai (mod bi ) and n ≡ aj (mod bj ), then r ≡ ai (mod bi ) and r ≡ aj (mod bj ), hence i = j.

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(b) The first identity is obvious, so we omit its proof. We have     N/bj −1 N/bj −1 k k X X X X  (αbj )s  αaj +sbj  = αaj  j=1

s=0

s=0

j=1

=

k X j=1

αaj (αN − 1)/(αbj − 1),

which proves the second identity. Suppose that (ai , bi ), for i = 1, . . . , k, is a disjoint covering system, then   N/bj −1 N −1 k X X X  αt = αaj +sbj  , t=0

s=0

j=1

hence using the two identities, we get the following relation. αa1 /(αb1 − 1) + · · · + αak /(αbk − 1) = 1/(α − 1).

Conversely suppose that this relation holds, then according to the two identities, we obtain   N/bj −1 N −1 k X X X  αt = αaj +sbj  . t=0

s=0

j=1

This shows that α is a root of the following polynomial with integral coefficients   N/bj −1 k N −1 X X X  xaj +sbj  . f (x) = xt − t=0

s=0

j=1

This polynomial is the zero polynomial or its degree is less than N . The assumptions made on α implies that f (x) = 0, thus   N/bj −1 N −1 k X X X  xt = xaj +sbj  . t=0

j=1

s=0

The polynomial on the left hand side of this identity is of degree N − 1 and contains N non zero terms. The same must be true for the polynomial appearing on the right hand side. It follows that for any t ∈ {0, . . . , N − 1}, there exists a unique j ∈ {1, . . . , k} such that t = aj + sbj .

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Exercise 10.66. Let K be a number field, A be its ring of integers, F, P be the sets of fractional ideals of K and principal fractional ideals respectively. Let H = F/P. For any ideal I ∈ F, denote by [I] be its class modulo P. (1) Let A be a fractional ideal of K and let d be the order of [A] in H. Let ω ∈ K such that Ad = ωA, ρ = ω 1/d , E = K(ρ) and B be the ring of integers of E. Show that (a) AB = ρB. (b) If for some number field F containing K, with ring of integers C, we have AC = αC, with α ∈ F , then α = µρ, where µ is an algebraic unit. (2) Let h be the class number of K and let A1 , . . . , Ah be a complete set of representatives of the classes of ideals modulo P. Let d1 , . . . , dh be the respective orders of [A1 ], . . . , [Ah ]. For i = 1, . . . , h let ωi ∈ K such 1/d that Adi i = ωi A and let ρi = ωi i . Let E = K(ρ1 , . . . , ρh ) and B be its ring of integers. Show that for any fractional ideal I of K, IB is principal. (3) Let γ and η be algebraic numbers. Show that there exists an algebraic number φ such that the following relation holds aγ + bη = φ, where a and b are algebraic integers. Show that φ is unique up to multiplication by an algebraic unit. The number φ will be called the gcd of γ and η. Solution 10.66. (1)(a) We have (AB)d = Ad B = ωB = (ρB)d , hence by the uniqueness of the factorization in E into a product of prime ideals, we obtain AB = ρB. (b) It is known that any fractional ideal of K may be generated by two elements. So let a1 , a2 be elements of K such that A = a1 A + a2 A, then αC = a1 C +a2 C, thus α = a1 c1 +a2 c2 with c1 , c2 ∈ C. We have a1 = ρb1 and a2 = ρb2 , where b1 , b2 ∈ B, hence α = ρ(b1 c1 + b2 c2 ). We obtain a similar relation by permuting ρ and α. Thus b1 c1 + b2 c2 is an algebraic unit. (2) For any i ∈ {1, . . . , h}, let Ei = K(ρi ) and Bi be its ring of integers. According to (1), Ai Bi = ρi Bi , hence Ai B = ρi B, that is Ai B is principal. Since I is equivalent to some Ai , say Ai0 , then I = Ai0 (αA),

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where α ∈ K. We deduce that

IB = Ai0 B(αB) = ρi0 BαB = ρi0 αB,

thus IB is principal.

Exercise 10.67. √ Let d be a non zero square free integer, K = Q( d) and A be the ring of integers of K. Recall that A = Z[ω], where (√ d if d ≡ 2, 3 (mod 4) √ . ω= (1 + d)/2 if d ≡ 1 (mod 4) Let I be a non zero ideal of A.

(1) Let δ = gcd{b, there exists a ∈ Z such that a + bω ∈ I}. Show that there exists a0 ∈ Z such that a0 + δω ∈ I. (2) For any α = a + bω ∈ I, let c(α) = a + bω − (b/δ)(a0 + δω). Show that c(α) ∈ Z ∩ I and I is generated over Z by a0 + δω and {c(α), α ∈ I}. (3) Let c = gcd{c(α), α ∈ I}. Show that I is generated over Z by c and a0 + δω. Show that c is the smallest positive integer which belongs to I. (4) Let q and r ∈ Z such that a0 = cq + r with 0 ≤ r < c. Show that I is generated by c and r + δω. Show that δ | c and δ | r. (5) Let P be a prime ideal of A lying over some rational prime p such that A/P ' Fp . Show that P = (p, r + ω) with 0 ≤ r < p. Solution 10.67. Pn (1) Let (ai , bi ) ∈ Z2 , i = 1, . . . , n such that ai + bi ω ∈ Z and δ = i=1 xi bi Pn with xi ∈ Z. Since i=1 xi (ai + bi ω) ∈ I, then a0 + δω ∈ I with Pn a0 = i=1 xi ai . (2) We have c(α) = a − (b/δ)a0 , hence c(α) ∈ Z. Obviously c(α) ∈ I, thus c(α) ∈ I ∩ Z. Any α = a + bω ∈ I may be written in the form α = c(α) + (b/δ)(a0 + δω),

hence a0 + δω with the set {c(α), α ∈ I} generate I. (3) The first part is obvious. Let d ∈ I be a positive integer, then d = c(d), hence c | d, that is c is the smallest positive integer which belongs to I. (4) We have r + δω = (a0 − cq) + δω = a0 + δω − cq ∈ I

and

a0 + δω = r + δω + qc,

hence c and r + δω generate I over Z. Since c ∈ I and r + δω ∈ I, then δ | c and δ | r.

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(5) Since NK/Q (P) = p, then p ∈ P. Since 1 6∈ P, then the element c defined in (2) is equal to p. Hence I has the form I = (p, r + δω), with 0 ≤ r < p. Since δ | r and δ | p, then δ = 1. Thus I = (p, r + ω).

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Chapter 11

Derivations

Exercise 11.1. Let p be a prime number, K be a perfect field of characteristic p and E be an Pn extension of K. Let α1 , . . . , αn be elements of E such that αi = j=1 cij αjp for i = 1, . . . , n, where cij ∈ K for any (i, j) ∈ {1, . . . , n}2 . Show that α1 , . . . , αn are algebraic over K. Solution 11.1. Suppose that there exists i ∈ {1, . . . , n} such that αi is transcendental over K. Let r ≥ 1 be the transcendence degree of K(α1 , . . . , αn ) over K. We may suppose that {α1 , . . . , αr } is a transcendence basis and αr+1 , . . . , αn are algebraic over K(α1 , . . . , αr ). Let d : K(α1 , . . . , αr ) → K(α1 , . . . , αn ) 1 ,...,αr ) . Since K is be the K-derivation such that d(f (α1 , . . . , αr )) = ∂f (α∂α 1 perfect and K(α1 , . . . , αn ) is algebraic and separable over K(α1 , . . . , αr ), then d has one and only one extension dˆ : K(α1 , . . . , αn ) → K(α1 , . . . , αn ) Pn [Lang (1965), Th. 7, Chap. 10.7]. Since α1 = j=1 c1j αjp , then ˆ 1) = 1 = d(α

n X

ˆ j ) = 0, cij pαjp−1 d(α

j=1

which is a contradiction. Therefore α1 , . . . , αn are algebraic over K. Exercise 11.2. Let K be a field of characteristic 0, f (x, y) be an irreducible polynomial with coefficients in K. Let α be an element of an algebraic closure of K(x), E = K(x, α) and K0 be the algebraic closure of K in E. Let d be the derivation of K(x) such that d(u(x)) = u0 (x) and let d˜ be the unique extension of d to E. Show that Ker d˜ = K0 . 411

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Solution 11.2. Let β ∈ K0 and let g(x) be its minimal polynomial over K. We have ˜ ˜ 0 = d(g(β)) = g 0 (β)d(β). If g 0 (β) = 0, then g 0 (x) = 0, hence g(x) is a constant, which is excluded. ˜ ˜ Let β ∈ Ker d˜ and It follows that d(β) = 0 which means β ∈ Ker d. let h(x, y) ∈ K[x, y] be the unique (up to a multiplication by a non zero constant in K) irreducible polynomial satisfying h(x, β) = 0. We have ˜ 0 = d(h(x, β)) =

∂h(x, β) ˜ ∂h(x, β) . + d(β) ∂x ∂y

= 0, hence h(x, y) | ∂(h(x,y)) in K(x)[y]. Taking We deduce that ∂h(x,β) ∂x ∂x ∂h into account of the respective degrees in y of h and ∂x , we conclude that there exists c(x) ∈ K(x) such that ∂(h(x,y)) = c(x)h(x, y). Set ∂x h(x, y) = hm (x)y m + · · · + h0 (x) and c(x) = a(x)/b(x), where h0 (x), . . . , hm (x), a(x) and b(x) ∈ K[x], hm (x) 6= 0, b(x) 6= 0 and gcd(a(x), b(x)) = 1. Using the above relation 0 between h(x, y) and ∂h(x,y) ∂x , we get the equations: b(x)hj (x) = a(x)hj (x) for j = 0, . . . , m. This implies that b(x) divides all the polynomials hj (x) hence b(x) is a constant in K. It follows that a(x) = 0 and h0j (x) = 0 for j = 0, . . . , m. Since the characteristic of K is 0, we conclude that hj (x) is a constant in K for j = 0, . . . , m which means β ∈ K0 . Exercise 11.3. Let E be a field of characteristic 6= 2. Suppose that there exists a map σ : E ⇒ E satisfying the following conditions. (1) For any x, y ∈ E, σ(x + y) = σ(x) + σ(y). (2) For any x ∈ E, σ(x2 ) = 2x(σ(x)). Show that σ is a derivation of E. Solution 11.3. The proof is similar to the one given in Exercise 1.13. Exercise 11.4. ∂f Let d be the K-derivation of K[x, y] defined by d(f ) = x ∂f ∂x + y ∂y . Show that Ker d = K.

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Solution 11.4. Let f (x, y) = an (y)xn + · · · + a0 (y) be an element of Ker d with an (y) 6= 0 and let m = deg an (y), then the coefficient of xn of d(f ) is equal to nan (y) + ya0n (y), hence nan (y) + ya0n (y) = 0. Set an (y) = bm y m + · · · + b0 , where bm 6= 0, then the coefficient of y m in the equation relating a and a0 is equal to (n + m)bm , hence n + m = 0 which implies n = m = 0 and f ∈ K. Exercise 11.5. Let K be a field of characteristic 0, ~x = (x1 , . . . , xn ) and f (~x) ∈ K[~x] irreducible. Let δ be a K-derivation of K[~x] such that Ker δ contains n − 1 algebraically independent polynomials over K. Show that there exists T (~x) ∈ K[~x] such that δ(f ) = T f if and only if for any i ∈ {1, . . . , n}, f | δ(xi ) or there exists P (~x) ∈ Ker δ such that f | P . Solution 11.5. • Necessity of the conditions. Suppose that there exists some i ∈ {1, . . . , n}, say i = n, such that f - δ(xi ) and let P1 , . . . , Pn−1 be algebraically independent elements over K of Ker d. Let A = K[~x]/(f ) and E the fraction field of A. Since the transcendence degree of E over K is equal to n − 1, then the elements of A: x ¯n , P1 , . . . , Pn−1 are algebraically dependent over K, hence there exists R(y1 , . . . , yn ) ∈ K[y1 , . . . , yn ] such that R(P1 , . . . , Pn−1 , x ¯n ) = ¯0. We deduce that there exists q(~x) ∈ K[~x] such that R(P1 , . . . , Pn−1 , xn ) = f q. We may write this identity in the form ak (P1 , . . . , Pn−1 )xkn +· · ·+a1 (P1 , . . . , Pn−1 )xn +a0 (P1 , . . . , Pn−1 ) = f q, where k is supposed to be minimal. Suppose that k ≥ 1, then applying δ to both sides of this identity, yields
δ(xn ) kak (P1 , . . . , Pn−1 )xnk−1 + · · · + a1 (P1 , . . . , Pn−1 ) = f δ(q) + qδ(f ) = f (δ(q) + T q).

Since f - δ(xn ), then there exists g(~x) ∈ K[~x] such that

kak (P1 , . . . , Pn−1 )xk−1 + · · · + a1 (P1 , . . . , Pn−1 ) = f g, n

which contradicts the minimality of k. Therefore k = 0 and a0 (P1 , . . . , Pn−1 ) ≡ 0 (mod f ) and the proof is complete. • Sufficiency of the conditions. P ∂f Suppose that f | δ(xi ) for i = 1, . . . , n. We have δ(f ) = ∂xi δ(xi ), hence f | δ(f ) in K[~x], thus δ(f ) = T f for some T ∈ K[~x]. If

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f (~x) | P (x) for some P (x) ∈ K[~x], let g(~x) ∈ K[~x] such that P (~x) = f (~x)e g(~x), where g(~x) ∈ K[~x], e is a positive integer and gcd(f (~x), g(~(x)). Then 0 = δ(P ) = ef (~x)e−1 g(~x)δ(f ) + f (~x)δ(g). We deduce that eg(~x)δ(f ) + f (~x)δ(g) = 0, which shows that f (~x) | δ(f ) in K[~x]. Therefore, there exists T (~x) ∈ K[~x] such that δ(f ) = T f . Exercise 11.6. (1) Let A be an integrally closed domain, d : A → A a derivation and d˜ its extension to Frac(A). Show that Ker d˜ = Frac(Ker d) if and only if Ker d˜ is algebraic over Frac(Ker d). (2) Let K be a field of characteristic 0, A = K[x1 , . . . , xn ], d a K-derivation of A and d˜ its extension to Frac(A) = K(x1 , . . . , xn ). If Ker d contains n − 1 algebraically independent polynomials over K, show that Ker d˜ = Frac(Ker d). (3) Let K be a field of characteristic 0, d : K[x1 , x2 ] → K[x1 , x2 ] the ∂ ∂ + x2 ∂x . Show that Ker d = K and derivation such that d = x1 ∂x 1 2 ˜ x2 /x1 ∈ Ker d. Conclude that the assumption on the transcendence degree in (2) cannot be omitted. Solution 11.6. (1) • Necessity of the condition. ˜ then α = u/v, where u, v ∈ Ker d and v 6= 0, hence Let α ∈ Ker d, α − u/v = 0 which shows that α is algebraic over Frac(Ker d). • Sufficiency of the condition. The inclusion Frac(Ker d) ⊂ Ker d˜ is obvious. We prove the reverse ˜ then there exists a positive integer n and inclusion. Let α ∈ Ker d, u0 /v0 , u1 /v1 , . . . , un−1 /vn−1 ∈ Frac(Ker d) such that αn + (un−1 /vn−1 )αn−1 + · · · + (u0 /v0 ) = 0. Let v = v0 · · · vn−1 , then v ∈ Ker d and vα is integral over A, which is integrally closed. Hence vα ∈ A. Set vα = a ∈ A, then d(a) = ˜ = 0. Therefore α = a/v ∈ Frac(Ker d). d(a)

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(2) We apply (1) for A = K[x1 , . . . , xn ]. Here A is a unique factorization domain, hence integrally closed. From the assumptions, we conclude that the transcendence degree of Ker d˜ over K is equal to n or n − 1. ˜ hence In the first case, K(x1 , . . . , xn ) is algebraic separable over Ker d, ˜ ˜ d = 0. Therefore d = 0, Ker d = K[x1 , . . . , xn ], Ker d = K(x1 , . . . , xn ) and the result is proved in this case. In the second case, let t be the ˜ transcendence degree of Frac(Ker d) over K. Since Frac(Ker d) ⊂ Ker d, then n−1 ≤ t ≤ n−1, hence t = n−1. It follows that Ker d˜ is algebraic over Frac(Ker d). By (1), we conclude that Ker d˜ = Frac(Ker d). (3) Let f (x1 , x2 ) ∈ K[x1 , x2 ], X f (x1 , x2 ) = ai1 i2 xi11 xi22 , then

d(f ) =

X (i1 + i2 )aij xi11 xi22 ,

hence d(f ) = 0 if and only if f ∈ K. Therefore Ker d = K. It is easy ˜ 2 /x1 ) = 0. This shows that Ker d˜ 6= Frac(Ker d). It is to check that d(x obvious that, in this example, the assumptions of (2) are not satisfied. Exercise 11.7. ~ = (x1 , . . . xn ) be an n-tuple Let K be a field of characteristic p ≥ 0, X of variables, y another variable such that x1 , . . . , xn , y are algebraically ~ y) ∈ K[X, ~ y] be irreducible and such that independent over K. Let f (X,

~ such that degy f ≥ 1 and degxi f ≥ 1 for i = 1, . . . , n. Let α ∈ K[X] ~ ~ f (X, α) = 0 and let E = K(X, α). Let K0 be the algebraic closure of K in E. If p > 0, we assume that f 6∈ K[xp1 , . . . , xpn , y p ].

(1) Show that [K0 : K] | degy f . (2) If f is absolutely irreducible, show that K0 = K. ~ y) ∈ K[ ¯ X, ~ y] be an irreducible factor of f such that (3) Let f1 (X, ~ α) = 0. Suppose that the leading coefficient of f1 for the lexf1 (X, icographic order is equal to 1. Let K1 be the extension of K generated by the coefficients of f1 . (a) If p > 0, let K1s be the separable closure of K in K1 and let pe = [K1 : K]i = [K1 : K1s ]. Let k be the smallest non negk ~ y) ∈ K s [X, ~ y]. Let ative integer such that k ≤ e and f1p (X, 1 k p ~ ~ g1 (X, y) = f1 (X, y). • Show that g1 is irreducible over K1s .

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¯ be the • • Let d1 = [K1s : K] and let σ1 , . . . , σd1 : K1s → K Qd1 σi ~ ~ distinct embeddings. Let g(X, y) = i=1 g1 (X, y). Show ~ y) = cg(X, ~ y) for some c ∈ K ? . that f (X,

• • • Deduce that k = 0, K1 /K is separable and g1 = f1 .

(b) In any case p = 0 or p > 0, show that K1 = K0 . (c) Deduce that f is absolutely irreducible if and only if K0 = K. ~ in E. Show that K0 ⊂ F . (4) Let F be the separable closure of K(X) (5) From now on, we suppose that n = 1 and p = 0. We set x1 = x and we suppose that f (x, y) 6∈ K[xp , y] and f (x, y) 6∈ K[x, y p ]. Let d : K(x) → K(x) be the derivation such that for any u(x) ∈ K(x), we have d(u(x)) = u0 (x). • Show that d extends, in a unique way, as a derivation d˜ of E and ˜ compute d(α). • • Show that Ker d˜ = K0 .

˜ (6) Let m = degy f . For any β ∈ E we denote by β 0 the element d(β) and generally β (k) the k-th derivative of β. Show that there exist a0 (x), a1 (x), . . . , am (x) ∈ K[x] not all 0 such that a0 (x)α + a1 (x)α0 + · · · + am (x)α(m) = 0.

(7) • Let β be a conjugate of α over K(x) and let σ : E → K(x) be the K(x)-embedding such that σ(α) = β. Show that σ(α(k) ) = β (k) . • • Let k ≤ m be the smallest positive integer such that b0 (x)α + b1 (x)α0 + · · · + bk (x)α(k) = 0,

where bi (x) ∈ K[x] for i = 0, . . . , k and bk (x) 6= 0. Show that all the conjugates of α over K(x) satisfy this differential equation. Solution 11.7. (1) Consider the diagram.  K0 (X)

 K(X)

K0

K

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~ K0 /K is algebraic while K(X)/K is purely transcendental, hence K0 ~ are linearly disjoint over K [Lang (1965), Pro. 3, Chap. 10.5]. and K(X) ~ : K(X)] ~ = [K0 : K]. It follows that Therefore [K0 (X) ~ = [E : K0 (X)][K ~ ~ ~ degy f = [E : K(X)] 0 (X) : K(X)] ~ = [E : K0 (X)][K 0 : K], which proves that [K0 : K] | degy f . ~ y) is absolutely irreducible. We have (2) Suppose that f (X, ~ α) : K0 (X)] ~ = degy f = [K(X, ~ α) : K(X)]. ~ [K(X, ~ = K(X) ~ and then K0 = K. Therefore K0 (X) (3)(a) • Suppose that g1 is reducible over K1s , then there exists h < k such h that f p ∈ K1s [x, y], which contradicts the minimality of k. Hence g1 is irreducible over K1s . ~ α) = g1 (X, ~ α) = 0, the two polynomials f and g1 • • We have f (X, have their coefficients in K1s and the second is irreducible over this ~ α) | f (X, ~ α) in K s [x, y]. Therefore, field. It follows that g1 (X, 1 ¯ g σ (X, ~ α) | f (X, ~ α) in K s [x, y]. for any K-embedding σ : K1s → K, 1 1 Qd1 σi ~ σ τ ~ ~ y]. Since g1 6= g1 for σ 6= τ , then i=1 g1 (X, y) | f (X, y) in K1s [X, The first polynomial has its coefficients in K and the second is ~ y) = irreducible over K, hence there exists c ∈ K ? such that f (X, ~ y). cg(X, • • • We have ~ y) = c f (X,

d1 Y

k

~ y))σi = c (f1p (X,

i=1

d1 Y

˜ (X, ~ y))pk , (f1,σ i

i=1

˜ (X, ~ y) is the polynomial obtained from f σi (X, ~ y) by where f1,σ i 1 1/pk replacing any of its coefficient a by a . If k ≥ 1 then f ∈ K[xp1 , . . . , xpn , y p ], which contradicts the assumptions. Hence k = 0, K1s = K1 , K1 /K is separable and g1 = f1 . (b) We have shown that if p > 0, then K1 /K is separable. This claim is ~ y) = c Qd1 (f1 (X, ~ y))σi , also true if p = 0. We have in any case f (X, i=1 where the product runs over the distinct K-embeddings σi : K1s → ¯ We deduce that degy f = degy f1 ˙[K1 : K]. Consider the diaK. gram.

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 K1 (X)

 K(X)

K1

K

On the one hand, we have ~ α) : K(X)] ~ = [K1 (X, ~ α) : K1 (X)][K ~ ~ ~ [K1 (X, 1 (X) : K(X)] ~ α) : K1 (X)][K ~ = [K1 (X, 1 : K] ˙ = deg f1 [K1 : K] y

= degy f. On the other hand, we have ~ α) : K(X)] ~ = [K1 (X, ~ α) : K(X, ~ α)][K(X, ~ α) : K(X)] ~ [K1 (X, ~ α) : K(X, ~ α)] degy f. = [K1 (X, ~ α) = K(X, ~ α), which implies K1 ⊂ K(X, ~ α) and then Hence K1 (X, K1 ⊂ K0 . Since f1 is absolutely irreducible, then by (2), we have K10 = K1 . Since K1 ⊂ K0 , then K 0 ⊂ K10 = K1 ⊂ K 0 , hence K1 = K0 . (c) We have proved, in (2), that if f is absolutely irreducible, then K 0 = K. We prove the converse. Suppose that f is not absolutely ¯ y] of f . Let irreducible and let f1 be an irreducible factor in K[x, K1 be the field generated by the coefficients of f1 over K, then K ( K1 = K0 by (3) (b), which is a contradiction. (4) Let a ∈ K0 and g(x) = Irr(a, K), then g is separable over K because ~ = Irr(a, K), then a is separable K0 /K is separable. Since Irr(a, K(X)) ~ over K(X). Therefore a ∈ F . (5) • Suppose that an extension d˜ exists, then since f (x, α) = 0, we have ∂f ∂f ˜ (x, α) + (x, α)d(α) = 0. ∂x ∂y Since

∂f ∂y (x, α)

∂f

∂x (x,α) ˜ 6= 0, then d(α) = − ∂f . Since d˜ is completely (x,α) ∂y

˜ determined by d(α), then d˜ is unique. Moreover, this formula for ˜ ˜ d(α) allows one to define d(β) for any β ∈ E.

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˜ We prove the reverse inclusion. Let • • It is clear that K0 ⊂ Ker d. β ∈ Ker d˜ and g(x, y) = y q + bq−1 (x)y q−1 + · · · + b0 (x) be its minimal polynomial over K(x). Since g(x, β) = 0, then ∂g ∂g ˜ (x, β) + (x, β)d(β) = 0. ∂x ∂y ∂g ∂g Hence ∂x (x, β) = 0. Therefore, g((x, y) | ∂x (x, y) in K(x)[y]. It ∂g (x, y) = 0 and then b0 (x) = 0 for i = 0, . . . , q − 1. We follows that ∂x deduce that b(x) ∈ K for i = 0, . . . , q − 1, which implies that β is algebraic over K. Therefore β ∈ K0 . (6) Since E is a vector space over K(x) of dimension m, then any family of m + 1 elements is linearly dependent. Therefore, there exist a0 (x), a1 (x), . . . , am (x) ∈ K[x] not all 0 such that

a0 (x)α + a1 (x)α0 + · · · + am (x)α(m) = 0. (7) • It is sufficient to prove that if σ(α) = β, then σ(α0 ) = β 0 . From (5), we have ∂f

∂x (x, α) ˜ α0 = d(α) = − ∂f , ∂y (x, α)

hence ∂f ∂x σ(α0 ) = − ∂f

(x, σ(α))

∂y (x, σ(α))

= β0.

• • It is clear that the integer k satisfying the given properties exists. Applying σ to the differential equation satisfied by α: b0 (x)α + b1 (x)α0 + · · · + bk (x)α(k) = 0 and using (7) •, we conclude that β satisfies the same equation. Exercise 11.8. Let K be a field and A be a K-algebra without zero divisors. Let ∆ be a set of K-derivations of A and A∆ = {a ∈ A, d(a) = 0, for any d ∈ ∆}. Show that Frac(A∆ ) ∩ A = A∆ .

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Solution 11.8. Let ¯ = {δ, derivation of Frac(A) extending some derivation of A} ∆ and ¯ ¯ Frac(A)∆ = {x ∈ Frac(A), δ(a) = 0, for any δ ∈ ∆}. ¯

¯

We have Frac(A∆ ) ⊆ Frac(A)∆ and Frac(A)∆ ∩ A = A∆ , hence ¯

A∆ ⊆ Frac(A)∆ ∩ A ⊆ Frac(A)∆ ∩ A = A∆ . Therefore Frac(A∆ ) ∩ A = A∆ . Exercise 11.9. Let R and S be integral domains such that S ⊂ R and let F and E be their respective fraction fields. Suppose that E is a Galois extension of F , R is integrally closed and that R ∩ F = S. Let D : E → E be a derivation such that D(F ) ⊂ F and D(R) ⊂ R. Let S¯ be the integral closure of S in E. (1) Show that for any σ ∈ Gal(E, F ), σDσ −1 is a derivation of E and the restrictions to F of σDσ −1 and D are equal. (2) Deduce that for any σ ∈ Gal(E, F ), σDσ −1 = D. ¯ = S. ¯ (3) Show that for any σ ∈ Gal(E, F ), σ(S) ¯ ¯ (4) Show that D(S) ⊂ S. Solution 11.9. (1) Obvious. (2) DF is a derivation of F and D, σDσ −1 are extensions of DF to E, where DF is the restriction of D to F . Since E/F is algebraic, this extension is unique, hence σDσ −1 = D. ¯ ⊂ S¯ for any σ ∈ Gal(E, F ). Since (3) It is sufficient to prove that σ(S) ¯ ¯ ¯ ⊂ S. ¯ σ(S) ⊂ E and σ(S) is integral over S, then σ(S) ¯ then its conjugates over F , σ(¯ (4) Let s¯ ∈ S, s), σ ∈ G, belong to S¯ by (3), hence any elementary symmetric function of these elements belongs to ¯ Since S¯ ⊂ R, ¯ where R ¯ denotes the integral closure of R in E and S. ¯ = R, hence all these elementary since R is integrally closed, then R symmetric functions belong to R. Since they also belong to F , they are elements of R ∩ F , which is equal to S. It follows that D(¯ s) is integral ¯ thus D(S) ¯ ⊂ S. ¯ over S, that is D(¯ s) ∈ S,

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Exercise 11.10. Let K be a perfect field, E be an algebraic extension of K, d : K → K be a derivation and dˆ : E → E be its unique extension. Suppose that Ker d is algebraically closed in E. Show that Ker dˆ = Ker d. Solution 11.10. ˆ For the reverse inclusion, let α ∈ Ker dˆ Obviously, we have Ker d ⊂ Ker d. and let f (x) = Irr(α, K) = xn + an−1 xn−1 + · · · + a0 . Since f (α) = 0, then n−1 X

0 ˆ i )αi + d(α)f ˆ d(a (α) = 0,

i=0

hence

n−1 X

ˆ i )αi = d(a

i=0

n−1 X

d(ai )αi = 0.

i=0

It follows that ai ∈ Ker d for any i ∈ {0, . . . , n − 1} which implies that α is algebraic over Ker d. Since Ker d is algebraically closed in E, then α ∈ Ker d. Exercise 11.11. (1) Let K be a field, A be a K-algebra, ∆ be a set of K-derivations of A and A∆ = {a ∈ A, d(a) = 0 for any d ∈ ∆}. ¯

Show that Frac(A∆ ) ⊂ (Frac A)∆ , where

¯ = {δ derivation of Frac A, δ is a prolongation of some d ∈ ∆}. ∆

(2) Show that the reverse inclusion does not hold in the following case: K is a field of characteristic 0, A = K[x, y], ∆ = {d}, where d is the unique K-derivation of A such that d(x) = x and d(y) = y. Solution 11.11. ¯

¯

(1) Since (Frac A)∆ is a field, it is sufficient to prove that A∆ ⊂ (Frac A)∆ . ¯ then a = a ∈ Frac A and Let a ∈ A∆ and δ ∈ ∆, 1 δ(a) =

δ(a) · 1 − aδ(1) d(a) − ad(1) = = 0, 12 12

¯

hence a ∈ (Frac(A))∆ .

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x y

∈ (Frac A)∆ but

x y

6∈ Frac(A∆ ). It is easy to see δf ˆ that for any f (x, y) ∈ K[x, y], d(f ) = δf δx x + δy y. Let d be the unique extension of d to K(x, y) = Frac A, then

(2) We show that

ˆ ˆ ˆ x ) = y d(x) − xd(y) = yd(x) − xd(y) = 0, d( 2 y y y2 thus

x y

¯

∈ (Frac A)∆ . Let f (x, y) ∈ A∆ , then x

hence x| δf δy . Set

δf δy

δf δf +y = 0, δy δx

= xg(x, y), where g(xy) ∈ K[x, y], then

−yg(x, y). It follows that

δf δy /x

=

− δf δx /y.

(Eq 1) δf δx

=

Set

f (x, y) = an (x)y n + · · · + a1 (x)y + a0 (x). Since y | δf δx , then a0 (x) = 0. We deduce that n

a1 (x) an (x) n−1 y + ··· + = −[a0n (x)y n−1 − · · · − a01 (x)] x x

and then n anx(x) = −a0n (x). Set an (x) = bm xm + bm−1 xm−1 + · · · + b1 x with bm 6= 0, then nbm xm−1 + nbm−1 xm−2 + · · · + nb1

= −mbm xm−1 − (m − 1)bm−1 xm−2 − · · · − b1 .

It follows that (n + m)bm = 0, which implies n = m = 0. Therefore, f (x, y) ∈ K[x] and then from (1), δf δx = 0 which means f (x, y) ∈ K. We have proved that A∆ = K, hence Frac(A∆ ) = K and xy ∈ K. It follows that Frac(A∆ ) ( (Frac A)∆ . Exercise 11.12. Let K be a field of characteristic 0 and ∆ be a K-algebra without zero divisors. Let ∆ be a set of derivations of A and A∆ = {a ∈ A, d(a) = 0 for any d ∈ ∆}. Show that A∆ is integrally closed in A.

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Solution 11.12. Let B = A∆ and let a ∈ A be integral over B. Then there exist b0 , b1 , . . . , bn − 1 ∈ B such that an + bn−1 an−1 + · · · + b0 = 0 and we may suppose that n is minimal. We deduce that for any d ∈ ∆,
d(a) nan−1 + (n − 1)bn−1 an−2 + · · · + b1 = 0,

hence d(a) = 0, that is a ∈ A∆ . Therefore, A∆ is integrally closed in A.

Exercise 11.13. Let K be a field, n be an integer at least equal to 2, A = K[x1 , . . . , xn ] and B the integral closure of K[x1 , x1 x2 ] in A. Show that for any non empty set ∆ of derivations of A, we have B 6= A∆ , where A∆ is the set of elements a ∈ A such that d(a) = 0 for any d ∈ ∆. Show that B ⊂ K[x1 , x2 ]. Solution 11.13. Suppose that there exists a non empty set ∆ of derivations of A such that B = A∆ . For any d ∈ ∆, we have d(x1 ) = 0 = d(x1 x2 ) = x1 d(x2 ), hence d(x2 ) = 0, which implies x2 ∈ B. We show that this leads to a contradiction. There exist a0 , a1 , . . . , am−1 ∈ K[x1 , x1 x2 ] such that m−1 xm + · · · + a0 = 0. Write the left side of this identity as 2 + am−1 x2 a sum of monomials bxi1 xj2 . Then it is seen that the coefficient of xm 2 is equal to 1. This means that the polynomial on the left side of the identity is non zero, hence x1 , x2 are algebraically dependent over K, which is a contradiction. We have K[x1 , x1 x2 ] ⊂ K[x1 , x2 ] ⊂ A, hence B ⊂ K[x1 , x2 ], where K[x1 , x2 ] is the integral closure of K(x1 , x2 ) in A. It is easy to show that K[x1 , x2 ] is integrally closed in A, hence B ⊂ K[x1 , x2 ]. Exercise 11.14. Let K be a field of characteristic 0, d be a K-derivation of K[x1 , . . . , xn ]. Suppose that d 6= 0. (1) Show that the number of polynomials algebraically independent over K and belonging to Ker d is at most equal to n − 1. (2) Show that the condition on the characteristic of K cannot be omitted. Solution 11.14. (1) Suppose that Ker d contains n polynomials f1 (x1 , . . . , xn ), . . . , fn (x1 , . . . , xn )

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algebraically independent over K and let g(x1 , . . . , xn ) ∈ K[x1 , . . . , xn ]. Then clearly f1 , . . . , fn , g are algebraically dependent over K. Therefore, there exists φ(y1 , . . . , yn , z) ∈ K[y1 , . . . , yn , z] such that φ is irreducible over K and φ(f1 , . . . , fn , g) = 0. Set m X φ(y1 , . . . , yn , z) = φi (y1 , . . . , yn )z i , i=0

where m is a positive integer, φi ∈ k[y1 , . . . , yn ] for i = 0, . . . , m and φm 6= 0. Pm Since i=0 φi (f1 , . . . , fn )g i = 0, then ! m m X X i i−1 d(φi (f1 , . . . , fn ))g + iφi (f1 , . . . , fn )g d(g) = 0, i=0

i=1

hence

m X

iφi (f1 , ..., fn )g

i−1

i=1

!

d(g) = 0.

Since K is of characteristic 0, f1 , . . . , fn are algebraically independent over K and φm (y1 , . . . , yn ) 6= 0, then mφm (f1 , . . . , fn ) 6= 0, hence Pm Pm i−1 6= 0. Therefore i=1 iφi (f1 , . . . , fn )g i−1 6= 0 i=1 iφi (y1 , . . . , yn )y and then d(g) = 0. Since g was arbitrary, then d = 0, which is a contradiction to our assumptions. We conclude that the number of elements of Ker d, algebraically independent over K is at most equal to n − 1. (2) Let K be a field of characteristic p and let d : K[x] → K[x] such that d(f (x)) = f 0 (x). Then clearly d(xp ) = pxp−1 = 0. Therefore Ker d contains a transcendental element over K. Exercise 11.15. Let K be a field of characteristic 0 and P (x, y) be a non constant polynomial with coefficients in K. Let D : K(x, y) → K(x, y) be the map such that for any f (x, y) ∈ K(x, y), ∂p ∂p ∂y ∂x D(f ) = ∂f ∂f ∂x

∂y

(1) Show that D is a K-derivation. (2) Let f (x, y) ∈ K(x, y). Show that the following propositions are equivalent. (i) f ∈ Ker D. (ii) f and P are algebraically dependent over K.

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Solution 11.15. (1) Let f, g ∈ K(x, y) and λ, µ ∈ K. Then ∂p ∂p ∂x ∂y D(λf + µg) = ∂f ∂g ∂g λ ∂x + µ ∂x λ ∂f + µ ∂y ∂y     ∂p ∂f ∂g ∂p ∂f ∂g = λ +µ − λ +µ ∂x ∂y ∂y ∂y ∂x ∂x     ∂p ∂f ∂p ∂f ∂p ∂g ∂p ∂g =λ − +µ − ∂x ∂y ∂y ∂x ∂x ∂y ∂y ∂x = λD(f ) + µD(g),

hence D is K-linear. We have ∂p ∂p ∂x ∂y D(f g) = ∂f ∂g ∂g g ∂x + f ∂x g ∂f + f ∂y ∂y     ∂p ∂f ∂g ∂p ∂f ∂g = g +f − g +f ∂x ∂y ∂y ∂y ∂x ∂x     ∂p ∂f ∂p ∂g ∂p ∂g ∂p ∂f − +g − =f ∂x ∂y ∂y ∂x ∂x ∂y ∂y ∂x = f D(g) + gD(f ).

Therefore D is a derivation. Obviously D(a) = 0 for any a ∈ K, hence D is K-derivation. ∂p ∂ ∂p ∂ Remark. We may write the map D in the form D = ∂x ∂y − ∂y ∂x , therefore D is a K-derivation. (2) • (i) ⇒ (ii). Suppose that f and P are algebraically independent over K. Let Q(x, y) ∈ K(x, y)\K, then f, P and Q are algebraically dependent over K. Let R(x, y, z) ∈ K[x, y, z]\{0} such that R(P, f, Q) = 0. We may write this identity in the form: Pn i = 0, where n ≥ 1, R1 (x, y) ∈ K[x, y] and i=0 Ri (P, f )Q Rn (x, y) 6= 0. Moreover, we may suppose that n is minimal. We have ! n n X X i i−1 0 = D(R(P, f, Q)) = D(Ri (P, f ))Q + Ri (P, f )iQ D(Q). i=0

i=1

Since D(P ) = D(f ) = 0, then n X D(Q) Ri (P, f )iQi−1 = 0. i=1

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If n ≥ 2, then since n is minimal, we conclude that D(Q) = 0. If n = 1, then R1 (P, Q)D(Q) = 0 and since R1 (P, Q) 6= 0, we conclude that D(Q) = 0. In any case we obtain D(Q) = 0 for any Q ∈ K[x, y]\K. Therefore D = 0. We deduce that 0 = D(x) = −

∂P ∂y

and

0 = D(y) =

∂P . ∂x

It follows that P (x, y) ∈ K, a contradiction. • (ii) ⇒ (i). Obviously if f ∈ K, then f ∈ Ker D, hence we may suppose that f 6∈ K. Since P (x, y) and f (x, y) are algebraically dependent over K, there exist a positive integer n and a0 (x), a1 (x), . . . , an (x) ∈ k[x] such that an (x) 6= 0, Pn Pn i i i=0 ai (f (x, y))P (x, y) = 0 and the polynomial i=0 ai (x)y is irreducible over K. We deduce that n X

D(ai (f ))P i + D(P )

i=0

n X

iai (f )P i−1 = 0.

i=1

Since D(P ) = 0, then D(f )

n X

a0i (f )P i = 0.

i=0

Pn

Pn 0 i i Suppose that = 0, then divides i=0 ai (f )P i=0 ai (x)y Pn 0 i 0 a (x))y . It follows that a (x) = 0 for i = 0, . . . , n, hence i i=0 i ai (x) ∈ K for i = 0, . . . , n. Therefore P (x, y) ∈ K which is a contradiction. We conclude that D(f ) = 0, that is f ∈ Ker D. Exercise 11.16. Let R and S be integral domains such that S ⊂ R and let E and F be their fraction fields respectively. Suppose that R is integrally closed, E is a finite Galois extension of F and R ∩ F = S. Let R0 be the integral closure of S in E. Let d be a derivation of E such that d(F ) ⊂ F and d(R) ⊂ R. (1) Show that R0 ⊂ R. (2) Show that d(R0 ) ⊂ R0 . Solution 11.16. (1) The following diagram may be helpful.

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E

R

R0

F

S Let R1 be the integral closure of R in E. Since R is integrally closed, then R1 = R. On the other hand, since S ⊂ R, then R0 ⊂ R1 , hence R0 ⊂ R. (2) Let G = {σ1 , . . . , σn } be the Galois group of E over F and let α ∈ R0 . We must show that d(α) ∈ R0 . This is equivalent to show that the eleP ments sk := σi1 (d(α)) · · · σik (d(α)), (k = 1, . . . , n), which represent, up to multiplication by −1, the coefficients of the characteristic polynomial of d(α) over F , except the leading one, belong to S = R ∩ F . Obviously, they belong to F . We show that they belong to R. For −1 any σ ∈ G, σdσ −1 is a derivation of E and σdσIF = dIF . Since E/F is algebraic and separable, then any derivation of F has one and only one extension to E. It follows that σdσ −1 = d, hence σd = dσ. Since α ∈ R0 , then σ(α) ∈ R0 and σ(α) ∈ R by (1). We conclude that sk ∈ R. Exercise 11.17. Let A be an integral domain containing Q and d : A[x] → A[x] be the derivation defined by d(f (x)) = f 0 (x) for any f (x) ∈ A[x]. (1) Show that d is surjective. (2) Let I and J be ideals of A[x] such that I ⊂ J. Show that J + d(I) is an ideal of A[x]. Show that d(I) is not always an ideal of A[x]. (3) Let I be an ideal of A[x] and let c(I) be the ideal of A generated by all the coefficients of the polynomials f (x) ∈ I. Show that d−1 (I) is a subring of A[x] and A ⊂ d−1 (I) ⊂ A + c(I)A[x]. Solution 11.17. Pn (1) Let g(x) = i=0 ai xi be an element of A[x], then ! n X 1 i+1 ai x , g(x) = d i+1 i=0 hence d is surjective.

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(2) Let g1 + f10 , g2 + f20 be elements of J + d(I), where f1 , f2 ∈ I and g1 , g2 ∈ J and let h ∈ R[x], then g1 + f10 + g2 + f20 = (g1 + g2 ) + (f1 + f2 )0 ∈ J + d(I) h(g1 +

f10 )

= hg1 +

hf10

0

and

0

= (hg1 − h f1 ) + (f h) ∈ J + d(I),

hence J + d(I) is an ideal of A[x]. Let A = Q and I = (x2 + x + 1)A[x], then

d(I) = {[(x2 + x + 1)f (x)]0 , f (x) ∈ A[x]},

hence [x(x2 + x + 1)]0 ∈ d(I), that is 3x2 + 2x + 1 ∈ d(I). We show that x(3x2 + 2x + 1) 6∈ d(I) which will prove that d(I) is not an ideal of Q[x]. Suppose that x(3x2 + 2x + 1) ∈ d(I), then 0

3x3 + 2x2 + x = [(x2 + x + 1)f (x)]

for some f (x) ∈ Q[x]. Since



2x2 x2 3x4 + + 3x + 2x + x = 4 3 2 3

2

then

0

,

3 4 2 3 1 2 x + x + x + a, 4 3 2 where a ∈ Q. We deduce that x2 + x + 1 divides 43 x4 + 23 x3 + 21 x2 + a in Q[x]. Performing an Euclidean division, we obtain   3 4 2 3 1 2 3 2 1 2 1 1 1 2 x + x + x + a = (x + x + 1) x − x − x + x + + a. 4 3 2 4 12 12 4 6 (x2 + x + 1)f (x) =

Since for any a ∈ Q, x2 + x + 1 - 34 x4 + 32 x3 + 12 x2 + a, then we get a contradiction. We conclude that x(3x2 + 2x + 1) 6∈ d(I). (3) Let f1 , f2 ∈ d−1 (I), then 0

0

0

d(f1 − f2 ) = (f1 − f2 ) = f1 − f2 ∈ I 0

and

0

d(f1 · f2 ) = f1 f2 + f2 f1 ∈ I,

hence f1 − f2 and f1 f2 ∈ d−1 (I). Therefore d−1 (I) is a subring of A[x]. Let a ∈ A, then d(a) = 0 ∈ I, hence a ∈ d−1 (I). We deduce that 0 A ⊂ d−1 (I). Let f (x) ∈ d−1 (I) and let a = f (0), then d(f ) = f ∈ I. P 0 n Set f (x) = i=0 ai xi , then ai ∈ c(I) and n X ai i+1 f (x) − f (0) = f (x) − a = x ∈ c(I)A[x]. i + 1 i=0 Therefore

f (x) = a + (f (x) − a) ∈ A + c(I)A[x].

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Exercise 11.18. Let A be an integral domain containing Q, K be its fraction field, R be a subring of some extension E of K satisfying the condition R ∩ K = A. Suppose that there exist t, u ∈ R and an A-derivation d of R such that R ⊂ K[u] and d(t) = 1. (1) (2) (3) (4)

Show that u is transcendental over K. Show that for any x, y ∈ R, xy ∈ A ⇒ x ∈ A and y ∈ A. Show that R ⊂ K[t]. Deduce that R = A[t].

Solution 11.18. (1) The following diagram may be useful. E K[u] K

R A

Suppose that u is algebraic over K and let f (x) = xn + an−1 xn−1 + · · · + a0

be its minimal polynomial over K. Since f (u) = 0, then f 0 (u)d(u) = 0. If f 0 (u) = 0, then f (x)|f 0 (x) in K[x] and then f 0 (x) = 0 which contradicts the fact that the characteristic of K is equal to 0. Therefore d(u) = 0. We deduce that for any g(u) ∈ R ⊂ K[u], we have d(g(u)) = g 0 (u)d(u) = 0, which contradicts the existence of t ∈ R, such d(t) = 1. We conclude that u is transcendental over K. (2) Let x, y ∈ R such that xy ∈ A. We may write x and y in the form x = f1 (u)/c1 and y = f2 (u)/c2 , where c1 , c2 ∈ A and f1 (u), f2 (u) ∈ A[u]. We have f1 (u)f2 (u) = c1 c2 xy ∈ A. Since u is transcendental over K, then f1 (u) and f2 (u) ∈ A. We deduce that x and y ∈ R ∩ K = A. (3) Since R ⊂ K[u], it is sufficient to show that u ∈ K[t]. Since t ∈ R, then t ∈ K[u], hence there exist a ∈ A\{0} and F (u) ∈ A[u] such that at = F (u). Since d is an A-derivation, we deduce that d(at) = ad(t) = a = F 0 (u)d(u) ∈ A.

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By (2), we conclude that F 0 (u) and d(u) ∈ A. Let b ∈ A such that F 0 (u) = b, then F (u) = bu + c with c ∈ A. We deduce that at = bu + c and this shows that b 6= 0. We now have a = F 0 (u)d(u) = bd(u), hence ab−1 = d(u) ∈ R ∩ K, thus ab−1 ∈ A. We deduce that cb−1 = ab−1 t − u ∈ R ∩ K. Since R ∩ K = A, then u = ab−1 t − cb−1 ∈ A[t]. We conclude that R ⊂ K[u] ⊂ K[t]. (4) The inclusion A[t] ⊂ R is obvious. We prove the reverse inclusion. Let x ∈ R\{0}. Since R ⊂ K[t], then x ∈ K[t], hence there exists a ∈ A such that ax ∈ A[t]. Set ax = b0 tn + b1 tn−1 + · · · + bn , where n is a positive integer, b0 , b1 , . . . , bn ∈ A and b0 6= 0. Let dn be the n-th iterate of d, then dn (ax) = adn (x) = n!b0 . Since Q ⊂ A, then n! is a unit of A, hence a|b0 in A. We complete the proof that a|bi for i = 0, . . . , n by induction. Suppose that a|bi for i = 0, . . . , m for m ≥ 1, then   bm n−m b0 n t = bm+1 tn−m−1 + · · · + bn . a x − t − ··· − a a Let y = x − ba0 tn − · · · − bam tn−m , then y ∈ R and we have adn−m−1 (y) = (n − m − 1)!bm+1 . This shows that dn−m−1 (y) ∈ K ∩ R = A, hence a | bm+1 in A and then x ∈ A[t] as desired. Exercise 11.19. Let K be a field, d be a K-derivation of K(x, y) such that d(x) and d(y) ∈ K[x, y]. (1) Show that the map τ : K(x, y) \ {0} → K(x, y) defined by τ (f ) = d(f )/f for any f (x, y) ∈ K(x, y)∗ \ {0} is a homomorphism of the multiplicative group into the additive group. (2) Let f1 , f2 ∈ K[x, y] \ {0} such that gcd(f1 , f2 ) = 1. Suppose that τ (f1 f2 ) ∈ K[x, y] or τ (f1 /f2 ) ∈ K[x, y]. Show that τ (f1 ) and τ (f2 ) ∈ K[x, y]. Solution 11.19. (1) Let f1 , f2 ∈ K(x, y) \ {0}, then τ (f1 f2 ) = d(f1 f2 )/f1 f2 = and

f1 d(f2 ) + f2 d(f1 ) = τ (f1 ) + τ (f2 ) f1 f2

d(f1−1 ) = f1 d(1/f1 ) = f1 (−d(f1 )/f12 ) = −d(f1 )/f1 = −τ (f1 ), f1−1 hence τ is a homomorphism of groups.

τ (f1−1 ) =

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(2) If τ (f1 f2 ) ∈ K[x, y], then from the identity τ (f1 f2 ) · f1 f2 = f1 d(f2 ) + f2 d(f1 ), we see that f1 | d(f1 ) and f2 | d(f2 ). Hence τ (f1 ) ∈ K[x, y]. If τ (f1 /f2 ) ∈ K[x, y], we use the identity f1 f2 d(f1 /f2 ) = d(f1 )f2 − d(f2 )f1 and we obtain the same conclusions as above: f1 | d(f1 ) and f2 | d(f2 ). Therefore τ (f1 ) ∈ K[x, y] and τ (f2 ) ∈ K[x, y]. Exercise 11.20. Let K be a field of characteristic 0, d be a derivation of K and K0 be a subfield of K containing Ker d and satisfying the condition d(K0 ) ⊂ K0 . (1) Let {f1 , f2 , f3 , . . .} be a subset of K such that d(fi ) ∈ K0 for any i and let ( n ) X n T0 = ci fi , n ≥ 1, (c1 , . . . , cn ) ∈ (Ker d) \{0} . i=1

If T0 ∩ K0 = φ, show that 1, f1 , f2 , . . . are linearly independent over K0 . (2) Let F be a subfield of K such that K0 ⊂ F and d(F ) ⊂ F . Let L ∈ T0 such that 1, L are linearly independent over F . (3) Deduce that if K0 ∩ T0 = φ, then for any L ∈ T0 and any integer n ≥ 1, 1, L, . . . , Ln are linearly independent over K0 . (4) Let F be as in (2) and let L ∈ T0 . If λ + µL = 0, where λ, µ ∈ F , show that d(λ)µ − λ(µ) + µ2 d(L) = 0. (5) Let A0 = K0 and Am = K0 [f1 , . . . , fm ] for any integer m ≥ 1. Let Km be the fractions field of Am . (a) Show that d(Am ) ⊂ Am and d(Km ) ⊂ Km for any m ≥ 0. (b) For any m ≥ 1 let ( n ) X n−m Tm = ci fi , n ≥ m + 1, (cm+1 , . . . , cn ) ∈ (Ker d) \{0} . i=m+1

Suppose that T0 ∩ K0 6= φ and let L ∈ Tm for some m ≥ 0. Show that 1, L, . . . , Ln are linearly independent over Km for any n ≥ 1.

(6) Let f1 , f2 , . . . , and T0 as in (1). Show that f1 , f2 , f3 , . . . are algebraically independent over K0 if and only if T0 ∩ K0 6= φ. (7) Let F be a field of characteristic 0, K = F ((x)), K0 = F (x) and d be d (u(x)) for any u(x) ∈ K. the derivation of K such that d(u(x)) = dx

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(a) Let f (x) ∈ K\K0 such that d(f ) ∈ K0 . Show that f (x) is transcendental over K0 . P∞ n (b) Show that f (x) = n=1 xn satisfies the conditions of (a) and then it is transcendental over F (x). (8) Let F, d, K and K0 as in (7). (a) For g(x) ∈ K0 and α ∈ F¯ write g(x) in the form g(x) = P∞ i ¯ i=−m ai (x − α) , where m is a non negative integer and ai ∈ F for any. The coefficient a−1 is called the residue of g(x) at α. Show that the residue of g 0 (x) at α is equal to 0. (b) Let f1 (x), f2 (x), . . . be elements of K such that d(f (x)) ∈ K0 . Suppose that there exist α1 , α2 , . . . ∈ F¯ such that for any integer n ≥ 1, Det(rij ), where rij is the residue of d(fi (x)) at αj . Show that f1 (x), f2 (x), . . . are algebraically independent over K0 . (9) Let F be a field of characteristic 0, g1 (x), g2 (x), . . . be pairwise relatively prime and non constant polynomials with coefficients in F . Suppose that gi (0) = 1 for any i ≥ 1. Let ∞ X (1 − gi (x))n . fi (x) = Log(gi (x)) := n n=1 Show that the gi (i ≥ 1) are algebraically independent over F (x).

Solution 11.20. (1) We first show that 1, f1 are linearly independent over K0 . Suppose the contrary. Then there exist (λ, µ) ∈ K02 \{0} such that λ + µf1 = 0, hence λ = −µf1 ∈ K0 ∩ T0 , which is a contradiction. Therefore 1, f1 are linearly independent over K0 . Let n ≥ 2, we show by induction on n that 1, f1 , . . . , fn are linearly independent over K0 . Suppose that n X λi fi = 0, (Eq 1) λ0 + i=1

where λj ∈ K0 for j = 0, . . . , n. We assume that λn 6= 0 and we will get a contradiction. Differentiating the above identity we obtain: n n X X d(λ0 ) + d(λi )fi + λi d(fi ) = 0. (Eq 2) i=1

i=1

Eliminating fn from the relations (1) and (2), yields to the equation: µ0 +

n−1 X i=1

(λn d(λi ) − d(λn )λi )fi = 0,

(Eq 3)

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where µ0 = λn d(λ0 ) − d(λn )λ0 + λn

n X

λi d(fi ).

i=1

The assumptions on the λj and on the fi show that µ0 and λn d(λi ) − d(λn )λi = 0 for i = 1, . . . , n − 1. These equations may be written in the form: µ0 = 0 and d( λλni ) = 0 for i = 1, . . . , n − 1. We deduce that for any i = 1, . . . , n − 1, there exists ci ∈ Ker d such that λi = ci λn . Notice that we also have λn = cn λn with cn = 1. Replacing these values of the λi in the equation µ = 0 and dividing by λ2n , we obtain: n

λn d(λ0 ) − d(λn )λ0 X + ci d(fi ) = 0, λ2n i=1

Pn hence d( λλn0 + i=1 ci fi ) = 0. This implies that there exists c ∈ Ker d Pn such that λλn0 + i=1 ci fi = c. Pn It follows that c − λλn0 = i=1 ci fi ∈ K0 ∩ T0 , which is a contradiction. (2) Let n ≥ 2 be an integer. Suppose that n X

λi Li = 0,

(Eq 4)

i=0

where λi ∈ F for i = 0, . . . , n and λn 6= 0. We will get a contradiction. Differentiating the above relation, we obtain: ! n n X X i i−1 d(λi )L + iλi L d(L) = 0. i=0

i=0

We may write this identity in the form: d(λn )Ln +

n−1 X

(d(λi ) + (i + 1)d(L)λi+1 )Li = 0.

(Eq 5)

i=0

Eliminating Ln from (4) and (5), we get the following identity: n−1 X i=0


λn [d(λi ) + (i + 1)d(L)λi+1 ] − d(λn )λi Li = 0.

Notice that d(L) ∈ F since d(fi ) ∈ K0 ⊂ F for any i so that the coefficients of Li belong to F . Using the inductive hypothesis, we conclude that, in particular, λn d(λn−1 ) − d(λn )λn−1 + nd(L)λ2n = 0.

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Dividing by λ2n , we get: λn d(λn−1 ) − d(λn )λn−1 + nd(L) = 0, λ2n that is d( λλn−1 + nL) = 0. n We deduce that there exists c ∈ Ker d such that λn−1 λn

λn−1 λn

+ nL = c. There-

fore − c + nL = 0, which shows that 1, L are linearly independent over F , a contradiction. (3) We apply (2) for the particular case F = K0 . For this aim, we must Pn show that 1, L are linearly independent over K0 . Set L = i=1 ci fi , where ci ∈ Ker d for i = 1, . . . , n and (c1 , . . . , cn ) 6= 0 and suppose that Pn there exist λ, µn ∈ K0 such that λ·1+µL = 0. Then λ+ i=1 µci fi = 0, hence by (1), λ = 0 and µc1 = 0 for i = 1, . . . , n. Since one of the ci is non zero, then λ = µ = 0. Now the condition in (2) is fulfilled and we conclude that 1, L, . . . , Ln are linearly independent over K0 . (4) Differentiating the identity λ + µL = 0, we obtain d(λ) + d(µ)L + µd(L) = 0. We eliminate L from these two relations and we get the result. (5)(a) Since d(fi ) ∈ K0 for any i ≥ 1, then d(Am ) = d(K0 )[d(f1 ), . . . , d(fm )] ⊂ K0 ⊂ Am and

d(Km ) = d(K0 )[d(f1 ), . . . , d(fm )] ⊂ K0 ⊂ Km . (b) We proceed by induction on m. If m = 0, the result follows from (3). Let m ≥ 1 and suppose that the result is true for m − 1. According to (2) with F = Km it is sufficient to prove that 1, L are linearly independent over Km or equivalently over Am . Suppose that there exist λ, µ ∈ Am \{0} such that λ + µL = 0. If none of λ and µ depends of fm , then we may apply the inductive hypothesis since Tm ⊂ Tm−1 , and we get a contradiction. In the other case Pn Pn i i we set λ = i=0 ai fm and µ = i=0 bi fm , where n is an integer ≥ 1, ai , bi ∈ Am−1 for i = 0, . . . , n − 1, an 6= 0 or bn 6= 0. By (4), we have d(λ)µ − λd(µ) + µ2 d(L) = 0, hence n n n n X X X X i i i i bi f m · d(ai )fm − ai fm d(bi )fm i=0

i=0

+ d(fm )

" n X i=0

+

n X i=0

i bi fm

i=0

i bi fm ·

!2

n X i=1

i=0

i−1 iai fm −

d(L) = 0.

n X i=0

i ai fm ·

n X i=1

i−1 bi fm

#

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P2n i We consider this relation as an equation of the form i=0 ci fm , where ci ∈ Km−1 for i = 0, 1, . . . , 2n. Since fm ∈ Tm−1 , then we apply the inductive hypothesis and conclude that ci = 0 for i = 0, . . . , n and this will lead to a contradiction. We have c2n = bn d(an ) − an d(bn ) + d(L) · b2n = 0. Suppose that bn 6= 0, then c2n =

bn d(an ) − an d(bn ) + d(L) = 0, b2n

hence d( abnn + L) = 0. Therefore, there exists c ∈ Ker d such that an bn + L = c. Since L ∈ Tm ⊂ Tm−1 and abnn − c ∈ Km−1 , this equality contradicts our inductive hypothesis. Thus bn = 0 and then an 6= 0. We have: c2n+1 = bn−1 d(an ) − an d(bn−1 ) = 0,

bn−1 hence d( bn−1 an ) = 0. Therefore an ∈ Ker d. Thus bn−1 = ean with e ∈ Ker d. Using the relations bn = 0 and c2n−2 = 0, we obtain the following equation:

bn−2 d(an ) − an d(bn−2 )

+ bn−1 d(an−1 ) − an−1 d(bn−1 )

+ d(fm )bn−1 an + d(L)b2n−1 = 0. Substituting ean for bn−1 , we get: bn−2 d(an ) − an d(bn−2 )

+ ean d(an−1 ) − ean−1 d(an )

+ ed(fm )a2n + e2 a2n d(L) = 0.

Dividing by a2n , we obtain: bn−2 d(an ) − an d(bn−2 ) an d(an−1 ) − an−1 d(an ) +e +ed(fm )+e2 d(L) = 0, 2 an a2n hence

  bn−2 an−1 d − +e + efm + e2 L = 0. an an

We conclude that there exists h ∈ Ker d such that −bn−2 + ean−1 + e(fm + eL) = h. an

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Thus −bn−2 + ean−1 − han + ean (fm + eL) = 0.

This equation has the form: λ + µ(fm + eL) = 0, where λ and µ ∈ Km−1 . Since fm + eL ∈ Tm−1 , the inductive hypothesis implies λ = µ = 0. Therefore ean = 0. Since bn−1 = ean , then bn−1 = 0. Suppose that bn = bn−1 = . . . = bn−(r−1) = 0, for r > 1, we show that bn−r = 0. We have hence

d( bn−r an )

c2n−1 = bn−r d(an ) − d(bn−r )an = 0,

= 0. Thus bn−r = kan for some k ∈ Ker d. We have

c2n−(r+1) = bn−(r+1) d(an ) − d(bn−(r+1) )an + bb−r d(an−1 )

− d(bn−r )an−1 + rd(fm )bn−r an = 0.

We substitute kan for bn−r and we obtain:

bn−(r+1) d(an ) − d(bn−(r+1) )an

+ kan d(an−1 ) − kan−1 d(an ) + rka2n d(fm ) = 0.

Dividing by a2n , we get bn−(r+1) d(an ) − d(bn−(r+1) )an an d(an−1 ) − an−1 d(an ) +k +rkd(fm ) = 0, 2 an a2n hence   bn−(r+1) an−1 d +k + rkfm = 0. an an It follows that there exists l ∈ Ker d such that bn−(r+1) − lan + kan−1 + rkan fm = 0.

This identity has the form λ + µfm = 0, where λ, µ ∈ Km−1 and fm ∈ Fm−1 . The inductive hypothesis implies, in particular that rkan = 0. Since bb−r = kan and since the characteristic of K is 0, then bn−r = 0. (6) Suppose that T0 ∩ K0 6= φ and let g be an element of this intersection, then g ∈ K0 , and there exist n ≥ 1 and (c1 , . . . , cn ) ∈ Ker dn such that Pn i=1 ci fi − g = 0. Therefore f1 , . . . , fn are algebraically dependent over K0 . Suppose that T0 ∩ K0 6= φ. To conclude for the algebraic independence over K0 of the fi , it is sufficient to prove that for any n m ≥ 1 fm is transcendental over Km−1 , that is 1, fm , . . . , fm are linearly independent over Km−1 for any n ≥ 1. By (5) this claim is true, hence the fi are algebraically independent over K0 .

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(7)(a) Here Ker d = F and T0 = {λf (x), λ ∈ F ∗ }. We use (6). It is clear that for any λ ∈ F ∗ that λf (x) 6∈ K0 since f (x) ∈ K\K0 . Therefore T0 ∩ K0 = φ and the conclusion follows. (b) We must show that f (x) 6∈ K0 and d(f (x)) ∈ K0 . We have d(f (x)) =

∞ X

xn =

n=0

1 , 1−x

hence d(f (x)) ∈ F (x). Suppose that f (x) ∈ F (x) and let u(x), v(x) ∈ F [x] such that f (x) = u(x) v(x) and gcd(u(x), v(x)) = 1, then d(f ) =

1 d(u(x))v(x) − d(v(x))u(x) = . 1−x v 2 (x)

We deduce that v 2 (x) = (1 − x)(d(u(x))v(x) − d(v(x))u(x)). Suppose that there exists P (x) ∈ f [x] such that P (x)|v(x), P (x) 6= 1 − x and P (x) is irreducible over F , then from the above identity, we conclude that P (x)|d(v(x)). Denote by νp , the valuation at p and let e = νp (v(x)), then νp [d(u(x))v(x) − d(v(x))u(x)] = e − 1. But νp (v 2 (x)) = 2e and we have reached a contradiction. From the identity, we see that 1 − x|v(x). Therefore v(x) = λ(1 − x)e with λ ∈ F and e ≥ 1. Using valuations at 1 − x as previously we obtain a contradiction, hence f (x) 6∈ F (x). (8)(a) We may suppose that α = 0. Then g(x) has the form: g(x) = axk

∞ X

bj x j ,

j=0

where bj ∈ F and b0 = 1 and k an integer. If k ≥ 0, then obviously the residue of g 0 (x) at 0 is 0. If k < 0, then   ∞ X 0 k−1 k+j−1 . g (x) = a b0 x + (k + j)bj x j=1

If k + j − 1 = then j = −k and a−1 = (k + j)bj = 0.

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(b) We apply (6). Suppose that for some positive integer n, there exist c , c , . . . , cn ∈ Ker d = F and g(x) ∈ K0 = F such that Pn 1 2 Pn i=1 ci fi (x) = g(x), then i=1 ci d(fi (x)) = d(g(x)). By (8) (a) For any j = 1, . . . , n the residue of d(g(x)) at αj is Pn zero, hence i=1 ci rij = 0 for j = 1, . . . , n. This is a homogeneous linearly system of equations in c1 , . . . , cn , whose determinant is by assumption, non zero. Hence c1 = · · · = cn = 0. We deduce that T0 ∩K0 = φ and then by (6), f1 , f2 , . . . are algebraically independent over F (x). (9) We show that the assumptions of (8) (b) are fulfilled. We have d 1 d (gi (x)) = fi (x), dx 1 − gi (x) dx

d hence dx (fi (x)) ∈ F (x). For any i ≥ 1 select a root αi of gi (x) in ¯ F , then the preceding computations show that αi is a simple pole of d d dx (fi (x)) and αi is not a pole of dx (fj (x)) for any j 6= i. Therefore the d residue of dx (fj (x)) at αi is 0 if i 6= j. We show that the residue of d dx (fi (x)) at αi is equal to the multiplicity of αi as a root of gi (x). Set g1 (x) = (x − αi )mi Q(x), where mi is a positive integer, Q(x) ∈ F¯ [x] and Q(αi ) 6= 0. Then

d d g1 (x) = mi (x − αi )mi −1 Q(x) + (x − αi )mi Q(x), dx dx

hence d mi Q(x) + (x − α) dx Q(x) d d (f1 (x)) = (gi (x))/gi (x) = . dx dx (x − α)Q(x)

d fi (x) is equal to mi and Therefore the value at x = α of (x − α) dx d then the residue of dx (fi (x)) at αi is equal to mi . It follows that Qn Det(rij ) 1≤i≤n = i=1 mi 6= 0, and the hypotheses of (8) are satisfied. 1≤j≤n

Therefore the fi (x)(i ≥ 1) are algebraically independent over F (x).

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Notations

a|b a-b pe || a νp (a),

νP (I)

gcd(a1 , . . . , an ) lcm(a1 , . . . , an ) a ≡ b (mod q) a ¯, f (x)


n m q (p)

A\B max(A) |A| N Z Q R C Q h(K) K[x] K(x)

a divides b a does not divide b pe is the exact power of p dividing a p-adic valuation of the integer a and P-adic valuation of the ideal I respectively greatest common divisor of the integers a1 , . . . , an least common multiple of the integers a1 , . . . , an a is congruent to b modulo q class modulo an ideal of the integer a and class of the polynomial f (x) respectively binomial coefficient quadratic character of q modulo p complementary set of B in A supremum of the elements of A cardinal of the set A set of non negative rational integers set of rational integer set of rational numbers set of real numbers set of complex numbers set of algebraic numbers class number of the number field K ring of polynomials with coefficients in K field of rational functions over K

439

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Galois Theory and Applications: Solved Exercises and Problems

K[[x]] K((x)) Sn An ∅ En Ker φ, Im φ, Rank φ E'F Det(A) At Mn (K) GLn (K) DimK E Dim R Frac (R) [E : K] [E : K]s [E : K]i Trdeg E/K (G : H) TrE/K (α) NE/K (α) Gal(E, K), Gal(E/K) Inv(H) K(α) K(A) E.F deg f Disc(f ) degy f cont(f )

10939-main

ring of formal power series with coefficients in K field of formal series with coefficients in K symmetric group operating on a set of n elements alternating group operating on a set of n elements empty set Cartesian product of the set E with itself n times or set of xn when x runs in E kernel, image and rank of φ respectively E is isomorphic to F determinant of A transpose of the matrix A set of all n × n matrices with coefficients in K linear group of K n dimension over K of E Krull dimension of R fraction field of the integral domain R degree of the extension E/K degree of separability of the extension E/K degree of inseparability of the extension E/K transcendence degree of E/K index in G of the subgroup H trace of α in E/K norm of α in E/K Galois group of the extension E/K field or ring of invariants of the group H field obtained by adjoining α to K field obtained by adjoining the elements of A to K composite field of E and F degree of the polynomial f discriminant of the polynomial f degree of the multivariate polynomial f relatively to y content of the polynomial f

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Notations 0

f (x) f (k) (x) ∂φ ∂x

Resx (f (x), g(x)) Irr(α, K, x) Char(α, K, x) I(α) DiscL/K (α1 , . . . , αn ) N E KF

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derivative of f derivative of order k of f partial derivative relatively to x of φ resultant of the polynomials f (x) and g(x) minimal polynomial of α over K characteristic polynomial of α over K index of α discriminant of the elements of L, α1 , . . . , αn tensor product of E and F over K

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Bibliography

Alaca, S. and Williams, K. S. (2003). Introductory Algebraic Number Theory, (Cambridge University Press). Ayad, M. (1997). Th´eorie de Galois 122 Exercices corrig´es, Niveau I, (Ellipses, Paris). Ayad, M. (1997). Th´eorie de Galois 115 Exercices corrig´es, Niveau II, (Ellipses, Paris). Baker, A. (2012). A Comprehensive Course in Number Theory, (Cambridge University Press). Bewersdorf, J. and Kramer, D. (2006). Galois Theory for Beginners: A Historical Perspective, (American Mathematical Society, Providence). Borevich, Z. I. and Shafarevich I. R. (1966). Number Theory, (Academic Press). Bourbaki, N. (1950). Alg`ebre Chap. V (Corps commutatifs), (Hermann, Paris). Clark, P. L. (1993). ftp://math.uga.edu/~pete/FieldTheory.pdf. Cox, D. A. (2012). Galois Theory, 2nd edn. (Wiley). Escofier, J. P. and Schneps, L. (2012). Galois Theory, (Graduate Texts in Math.) (Springer). Flath, D. F. (1989). Introduction to Number Theory, (Wiley, New York). Hancock, H. (1932). Foundations of the Theory of Algebraic Numbers, Vol. 2, (Dover Pub. Inc., New York). Hasse, H. (1980). Number Theory, (Springer-Verlag). Howie, J. M. (2010). Fields and Galois Theory, (Springer). Hua, L. K. (1982). Introduction to Number Theory, (Springer-Verlag, New York). Ireland, K. and Rosen I. M. (1990). Classical Introduction to Modern Number Theory, (Springer-Verlag). Janusz, G. J. (1973). Algebraic Number Fields, (Academic Press, New York). Jarvis, F. (2014). Algebraic Number Theory, (Springer). Lang, S. (1965). Algebra, (Addison-Wesley). Lang, S. (1994). Algebraic Number Theory, (Springer-Verlag, New York). Lidl, R. and Niederreiter H. (2008). Finite fields (Encyclopedia of Mathematics and its Applications), (Cambridge University Press). Marcus, D. A. (1977). Number fields, (Springer-Verlag).

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Mollin, R. A. (2011). Algebraic Number Theory, 2nd edn. (Chapman and Hall CRC). Murty, M. R. and Esmonde J. (2006). Problems in Algebraic Number Theory, 2nd edn. (Springer). Narkievicz, K. (1990). Elementary and Analytic Theory of Algebraic Numbers, (Springer-Verlag). Postnikov, M. M. (2004). Foundations of Galois Theory, (Dover Publications). Ribenboim, P. (2001). Classical Theory of Algebraic Numbers, (Springer). Roitman, M. (1997). On Zsigmondy primes, (Proc. A. M. S., 125, 1913–1919). Rotman, J. (2013). Galois Theory, 3rd edn. (Springer). Samuel, P. (1970). Algebrai Theory of Numbers, (Houghton-Mifflin). Schinzel, A. (2000). Polynomials with special regard to reducibility, (Cambridge University Press). Small, C. (1991). Arithmetic of finite fields, (Marcel Dekker). Stewart, I. N. (2015). Galois Theory, (CRS Press). Stewart, I. and Tall D. (2015). Algebraic Number Theory and Fermat’s Last Theorem, 4th edn. (Chapman and Hall CRC). Swallow, J. (2004). Exploratory Galois Theory, (Cambridge University Press). Weil, A. (1967). Basic Number Theory, (Springer-Verlag). Weintraub, S. (2007). Galois Theory, (Springer).

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