Radical extensions and Galois groups

een wetenschappelijke proeve op het gebied van de Natuurwetenschappen, Wiskunde en Informatica

Proefschrift

ter verkrijging van de graad van doctor aan de Radboud Universiteit Nijmegen op gezag van de Rector Magnicus prof. dr. C.W.P.M. Blom, volgens besluit van het College van Decanen in het openbaar te verdedigen op dinsdag 17 mei 2005 des namiddags om 3.30 uur precies

door Mascha Honsbeek

geboren op 25 oktober 1971 te Haarlem Promotor: prof. dr. F.J. Keune Copromotores: dr. W. Bosma dr. B. de Smit (Universiteit Leiden) Manuscriptcommissie: prof. dr. F. Beukers (Universiteit Utrecht) prof. dr. A.M. Cohen (Technische Universiteit Eindhoven) prof. dr. H.W. Lenstra (Universiteit Leiden) Dankwoord

Zoals iedereen aan mijn cv kan zien, is het schrijven van dit proefschrift niet helemaal volgens schema gegaan. De eerste drie jaar wilde het niet echt lukken om resultaten te krijgen. Toch heb ik in die tijd heel jn samengewerkt met Frans Keune en Wieb Bosma. Al klopte ik drie keer per dag aan de deur, steeds maakten ze de tijd voor mij. Frans en Wieb, dank je voor jullie vriendschap en begeleiding. Op het moment dat ik dacht dat ik het schrijven van een proefschrift misschien maar op moest geven, kwam ik in contact met Bart de Smit. Hij keek net een beetje anders tegen de wiskunde waarmee ik bezig was aan en hielp me daarmee weer op weg. Samen met Hendrik Lenstra kwam hij met het voorstel voor het onderzoek dat ik heb uitgewerkt in het eerste hoofdstuk, de kern van dit proefschrift. Daarnaast heeft Bart mij ook vanaf dat moment als copromotor begeleid. Bart, Hendrik, dank je wel voor jullie steun en goede ideeen. Daarnaast wil ik ook graag mijn leescommissie bedanken. Hendrik, Arjeh en Frits, dank je wel voor jullie inzet en goede suggesties. Ook vanuit mijn vrienden en ouders heb ik veel steun ontvangen. Voorzichtig in- formeerden ze soms of het eind al in zicht was. Speciaal wil ik Arnoud, Kirsten en Timo noemen. Pap, mam en al die anderen die er met mij in zijn blijven geloven, dank je wel. Verder ontwierp Jan een schitterende kaft voor dit proefschrift. Helaas kreeg ik geen toestemming om deze Nijntje-afbeelding te gebruiken.

Tot slot, Richard, wil ik jou bedanken voor alle steun en hulp. Zonder jou had het proefschrift er niet zo mooi uitgezien. Mascha

III IV Contents

Dankwoord III

Table of Contents V

Introduction VII

1 Maximal radical extensions 1 1.1 Preliminaries ...... 2 1.2 Kummer theory ...... 5 1.3 Maximal Kummer extensions ...... 9 1.4 Separating the roots of unity ...... 11 1.5 Galois action on radicals ...... 15 1.6 The maximal Kummer subextension in L ...... 19 1.7 Determining the Galois group ...... 21 1.8 Computing the eld D ...... 27

2 Subelds of radical extensions 33 2.1 Known results ...... 34 2.2 First implication ...... 37 2.3 The Galois group ...... 41 2.4 Subgroups ...... 46 2.5 Conclusion of the proof ...... 48

3 Ramanujan’s nested radicals 57 3.1 The problem and its history ...... 58 3.2 Denesting condition ...... 59 3.3 A special polynomial ...... 62 3.4 Comparison to Ramanujan’s method ...... 65

4 Nested radicals of depth one 69 4.1 Counterexample to Zippel’s conjecture ...... 70 4.2 Denesting without roots of unity ...... 71

V VI CONTENTS

4.3 Denesting allowing roots of unity ...... 73

5 Rogers-Ramanujan continued fraction 77 5.1 Introduction ...... 79 5.2 The modular function eld of level 5 ...... 80 5.3 Galois theory ...... 85 5.4 The ray class eld H5 ...... 90 5.5 Nested radicals ...... 94

Bibliography 97

Nederlandse samenvatting 101

Curriculum Vitae 105 Introduction

The use of symbolic computing is one of the characteristics of a computer algebra package. For example, the number √2 is represented as a symbol with the property that its square is 2. This enables us to do exact calculations. Compared to numerical calculations, there also are some drawbacks concerning computations with radical expressions. For example, algebraically, one will not be able to distinguish between the positive and the negative square root of 2. If one wants to do a computation for one specic choice of these roots sometimes a numerical estimate is necessary. Moreover, we often get complicated expressions although there is a simpler expression for the number that we are interested in. The simplication of radical expressions was an important motivation for the research in this thesis. Look at 3 3 = √5 + 2 √5 2, where, as in the rest of this introqduction, whenq we take a root of a positive real number, we mean the real positive root of this number. A numerical estimate of is given by

3 3 √5 + 2 √5 2 = 1.618033988 . . . 0.618033988 . . . . q q We see that is approximately equal to 1. Using symbolic computations we prove that actually equals 1. One can check that is a root of the polynomial x3 + 3x 4 = (x 1)(x2 + x + 4), which has only a single real root. Therefore has to equal 1. Intuitively the notation 1 is simpler than 3 √5 + 2 3 √5 2. To make this precise we introduce the notion of radicals and nesting depth. p p Let K be a eld of characteristic 0 and let K be some xed algebraic closure of K. We say that K is a radical over K if there exists some n Z>0 with n K. We denote by∈K(0) the eld K itself and inductively dene the∈elds K (k) for ∈k 1 as (k) (k 1) n (k 1) K = K ( K with K for some n Z ). { ∈ ∈ ∈ >0} We say that K is a nested radical over K if there exists some k Z>0 with (k) ∈ (k) (k ∈1) K . We say that has nesting depth k over K if K K . ∈ ∈ \ VII VIII

It is desirable for the computer to produce the simplest possible expression for the outcome of a computation with nested radicals. Richard Zippel [40] gave the following identity

6 7√3 20 19 = 3 5/3 3 2/3. (1) q Following [5], [20], [21], [28], [40] we say thatp the righp t hand side of this equation is a denesting of the left hand side; the equality shows that some element that is clearly contained in K(2) is even contained in K(1). The element K is a nested radical if and only if there exists a Galois extension L/K with L and∈ a chain of eld extensions ∈ K = K K . . . K = L, 0 1 t such that for 1 i t the eld Ki is generated by radicals over Ki 1. For example, if we take the nested radical = 6 7√3 20 19 from equality (1) over Q, then an obvious choice for the rst elds in this chain would be p

6 6 Q Q √3 20 Q √3 20, 7√3 20 19 = Q 7√3 20 19 . ³ ´ µ q ¶ µq ¶ In general we have to adjoin all conjugates of to obtain a Galois extension. In this case however, it turns out that Q 6 7√3 20 19, is Galois over Q. One can 6 check this using a computer algebra pac³pkage or by using´equality (2) below. If we adjoin suciently many roots of unity in the rst step of the chain, then we see that Ki/Ki 1 is abelian for all i 1. In our example:

6 Q Q( ) Q √3 20, Q 7√3 20 19, . 6 6 6 ³ ´ µq ¶ As K is a eld of characteristic 0, an element is a nested radical if and only if the Galois group of the normal closure of K() over K is solvable ([22] Chapter VI). Given a chain of eld extensions K = K0 K1 . . . Ks, for which Ki/Ki 1, for i Z>0, is generated by a radical, and an element Ks of nesting depth t s Susan∈ Landau [20] provides an algorithm that computes∈ a symbolic representation for in K(d), where d is the length of the derived series of the Galois closure of Ks/K. We have t d t + 1. Carl Cotner [7]impro ved this result. He showed that the nesting depth of is computable and gave an algorithm to nd a symbolic representation of minimal nesting depth. For specic types of nested radicals it is possible to give a simple algorithm to compute a symbolic representation of minimal depth. This we will do in chapters 3 and 4 for certain elements of K(2). Given a subeld L of K(1), an element of L, and n (1) some n Z 2 we will prove that √ K implies that satises a strong condition ∈ ∈ INTRODUCTION IX which may be phrased as ‘ is almost an n-th power in L’. Indeed, equality (1) holds because ‘7√3 20 19 is almost a sixth power in Q(√3 20 )’; we have 1 6 7√3 20 19 = √3 20 2 (2) 144 and ³ ´ 1 6 √3 20 2 = 3 1/12 √3 20 √3 8 = 3 5/3 3 2/3. r144 To prove our claim³ in general´ wpe use Galois³ groups´ ofpradical extensions.p If the ground eld K contains the appropriate roots of unity and the extension is nite then we have a Kummer extension and the Galois group is well known. We discuss Kummer theory in chapter 1. Moreover, we discuss non-nite Kummer extensions and describe the Galois group for certain radical extensions for which the ground eld K does not contain the appropriate roots of unity. We show that the Galois group of n L = K( K with K for some n Z ) { ∈ ∈ ∈ >0} over K can be embedded in a semidirect product of two explicitly described groups. An important reason to study Galois groups of radical extensions is to gain insight in the structure of subelds of the extension. Let L/K be a Kummer extension generated by radicals 1, 2, . . . over K. By Kummer theory all subelds of L = K(1, 2, . . .)/K are generated by a subgroup of W = K, 1, 2, . . . . This also holds for pure radical extensions. (A eld extension L/K his pure if everyiprimitive p-th root of unity contained in L is in fact contained in K, for prime numbers p and for p equal to 4.) However, this is not true in general. 1 For example, the Galois extension Q(7)/Q has two non-trivial subelds, Q(7 +7 ) and Q(√ 7 ), which are not generated by a power of 7. In chapter 2 we study the extension Q()/Q for some radical C and give a both necessary and sucient condition for the subelds of Q() to∈be generated by a subgroup of Q, . This gives us also some non-pure extensions for which all h i subelds are generated by a subgroup of Q, , like Q()/Q where is a complex number with 4 = 3. h i Above we considered the simplication of nested radicals. Another problem con- cerning nested radicals is nding a representation of some element as a radical expres- sion. The element can, for example, be given as the root of a solvable polynomial or can be given by an analytic expression. For singular values of the Rogers-Ramanujan continued fraction we determine radical expressions in chapter 5. Below we give a more detailed description of this thesis in an overview per chap- ter.

Chapter 1 For ease of exposition we consider a perfect eld K in this introduction. In chapter 1 we will have a slightly weaker condition on K. Let K be a xed algebraic closure X of K. We dene the group of radicals over K as

n A = K : K for some n Z . { ∈ ∈ ∈ >0} In chapter 1 we describe the Galois group of the extension K(A)/K. In the case K = Q and K C , the group A is the direct product of (C ), the group of roots of unity inC , and the group R of real positive radicals in A. The group AutQ (A) of group automorphisms of A that are the identity on Q is isomorphic to Hom(Q>0, ˆ) o Aut((C )), where ˆ denotes limn n(C ), the inverse limit of the groups of n-th roots of unity in C for all n Z>0, with respect to the maps (C ) (C ): n/m, for all m n Z ∈. The map n → m 7→ | ∈ >0

AutQ (A) = Hom(Q , ˆ) o Aut((C )) → >0 is given by ( √n a ) a n , (C ) , 7→ 7→ √a n Z | ³ µ ¶ ∈ >0 ´ where √n a denotes the positive real root of a. The Galois group Gal(Q(A)/Q) can be embedded in this group . We know that √x Q((C )) Q(R) for all x ∈ ∩ ∈ Q>0. This gives a condition that turns out to determine the image of Gal(Q(A)/Q) completely. Any homomorphism f from Q>0 to ˆ induces a map f˜ from Q>0 to 1 by composition with the projection map ˆ 1 . In section 1.7 we nd h i → h i a Gal(Q(A)/Q) (f, z) with f˜(a) = for all a Q , ' ∈ z ∈ >0 n ³ ´ o a where z denotes a generalised Jacobi symbol. In chapter 1, we also give such a description for Gal(K(A)/K), where K is a perfect¡eld¢ and n A = K : K for some n I , h ∈ ∈ ∈ i with I = an arbitrary subset of Z>0. In general we cannot give a splitting of A, but we 6 can∅ give a splitting of A/(K). We x a choice for this splitting and follow the lines of the proof sketched above to give a similar description of Gal(L/K).

Chapter 2 In chapter 2 we determine all radicals C for which the subelds of Q() are all generated over Q by a power of . ∈ For instance, if we take for a primitive fth root of unity, 5, then there exist a subeld Q(√5 ) of Q() that is not generated by a power of . If we take = 5 √5, then all subelds of Q()/Q are generated by a power of . To nd the subelds of Q() we study the Galois group of Q(n, ) over Q, where n is the least positive integer with n Q. The group Gal(Q( , )/Q) can ∈ n be embedded in Z/nZ o (Z/nZ) and its image G is determined by the intersection INTRODUCTION XI

d eld Q(n) Q(). For every radical C this eld is of the form Q( ) for some divisor∩ d of n. The number of n-th∈roots of unity in this intersection eld, #n Q(n) Q() , splits the problem in dierent cases. We identify some radicals for which eviden∩ tly subelds exist that are not generated by a power of . It turns out that¡ there are¢no other radicals for which such subelds exist. This gives us the following necessary condition on for all subelds of Q() to be generated by a power of :

(i) # Q( ) Q() 2 and we have 6 - n or √ 3 / Q, or n n ∩ ∈ h i (ii) # ¡Q( ) Q()¢ = 3 or n n ∩ (iii) # ¡Q( ) Q()¢ = 4 and 1 + i Q, or n n ∩ ∈ h i (iv) # ¡Q( ) Q()¢ = 6 and √ 3 Q, or n n ∩ ∈ h i

(v) # ¡Q( ) Q()¢ = 10 and both 4 - n and √5 Q, . n n ∩ ∈ h i H If H G is¡ such that Q(¢n, ) = Q(), then there is a one-to-one correspondence between subgroups of G containing H and subelds of Q(). We prove in each of the ve cases above that all subgroups of G containing H are of the form G dZ/nZ o ∩ (Z/nZ) for some divisor d of n. These groups correspond to the elds Q(n/d), ¡ which showes that the above condition is also sucient. ¢ Chapter 3 At the beginning of the 20th century Ramanujan ([1]) gave the following denesting formula:

m 3 4(m 2n) + n√3 4m + n q p = 1 3 (4m + n)2 + 3 4(m 2n)(4m + n) 3 2(m 2n)2 . 3 ³p p p ´ Let and be integers and dene = √3 + √3 . In chapter 3 we prove, using the normal closure of Q()/Q, that can be denested if and only if √3 + √3 equals p 3 k 3 l 2 3 3 √ √ f e for some f Q, k, l Z>0 and e in Q(√, √ ). We use this to showthat can only be denested∈ if there∈ exist integers m and n with (4m + n)n3 = (3) 4(m 2n)m3 or if / is a cube in Q. 5 For example, if we take m = n = 1, then the quotient above equals 4 . Hence we have the denesting

√3 5 √3 4 = 1 (√3 2 + √3 20 √3 25 ). 3 q XII

It can be shown that there do not exist integers m, n satisfying equation (3) if, for instance, we take = 2 and = 3. Hence √3 2 + √3 3 cannot be denested. p Chapter 4

Let K be a eld, let 1, . . . , t be radicals over K and dene L = K(1, . . . , t). In this chapter we give a necessary condition for a nested radical of the form √n with n Z and L K to be contained in K(1). ∈ >0 ∈ \ We dene the eld K K as the smallest extension over K containing all roots of unity in K . At the∞beginning of this introduction we described the Kummer correspondence between subelds and subgroups of the multiplicative group of gen- erating radicals. We use this correspondence to show that √n K(1) implies that ∈ n si there exist w K L, e L and integers s1, . . . , st with = w e i . ∈ ∞ ∩ ∈ i (1) √n If both the eld L and the nite subextension of K containing Q are pure then we even have w K. In general w is not contained in K. We provide an example where w is con∈tained in (K L) K. ∞ ∩ \

Chapter 5

This chapter consists of an article co-authored with Alice Gee. It was accepted by the Ramanujan Journal in May 2001 and can also be found in [10]. We determine nested radicals for singular values of the Rogers-Ramanujan continued fraction

n 1 ∞ n ( ) R(z) = q 5 1 q 5 , n=1 Y ¡ ¢ 2iz n where z is an element of the complex upper half plane, q is e and 5 denotes the Legendre symbol. ¡ ¢ First we prove that R(z) is a modular function of level 5. In fact, the eld F5 of functions of level 5 over Q(5) equals Q(R, 5). Let be an element of the upper half plane with [1, ] a Z-basis of some imaginary quadratic order. We denote by H5 the eld of function values in of elements of F5. The rst main theorem of complex multiplication states that H5 is abelian over Q(). Therefore, all elements of H5, and especially R(), are nested radicals over Q() and consequently over Q since [Q() : Q] equals 2. To compute the radical expression for these numbers we use Lagrange resolvents. For example, let i be a primitive 4-th root of unity and assume that K is a eld containing i. If = R() generates a cyclic extension of degree 4 over K, then we use the Galois group of K()/K to compute conjugates 1 = , 2, 3 and 4 of . INTRODUCTION XIII

We dene

l0 = 1 + 2 + 3 + 4, l = + i i , 1 1 2 3 4 l = + , 2 1 2 3 4 l = i + i . 3 1 2 3 4 2 4 4 Then we have 4 1 = l0 + l1 + l2 + l3 and l0, l2, l1 and l3 are elements of K. A radical expression for over K is

1 l + l2 + 4 l4 + 4 l4 . 4 0 2 1 3 µ q q q ¶ In general we form a chain of subextensions of H5/Q. A radical expression for over Q is now derived by doing similar computations as above recursively in every subextension in the chain until we end up in Q. In chapter 5 we use that, for our choice of , the powers are elements of Z, the ring of integers of Q. We compute a suciently precise numerical estimate for = R() to nd these integers. In every step the nesting depth increases by 1. We give radical expressions that can be uniquely interpreted: we only take roots of positive real numbers for which we choose the unique positive real root. XIV Chapter 1

Galois groups of maximal radical extensions

Kummer theory gives a description of the Galois group of a radical extension over a ground eld containing the appropriate roots of unity. In this chapter we give a similar description for certain radical extensions of a ground eld that is not required to contain those roots of unity. For a eld K we x an algebraic closure K . We denote by K the group of invertible elements in K and by (K) the group of roots of unity in K. For a subgroup A of K we denote by (A) the subgroup of roots of unity in A. Let K be a eld of characteristic char(K). Then the group of radicals over K is given by

n A = K : K for some n Z with char(K) - n . { ∈ ∈ ∈ >0 } This set A is a multiplicative group and the radical extension L = K(A) over K is a Galois extension. In this chapter we describe its Galois group Gal(L/K). In sections 1.1, 1.2 and 1.3 we give the necessary denitions and recall the main results from Kummer theory. In section 1.4 we show that there exists a subgroup C of A with A/(K) = (K )/(K) C/(K). The group C is not uniquely determined. We x a choice for C and dene the elds M = K((A)) and F = K(C). In section 1.5 we study how Gal(L/K) acts on elements of M and F . This gives a subgroup Z of Zˆ isomorphic to Gal(M/K) and a Kummer-like homomorphism group J. We will show that Gal(L/K) can be embedded in a semidirect product = J o Z of these groups. In order to determine the image of Gal(L/K) in we determine the intersection D of the elds M and the maximal Kummer extension inside F over K. In sections 1.6

1 2 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS and 1.7 we construct the following diagram.

= J o Z J

Z Gal(D/K)

The image of Gal(L/K) in consists of the elements (f, z) J o Z for which the images of f and z in Gal(D/K) are equal. Finally, in section∈ 1.8 we prove some properties of the eld D and compute it in some simple examples. Throughout the chapter we use the case K = Q as a standard example. The group Gal(Q(A)/Q) was described before by H.W. Lenstra in the exercises of the lecture notes ‘Galois Theory for Schemes’ [24].

1.1 Preliminaries

In this section we give some basic denitions and results. We start with the basics about pronite groups. For details we refer to [12]. Given a directed set I and an inverse system of nite groups (Gi)i I we obtain a pronite ∈ group G = limi I Gi. This is a topological group under the relative topology inside ∈ the product i I Gi, with the discrete topology on the groups Gi. We denote an ∈ element g of G by g = (gi)i I . A pronite group is compact and Hausdor. If the Q ∈ groups Gi are all abelian, then G is abelian as well. For all n, n0 Z with n n0 we consider the reduction maps of the abelian ∈ >0 | groups Z/n0Z Z/nZ. We denote by Zˆ the projective limit of Z/nZ for all n Z>0 with respect to→these homomorphisms. Moreover, we dene for a prime num∈ber p the group n Zp = lim Z/p Z. n Z ∈ >0

In fact Zˆ and Zp, for all primes p, are rings and we have a canonical isomorphism of rings Zˆ Z , where p ranges over the prime numbers. Similarly the group Zˆ ' p p is the projective limit limn Z (Z/nZ) with reduction maps (Z/n0Z) (Z/nZ) Q ∈ >0 → for all n, n0 with n n0 in Z . | >0 Let (Gi)i I be an inverse system of nite abelian groups for some index set I. ∈ ˆ ˆ We show that G = limi I Gi is a Z-module. Let a be an element of Z; we write ∈ a = (an)n Z . For all i we dene ni = #Gi. If h is an element of the group Gi for ∈ >0 a an some i I, then we dene h = h i . With this action Gi is a topological Zˆ-module with discrete∈ topology. As G inherits the action on its components we have

a ani g = gi i I ∈ ¡ ¢ for all (gi)i I G. Hence G is a topological Zˆ-module as well. ∈ ∈ 1.1. PRELIMINARIES 3

Denition 1. By P we denote the set of prime numbers. A Steinitz number is a formal expression m = pm(p), p P Y∈ where m(p) is an element of 0, 1, 2, . . . , for all p P . { ∞} ∈ The Steinitz numbers for which we have m(p) < for all primes p and m(p) = 0 for almost all p are identied with positive integers∞by multiplying out the formal expression. m(p) n(p) Let m = p P p and n = p P p be Steinitz numbers. We say that m divides n and∈write m n if m(p) ∈ n(p) for all p P . Moreover, we dene for a set I of QSteinitz numb| ers the greatestQ common divisor∈ and the least common multiple as

minm∈I m(p) supm∈I m(p) gcdm I (m) = p and lcmm I (m) = p . ∈ ∈ p P p P Y∈ Y∈ ˆ ˆ Let m be a Steinitz number, then we dene mZ = n m nZ, where the intersec- tion ranges over the integers n dividing m. | T Proposition 2. The map m mZˆ from the set of Steinitz numbers to the set of closed subgroups of Zˆ is bijective.7→

Proof. For every integer n, the subgroup nZˆ of Zˆ is closed because it is the kernel of the continuous projection map Zˆ Z/nZ. As an intersection of closed groups, → also mZˆ is a closed subgroup of Zˆ for every Steinitz number m. Let S be a closed subgroup of Zˆ and denote, for every integer n, the projection 1 map Zˆ Z/nZ by . Let x be an element of ( (S)). As the maps → n n n n n : S n(S) are surjective, there exists for every n Z>0 an element sn S → T ∈ ∈ with n(x) = n(sn). It is easy to see that Zˆ equals lim Z/n!Z. Since S is a n Z>0 ˆ ∈ closed subgroup of Z, the limit x of the sequence (sn!)n is also contained in S. It 1 follows S = ( (S)) . Writing (S) = d Z/nZ for some integer d n one n n n n n n | obtains S = mZˆ for the Steinitz number m that equals lcmn Z (dn). T ∈ >0 As a pronite group G is a continuous topological Zˆ-module, the annihilator ˆ AnnZˆ(G) is closed. Hence, by proposition 2, it equals mZ for some Steinitz num- ber m.

Denition 3. The exponent of a pronite group G is the Steinitz number exp(G) for which we have

exp(G) Zˆ = Ann (G) = a Zˆ : ga = 1 for all g G . Zˆ { ∈ ∈ } 4 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Proposition 4. Let G = limi I Gi for nite abelian groups Gi. If the projection map G G is surjective for ∈all i I then we have → i ∈

exp(G) = lcmi I (exp(Gi)) . ∈ Proof. Let a Zˆ. By denition we have a exp(G) Zˆ if and only if ga = 1 holds for all g G.∈As G maps surjectively to the∈G this is also equivalent to ∈ i ga = 1 for all g G and all i I. ∈ i ∈ So, we have the following equivalent statements:

a exp(G) Zˆ a exp(G ) Zˆ for all i I ∈ ⇐⇒ ∈ i ∈ a exp(G ) Zˆ ⇐⇒ ∈ i i I \∈ a lcmi I (exp(Gi)) Zˆ. ⇐⇒ ∈ ∈ This concludes the proof of the proposition.

In the next section we study Kummer extensions. The eld extensions that we consider do not have to be nite. We state two main theorems of innite Galois theory ([12]).

Denition 5. Let K L be an algebraic eld extension. For a group G AutK (L) we write LG for the eld of elements of L that are invariant under G. More explicitly we have LG = l L : (l) = l for all G . { ∈ ∈ } We call L/K a Galois extension if there exists a group G AutK (L) such that K = LG. If L/K is a Galois extension then we dene the Galois group Gal(L/K) to be AutK (L).

Theorem 6. Let K be a eld and L a subeld of K containing K. Denote by I the set of subelds E of L for which E is a nite Galois extension of K. Then I, when partially ordered by inclusion, is a directed set. Moreover, the following four assertions are equivalent:

The eld L is a Galois extension of K. The eld L is normal and separable over K. There is a set F K[x] of separable non-constant polynomials such that L is the splitting eld of F over K in K .

The composite of the elds E in I equals L. 1.2. KUMMER THEORY 5

If L/K is a Galois extension, then there is a group isomorphism

Gal(L/K) lim Gal(E/K) ' E I ∈ mapping Gal(L/K) to ( )E I . ∈ |E ∈ For a Galois extension L/K, this gives Gal(L/K) the structure of a pronite group. For nite normal subextensions E/K of L/K the projection maps to the Galois groups Gal(E/K) are surjective.

Theorem 7. Let L/K be a Galois extension of elds with Galois group G. The maps

ϕ E : E is a subextension of L/K H : H is a closed subgroup of G , { } { }

H dened by ϕ(E) = AutE(L) and (H) = L , are bijective and inverse to each other.

1.2 Kummer theory

In this section we give the denition of a Kummer extension of exponent m, for a Steinitz number m, and give some basic results from classical Kummer theory, following [22] chapter VI.

Denition 8. Let m be a Steinitz number and let K be a eld. We dene the multiplicative group

n (K) = K : = 1 for some n Z with n m . m { ∈ ∈ >0 | } When we consider a xed eld K within some algebraic closure K we will also use the notation m for m(K ). Similarly, we dene for a multiplicative group A the group m(A). Denition 9. For n Z we write w (K) = # (K). For a Steinitz number m ∈ >0 n n we dene the Steinitz number wm(K) as lcmn m wn(K) , where n ranges over the positive integers dividing m. | ¡ ¢ Denition 10. Let m be a Steinitz number. A eld extension L over K is called a Kummer extension of exponent m if it is a Galois extension, wm(K) equals m, and the group Gal(L/K) is a pronite abelian group of exponent dividing m.

m(q) If L/K is a Kummer extension of exponent m = q P q and K has charac- ∈ p teristic p for some prime p then p(K) consists of one single element as x 1 = (x 1)p K[x] holds. Therefore we have p - w (K).QWe conclude m(p) = 0. ∈ m 6 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Proposition 11. A eld extension L/K is a Kummer extension of exponent m if and only if all nite extensions M/K with K M L are Kummer extensions of exponent m. Proof. Let I be the set of all nite subextensions of L over K Assume that L/K is a Kummer extension of exponent m with Galois group G. As L/K is abelian, also M/K is a Galois extension with abelian Galois group GM for all M I and, by theorem 6, we have G = limM I GM . Let M be an element of I. As G∈ maps surjectively to G we have exp(G∈ ) = n for some n N with M M ∈ n m. Using wm(K) = m, we see that the extension M/K is a Kummer extension of|exponent m over K. Now assume that all nite subextensions of L/K are Kummer extensions of exponent m. Then, as L is the composite of the elds M for M I by theorem 6, ∈ also L/K is a Galois extension. Its Galois group is G = limM I Gal(M/K) and thus ∈ L/K is abelian. By proposition 4 the group G is of exponent m and as wm(K) = m holds we conclude that L/K is a Kummer extension of exponent m. Corollary 12. A eld extension L/K is a Kummer extension of exponent m if and only if all extensions M/K with K M L are Kummer extensions of exponent m. Denition 13. Let n be a positive integer and let K be a eld containing a primitive n-th root of unity. Then within a xed algebraic closure K of K we dene K a1/n n for a K as the splitting eld of x a over K. For a subset W of K we write K(W 1∈/n) for the composite of the elds K(a1/n) for all a W . ¡ ¢ ∈ For a Steinitz number m and a eld K with wm(K) = m we dene for all 1/m 1/n W K the eld K(W ) as the composite of the elds K(W ) for all n Z>0 with n m. ∈ | In classical Kummer theory we have the following theorem ([22], VI, 8.2). Theorem 14. Let n be a positive integer and K a eld containing a primitive n-th root of unity. There is a bijection from the set of all subgroups W of K containing n K to the set of all Kummer extensions of exponent n over K inside K , given by W K W 1/n . 7→ It is not¡ hard¢ to check that the inverse of this map is given by sending a Kummer n extension L/K of exponent n to the subgroup L K of K. Using this correspondence we can give a nice description∩ of the Galois group of a Kummer extension of exponent n Z . ∈ >0 Theorem 15. Let n be a positive integer and let L/K be a Kummer extension of n exponent n. If W is the subgroup of K containing K corresponding to L/K as in theorem 14, then there is an isomorphism of topological groups

n ϕ: Gal(L/K) Hom W/K , (K) , → n ¡ ¢ 1.2. KUMMER THEORY 7

n n such that ϕ()(aK ) = ()/ for all L and all a W with = a. Here the n ∈ ∈ group Hom W/K , n(K) has the relative topology in n(K), where the product n ranges over the elements of W/K , and n(K) has the discrete topology. ¡ ¢ Q Proof. Write WL = W and dene for all nite subextensions E/K of L/K the unique n group WE such that K WE WL and E = K(WE). Then it is shown in [22], chapter VI theorem 8.1, that for every nite subextension E/K of L/K the map ϕ n induces an isomorphism from Gal(E/K) to Hom WE/K , n(K) . As we have n n WL/K = lim→E L WE/K this gives the following commuting diagram ¡ ¢

ϕ n Gal(L/K) Hom(WL/K , n(K))

n limE Gal(E/K) limE Hom(WE/K , n(K)).

It follows that ϕ is an isomorphism as well. As Gal(L/K) = limE L Gal(E/K) is n compact and the group Hom W/K , n(K) is Hausdor it remains to prove that the map ϕ is continuous. ¡ ¢ n Dene for all a W the map ϕa : Hom W/K , n(K) n(K) by ϕa(f) = n ∈ →1/n f(a K ) and the projection map : Gal(L/K) Gal K(a )/K . Then ϕ is a ¡ → ¢ continuous if and only if ϕa ϕ is continuous for all a W . Let a in W and let K n 1/n ∈ ¡ ¢ ∈ with = a. We dene a map fa : Gal(K(a )/K) n(K) by fa() = ()/. Then the map ϕ ϕ: Gal(L/K) (K) equals f→ . a → n a a

ϕa ϕ Gal(L/K) n(K)

a fa Gal K(a1/n)/K ¡ ¢ Both a, by denition of a projection map, and fa, as a map between nite groups, are continuous and thus also ϕ ϕ is continuous. a Let K be a eld and let n be a positive integer. In theorem 14 we saw that there is a one-to-one correspondence between Kummer extensions of exponent n n and subgroups W of K containing K . For a Steinitz number m, we can use this correspondence to show that the intersection of Kummer extensions of exponent m can be expressed in terms of the generating radicals of the extensions. Denition 16. Let K be a eld, K an algebraic closure of K and m a Steinitz number not divisible by the characteristic of K. If m is contained in K we dene the multiplicative group of radicals of exponent m over K as

1/m n (K) = K with K for some n Z with n m . { ∈ ∈ ∈ >0 | } 8 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Proposition 17. Let m be a Steinitz number not divisible by the characteristic of K. If K1 = K(W1) and K2 = K(W2) are Kummer extensions of exponent m, for 1/m subgroups W1, W2 of (K) containing K, then we have K K = K(W W ). 1 ∩ 2 1 ∩ 2

Proof. Let n be a positive integer dividing m. As n K holds, there is a bijection n between the set of subgroups of K containing K and the set of subgroups of 1/n n 1/n (K) containing K given by sending K W K to W = a K : an W . Hence, theorem 14 gives a bijection between the set of subgroups{ ∈ of ∈1/n } (K) containing K and the set of Kummer extensions of exponent n over K 1/n inside K . As both these sets are partially ordered we have for K U (K) 1/n and K V (K) that K(U) K(V ) = K(U V ). ∩ 1/n ∩ Dene for i = 1, 2 groups Wi,n = Wi (K) for all n Z>0 dividing m. Then we have ∩ ∈

K(W W ) = K(W W ) 1 ∩ 2 1,n ∩ 2,n n m [| = K(W ) K(W ) 1,n ∩ 2,n n m [| ³ ´ = K(W ) K(W ) = K K . 1 ∩ 2 1 ∩ 2

In chapters 3 and 4 we study denestings of nested radicals. We will use the following reformulation of theorem 14.

Corollary 18. Let n Z>0 and let K be a eld containing a primitive n-th root of ∈ n n n unity. Let , , . . . , K such that we have , , . . . , K. Then is an 1 k ∈ 1 k ∈ element of K 1, . . . , k if and only if there are b K and l1, . . . lk N such that can be written in the form ∈ ∈ ¡ ¢ k li = b i . i=1 Y Proof. We only prove the implication from left to right as the other one is clear. Let L be the eld K(1, . . . , k ) and assume that is contained in L. Let W denote n n n 1/n the subgroup 1 , . . . , k K of L. Then we have K(W ) = L. Hence, by the h i n n remark following theorem 14, we have L K = W . Therefore we nd ∈ ∩ k n n nli = d i , i=1 Y for some d K and l1, . . . lk N. As K contains a primitive n-th root of unity, taking n-th ∈roots and multiplying∈ d by some n-th root of unity gives the corollary. 1.3. MAXIMAL KUMMER EXTENSIONS 9 1.3 Maximal Kummer extensions

In this section K is a eld and K is a xed algebraic closure of K. Furthermore m is a Steinitz number not divisible by the characteristic of K such that wm(K) = m and L/K is the maximal Kummer extension of exponent m over K:

1/m n L = K((K) ) = K K : K for some n Z with n m . { ∈ ∈ ∈ >0 | } In this section we describe ¡the Galois group of the eld extension L/K. ¢ For all n Z we dened the group as the multiplicative group of n-th ∈ >0 n roots of unity in K . For all n, n0 Z>0 with n n0 there is a reduction map ∈n0/n | f 0 : 0 dened by f(x) = x . For this system of groups and maps we n n n → n write ˆ = limn Z n. ∈ >0 The group ˆ is isomorphic to Zˆ as a Zˆ-module if the characteristic of K is 0. It is isomorphic to p P,p=q Zp if the characteristic of K is q > 0. One checks that ∈we ha6 ve, for all Steinitz numbers m, a canonical isomorphism of Zˆ-modules Q m /ˆ ˆ lim n, → n m | where ˆm = ˆ(mZˆ) and where n ranges over the positive integers. ˆ ˆ ˆ ˆ Similarly, we have Z/mZ limn m Z/nZ and (Z/mZ) limn m(Z/nZ). → | → | Theorem 19. Let L/K be the maximal Kummer extension of K of exponent m. There is an isomorphism of topological groups

m Gal(L/K) Hom(K, ˆ/ˆ ), ' given by ( ) a n , 7→ 7→ Ã n n Z>0,n m! µ ¶ ∈ | n where for all a K and all n Z with n m we choose L with = a. ∈ ∈ >0 | n ∈ n 1/n Proof. Dene for all n Z>0 with n m the eld Ln = K((K) ). As L is the composite of the L for∈all n Z with| n m, an application of theorem 6 shows n ∈ >0 | that the group G = Gal(L/K) is topologically isomorphic to limn m Gal(Ln/K). Let n be a positive integer with n m. Using theorems 14 and| 15 we see that | n Gal(Ln/K) is isomorphic to Hom K/K , n(K) under the isomorphism sending n Gal(Ln/K) to the map (aK ()/), where is an element of L with n∈= a. ¡ 7→ ¢ n As every homomorphism from K to n annihilates K we have

Gal(L /K) Hom(K, ). n ' n 10 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Let n, n0 Z such that n n0 then we get the following diagram. ∈ >0 |

Gal(L 0 /K) Hom(K, 0 ) n → n ↓ ↓ Gal(L /K) Hom(K, ) n → n

The map from Hom(K, n0 ) to Hom(K, n) is induced by the identity on K and n0/n the homomorphisms fn0n : n0 n are dened by fn0n(x) = x . It is easy to check that the diagram commutes.→ As the maps fn0n are the same as in the denition of ˆ, we get the following isomorphisms of topological groups:

Gal(L/K) lim Gal(Ln/K) ' n m |

lim Hom(K, n) ' n m | ( ) Hom(K, lim n) ' n m | m = Hom(K, ˆ/ˆ ).

The canonical map ( ) is a homeomorphism as the product topologies on

n and n n m a K a K n m Y| ³ Y∈ ´ Y∈ ³ Y| ´ are the same.

Now we state a lemma that we use in section 1.7. Dene m∞ = p m p∞, where | the product ranges over the prime numbers dividing m and write w = w ∞ (K). Q m Then w is contained in K, the eld K(wm) is contained in L and Gal(L/K(wm)) is a subgroup of Gal(L/K).

Lemma 20. If L/K is the maximal Kummer extension of K of exponent m, then the isomorphism of theorem 19 induces an isomorphism

m Gal L/K( ) Hom K/ , ˆ/ˆ . wm ' w Proof. Let ϕ be the isomorphism¡ from¢ theorem¡ 19. We dene¢ the homomorphism

m ϕ0 : Gal L/K( ) Hom K/ , ˆ/ˆ wm → w ¡ ¢ ¡ ¢ by ϕ0 (a ) = ϕ((a)) for all a K. We show that ϕ0 is well-dened. Let w ∈ a, b K with a/b . Let n Z with n m and let Gal L/K( ) . ∈¡ ¢ ∈ w ∈ >0 | ∈ wm ¡ ¢ 1.4. SEPARATING THE ROOTS OF UNITY 11

Then, for all , L with n = a and n = b we have ()/ = ()/ as ∈ (/) = / K(wm). This gives the∈ following commuting diagram

m Gal(L/K) ϕ Hom(K, ˆ/ˆ )

7→ 0 ϕ m Gal L/K(wm) Hom K/w, ˆ/ˆ , ¡ ¢ ¡ ¢ where (f)(a) = f a . The homomorphism ϕ0 is injective as ϕ is injective. Let w f Hom K / , ˆ/ˆm and dene = ϕ 1((f)). Let a . As is contained w ¡ ¢ w w in∈the kernel of f we have f(a) = 1 /ˆ ˆm. Therefore also∈ (f)(a) = 1 /ˆ ˆm ¡ ¢ ∈ ∈ and consequently for all integer divisors n of m we have (n)/n = 1 for all n L with n = a. Hence it follows () = for all L with n = 1 and n wm.∈So n ∈ | Gal(L/K) is contained in Gal(L/K( )). As we have ϕ0() = f, we conclude ∈ mw that ϕ0 is surjective.

1.4 Separating the roots of unity

We x the following notation for the rest of this chapter: K is a eld; K is an algebraic closure of K; m is a Steinitz number not divisible by the characteristic of K; w is wm∞ (K); A is the multiplicative group of radicals of exponent m over K n dened by A = K : K for some n Z>0 with n m ; L is the eld K(A{). ∈ ∈ ∈ | }

The torsion elements in the groups K and A are precisely the roots of unity. In particular the groups of m∞-torsion in K and A are m∞ (K) = w respectively m∞ (A) = mw. We would like to write A as a direct product with its torsion subgroup as one of the factors. Unfortunately, as we will see in example 22, this is not possible for every abelian group. However, we can write A/w as a direct product with its torsion subgroup as one of the factors. To show this we consider the inclusion map

: / A/K. mw w → Theorem 21. There exists a homomorphism : A/K mw/w such that = 1 holds. → We will use the rest of this section to prove this theorem and to construct the desired splitting of A/w. But rst we show that there exist abelian groups that are not a direct product with the torsion subgroup as a factor. 12 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Example 22. Consider the group p Fp, where p ranges over the prime numbers. The torsion subgroup of F is F . For H = F / F the sequence p p Qp p p p p p Q L ¡ Q ¢ L 0 F F H 0 → p → p → → p p M Y is exact. First we show that H is divisible. Let x = (xp)p P be an element of Fp ∈ p and let n be a positive integer. We dene y = (yp)p P by ∈ Q

xp/n if p - n yp = 0 if p n. ( |

Then we have ny x p Fp. So, for every x H and every n Z>0 there is an element y Hwith∈ x = ny. This shows that∈ H is a divisible∈group. If the sequence splits,∈H is isomorphicL to the kernel of the splitting map F F p p → p p and consequently H is isomorphic to a subgroup of Fp. But this is not possible p Q L as Fp has no non-zero divisible subgroups. We conclude that the sequence above p Q does not split. It is therefore not possible to give a splitting of p Fp with its torsion subgroupQ as one of the factors. Q Inspired by the previous example we also construct a Kummer extension gener- ated by a group of radicals for which the torsion subgroup is not a direct factor. Example 23 is based on a suggestion by B. Poonen.

Example 23. Denote by p a primitive p-th root of unity. Consider the elds K = Q( : p P ) and K( p 2 : p P ) and let R be the generating group of p ∈ p ∈ radicals K , p 2 : p P . Below we will show that R/(R) is not a direct factor p p of R. It hfollows immediately ∈ ithat also (R) is not a direct factor of R. p First we determine (R). Suppose is a root of unity contained in R. Then is n of the form k p 2 p for some element k of K and integers 0 n < p. Let r p p be the product of all primes for which we have np = 0 and let n be the least common multiple of r2Qandpthe order of . Then we have 6

n n n n/p = 1 = k 2 p K. ∈ p P Y∈ Let p be a prime above 2 in Q(k). We normalise the p-valuation in Q(k)/Q by vp(2) = 1 and nd

n vp(k) = vp(2) n n/p = n n/p. p p p P p P X∈ X∈

Suppose that p is a prime with np = 0. Then p divides n. Let l be the largest integer with pl n. We have pl n 6 n/q for all primes q = p. As pl is not a | | q 6 1.4. SEPARATING THE ROOTS OF UNITY 13 divisor of n n/p, we have pl - n n/p. But pl is a divisor of n and we have a p p p contradiction. We conclude that np equals 0 for all primes p and that (R) equals P (K) = p : p P . If R/h(R) is∈a directi factor of R then there exists a splitting of the projection map in the exact sequence

0 (R) R R/(R) 0. → → → → Clearly, the element 2 (R) is a p-th power in R/(R) for all primes p. Hence, the image of 2 (R) under the splitting homomorphism is also a p-th power for all primes p. By denition of this image is of the form 2 for some root of unity . Let q be a prime that does not divide the order of . Then each of the elements , 2 and 2 is contained in Rq. It follows that also is contained in Rq. But this q q is impossible as (R) equals p : p P . Therefore the sequence does not split and the group R is not a direct proh duct∈withi (R) as one of the factors.

To construct a splitting of the inclusion map from the beginning of this section, we remark that A/K is a torsion group. All elements are of nite order and we dene

n A = K : K for some n Z with n gcd(m, p∞) . p { ∈ ∈ ∈ >0 | } We have A/K = Ap/K. p P M∈ Also mw/w is a torsion group, hence we have

mw/w = Up, where Up = gcd(p∞,m)w/w. p P M∈

The p-part of is the restriction of to U , so we have : U A /K. It p p p → p suces to show that there exists a splitting of p for every prime p dividing m. Denition 24. Let R be a ring. An R-module Q is called injective, if given any R-module N, a submodule N 0 and a homomorphism N 0 Q, there exists an extension of this homomorphism to N. →

Lemma 25. Let p be a prime dividing m. Let R be a ring and suppose we have R-module structures on U and A /K such that : U A /K is an R-module p p p p → p homomorphism and such that the group Up is an injective R-module. Then there exists a homomorphism : A /K U with = 1. p p → p p p Proof. When we take Q = N 0 = Up and N = Ap/K, in the notation of denition 24, then there exists an extension p of the identity on Up mapping Ap/K to Up, as Up is injective. For this homomorphism we have pp = 1. 14 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

First we prove that U is an injective Z-module in the case that p∞ m. p | Lemma 26. An abelian group is injective if and only if it is divisible.

Proof. This is proved in [22], chapter XX, lemma 4.2.

Lemma 27. If p is a prime with p∞ m then the group U is an injective Z-module. | p

Proof. It suces to prove that Up = p∞w/w is divisible; to do so, we show that for q every a Up and for every prime q the group Up contains an element b with a = b . ∈ n Let a be in Up and let q be a prime. Then a is the class of a p -th root of unity n+1 pn for some n N. If p equals q there exists a primitive p -th root of unity in p ∈ Up with = pn ; we take b = w Up. If p = q then gcd(p, q) = 1 and thus there n ∈ y 6 q are x, y Z with xp +yq = 1. Let b = n U , then we have b = n . ∈ p w ∈ p p w

Now we will consider the case that p is a prime dividing m such that p∞ - m. Again we will show that Up is an injective module, only this time, the group is not an injective Z-module, but it is an injective Z/pnZ-module.

Lemma 28 (Baer’s Criterion). Let R be a ring and E an R-module. Then E is an injective R-module if and only if for all left ideals J R, every R-module homomorphism J E can be extended to a homomorphismR E. → → Proof. This is theorem 2.3.1 in [39].

n Lemma 29. Let p be a prime dividing m with p∞ - m and let n Z>0 with p m n+1 n ∈ | and p - m. Then the group Up is an injective Z/p Z-module.

Proof. If p∞ divides w then Up is the trivial group and hence Up is an injective Z/pnZ-module. n Assume p∞ - w. Then Up is isomorphic to Z/p Z. We apply Baer’s criterion. Let J be an ideal of Z/pnZ, then there exists some k N with k n such that J = pk Z/pnZ holds. When : J Z/pnZ is a homomorphism∈ then we have (J) p k Z/pnZ. As both J and Z/p→nZ are additive cyclic groups generated by k n n p respectively 1, we dene an extension 0 : Z/p Z Z/p Z of as follows. Let n k k → Z/p Z such that (p ) = p holds. We dene 0(1) = . Then for all ∈ k = p 0 J we have ∈ k k () = 0 (p ) = 0 p = 0(1) = 0(), n so, 0 is an extension of . We conclude that Z/p Z, and consequently Up, is an injective Z/pnZ-module.

Proof of theorem 21. Let p be a prime with p∞ m. Then by lemma 27 the group | Up is an injective Z-module. As an abelian group also Ap/K is a Z-module and the inclusion map p is a Z-module homomorphism. 1.5. GALOIS ACTION ON RADICALS 15

n n+1 Let p be a prime dividing m and let n Z>0 with p m and p - m. Then ∈ n | by lemma 29 the group Up is an injective Z/p Z-module. As an abelian group n n annihilated by p also Ap/K is a Z/p Z-module and the inclusion map p is a Z/pnZ-module homomorphism. Therefore, by lemma 25, for every prime p dividing m there exists a homomor- phism : A /K U with = 1. Since A/K is the direct sum of the groups p p → p p p A /K for p P the maps on the p-part together give us a splitting of . p ∈ p Corollary 30. There exists a subgroup C of A with K C such that A/ = / C/ . w mw w w

Proof. By theorem 21 there exists a homomorphism : A/K / such that → mw w = 1. We dene : A/ / by (a) = (a K) for all a A/ . Let w → mw w ∈ w ϕ: mw/w A/w be the inclusion map. Then we have the following splitting exact sequence:→

0 mw/w ϕ A/w ker() 0 .

We conclude A/ = / ker(). w mw w Let C be the set of all elements c of A with c ker(), then C is a subgroup w ∈ of A and by construction we have K/ ker(). w Example 31. We consider Q C . We set K = Q and m = p P p∞. We have L = K(A) with ∈ Q n A = C with Q for some n Z . { ∈ ∈ ∈ >0}

There exists a splitting of A with mw(A) = (C ) as a component:

A = (C ) R with n Q for some n Z . { ∈ >0 ∈ ∈ >0}

Taking both factors modulo (Q) = 1 gives a splitting of A/ . The corre- w { } w sponding maps and are given by (a Q) = 1 1 for all a A R and { } ∈ ∩ >0 ( Q) = 1 for every root of unity and (a w) = (a Q) for all a A. Note that man {y other} splittings can be found. ∈

1.5 Galois action on radicals

In this section we embed the group Gal(L/K) in a semidirect product

m = Hom(K/w, ˆ/ˆ ) o Z 16 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

where Z is a particular subgroup of (Zˆ/wmZˆ). Let from now on C be a xed subgroup of A with K C such that A/ = / C/ . w wm w w The existence of C follows from corollary 30. We dene the following two subelds of L = K(A): M = K(wm) and F = K(C).

First we will consider the action of Gal(L/K) on the roots of unity in wm, then the action on C. Let be an element of Gal(L/K), then for every n Z with n wm ∈ >0 | there exists a natural number k,n such that for every primitive n-th root of unity, k,n n, we have (n) = n . Dene the homomorphism ˆ ˆ ω : Gal(L/K) (Z/wmZ) lim (Z/nZ) → ' n wm | by

ω() = (k,n)n wm . | We show that ω is well-dened. Let n, n0 Z>0 with n n0. For every primitive ∈ | n0/n 0 n0-th root of unity n there exists a primitive n-th root of unity n with n0 = n. Now for all Gal(L/K) the elements k (Z/nZ) and k 0 (Z/n0Z) satisfy ∈ ,n ∈ ,n ∈ 0 k 0 0 k 0 k,n n /n ,n n /n ,n n = (n) = (n0 ) = (n0 ) = n .

Hence, we have k k 0 modulo n for all n n0 in Z and all in Gal(L/K). ,n ,n | >0 It follows that ω() is an element of (Zˆ/wmZˆ) for all . Lemma 32. The homomorphism ω induces an isomorphism of topological groups between the Galois group Gal(M/K) and a closed subgroup Z of the kernel of the projection map : (Zˆ/wmZˆ) (Zˆ/wZˆ). w → Proof. Since we have M = K(wm) the map ω is injective on Gal(M/K). As w is contained in K we have k,n = 1 for all n Z>0 with n w and all Gal(L/K). Therefore, the image Z = ω(Gal(M/K)) is a∈subgroup of the| kernel of the∈ projection map w. The map ω is continuous if and only if for all n Z with n wm the compo- ∈ >0 | sitions ωn of ω with the projection on (Z/nZ) are continuous. The maps ωn factor over Gal(K(n)/K):

ωn Gal(M/K) (Z/nZ)

Gal K(n)/K . ¡ ¢ 1.5. GALOIS ACTION ON RADICALS 17

Being the composition of a projection map and an isomorphism of nite groups, the homomorphisms ωn are continuous. As both Gal(M/K) and (Zˆ/wmZˆ) are pronite groups, the map induced by ω is a continuous bijection from a compact space to a Hausdor space and therefore it is a homeomorphism. We x the following notation:

Z = ω Gal(M/K) (Zˆ/wmZˆ), and we dene ¡ ¢ m = Hom(K/w, ˆ/ˆ ) o Z, m where the action of Z on Hom(K/w, ˆ/ˆ ) is induced by the action of (Zˆ/wmZˆ) on /ˆ ˆm. Now we consider the more dicult part of the action of Gal(L/K) on A: the action on the elements of C. The extension F = K(C) over K is, in general, not a Galois extension, but K(m)(C)/K(m) is. Inspired by the homomorphism from lemma 20 we dene a map

m : Gal(L/K) Hom(K/ , ˆ/ˆ ), → w describing the action of Gal(L/K) on the elements of C. First we dene the map in our standard example. Example 33. We look at the situation of example 31 again. We assume that the xed group C for this example is the group of real radicals in C . Let the elements of C/ 1 be represented by the radicals in the set { } n R : Q for some n Z . { ∈ >0 ∈ ∈ >0} The eld F is dened as

n F = K(C) = Q R : Q for some n Z . { ∈ >0 ∈ ∈ >0} n Dene for every a Q>0 and¡ for all n Z>0 the element √a as the¢ positive real n-th root of a. In analogy∈ with Kummer∈theory we dene the map by

√n √n () = a ( a )/ a n Z Hom(Q>0, ˆ) 7→ ∈ >0 ∈ ³ ´ for all Gal(L/Q) and for¡all a Q .¢ We now show that is a well-dened ∈ ∈ >0 map. Take a Q>0 and dene for n Z>0 and Gal(L/Q) the root of unity ,n n n ∈ ∈ ∈ by ( √a )/ √a. Then we have for all n, n0 Z with n n0 ∈ >0 | 0 0 0 n n n n /n n /n ,n √a = ( √a ) = ( √a ) = ,n0 and thus, by denition of ˆ, we have () Hom(Q , ˆ). As Q is a system of rep- ∈ >0 >0 resentatives of Q/ 1 this also denes a map from Gal(L/Q) to Hom(Q/ 1 , ˆ). { } { } 18 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

To dene the map in the general case we use the following lemma.

Lemma 34. For every a K and every n Z>0 with n m there exists a unique n-th root of a in C/∈. ∈ | w w Proof. Let a K and let n Z>0 with n m. There is an element b A such that bn = a. Therefore∈ we have ∈ | ∈ (b )n = a = (1 )(a ) / C/ . w w w w ∈ wm w w As b w is an element of A/w there exist unique wm/w and c C/w such that b = c holds. For these and c we have n∈= 1 and cn =∈ a , this w w w proves the existence of an n-th root of a w in C/w. n n Let x, y be elements of C/ with x = y K/ for some n Z . The w ∈ w ∈ >0 quotient x/y is contained in C/w mw/w = 1 , hence x and y are equal. Therefore the n-th root of a in C∩/ is unique.{ } w w Note that Gal(L/K) acts on A. As w is contained in K we have (c)/c = (c0)/c0 for all c, c0 C with c w = c0 w and for all Gal(L/K). Now we∈are ready to dene the map . We assign∈ to Gal(L/K) a homomor- ∈ phism that for every n Z>0 with n m sends a K/w to (xn)/xn n where x is an element of C with∈ xn =|a. That is, ∈we dene ∈ n n w m : Gal(L/K) Hom(K/ , ˆ/ˆ ) → w by (xn) n () = a , where xn C with xn w = a. 7→ xn n m ∈ µ | ¶ ³ ´ n By lemma 34 we see that for all n, n0 Z>0 we have xnn0 w = xn0 w. In exactly the same way as we did in example 33∈one proves that is a well-dened map. This map in general is not a homomorphism. Let be the map given by : Gal(L/K) , where () = (), ω() . → Proposition 35. The map is an injective homomorphism.¡ ¢

Proof. Let a be an element of K/w and dene x = (xn)n m n m C, where n | ∈ | xn w = a for all n Z>0 dividing m and where again for all n, n0 Z>0 we n ∈ Q ∈ have x 0 = x 0 . For all Gal(L/K) we denote by (x)/x the element nn w n w ∈ (xn) of /ˆ ˆm. For all , Gal(L/K) we have xn n m | ∈ ¡ ¢ (x) (x) (x) ()(a) = = x x (x) ³ (x)´ ³ ´ = ()(a) x = ()(a) ³()ω()´(a). 1.6. THE MAXIMAL KUMMER SUBEXTENSION IN L 19

So, as ω is a homomorphism, is a homomorphism as well. Let , be elements of Gal(L/K) with () = (). We have ω() = ω() and () = (). It follows that (c) = (c) holds for all c mw and that (c) = (c) holds for all c C. Hence equals and is injectiv∈e. ∈

Proposition 36. The map is continuous.

Proof. As we dened () = (), ω() we see that is continuous if and only if both and ω are continuous. In lemma 32 we saw that ω is continuous. ¡ ¢ The map is continuous if and only if for all n m the compositions of | n and the projection on n are continuous. These factor over the Galois groups of the 1/n subextensions K(n)(a )/K of L/K:

n Gal(L/K) n .

1/n Gal K(n)(a )/K . ¡ ¢ As the composition of a projection map and a homomorphism of nite groups the homomorphisms n are continuous.

In this section we dened the maps in the following commuting diagram.

Gal(L/K) → res projection ↓ ↓ Gal(M/K) ω Z → As is injective, the group Gal(L/K) is isomorphic to a subgroup of .

1.6 The maximal Kummer subextension in L

In the previous section we introduced elds F and M such that the composite of F and M is L. In general, if a Galois extension is the composite of two normal subextensions, one can express its Galois group in terms of the Galois groups of the subextensions. Unfortunately, the eld F may not be a normal extension. In order m to identify the image of Gal(L/K) in = Hom(K/w, ˆ/ˆ ) o Z we introduce a normal subeld of F . Fix the notation v = gcd(m, w). We dene a Kummer subextension Fv/K of F/K by

n F = K ( c C with c K for some n Z with n v ) . v { ∈ ∈ ∈ >0 | } 20 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

The maximal Kummer extension of exponent v inside L is

n L = K ( a A with a K for some n Z with n v ) . v { ∈ ∈ ∈ >0 | } Below we prove

L = F K( ) and K( ) F = K. v v vw vw ∩ v This enables us to extend the diagram of subextensions of L/K. Let D be the intersection eld M F , then we have the following diagram. ∩ v L

M = K(wm) Lv F

D(vw) Fv

K(vw) D

K

Theorem 37. The eld L is F K( ) and we have K( ) F = K. v v vw vw ∩ v

Proof. Dene groups of generating radicals for the elds K(vw) and Fv by W1 = n 1/v vw, K and W2 = c C with c K for some n v . As the group (K) hequals Wi , W we see{ that∈ L is the ∈composite of K(| }) and F . h 1 2i v vw v By proposition 17, the intersection of the elds K(vw) and Fv is the eld K(W1 W ). As A/ is the direct product of / and C/ and we have ∩ 2 w wm w w vw wm and W C, we see that W W is K. It follows that K( ) F is K. 2 1 ∩ 2 vw ∩ v

In the next section we use one more property of the eld Fv. Below we will show that for all c C with K(c)/K abelian we have c Fv. To prove this we use the following result∈ ([29], theorem 2). ∈

Theorem 38 (Schinzel). Let R be a eld, let n Z>0 not divisible by char(R) and let u be the number of n-th roots of unity in R. L∈et S be the splitting eld of xn a over R for some a R. Then S/R is abelian if and only if au = bn for some b R. ∈ ∈ Proof. (cf. [35], theorem 4.1) Let a be an element of R. Suppose that au = bn holds for some b R. Let be in R with u = b. Then we have un = au and ∈ 1.7. DETERMINING THE GALOIS GROUP 21 hence n a . It follows that S is a subeld of R(, ). As u = b R and ∈ u un ∈ u R hold, the extension R()/R is cyclic. It follows that R(, un) is abelian andconsequently S/R is abelian. For the converse, suppose that S/R is abelian with Galois group G. Take G k∈ and suppose that it acts on a primitive n-th root of unity n in S as (n) = n . If is an element of S with n = a, then for all G we have ∈ k () () () (k ) = = = , () k µ ¶ µ ¶ from which we deduce that (k )/() is in R as it is xed by every G. Its n-th k 1 n ∈ power a therefore is in R . Let g denote the greatest common divisor of n and the numbers k 1 for all G. Then we have ag Rn. If is an n-th root of unity contained in S, then we ∈ ∈ k 1 have () = if and only if = 1 holds for all G. That is, if and only if ∈ is a g-th root of unity. As the group of n-th roots of unity in R is u the equality g = u follows.

Proposition 39. Let c C. If K(c)/K is abelian then c is an element of F . ∈ v Proof. Let c be an element of C and assume that K(c)/K is abelian. There exists n a positive integer n with n m such that c K holds. Denote by v the greatest | ∈ n common divisor of v and n. By theorem 38 there exists an element b K with n vn n vn ∈ (c ) = b . It follows c = n b for some n-th root of unity n. As we have b K vn ∈ and K C we have b C. As also c C holds, n is an element of C. We ∈ ∈ vn conclude that n is an element of K and therefore also c is contained in K.

1.7 Determining the Galois group

Denote by D the intersection of the elds M and Fv. In this section we will prove the following theorem for K, L, m, , Fv and Z as before.

Theorem 40. The homomorphism gives an isomorphism between Gal(L/K) and m the subgroup H of = Hom K/w, ˆ/ˆ o Z dened by ¡ ¢ H = (f, z) with (f) = (z) Gal(D/K) , { ∈ 1 2 ∈ } 1 where 2 : Z Gal(D/K) is dened by 2(z) = ω (z) D for all z Z and → m | ∈ : Hom K/ , ˆ/ˆ Gal(D/K) is the composition of the canonical Kum- 1 w → mer map from Hom K / , ˆ/ˆm to Gal(F /K) and the restriction map from ¡ ¢ w v Gal(F /K) to Gal(D/K). v ¡ ¢ 22 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

First we show that the maps 1 and 2 as dened in theorem 40 give a commuting diagram. m Gal(L/K) Hom K/w, ˆ/ˆ

ω ¡ 1 ¢

2 Z Gal(D/K) Recall that in general is not a group homomorphism.

Proposition 41. The map

v 0 : Gal(F /K) Hom(K/ , /ˆ ˆ ) v → w dened by

(xn) n 0() = a , where xn C with xn w = a 7→ xn n v ∈ µ ³ ´ | ¶ m v is an isomorphism of abelian groups. Moreover, if ϕv : /ˆ ˆ /ˆ ˆ is the pro- jection map, then the following diagram commutes. →

m Hom K/w, ˆ/ˆ Gal(L/K)

¡ f ϕv f ¢ res 7→ 0 v Hom K/w, ˆ/ˆ Gal(Fv/K) ¡ ¢ Proof. From the previous section we know that F K( ) equals K and that L v ∩ vw v is Fv(vw). Hence, there is an isomorphism

Gal L /K( ) Gal(F /K) v vw ' v ¡ ¢ given by restricting the automorphisms of Lv/K(vw) to Fv. In lemma 20 we proved for the maximal Kummer extension Lv of K that

v Gal L /K( ) Hom K/ ∞ (K), ˆ/ˆ v vw ' v v ¡ ¢ = Hom ¡K/w(K), ˆ/ˆ .¢ ¡ ¢ We now dene the isomorphism 0 as the composition

v Gal(F /K) Gal L /K( ) Hom(K/ , ˆ/ˆ ). v → v vw → w ¡ ¢ Let a be an element of K/w. We proved in lemma 34 that for every n v there exists a unique element C/ with n = a. We also showed that the |quotient n ∈ w n 1.7. DETERMINING THE GALOIS GROUP 23

(x )/x does not depend on the choice of x C with x = . So, it follows n n n ∈ n w n from theorem 19 that the isomorphism 0 is given by

(xn) n 0() = a , where xn C with xn w = a. 7→ xn n v ∈ µ ³ ´ | ¶ It is easy to verify that for this map 0 the diagram above commutes.

The map 1 from theorem 40 is the composition of the homomorphisms in the following diagram

m Hom K/w, ˆ/ˆ

¡ f ϕ f ¢ 7→ v 01 v Hom K/w, ˆ/ˆ Gal(Fv/K) ¡ ¢ res

Gal(D/K).

From the commuting diagram in proposition 41 we conclude that for all Gal(L/K) we have ∈ 1(()) = |D and 1 2(ω()) = ω (ω()) = . |D |D We get (()) = (ω()) for all Gal(L/K) and thus the diagram 1 2 ∈ m Gal(L/K) Hom K/w, ˆ/ˆ

ω ¡ 1 ¢

2 Z Gal(D/K) commutes. Now we can prove theorem 40. Proof of theorem 40. Let G be the image of Gal(L/M) under in H. We have canonical maps G , H and H Z. In lemma 32 we showed that ω induces an isomorphism of Gal(→M/K) to Zand→ in section 1.5 we dened the homomorphisms and . This gives the following commuting diagram.

0 Gal(L/M) Gal(L/K) Gal(M/K) 0

ω ' ' ϑ G H Z 24 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

The rst row is exact because of Galois theory. The map is an injective continuous homomorphism by propositions 35 and 36. The map is injective and because the diagram commutes and ω is an isomorphism we see that ϑ is surjective. We will prove exactness of the sequence G H ϑ Z in H. For all g G we have ϑ((g)) = 1 so all there is left to prove→is that→for (f, 1) H we have (∈f, 1) G. ∈ ∈m Let (f, 1) be an element of H. Then we have f Hom(K / , ˆ/ˆ ) and ∈ w 1(f) = 2(1) = 1. For all n Z>0 with n m we dene the elds Ln = M n ∈ | { ∈ L : K . Then, by theorem 15, we have ∈ } ¡ n n ¢ Gal(L /M) Hom (K M )/M , n ' n n Hom K/(M K), . ' ¡ ∩ n ¢ m Let ϕn : /ˆ ˆ n, for all n Z>0 with¡ n m, be the pro¢jection map; we prove → n ∈ | that ϕn f Hom(K/(M K), n). ∈ ∩ n Let n be some positive integer and let a be an element of K M . Because K n n ∩ is contained in C we have a = c for some c C M . As c is an element of M the extension K(c)/K is abelian. Moreover, as c ∈ C ∩we have c F , by proposition 39. ∈ ∈ v It follows that c Fv M = D. We saw that 1(f) = 1, that is 1(f) acts trivial on the elements of∈ the∩eld D. Because c is an element of D with cn = a we have n f(a)n = 1. We conclude that K M is contained in ker(ϕn f) and thus ϕn f n ∩ is an element of Hom(K/(M K), ). ∩ n Now, for all n, n0 N with n n0 we get the following commuting diagram. ∈ | n0 Gal(Ln0 /M) Hom(K/(M K), 0 ) ∩ n 0 res g gn /n 7→ n Gal(L /M) Hom(K/(M K), ) n ∩ n

The eld L is the composite of the elds Ln for all positive divisors n of m and thus we have

G = Gal(L/M) = lim Gal(Ln/M). n m | n For all n m we have ϕn f Hom(K/(M K), n) and for all a K/w we | n0/n ∈ ∩ ∈ have ϕ 0 f(a) = ϕ f(a) for all n, n0 Z with n n0. It follows that n n ∈ >0 | n ¡ ¢f = (ϕn f)n m lim Hom(K/(M K), n). | ∈ n ∩ Hence (f, 1) is an element of G. We have a commuting diagram consisting of two exact sequences. As the homo- morphisms Gal(L/M) G and Gal(M/K) Z are isomorphisms, the injection also has to be an isomorphism.→ The groupGal→ (L/K) is compact because it is a pronite group and as H is Hausdor the continuous isomorphism is a homeomor- phism and thus Gal(L/K) and H are isomorphic as topological groups. 1.7. DETERMINING THE GALOIS GROUP 25

We will use this theorem to give an explicit description of Gal(L/K) in the case that K equals Q and m is p P p∞. ∈ Q Lemma 42. Let a be a positive integer and let n be an element of Zˆ. Then there exist positive integers ni with gcd(ni, 2a) = 1 for all i Z>0 such that limi ni ∈ →∞ equals n in Zˆ.

Proof. Let , for all k Z , be the projection from Zˆ to (Z/kZ). Now dene n ∈ >0 i for every i Z as the integer obtained by lifting to Z. Then for all i Z ∈ >0 2a(i!) ∈ >0 we have gcd(ni, 2a) = 1 and n equals limi ni. →∞ Denition 43. Let p be a prime and let a be an integer. We denote the Legendre a symbol of a and p by p . For a positive odd³integer´ b and an integer a with gcd(a, b) = 1 the Jacobi symbol is dened as a a vp(b) = , b p p b µ ¶ ³ ´ Y| where for a prime p and an integer b we denote by vp(b) the p-valuation of b. Now let a be an element of Q and let n be an element of Zˆ. Write a = b/c for b, c Z 0 and write n = limi ni for a sequence (ni)i∞=0 of positive integers ni ∈ \{ } →∞ for which gcd(ni, 2bc) = 1 for all i > 0. We dene the Jacobi symbol

a b c = lim , n i ni ni ³ ´ →∞ µ ¶ Áµ ¶ where b and c are the ordinary Jacobi symbols. Below we show that this is ni ni well-dened.³ ´ Let¡a, b,¢c, n, ni be as above and let p be an odd prime number. Then we have by quadratic reciprocity

ni p p if p 1 mod 4 = ni1 ³ ni ´ ni ( 1) 2 if p 3 mod 4. µ ¶  p ³ ´ n21  1 ni 1 2 i By the supplementary laws we have = ( 1) 2 and = ( 1) 8 . As ni ni n21 n ni 1 i ³ ´ ³ ´ i , ( 1) 2 and ( 1) 8 stabilise for large i the limits p ³ ´ 1 1 p p = lim and = lim , for p prime, n i ni n i ni µ ¶ →∞ µ ¶ ³ ´ →∞ µ ¶ a exist and do not depend on the choice of the ni. By multiplicativity this gives n . ¡ ¢ 26 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Example 44. Let us return to the case where K = Q and m = p P p∞. We choose C as in example 33. ∈ 2 Q Remark that /ˆ ˆ 2. We have w = 2 and also v = 2. The eld Lv is 2 ' Q( C with Q ) and the eld K(vw) is Q(i). The elds Fv and D are equal:{ ∈ ∈ }

F = D = Q( C with 2 Q ). v { ∈ ∈ >0}

m Let, as before, ϕ2 : /ˆ ˆ = ˆ 2 be the projection map. Then the isomorphism : Hom(Q , ˆ) Gal(D/Q→) satises 1 >0 →

(f)(√a ) = ϕ (f(a)) √a for all f Hom(Q , ˆ) and all a Q . 1 2 ∈ >0 ∈ >0

(p 1)/2 Let p be an odd prime. Notice that Q(√p ), with p = ( 1) p, is the unique subeld of degree 2 of Q(p)/Q. If is an element of Gal Q(p)/Q and we have ( ) = e for some e 1, 2, . . . , p 1 then p p ∈ { } ¡ ¢

e √p = √p if e is a square in (Z/pZ) (√p ) = p e  √p =³ ´ √p if e is not a square in (Z/pZ).  p ³ ´  f Moreover, if 0 is an element of Gal(Q(8)/Q) and we have 0(8) = 8 for some f 1, 3, 5, 7 then ∈ { }

f1 f21 0(√ 1 ) = ( 1) 2 √ 1 and 0(√2 ) = ( 1) 8 √2.

Let a be an element of Z>0 with prime factorisation

a = 2v2(a) pvp(a) qvq (a) p 1 mod 4 q 3 mod 4 Y Y

Then we have

v2(a) v (a) x √a = √2 √pvp(a) √ q q √ 1 , p 1 mod 4 q 3 mod 4 Y Y with x = 0 if 2 q vq(a) and x = 1 if 2 - q vq(a). The homomorphism 2 : Z | 1 → Gal(D/K) maps z Z to the restriction of ω (z) Gal(M/K) to Gal(D/K). P ∈ P ∈ 1.8. COMPUTING THE FIELD D 27

Hence we have, for all z Z, ∈

v2(a) v (a) x (z)(√a ) = (z) √2 √pvp(a) √ q q √ 1 2 2   p 1 mod 4 q 3 mod 4 Y Y  v (a) vp(a) vq (a) z21 2 z z = ( 1) 8 √2 √p √ q p q p q ³ ´ Y µµ ¶ ¶ Y µµ ¶ ¶ z1 x ( 1) 2 √ 1 ³ v2(a) v´(a) v (a) 2 p p q z1 q z1 x = ( 1) 2 ( 1) 2 √a z z z p q µ ¶ Y ³ ´ Y ³³ ´ ´ a z1 P v (a)+x = ( 1) 2 ( q q ) √a z ³a´ = √a. z ³ ´ Now let a Q>0 with a = b/c for b, c Z 0 . Then Q(√a ) = Q(√bc ) and a b ∈c bc ∈ \ { } n = n n = n , therefore Gal(L/Q) is isomorphic to the group ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ a (f, z) Hom(Q , ˆ) o Zˆ with ϕ f(a) = for all a Q . ∈ >0 2 z ∈ >0 n ¡ ¢ ³ ´ o 1.8 Computing the eld D

In this section we will show that the eld D = M Fv equals the intersection M F . Moreover we give a method to compute the eld∩D in some specic cases. At∩the end of the section we provide some examples. Proposition 45. The intersection M F is the eld D. ∩ Proof. As D = M F and F F we have that D M F . Now let Gal(L/K) ∩ v v ∩ ∈ with = 1. We prove that also ∩ = 1. We have |D |M F

1 () = = 1 = 2 ω(1) |D |D and hence (), ω(1) ¡H. So,¢ by theorem 40 ¡there¢is an automorphism Gal(L/K) such that ∈ ∈ ¡ ¢ () = (), ω() = (), ω(1) . It follows () = () and therefore¡ we hav¢e (¡c) = (c) for¢ all c C. We conclude = . The equality ω() = ω(1) implies that = 1 holds. So,∈ we have |F |F |M

∩ = ∩ = 1 ∩ = 1. |M F |M F |M F Hence M F is contained in D. ∩ 28 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

We introduce some more notation.

Denition 46. Let p be a prime and let s be a Steinitz number. By Kp we denote the eld K(p∞ (K )) and by Kp,s we denote the maximal abelian subextension of Kp/K of exponent s. Note that if char(K) = p we get Kp = K.

Lemma 47. Let be the group Gal(M/K) and let p denote the Galois group Gal((M K )/K), for all primes p dividing m. If the natural map ∩ p g : /v ( /v) → p p p m Y| is injective, then the maximal abelian subextension of exponent v of M/K is the composite of the elds M K . ∩ p,v Proof. Consider the following commuting diagram.

v g v / p(p/p) Q p f As we have = Gal((M K )/K) it follows Gal((M K )/K) = /v. Dene p ∩ p ∩ p,v p p v ker(fp) for all primes p the map fp : p/p. Then we have M = M Kp,v. → v ∩ If g is injective then ker(f) = ker(p) = v. The eld M is the maximal abelian subextension of exponent v in M/K. As we will see in example 49 it is not true in general that the maximal abelian subextension of exponent v of M/K is the composite of the elds M Kp,v. But if this holds, then lemma 47 gives us a method to compute the eld D. ∩ Theorem 48. If the maximal abelian subextension of exponent v of M/K is the composite of the elds M Kp,v, then the eld D is the composite of the elds M K ( ) F for the∩ primes p m. ∩ p vw ∩ v | Pr¡ oof. We intro¢duce two elds. Let D0 be the composite of the elds M K ( ) ∩ p vw ∩ Fv for all primes p m and let E be the composite of the elds M Kp,v for all primes p m. | ¡ ∩ ¢ By lemma| 47 we have D0 D E. To prove that D = D0 we rst show that D(vw) = D0(vw). In section 1.6 we saw that the maximal Kummer extension of exponent v over K, the eld Lv, is the composite of Fv and K(vw). The groups of radicals of exponent v over K, generating the elds Lv, Fv and K(vw), are n A = a A with c K for some n Z with n v , v { ∈ ∈ ∈ >0 | } 1.8. COMPUTING THE FIELD D 29

n C = c C with c K for some n Z with n v v { ∈ ∈ ∈ >0 | } and . We have A = C , . vw v h v vwi Let p be a prime dividing m. The eld M K ( ) is contained in L , hence ∩ p,v vw v there exists a subgroup T of Av such that M Kp,v(vw) = K(T ). As T is a subgroup of C , and as is contained in∩T we have h v vwi vw (T C ) = T. ∩ v vw Therefore, by proposition 17, also the following equality of elds holds

K(T ) F ( ) = K(T ). (1.1) ∩ v vw ¡ ¢ As Kp(vw) Fv is a eld of exponent v over K we have, by corollary 18, Av K ( w) K∩ ( ). It follows ∩ p v p,v vw K ( ) F = K ( ) F . (1.2) p vw ∩ v p,v vw ∩ v

As K(T ) equals M Kp,v(vw) we can substitute equation 1.2 in the left hand side of equation 1.1. We∩obtain

(M K ( )) F ( ) = M K ( ). ∩ p vw ∩ v vw ∩ p,v vw ¡ ¢ Taking the composite over all primes p dividing m gives D0(vw) = E(vw). As we have D0 D E we derive that D( ) = D0( ). vw vw Now look at the following diagram:

D(vw)

D D0(vw)

D0 K(vw)

K.

All the extensions are Galois extensions and by theorem 37 we know that the inter- section D K( ) equals K. We conclude that ∩ vw

Gal(D( )/D) Gal(K( )/K) Gal(D0( )/D0), vw ' vw ' vw hence we have D = D0. 30 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

Example 49. Let p and q be distinct primes congruent to 1 modulo 4. Let m be 2pq. We construct a eld K for which the composite of the elds Kp,2 and Kq,2 is smaller than the maximal Kummer extension of exponent 2 over K in the eld M = K(pq). Dene elements rp, sp in Q(p) and elements rq, sq in Q(q) that generate subelds of Q(p) and Q(q) as indicated in the following diagrams.

Q(p) Q(q)

2 2

Q(rp) Q(rq)

2 2

Q(sp) Q(sq)

Q Q

As Q(pq)/Q(sp, sq) is a Galois extension with Galois group C4 C4 we have the following diagram of subelds.

Q(pq)

Q(p, rq) . . . Q(rp, q)

. . . Q(p, sq) . . . Q(rp, rq) . . . Q(sp, q) . . .

Q(rp, sq) K Q(sp, rq)

Q(sp, sq)

Let K be the quadratic subextension of Q(pq)/Q(sp, sq) that is not equal to Q(rp, sq) or Q(sp, rq). The maximal Kummer extension of exponent 2 in K(pq)/K is a degree 4 extension. However, the extension Kp,2 = Kq,2 = Q(rp, rq) is of degree 2 over K.

Below we compute the intersection eld D in some special cases. We start with our standard example. 1.8. COMPUTING THE FIELD D 31

Example 50. Let, as before, K = Q, m = p P p∞, and let F be the eld generated by the real radicals in C . In this case we ha∈ ve v = 2. As the Galois group ˆ Q of Q((C )) over Q is isomorphic to Z which is the direct product of the Zp for all primes p we can apply theorem 48. Let p be an odd prime. Then Q( )/Q is cyclic of degree p 1 and the unique p subeld of degree 2 of Q(p) is Q(√ p ). As vw = 4 we have Kp(vw) Fv = Q(√p ). ∩ Moreover, we have K ( ) F = K F = Q(√2). It follows that 2 vw ∩ v 2 ∩ v D = Q(√p: p P ). ∈

In the case that K = Q and m = p P p∞ we could also have made other choices for the subgroup C in the splitting of A∈(section 1.5) resulting in a non-real eld F . Also in these cases it is easy to computeQ the eld D. For every prime p precisely one of the elements √p and √ p is contained in Fv. When we denote this element by p then we have D = Q(p : p P ). ∈

Example 51. Let K = Q(√2 ) and m = p P p∞. The Galois group of Q((C )) ∈ over Q(√2 ) is the direct product of the groups Z for odd primes p with Q p

Gal(Q( ∞ )/Q(√2 )) Z. 2 2 Again we apply theorem 48. As in example 50 we have v = 2 and for all odd primes the intersection K ( ) p vw ∩ Fv equals Q(p), where p √p, √ p depending on the choice for C. ∈ { } Again we have K ( ) = K . For the intersection K F there are only few 2 vw 2 2 ∩ v possibilities. The only extensions over Q(√2 ) of exponent 2 in K2 are Q(8) Fv, 1 1 6 Q( + ) and Q( ). The following equalities hold: 16 16 16 16 1 2 1 2 + = 2 + √2 and = 2 + √2. 16 16 16 16 ¡ ¢ ¡ ¢ Moreover we have (2+ √2 )( 2+ √2 ) = 2 . By lemma 34 there is a unique w w square root of (2 + √2) w in C/w. As we have √2 C and i / Fv precisely one 1 1 ∈ ∈ of the elements + and is contained in F . 16 16 16 16 v Example 52. Let K be a real number eld and let m be a Steinitz number coprime to 2. Then v equals 1. Consequently, we have Fv = K = D.

Example 53. Let K be Q R and let m be 2. Then L = M = K(i) and the eld ∩ F equals the ground eld K. All square roots of the elements of K 0 are contained in K and the product of this group and the group generated by igives all square roots of elements of K.

Example 54. If L/K is a Kummer extension of exponent m for some Steinitz number m, then we have v = gcd(m, w) = m. As the eld M equals K(wm) we 32 CHAPTER 1. MAXIMAL RADICAL EXTENSIONS

have M = K(vw). In section 1.6 we showed that K(vw) Fv = K, so in this case we have D = M F = K. ∩ ∩ v Moreover we have L = MF = MFv and thus Gal(L/K) is the direct product m of Hom(K/w, ˆ/ˆ ) Gal(Fv/K) and Z Gal(M/K). As M/K is a Kummer ' ' m extension, the group Z is isomorphic to Hom(w, /ˆ ˆ ). We derive the result of m theorem 19 again: Gal(L/K) Hom(K, ˆ/ˆ ). ' Chapter 2

Subelds of radical extensions

In the main theorem in the previous chapter we computed the Galois group of a large eld generated by radicals. When we are able to compute all subgroups of this Galois group we know all subelds of the extension. As we will see in chapters 4 and 5, it is useful to know the subelds of a eld generated by radicals in order to denest radicals. In this chapter we study the subelds of the extension Q()/Q, where is an element of C with n Q for some positive integer n. ∈ Denition 55. Let L/K be a eld extension. A group C is called a radical group for L/K if K C L, if L = K(C) and if there exists a positive integer n with n C K. The extension L/K is called a radical extension of exponent n if there exists a n radical group C for L/K with C K. Let L/K be a radical extension of exponent n with radical group C. If K contains a primitive n-th root of unity then L/K is a Kummer extension of exponent n, the map

: subgroups of C containing K subelds of L containing K { } → { } given by (C0) = K(C0) is a bijection, and, if L/K is nite, we have #(C/K ) = [L : K]. Under special conditions such a correspondence also holds for other radical extensions. In this chapter we determine for which radicals the map

: subgroups of Q, containing Q subelds of Q() { h i } → { } given by (C) = Q(C) is surjective. This results in the following theorem. By n = n(C ) we denote, as in denition 8, the group of n-th roots of unity in C .

Theorem 56. Let C be a radical over Q with n Z>0 minimal such that n ∈ ∈ Q. Let D be the subgroup Q, of C . Every subeld of Q() is of the form ∈ h i 33 34 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Q(d) for some positive integer d n if and only if one of the following conditions holds: |

(i) D and we have 6 - n or √ 3 / D , ∩ n 2 ∈ (ii) D = , ∩ n 3 (iii) D = and 1 + i D , ∩ n 4 ∈ (iv) D = and √ 3 D , ∩ n 6 ∈ (v) D = and both 4 - n and √5 D . ∩ n 10 ∈ In section 2.1 we give the context of the results in this chapter. We state some known results for generalising Kummer theory to radical extensions over elds that do not contain the appropriate roots of unity. The proof of theorem 56 comprises the rest of the chapter. In section 2.2 we show the rst implication: if every subeld of Q() is generated by a power of then has to satisfy one of the conditions (i), (ii), (iii), (iv) or (v). In section 2.3 we embed the Galois group of Q(n, ) over Q in Z/nZo(Z/nZ). We use the intersection eld Q( ) Q() to give an explicit description of the image n ∩ G of Gal(Q(n, )/Q) in Z/nZ o (Z/nZ). In sections 2.4 and 2.5 we show, for the radicals satisfying one of the ve conditions of theorem 56, that all subgroups of the group G are of the form G dZ/nZ o (Z/nZ) for some divisor d of n. These ∩ subgroups correspond to the elds Q(n/d). This will nish the proof of theorem 56. ¡ ¢ As a corollary we prove the following theorem at the end of section 2.5.

Theorem 57. Let C be a radical over Q with n Z>0 minimal such that n ∈ ∈ Q. Let D be the group Q, . The map is bijective if and only if one of the ∈following properties holds:h i

(i) D and we have 6 - n or √ 3 / D , ∩ n 2 ∈ (ii) D = . ∩ n 3 In this chapter we x the following notation. Always C will denote a radical over Q and n will be the smallest positive integer for whic∈h n is contained in Q. By D we denote the multiplicative group Q, C . If m is a natural number then we denoteh a primitivi e m-th root of unity in C by , the group we denote by and the number ϕ(m) is the Euler totient of m. m h mi m 2.1 Known results

The rst results in nding non-Kummer radical extensions for which the map from the introduction is a bijection impose conditions on the group C to ensure 2.1. KNOWN RESULTS 35 that the group index and the degree of the eld extension are equal. In [16] Kneser generalised results of Besicovitch [3], Mordell [25] and Siegel [33]. He proved the following theorem.

Theorem 58. Let K be a eld of characteristic 0. Let L/K be a radical extension of exponent n with radical group C. If L/K is a nite extension then the degree [L : K] equals the index (C : K) if and only if R1 and R2 hold: R : for all odd primes p: if C then K, 1 p ∈ p ∈ R : if 1 + i C then i K. 2 ∈ ∈

We prove one implication: the conditions R1 and R2 are necessary. Assume that [L : K] = (C : K) holds. For every element c of C we have

[K(c) : K] ( K, c : K). (2.1) h i

As [L : K] is equal to [L : K(c)][K(c) : K] and (C : K) equals the product (C : K, c )( K, c : K) it follows from [L : K] = (C : K) that equality holds in 2.1.h Letip hbe aniodd prime with C. Then, in particular we have [K( ) : p ∈ p K] = ( K, : K). As the degree [K( ) : K] divides p 1 and the group index h pi p ( K, p : K) is 1 or p it follows that p is an element of K. The same argument whorks ifi1 + i C: the eld K(1 + i) = K(i) is of degree 2 over K if we have i / K; ∈ ∈ in this case the group index ( K, 1 + i : K) is 4, so i must be an element of K. h i Whether or not the conditions R1 and R2 from theorem 58 are satised does not only depend on the extension L/K but also on the choice for the radical group C.

Example 59. Take K = Q and L = Q( ). If we take C = Q, , then C satises 8 h 8i the conditions R1 and R2 from theorem 58 and we have 4 = [L : K] = (C : K). But, if we take C = Q, 8, √2 then (C : K) = 8 = [L : K]. And indeed, in this case √2 = 1 + i ishan elementi of C and as i is not 6contained in Q this group does 8 not satisfy R2. Given a radical extension L/K it also depends on the choice for the radical group C whether or not the map is bijective as we see in the following example.

Example 60. Take K = Q and let C be an element with 4 = 9. Dene ∈ L = Q(). When we take C = Q, , then the conditions R and R hold and h i 1 2 the degree of the extension L/K equals the group index (C : K ). But, as L equals Q(i, √6 ) and C/K is a cyclic group, the map is not a bijection. If we take C = Q, i, √6 , then again C satises the conditions in Kneser’s theorem. However, inh this caseithe map is a bijection.

The previous example also shows that conditions R1 and R2 from theorem 58 are not sucient for to be a bijection. 36 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

In [11] Greither and Harrison introduce the notion of cogalois theory. Given a radical extension L/K they x a choice for the radical group C. They give a condition on the extension L/K for the map to be a bijection. This condition implies that every radical group for L/K satises R1 and R2.

Denition 61. A eld extension L/K is called pure if the following holds: if p = 4 or p is prime then every element L satisfying p = 1 is contained in K. ∈

Let L/K be a eld extension and let C be a subgroup of L/K, then we denote by K(C) the subextension of L/K generated by all b L with b K in C. ∈ Proposition 62 (Greither and Harrison). Let K be a eld of characteristic 0 and let L/K be a pure radical extension. Then the maps

: subgroups of (L/K) subelds of L containing K , { tors} → { } dened by (C/K) = K(C) and

ϕ: subelds of L containing K subgroups of (L/K) , { } → { tors} dened by ϕ(M) = (M /K)tors are bijections that are inverse to each other.

Corollary 63. Let 1, . . . , t be real numbers for which there exist positive integers n , . . . , n with ni Q for all i. Dene L as the subeld Q( , . . . , ) of R. Then 1 t i ∈ 1 t every subeld of L is generated by monomials in the radicals i.

Proof. As L is a real eld the extension L/Q is pure and we can apply proposition 62. The map is a bijection, hence every subeld of L is generated by a subgroup C of the multiplicative group Q, , . . . , with Q C. h 1 ti In [18] Halter-Koch reformulates the problem. He remarks that for certain radi- cals there exist subelds of Q() that are conjugate to a eld of the form Q(d) but that are not generated by a power of , see example 65. Therefore he denes a new map 0 mapping subgroups of the radical group C of L/K to classes of conjugate subelds of L/K. He gives sucient conditions on C for 0 to be a bijection.

Theorem 64. Let K be a eld of characteristic 0 and let L/K be a radical extension of exponent n Z with radical group C. If the degree of the extension [L : K] ∈ >0 equals the index (C : K) and if C satises the condition

R : If 4 n, i L, y L and (1 + i)y C, then i K(y) or i K(iy), 3 | ∈ ∈ ∈ ∈ ∈ then every subeld K0 of L containing K is conjugate to a eld of the form K(C 0) for some subgroup C0 of C containing K. 2.2. FIRST IMPLICATION 37

Example 65. Let be an element of C with 6 = 3. One easily checks that the group C = Q, satises the conditions R1, R2 and R3. In the extensionh i Q()/Q there are subextensions, like Q( 2), that are not 3 generated by a subgroup C0 of C, but all subelds of Q() are conjugate to a eld 3 2 Q(C0) for such a subgroup C0. Namely, if we x the notation √ 3 = , then the subelds of Q() are

Q, Q(√ 3 ), Q(√3 3 ), Q( √3 3 ), Q(2 √3 3 ) and Q(). 3 3 These are all conjugate to Q, Q(3), Q(2) or Q().

We give an example that shows that condition R3 in theorem 64 is not necessary. Example 66 ([18]). Let K be Q(√ 2 ) and let be an element of R with 8 = 5. If we take L = Q(8, ) and we let C be K, i, , then all subelds of L are conjugate to a eld generated by a subgrouphof C, althoughi the group C does not 2 2 satisfy condition R3: let y be (1 + i) , then we have (1 + i)y = 2i C, but i is an element of neither K(y) nor K(iy). ∈

2.2 First implication

In this section we prove proposition 67, which gives one implication of theorem 56. Recall that C is a radical over Q with n Z minimal such that n Q and ∈ ∈ >0 ∈ that D is the subgroup Q, of C . h i Proposition 67. If every subeld of Q() is of the form Q(d) for some positive integer d n then one of the following conditions holds: | (i) D and we have 6 - n or √ 3 / D , ∩ n 2 ∈ (ii) D = , ∩ n 3 (iii) D = and 1 + i D , ∩ n 4 ∈ (iv) D = and √ 3 D , ∩ n 6 ∈ (v) D = and both 4 - n and √5 D . ∩ n 10 ∈ First we give two lemmas. Lemma 68. Let K be a eld and let a be an element of K. Let f be the polynomial xm a K[x] for some integer m 2. Then f is reducible over K if and only if at leastone∈ of the following two properties holds. There exist a divisor d > 1 of m and an element b in K such that a = bd. We have 4 m and there exists an element c K such that a = 4c4. | ∈ 38 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Proof. This is [22] Chapter VI, section 9, page 297, theorem 9.1. Lemma 69. Let d be a divisor of n. If Q(n/d)/Q is an abelian extension then we 2n/d have Q = Q, . d h i Proof. As Q(n/d)/Q is abelian and n/d is a root of xd n Q[x], theorem 38 shows that there exists some rational number b with 2n= bd.∈Taking d-th roots gives 2n/d = k b for some positive integer k. d 2n/d d/g Let g = gcd(k, d), then we have Q. Suppose that g is greater t ∈ than 2, then there is an integer t < n with Q, which gives a contradiction with the minimality of n. ¡ ¢∈ k 2n/d If g equals 1, then is a primitive d-th root of unity and thus Q, is d h i d Q. t Suppose that 4 d. Then there exist t Z>1 and l N with d = 2 l. Substituting this in the| identity 2n/d = k b ∈gives ∈ d n/(2t1 l) k = b 2t l. 2t2 l n k k As n is minimal with Q we have 2t l / Q. Therefore 4 is not contained in Q and hence k is odd ∈and so g equals 1. ∈ ¡ ¢ k/2 If g equals 2 then 4 - d and we have gcd(d/2, k/2) is 1. Write = d/2 , then

2n/d = k b = b, d where is a primitive d/2-th root of unity. We saw above that d/2 is odd. Therefore is a primitive d-th root of unity. We conclude that, also in this case, the group 2n/d Q, equals Q. h i d Corollary 70. Let d be a divisor of n. If Q(n/d)/Q is an abelian extension then we have Q(n/d). d For the rest of this section we assume that every subeld of Q()/Q is of the form Q(d) for some positive integer d n. First we show that the intersection | D n is contained in 60, then we prove that D n equals k for some k in 1, 2∩, 3, 4, 6, 10 . Finally we give a subeld of Q()/Q∩ that is not generated by a {power of for}the cases that are excluded in proposition 67. Lemma 71. If all subelds of Q() are generated by d for some divisor d of n, then Q() has at most one subeld of degree 2 over Q. Proof. Suppose that K is a quadratic subeld of Q(). Then there exists an integer t with K = Q(n/t). All extensions of degree 2 over Q are abelian. So, by corollary 70, we have t K. Hence ϕ(t) 2. We conclude that t is one of the numbers 1, 2, 3, 4, 6. For t = 1 we have Q(n/t) = Q. Moreover we know that Q(n/3) Q(n/6) and Q(n/2) Q(n/4), so there are at most two dierent subelds of Q() 2.2. FIRST IMPLICATION 39

n/3 n/4 of degree 2 over Q. If both Q( ) = Q(3) and Q( ) = Q(4) are quadratic subelds of Q(), then Q(√3 ) is another subeld of Q() of degree 2 over Q. This eld cannot be generated by a power of , so, Q() has at most one subeld of degree 2.

Proposition 72. If all subelds of Q() are generated by d for some divisor d of n, then we have D . ∩ n 60

Proof. Let k be the positive integer with D n = k. If p is a prime dividing k, ∩ 1 then Q( ) is a subeld of Q(). Let K be Q( + ) Q( ). The extension p p p p K/Q is abelian and n K is contained in 1 . As a subeld of Q() the eld K is generated by a power∩of . Let r Z bhe thei largest integer with K = Q(n/r). ∈ >0 As K is abelian we have by corollary 70 that r K and thus r 2. So, K is a subeld of degree 1 or 2 over Q. We conclude that the only primes dividing| k are in the set 2, 3, 5 . { } Suppose that p is an element of 3, 5 such that p2 divides k. Then the eld { } Q( 2 ) is abelian and it is of degree p(p 1) over Q. Therefore Q() contains an p abelian subeld K of degree p with n K 1 , generated by a power of . This gives a contradiction with corollary 70∩andthhus gcdi (9 25, k) is a divisor of 15. Suppose that 8 divides k. Then 8 is contained in Q(). Therefore Q() has three dierent subelds of degree 2 over Q, which gives a contradiction with lemma 71. We conclude that k divides 15 4 = 60. Corollary 73. If all subelds of Q() are generated by d for some divisor d of n, then we have D = for some k in 1, 2, 3, 4, 6, 10 . ∩ n k { } Proof. By proposition 72 we have k 60 and by lemma 71 there is at most one subeld of degree 2 over Q in Q(). If| 3 divides k then this unique quadratic eld is Q(3). If 4 divides k it is Q(i) and if 5 divides k it is the subeld Q(√5 ) of Q(5). Hence k cannot be 12, 15, 20, 30 or 60. In the case that 5 divides k the eld Q(√5 ) is generated by a power of . As Q(√5 )/Q is an abelian extension we see by corollary 70 that the eld Q(√5 ) equals Q(n/t) for some t 1, 2 . As Q(n) is Q we have t = 2 and 2 n. We conclude that also k is even; this∈ { excludes} k = 5. Hence, k is an element of |the set 1, 2, 3, 4, 6, 10 . { } We have shown that the assumption that all subelds of Q() are generated by a power of implies that D n equals k for some k in 1, 2, 3, 4, 6, 10 . Proposition 67 states that this assumption∩ leads to some additional{ restrictions on} n and D. In the following lemmas we prove, for each of the possible values for k, that the extra conditions are necessary.

Lemma 74. If D n is contained in 2 and all subelds of Q() are generated by a power of then∩we have 6 - n or √ 3 / D . ∈ 40 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Proof. We use lemma 68 to show that [Q() : Q] equals n. Suppose that the poly- nomial xn n is reducible over Q. We do not have 4 n and n = 4c4 for some c Q, as in that case n/2 = 2ic2 | ∈ and i D n. Therefore there exists a divisor d > 1 of n and an element b Q ∈with ∩n = bd. If d is even, then also n is even and we nd the equality (x∈n n) = (xn/2 bd/2)(xn/2 + bd/2) Q[x], which gives a contradiction with the ∈ n/d k minimality of n. So, d is odd and hence we have = d b for some k N with k ∈ gcd(k, d) = 1. Therefore d D n 2. It follows that d equals 1 and thus xn n is irreducible over Q∈. ∩ Suppose that we have 6 n and √ 3 D. As [Q() : Q] equals n, the degree of Q(n/d) over Q is d| for all divisors ∈ d of n, thus Q(n/3) is the unique subextension of Q()/Q of degree 3 generated by a power of . As √ 3 D, we n/3 ∈ have 3 Q() and therefore also Q(3 ) is a subeld of Q() of degree 3 over ∈ n/3 n/3 n/3 Q. As 3 Q( , 3 ) and [Q(3) : Q] = 2, the eld Q(3 ) cannot be generated∈by a power of .

n 2 4 Lemma 75. Suppose that 4 divides n. If is an element of Q 4 Q then Q(n/4) has three dierent subelds of degree 2 over Q. \

Proof. By lemma 68 the minimal polynomial of n/4 is x4 n = x4 + t2 Q[x] for 4 2 2 1 3 1 ∈3 some t Q. Let be a root of x + t . The elements , t + and t are roots of∈the polynomials

x2 + t2, x2 + 2t and x2 2t. These are, again by lemma 68, all irreducible as we have t2 / 4 Q4, so there are three dierent subelds of Q() of degree 2 over Q. ∈

Lemma 76. If D n is 4 and all subelds of Q() are generated by a power of then we have 1 +∩ i D . ∈ n/2 Proof. As we have D n = 4 we know that 4 divides n and that is an ∩ n 4 element of i Q. Suppose that is not an element of 4 Q . By the previous lemma we kno w that Q() had three dierent subeldsof degree 2 over Q. This n 4 gives a contradiction with lemma 71. Therefore we conclude 4 Q and n/4 n/4 ∈ hence we have (1 + i) Q (1 i) Q. If (1 + i) Q, then clearly ∈ n/4 ∪ ∈ 3 we have 1 + i D . If (1 i) Q, then (1 + i) = (1 i) /2 is an element ∈ ∈ of D as well.

2n/d Lemma 77. If d is a divisor of n with D then we have Q, = Q. d h i d Proof. As d is contained in D, the group Q, d /Q is the unique subgroup of h i 2n/d order d/ gcd(d, 2) of the cyclic group D/Q. As also Q, /Q has order 2n/d h i d/ gcd(d, 2) the subgroups Q, and Q, d = d Q of D containing Q must be equal. h i h i 2.3. THE GALOIS GROUP 41

Lemma 78. If D n is 6 and all subelds of Q() are generated by a power of then we have √∩ 3 D . ∈

Proof. As we have D n = 6 the integer n is divisible by 6. By lemma 77 we n/3 ∩ have Q( ) = Q(3). In lemma 71 we saw that there can only be one subeld of Q() of degree 2 over Q. So, the unique quadratic subeld of Q() is Q(n/2) = n/2 Q( ) = Q(√ 3 ). Thus, by corollary 18, we have √ 3 Q. 3 ∈

Lemma 79. If D n is 10 and all subelds of Q() are generated by a power of then we have 2 ∩n, 4 - n and √5 D . | ∈ Proof. In lemma 71 we showed that there can be at most one subeld of degree 2 over Q in Q(). As we have Q(√5 ) Q(5) Q() this subeld is Q(√5 ). We n/2 therefore have 2 n and √5 Q. |n/5 ∈ We have Q( ) = Q(5) by lemma 77. Hence, if 4 is a divisor of n then the elds Q(n/4) and Q(n/5) are subelds of Q() of degree 4. Both of these elds contain Q(n/2) = Q(√5 ). The eld Q(n/4) is isomorphic to Q(√4 5c2 ) for some n/2 c Q as is an element of √5 Q. As D does not contain i, the extension Q(∈n/4)/Q is not abelian and thus Q(n/5) is not equal to Q(n/4). Hence, there also exists some third subeld of Q() of degree 2 over Q(√5 ). We show that there exist at most two subelds of degree 4 of Q() generated by a power of . Let t be a divisor of n such that [Q(n/t) : Q] is equal to 4. For divisors n/r0 n/r r r0 of n we have Q( ) Q( ). So, for all prime divisors p of t we have [Q|(n/p) : Q] divides 4. As, lemma 68, xp n is reducible if and only if there exists some integer b with n = bp and n is minimal, we have [Q(n/p) : Q] p, p 1 and hence p 2, 3, 5 . Suppose 3 t, then, as x3 n is reducible,∈ { we hav}e n/3 ∈ { } | Q( ) = Q(3). But, this is impossible as Q(√5 ) is the unique subeld of degree 2 of Q()/Q. Therefore t is the product of a power of 2 and a power of 5. If 5 t then Q(n/t) contains the eld Q(n/5), if 5 - t, then 4 t and Q(n/t) contains the| eld Q(n/4). Thus Q(n/t) equals one of these two elds.|

2.3 The Galois group

In the previous section we showed one implication in theorem 56. In this section we start the proof of the other implication. First we show that Gal(Q(n, )/Q) can be viewed as a subgroup of Z/nZ o (Z/nZ). Then we prove that the intersection eld n/r Q(n) Q() is of the form Q( ) for some integer r. For the radicals that satisfy one of∩the conditions (i), (ii), (iii), (iv) or (v) of theorem 56 we explicitly determine this intersection eld. We compute its Galois group and use this to give an explicit description of Gal(Q(n, )/Q) as a subgroup of Z/nZ o (Z/nZ). For computing the subelds of Q() we use the Galois correspondence between subelds of Q(n, ) and subgroups of its Galois group over Q. To show that 42 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Gal(Q(n, )/Q) can be viewed as a subgroup of Z/nZ o (Z/nZ) we dene the map : Gal(Q( , )/Q) Z/nZ o (Z/nZ) n → by () = (k, l) where () = k and ( ) = l . n n n Before we prove that is an injective group homomorphism we recall the group- operation in the semidirect product Z/nZ o (Z/nZ). Let t, t0 be positive integers such that t0 t. The action of (Z/tZ) on Z/t0Z is given by multiplication: |

Z/t0Z (Z/tZ) Z/t0Z → (k, l) kl. 7→ This action induces the following group operation in Z/t0Z o (Z/tZ): for elements (k, l) and (k0, l0) Z/t0Z o (Z/tZ) we have ∈

(k, l)(k0, l0) = (k + lk0, ll0);

1 1 1 the inverse of the element (k, l) is (k, l) = ( l k, l ). Proposition 80. The map is an injective group homomorphism.

Proof. The map clearly is well-dened. It is injective because an element of Gal(Q(n, )/Q) is determined by the images of n and . Let and be elements of Gal(Q(n, )/Q) with () = (k, l) and () = (k , l ). Then we have

( ) = (l )l = l l and () = (k ) = l k +k . n n n n n Therefore we have

() = (k + l k , l l ) = (k , l )(k , l ) Z/nZ o (Z/nZ), ∈ hence is a homomorphism.

The following lemma shows that the intersection eld Q(n) Q() is of the form Q(n/r) for some r N. In corollary 82 we compute the degree∩ of some of the ∈ subextensions of Q(n, ) that we use later. Lemma 81. Let m be a positive integer and let K C be a eld containing a primitive m-th root of unity. Let be an element of C with m K. Then K() over K is a cyclic Galois extension of degree r for some r dividing∈m, and we have r K. ∈ Proof. This follows directly from theorem 14, the main theorem of Kummer theory (cf. [22], Chapter VIII, section 6, theorem 10).

Corollary 82. Let d Z be minimal such that d Q( ); then we have ∈ >0 ∈ n 2.3. THE GALOIS GROUP 43

[Q( , ) : Q)] = ϕ(n) d and n Q( ) Q() = Q(d). n ∩ Proof. By lemma 81 we have [Q(n, ) : Q(n)] = d. As the degree of n over Q is ϕ(n) we have [Q(n, ) : Q] = ϕ(n) d. d Because is an element of Q(n) the degree of over Q(n) Q() is at most d. Denote the degree of Q( ) Q() over Q by t. Then we have ∩ n ∩ ϕ(n) [Q( , ) : Q()] [Q( ) : Q( ) Q()] = . n n n ∩ t We get the following diagram

Q(n, ) ϕ(n)/t d

Q(n) Q()

ϕ(n)/t d Q(d) Q( ) Q() n ∩ t

Q. It follows that all inequalities in the diagram have to be equalities. As the degree d of over Q(n) Q() is d we conclude that Q( ) equals the intersection eld Q( ) Q(). ∩ n ∩ For the rest of this section we x the following notation. Let r denote the largest divisor of n with Q( ) Q() = Q(n/r); for the number d in corollary 82 we have n ∩ r = n/d. Let X denote the subgroup of (Z/nZ) for which the image under of Gal Q( , )/Q() is 0 o X and let G denote the image of the Galois group n { } Gal(Q(n, )/Q) in Z/nZ o (Z/nZ) under the map . ¡ ¢ n/r Now the Galois group of Q( ) over Q is isomorphic to (Z/nZ)/X. The action of X on Z/nZ induces an action on the subgroup (n/r)Z/nZ Z/rZ. By corollary 70 n/r ' we have r Q( ) Q(). So k X implies that r divides k 1. Therefore ∈ n/r the action of X on Z/rZ is trivial. Restricting Gal Q(n, )/Q to Gal(Q( )/Q) gives an embedding of the group Gal(Q(n/r)/Q) in Z/rZo(Z/nZ) /X such that 0 ¡ ¢ the following diagram commutes

n/r Gal Q(n, )/Q Gal(Q( )/Q)

¡ ¢ 0 proj Z/nZ o (Z/nZ) Z/rZ o (Z/nZ)/X. 44 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

n/r Proposition 83. Let G0 be the image of Gal(Q( )/Q) under 0 in Z/rZ o (Z/nZ)/X. Then G is the pre-image of G0 under the projection map proj. Proof. This simply follows from the degrees of the subextensions of Q()/Q. The number of elements in the kernel of the map proj is #(rZ/nZ o X) = n/r #X. By corollary 82 we have

n/r n/r #X = [Q(n, ) : Q(n)] [Q(n) : Q( )] n/r = [Q(n, ) : Q( )] #G = . #G0

The number of elements of the pre-image of G0 equals the number of elements in G, consequently these groups are equal. Proposition 84. Let be a radical satisfying one of the conditions (i), (ii), (iii), (iv) or (v) of theorem 56. Let k be dened by D = . ∩ n k If satises condition (i) then r is 1 or 2, If satises one of the conditions (ii), (iii), (iv) or (v) then r = k and Q( ) n ∩ Q() = Q(r). n/r n/r Proof. The extension Q( )/Q is abelian as Q( ) is contained in Q(n). There- 2n/r fore, by lemma 69, we have r Q = Q, . As r Q D, we have r k h i 2n/k n/r and hence r k. Lemma 77 gives k Q = Q, Q( ), as k D. This results in| the following chain of eld extensions.h i

Q(n/r)

2n/k Q( ) = Q(k)

2n/r Q( ) = Q(r)

By denition of r every odd divisor of k is also a divisor of r. Thus r k and k 2r. If k is 1 or 2, then clearly r 1, 2 . If satises condition (ii) |then n is |odd n/r 2n/r ∈ { } and hence Q( ) = Q( ) = Q(r) and r equals k. If satises condition (iii) n/4 n/2 then k = 4, so r 2, 4 . As 1 + i D we have Q( ) = Q( ) = Q(i). ∈ { } n/r ∈ Hence r equals k = 4 and Q( ) is Q(r). If satises condition (iv) then n/3 n/6 n/3 n/2 k = 6, so r 3, 6 . As 3 Q( ) Q, we have Q( ) = Q( , ) = ∈ { } ∈ \ n/r Q(3, √ 3 ) = Q(3). Hence r equals k = 6 and Q( ) is Q(r). Similarly, if n/10 n/5 n/2 satises condition (v) then Q( ) = Q( , ) = Q(5). We therefore have n/r r = k = 10 and Q( ) = Q(r). 2.3. THE GALOIS GROUP 45

Theorem 85. Let be a radical satisfying one of the conditions (i), (ii), (iii), (iv) or (v) of theorem 56. With notation as in proposition 83 and for a particular choice for 4, 6 and 10, we have

the trivial group if r = 1,

(0, 1), (1, ε) Z/2Z o (Z/nZ)/X if r = 2, { }   (0, 1), (1, 1) Z/3Z o (Z/3Z) if r = 3, G0 = { }  (0, 1), ( 1, 1) Z/4Z o (Z/4Z) if r = 4, { } (0, 1), ( 1, 1) Z/6Z o (Z/6Z) if r = 6, { }  (0, 1), (6, 1), (7, 7), (9, 3) Z/10Z o (Z/10Z) if r = 10, { }   where, for r =2, the element ε is the non-trivial element of the group (Z/nZ)/X. Proof. If r is 1, then the eld Q(n/r) is Q and its Galois group over Q is trivial, hence also G0 is trivial. If r = 2 then the Galois group Gal(Q(n/r)/Q) has two elements. The non-trivial element acts non-trivially on n/r and it is the restriction to Q(n/r) of an element of Gal(Q( )/Q) that acts non-trivially on . This gives us G0 = (0, 1), (1, ε) n n { } Z/2Z o (Z/nZ)/X, where ε denotes the non-trivial element of (Z/nZ)/X. n/r If r > 2 then by proposition 84, we see that Q( ) equals Q(r). In these cases the group (Z/nZ)/X is isomorphic to (Z/rZ). Calculating the action of n/r n/r Gal(Q( )/Q) on and on r gives the groups listed in the theorem. For n/2 example if r equals 10 then it follows from the fact √5 D that √5 Q. n/10 n/5 ∈ n/5 ∈k As Q( ) is abelian corollary 70 gives 10 Q( ) and thus 5 Q for some k 1, 2, 3, 4 . We therefore know thatn/10 is of the form l √5∈ c for some ∈ { } 5 c Q and for l = 3k mod 5. If, for example, is an element of Gal(Q(5)/Q) with ∈ 2 7 (5) = 5 , then (10) = 10 (as 7 is the unique element t of (Z/10Z) for which t is 1 2 2 2 modulo 5) and (√5) = 2( + ) + 1 = 2( + ) + 1 = √5. The element 5 5 5 5 automorphism corresponds under to the element (2l + 5, 7) Z/10Z o (Z/10Z) as we have ∈

(n/10) = (l √5)c = 2l√5c = 5 l l √5c = 2l+5 l √5c = 2l+5n/10. 5 5 10 5 5 10 5 10 n/10 If we compute the action of each element of Gal(Q(5)/Q) on and on 10, then we nd the group (0, 1), (2l + 5, 7), (4l + 5, 3), (6l, 9) . For the choice l = 1, this gives the group listed{ in the theorem. } In the next two sections we nish the proof of theorem 56. We determine the subgroups of (G) that correspond to the subelds of Q(). The next lemma gives the subgroups of (G) that correspond to subelds of Q() generated by a power of . Lemma 86. Let d be a divisor of n. Then the -image of the Galois group F = n/d Gal Q( , )/Q( ) equals (dZ/nZ o (Z/nZ)) G. n ∩ ¡ ¢ 46 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Proof. An element of Gal Q(n, )/Q corresponds under to a pair

¡(k, l) Z/n¢Z o (Z/nZ), ∈ k l n/d nk/d n/d with () = n and (n) = n. We have ( ) = n and thus F holds if and only if d is a divisor of k. That is, the image of F is the intersection∈ of G and dZ/nZ o (Z/nZ).

2.4 Subgroups

In this section we give a group theoretical theorem that we will use in the next section to nish the proof of theorem 56. Lemma 87. Let N be a group and let G be a group acting on N by group homomor- phisms. For every subgroup H of N oG containing 1 oG we have H = (H N)oG. { } ∩ Proof. Because we have 1 o G H it is clear that (H N) o G is contained in H. Let h be an element of H{. }Then we have h = n g for some∩ n N and some g G. As g is an element of H we have n H N, so H is contained∈in (H N) o G.∈ ∈ ∩ ∩ Theorem 88. Let r, n Z with r n. Consider the following groups: ∈ >0 | Z : a subgroup of (Z/nZ), X : a subgroup of Z contained in the kernel of the projection map (Z/nZ) (Z/rZ), Q: a subgroup of→Z/rZ o Z/X isomorphic to Z/X under projection on the second coordinate, G: the pre-image of Q under the projection map : Z/nZ o Z Z/rZ o Z/X, H : a subgroup of Gc→ontaining 0 o X that maps surjectively to Z/X under . { } These groups give the following commuting diagram

0 o X H G Z/nZ o Z { }

Q Z/rZ o Z/X

Z/X.

If for all divisors s of r with s = 1 we have Q sZ/rZ o Z/X and if for all f Z/nZ and all g Z the condition6 (f, g) H implies6 that (rf, 1) is an element of∈H, then we have ∈ ∈ H = (dZ/nZ o Z) G ∩ 2.4. SUBGROUPS 47 for some divisor d of n.

Proof. Let H0 be the kernel of the projection map H Z/X. Then clearly we have 0 o X H rZ/nZ o X. By lemma 87 we therefore→ see that { } 0

H0 = d0Z/nZ o X, for some divisor d0 of n with r d0. Let a be an element of Z/nZ such that there exists some b Z with (a, b) |H. Then we have by the second condition in the ∈ ∈ theorem (ra, 1) H0 and thus ra is an element of d0Z/nZ. It follows that a is d0 ∈ contained in r Z/nZ. Let d be the integer d0/r. We showed that H is contained in (dZ/nZ o Z) G. Below we prove that these groups are equal. We consider the chain of groups ∩

H = d Z/nZ o X H (dZ/nZ o Z) G dZ/nZ o Z. 0 0 ∩

As we have (dZ/nZ o Z : d0Z/nZ o Z) = d0/d = r we see that the index dZ/nZ o Z : H0 equals r (Z : X) = r #(Z/X). By denition of H0 we have (H : H ) = #(Z/X). ¡ 0 ¢ To determine the index of (dZ/nZ o Z) G in dZ/nZ o Z we consider the projection map Z/nZ o Z Z/rZ o Z/X. As∩ H dZ/nZ o Z maps surjectively to Z/X via Q we have Q gcd(→ r, d)Z/rZ o Z/X. By assumption we therefore have gcd(r, d) = 1. G Z/nZ o Z ∪ ∪ (dZ/nZ o Z) G dZ/nZ o Z ∩

Q Z/rZ o Z/X As G is the pre-image of Q also (dZ/nZ o Z) G is the pre-image of Q under the surjective projection map dZ/nZ o Z Z/rZ∩o Z/X. We conclude that → (dZ/nZ o Z : (dZ/nZ o Z) G) = (Z/rZ o Z/X : Q) = r. ∩ Summarising, we showed

dZ/nZ o Z : (dZ/nZ o Z) G H : H = r #(Z/X) = dZ/nZ o Z : H . ∩ 0 0 ¡ ¢¡ ¢ ¡ ¢ We conclude that H is equal to (dZ/nZ o Z) G. ∩ 48 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS 2.5 Conclusion of the proof

In this section we nish the proof of theorem 56 and prove theorem 57. We apply theorem 88 to show that all subelds of Q() are generated by a power of if satises one of the conditions (i), (ii), (iii), (iv) or (v) of theorem 56. We rst give a proposition and some lemmas. In proposition 94 we will use these results to show that the conditions of theorem 88 are satised for the groups that we are interested in.

Proposition 89. Let Z be a subgroup of (Z/nZ) and let G be a subgroup of Z/nZo Z consisting of the elements in (A X) (B Y ) for subsets A, B of Z/nZ and subsets X, Y of Z with ∪ X : a subgroup of Z of index 2, Y : the complement of X in Z, A: a subgroup of Z/nZ, B : some coset of A in Z/nZ. Let f be an element of Z/nZ and let g be an element of Z. For subgroups H of G with 0 o X H we have the following implication { } (f, g) H (kf, 1) H ∈ ⇒ ∈ for all k in the ideal y + 1: y Y Z/nZ. h ∈ i Proof. We use the group action in Z/nZ o Z, as described in section 2.3. First assume that f is an element of Z/nZ and that g is an element of Z with 1 (f, g) in H A X. As X is a subgroup of Z we have x X for all x X. Therefore w∩e have for all k N ∈ ∈ ∈ 1 k (kf, 1) = (f, g)(0, g ) H. ∈ Now assume that f is an elemen¡t of Z/nZ and¢ that g is an element of Z with (f, g) in H B Y . As X is a subgroup of Z of index 2 and X Y equals Z the 1∩ ∪ element z y is contained in X for all y, z Y . Therefore we see that ∈ 1 (f, y) = (f, g)(0, g y) H for all y Y. ∈ ∈ Squaring this element shows that (1 + y)f, y2 H for all y Y . As y2 is an element of X we see that ∈ ∈ ¡ ¢ 2 2 (1 + y)f, 1 = (1 + y)f, y (0, y ) H for all y Y. ∈ ∈ Let k be an elemen¡ t of y¢+ 1¡: y Y . Then¢ there are elements y Y and integers h ∈ i i ∈ ci with k = i ci(yi + 1). As (1 + y)f, 1 is contained in A X for all y in Y , we have ¡ ¢ P (kf, 1) = (1 + y )f, 1)ci H, i ∈ i X ¡ also in this case. 2.5. CONCLUSION OF THE PROOF 49

Lemma 90. Let p be a prime number and let t be a positive integer. Let X be t a subgroup of (Z/p Z) of index 2 and let Y be its complement. Then the ideal y + 1: y Y in Z/ptZ is the ideal generated by 2 if and only if one of the following hconditions∈holds:i

p is greater than 3, p is 2 or 3 and X is not equal to the kernel of the projection map t (Z/p Z) (Z/mZ), where m = 4 if p is 2 and m = 3 if p is 3. → Proof. The ideal y + 1: y Y is of the form psZ/ptZ for some 0 s t. The number of elemenhts of Y is∈ϕ(pit)/2. The number of elements of psZ/ptZis pt/ps. Therefore, for all primes p > 3, we see that

# y + 1: y Y #Y > #(pZ/ptZ) h ∈ i and hence in these cases the ideal y + 1: y Y must equal the full ring Z/ptZ = 2Z/ptZ. h ∈ i t t 1 t 1 t If p is 3 we have #Y = ϕ(p )/2 = 2p /2 = p = #(pZ/p Z). We conclude t t that y +1: y Y is 3Z/3 Z if Y equals y (Z/3 Z) with y = 2 mod 3 and that y + h1: y Y ∈equalsi Z/3tZ = 2Z/3tZ if{there∈ is some y Y with y = 1 }mod 3. h ∈ i ∈ If p is 2 clearly we have y + 1: y Y 2Z/2tZ. In this case it holds that t t 1 h t ∈ i t #Y = ϕ(p )/2 = 2 /2 = #(4Z/2 Z). The ideal y + 1: y Y is 4Z/2 Z if Y t h ∈ i t equals y (Z/2 Z) with y = 3 mod 4 and y + 1: y Y equals 2Z/2 Z if there is some{y ∈ Y with y = 1 mod 4. } h ∈ i ∈

Lemma 91. Let n Z and let X be a subgroup of index 2 of (Z/nZ) such that ∈ >0 the group X is not equal to the kernel of the projection map (Z/nZ) (Z/mZ) → for any element m 3, 4 dividing n. Let Y be the complement of X in (Z/nZ). Then the ideal y +∈1:{y }Y in Z/nZ is the ideal generated by 2. h ∈ i

Proof. For all divisors d of n we dene the projection map ϕ : (Z/nZ) (Z/dZ). d → As X is a subgroup of index 2 of (Z/nZ) the group ϕd(X) is a subgroup of index at r1 rs most 2 in (Z/dZ) and #ϕd(X) = #ϕd(Y ) for all divisors d of n. Let n = p1 ps be the prime factorisation of n. By lemma 90 the element 2 is in the image of r y + 1: y Y under the map ϕp i for all i. The Chinese remainder theorem h ∈ i i now tells us that 2 is contained in y + 1: y Y . If n is odd, it follows that y + 1: y Y is Z/nZ = 2Z/nZ. If hn is even all∈ elemeni ts of Y are odd, hence we hhave 2 ∈y + i1: y Y 2Z/nZ. We conclude that also in this case y + 1: y Y is 2Z/n∈Z.h ∈ i h ∈ i

Lemma 92. Let r be a divisor of n and suppose that r is even if n is even. Let Y be the pre-image of 1 under the projection map (Z/nZ) (Z/rZ). Then the ideal y + 1: y Y Z/nZ equals rZ/nZ. → h ∈ i 50 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Proof. Let p be a prime and assume that n = pk and r = pl for integers k, l with l k. l l If l = 0 then Y is the set 1+ p (Z/nZ), so p is an element of y+1: y Y and it follo6 ws y + 1: y Y = rhZ/ni Z. h ∈ i Now considerh the case∈ thati l is 0. By assumption p is odd and as the pre-image of 1 is the group (Z/nZ) we have 1 Y . Therefore 2 is an element of y + 1: y Y and as gcd(2, p) = 1 the ideal y + 1∈: y Y equals rZ/nZ. h ∈ i So the lemma holds for n ah prime p∈ower.i For general n, we rst determine the r1 rk ri prime factorisation of n = p1 pk . As the projection of y + 1: y Y in Z/pi Z ri h ∈ i is rZ/pi Z for all i we have y + 1: y Y = rZ/nZ by the Chinese remainder theorem. h ∈ i Lemma 93. Let n be a positive integer with 10 n and 4 - n. Let G be the pre-image | of the group (0, 1), (6, 1), (7, 7), (9, 3) Z/10Z o (Z/10Z) under the projection { }

Z/nZ o (Z/nZ) Z/10Z o (Z/10Z). → 1 Let be the projection map (Z/nZ) (Z/10Z) and dene X = (1). → Let f be an element of Z/10Z and let g be an element of (Z/nZ). Then for all subgroups H of G containing 0 o X we have the implication { } (f, g) H (10f, 1) H. ∈ ⇒ ∈ Proof. Dene one more projection map p: Z/nZ Z/10Z and dene subsets of 1 1 → 1 (Z/nZ) by Y = ( 1), U = (7) and V = (3) and subsets of Z/nZ by 1 1 1 1 A = p (0), B = p (6), C = p (7) and D = p (9). Then G is the group given by the elements

(A X) (B Y ) (C U) (D V ) in Z/nZ o (Z/nZ). ∪ ∪ ∪ Let H be a subgroup of G with 0 o X H. Let f be an element of Z/10Z { } and let g be an element of (Z/10Z) such that (f, g) is contained in H. As before (proposition 89) one shows that (f, g) in (A X) implies 1 k (kf, 1) = (f, g)(0, g ) H ∈ for all integers k, and (f, g) in (B ¡Y ) implies ¢ 1 2 2 ((1 + y)f, 1) = (f, g)(0, yg ) (0, y ) H ∈ for all y Y and thus (kf, 1) in H¡for all k y +¢ 1: y Y . Now ∈assume that (f, g) is contained in H∈ h(C U). ∈Firsti we remark that for all 1 ∩ u U we have g u X. Therefore we have ∈ ∈ 1 (f, u) = (f, g)(0, g u) H for all u U. ∈ ∈ 2.5. CONCLUSION OF THE PROOF 51

Then also (f, u)2 = (1 + u)f, u2) is an element of H (B Y ) and by multiplying ∩ by (0, yu 2) A X we have (1 + u)f, y) H for all y Y and for all u U. ¡ Squaring this∈elemen t gives ∈ ∈ ∈ ¡ 2 2 (1 + u)(1 + y)f, 1) = (1 + u)f, y) (0, y ) H for all u U and for all y Y, ∈ ∈ ∈ hence¡ (kf, 1) is an elemen¡ t of H for all k in (1 + u)(1 + y): u U, y Y . In lemma 92 we saw that y + 1: y Y equals 10h Z/nZ. Similarly one∈ shows∈ thati U h ∈ i equals the set u (Z/nZ) with u = 7 mod 10 and therefore u + 1: u U is 2Z/nZ. Because{ 4∈is not a divisor of n we have } h ∈ i

(1 + u)(1 + y): u U, y Y = 20Z/nZ = 10Z/nZ. h ∈ ∈ i The implication (f, g) H (C U) (kf, 1) H, ∈ ∩ ⇒ ∈ for all k in y + 1: y Y = 10Z/nZ follows. The proof for elements (f, g) of H (D V )his exactly ∈the samei as the proof for (f, g) in H (C U). We conclude ∩ ∩ that for all f Z/nZ and all g (Z/nZ) with (f, g) H we have (kf, 1) H for all k in y + 1∈: y Y . ∈ ∈ ∈ h ∈ i Before we conclude the proof of theorem 56 in the following proposition we x three homomorphisms. Let : Gal(Q(n, ) : Q) Z/nZ o (Z/nZ) be the map that we introduced in section 2.3. Denote by be the→ composition of the map and the projection Z/nZ o (Z/nZ) (Z/nZ) and let, as in section 2.3, → n/r 0 : Gal(Q( ) : Q) Z/rZ o (Z/nZ)/ Gal(Q( , )/Q()) → n be the composite of and the projection ¡ ¢

Z/nZ o (Z/nZ) Z/rZ o (Z/nZ)/ Gal(Q( , )/Q()) . → n ¡ ¢ Proposition 94. Let C be a radical over Q with n Z>0 minimal such that n ∈ ∈ Q. Let D be the group Q, . ∈If satises one of the folhlowingiconditions

(i) D and we have 6 - n or √ 3 / D , ∩ n 2 ∈ (ii) D = , ∩ n 3 (iii) D = and 1 + i D , ∩ n 4 ∈ (iv) D = and √ 3 D , ∩ n 6 ∈ (v) D = and both 4 - n and √5 D , ∩ n 10 ∈ then every subeld of Q()/Q is of the form Q(d) for some positive divisor d of n. 52 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Proof. Let be a radical satisfying one of the conditions (i), (ii), (iii), (iv) or (v). Let K be a subeld of Q() and let HK be the image of the group Gal(Q(n, )/K) under . Let r, as before, denote the maximal positive divisor of n with Q(n) Q() = Q(n/r). ∩ We consider three cases: rst we suppose that K contains the eld Q(n/r). Then we consider the case that the intersection K Q(n/r) is Q. Finally we suppose that K Q(n/r) is a eld not equal to Q(n/r)∩or Q. In each of these cases we prove ∩ that HK is a group of the form

dZ/nZ o (Z/nZ) Gal(Q( , )/Q) , ∩ n for some divisor d of ¡n. It follows by lemma¢ 86¡ that all subelds¢of Q() are generated by a power of . Suppose that K is a subeld of Q() containing Q(n/r). Let X be the group n/r Gal(Q(n, )/Q()) . As Q( ) is a subeld of K we have ¡ Gal¢ Q( , )/K Gal Q( , )/Q(n/r) . n n n/r By Galois theory the group¡ Gal Q(¢n)/Q(¡ ) is isomorphic¢ to a subgroup X of n/r (Z/nZ). As Q( ) K equals Q( ) Q() = Q( ) the inclusion n ∩ ¡ n ∩ ¢ Gal Q( , )/K Z/nZ o X n holds; suppose (a, y) ¡Gal(Q¡ (n, )/K) ¢¢with y (Z/nZ) X then (a, y) does not x Q( ) Q() ∈and we have a contradiction with∈ Q( ) \Q() K. As K is n ¡ ¢ n a subeld of Q∩() it follows ∩ 0 o X H Z/nZ o X. { } K By lemma 87 we see that HK is of the form dZ/nZ o X for some divisor d of n. As the group Gal Q( , )/Q(n/r) xes the eld Q(n/r) = Q( ) Q() and as n n ∩ n/r dZ/nZ since Q(n/r) is contained in K, we see that for this particular choice ¡ ¢ of d ∈we have dZ/nZ o (Z/nZ) G = dZ/nZ o X. ∩ By lemma 86 we conclude¡ that K equals¢ Q(n/d) for some divisor d of n. This completes the proof for subelds K of Q() containing Q(n/r). If r equals 1, then for every subeld K of Q() we have Q(n/r) K. For the rest of the proof we assume that r > 1 holds. Suppose that K is a subeld of Q() with K Q(n/r) = Q. We apply theorem 88 with the groups ∩

Z = (Z/nZ) Gal(Q( )/Q), ' n X = Gal Q( , )/Q() Z, n n/r Q = 0 Gal(Q( )/Q) Z/rZ o Z/X, ¡ ¡ ¢¢ G = Gal(Q( , )/Q) Z/nZ o (Z/nZ). ¡ n ¢ ¡ ¢ 2.5. CONCLUSION OF THE PROOF 53

First we check that the groups above satisfy the conditions of theorem 88. By denition of the map 0 we see that Q is isomorphic to Z/X and that G is the pre- image of Q under the projection map Z/nZo(Z/nZ) Z/rZoZ/X. theorem 85 shows that there does not exist a divisor s > 1 of rsuc→ h that Q is contained in n/r sZ/rZ o Z/X. As K Q( ) equals Q the group HK maps surjectively to Z/X = Gal(Q(n/r)/Q) .∩ Now let f be an element of Z/nZ and let g be an element of (Z/nZ) with (f¡, g) H . Assume¢ that satises condition (i). As r > 1 holds, the intersection ∈ K Q(n) Q() is of degree 2 over Q. By proposition 83 the group G is of the form (A X∩) (B Y ), with A, B, X, Y as in proposition 89. The group X is not the ∪ kernel of the projection map (Z/nZ) (Z/3Z) as otherwise we would have 3 X → n/r Q(n, ) = Q(), which implies that 3 is contained in Q(n) Q() = Q( ) and thus by lemma 69 we would have D . As is not ∩contained in D , 3 ∩ n 4 ∩ n also X is not the kernel of the projection map (Z/nZ) (Z/4Z). Therefore we can apply lemma 91 and we see that y + 1: y Y is 2→Z/nZ and proposition 89 proves that (rf, 1) H . Assume thath satises∈ ione of the conditions (ii), (iii) ∈ K or (iv). By proposition 84 we have Q(n) Q() = Q(r). Also in this case G is of the form (A X) (B Y ), with A, B∩, X, Y as in proposition 89, where X is ∪ isomorphic to Gal(Q(n, )/Q()). The group X is isomorphic to rZ/nZ and the set Y , by theorem 84, is y (Z/nZ) with y = 1 mod r . By lemma 92 we have { ∈ } (rf, 1) HK as r is even if n is even. If satises condition (v), then lemma 93 shows (∈rf, 1) H . ∈ K We conclude that all conditions in theorem 88 are satised and thus HK is of the form (dZ/nZ o (Z/nZ)) G for some divisor d of n. By lemma 86, we see that K is of the form Q(d) for some∩ divisor d of n. This completes the proof for subelds K of Q() with K Q(n/r) = Q. If satises one∩of the conditions (i), (ii), (iii) or (iv) we have either Q(n/r) K or K Q(n/r) = Q because [Q(n/r) : Q] = 2. It remains to show that K Q(n/r) is unequal∩ to both Q and Q(n/r) which can only happen if satises condition∩ (v). 54 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS

Assume that satises condition (v) and let K be a subeld of Q() with n/r n/r n/r K Q( ) not equal to both Q or Q( ) = Q(5). Then we have K Q( ) = Q(∩√5 ). Consider the following diagram. ∩

Q(n, )

Q(n) Q()

Q( ) Q() = Q( ) n ∩ 5

Q(√5 )

We apply theorem 88 with

Z Gal Q( )/Q(√5 ) , ' n X = (Gal Q( , )/Q() Z ¡ n ¢ n/10 Q = 0 Gal(¡ Q( )/Q(√¢5 )) ,

G = ¡Gal(Q(n, )/Q(√5 )) ¢ Z/nZ o (Z/nZ) ¡ ¢ By denition of 0 the group Q is a subgroup of Z/10Z o (Z/nZ)/X. One easily n/10 checks that the 0-image of Gal(Q( )/Q(√5 )) is contained in Z/10ZoZ/X. We will consider Q as a subgroup of Z/10Z o Z/X. Similarly as before Z, X, Q and G satisfy the denitions in theorem 88 and HK is a subgroup of G containing 0 o X that surjects to Z/X. By lemma 93 we see that (f, g) H implies that{(10} f, 1) H for all f Z/nZ and all g Z. It ∈ K ∈ K ∈ ∈ follows from theorem 88 that HK equals (dZ/nZ o (Z/nZ)) G for some divisor d of n in this last case as well. We can literally copy the proof∩of lemma 86 to show that this group corresponds to the eld Q(n/d). So, also all subelds K of Q()/Q with K Q(n/r) not equal to both Q or Q(n/r) are generated by a power of . ∩ Theorem 57 now easily follows:

Proof. The map is surjective if and only if satises one of the conditions (i), (ii), (iii), (iv) or (v) of theorem 56. In each of these cases we decide whether or not is also injective. Suppose that satises condition (i). As D n is contained in 2 it follows by minimality of n from lemma 68 that we have [∩Q() : Q] = n. For all divisors d n/d of n we have [Q( ) : Q] = d, therefore is injective. 2.5. CONCLUSION OF THE PROOF 55

Suppose that satises condition (ii). By corollary 82 the degree [Q() : Q] is d d 2n/3. We easily derive that [Q( ) : Q] is n/d if 3 is not contained in Q( ) and d d that [Q( ) : Q] is 2n/(3d) if 3 is an element of Q( ). Let d, t Z>0 such that d t d ∈ Q( ) = Q( ). If 3 is an element of Q( ) then we have 2n/3d = 2n/3t and thus d d equals t. If 3 is not contained in Q( ) then n/t = n/d and also in this case we have d = t. Hence the map is injective. Suppose that satises condition (iii). Then we have the equality Q(n/4) = n/2 Q( ) = Q(i), hence is not injective. Suppose that satises condition (iv). Then the elds Q(n/2) and Q(n/3) both equal Q(3). We conclude that is not injective. Suppose that satises condition (v). Then the elements n/5 and n/10 both generate the eld Q(5), and also in this case is not injective. 56 CHAPTER 2. SUBFIELDS OF RADICAL EXTENSIONS Chapter 3

Ramanujan’s nested radicals

An application of the theory from the previous chapters can be found in denesting radicals. Let K be a eld of characteristic 0 and let K be the algebraic closure of K. An element K can be represented by a radical expression if and only if the Galois group of the∈ normal closure of K() over K is solvable ([22], chapter VI, theorem 7.2). First we give some denitions. Denition 95. Let K be a eld and K some xed algebraic closure of K. We denote by K(0) the eld K itself and inductively dene the elds K (k) for k 1 as n (k 1) K( K with K for some n Z ). { ∈ ∈ ∈ >0} Denition 96. Let K be a eld and let K be some xed algebraic closure of K. We call K a nested radical if there exists some t N with K (t). ∈ ∈ ∈ Denition 97. A nested radical is of nesting depth n N over a eld K if n is the smallest integer for which this nested radical is contained∈ in K (n). Let K be a number eld. The nesting depth of in K is computable [7], although computing it is not easy in general. Susan Landau [20] gave an algorithm that constructs, for an element K with nite nesting depth, a representation of in K(r) with r 1 + nesting depth(∈ ). In this chapter we study nested radicals of the special form

√3 + 3 Q(2) ∈ q p for rationals and and give a necessary and sucient condition for √3 + √3 3 √3 to be denestable; that is, we show under what conditions √ + phas nesting depth smaller than 2. p We prove the following theorem.

57 58 CHAPTER 3. RAMANUJAN’S NESTED RADICALS

Theorem 98. Let , be elements of Q such that / is not a cube in Q. Then the following three statements are equivalent.

(1) The nested radical √3 + √3 is contained in Q(1).

(2) The polynomial t4 +p4t3 + 8 t 4 Q[t] has a rational root. ∈ (4m+n)n3 (3) There exist integers m, n such that = 4(m 2n)m3 holds.

Radicals are often not unambiguously dened; the expression √3 2, for example, can denote three dierent elements of C . In this chapter we will consider nested radicals dened over Q and only take square roots and cube roots of real numbers. We x the values for these expressions in C according to the following rules. Given a real number , the expression √3 will represent the unique real cube root of . When is a positive real number then √ represents the positive square root of . By i we denote a xed primitive fourth root of unity in C . If is a negative real number then √ represents i √ . 3.1 The problem and its history

As for many subjects in number theory the history of denesting nested radicals leads us to Srinivasa Ramanujan (1887-1920). One of the questions he sent to the Journal of the Indian Mathematical Society was Question 525 [28]:

Shew how to nd the square roots of surds of the form √3 A + √3 B and hence prove that

(i) √3 5 √3 4 = 1 √3 2 + √3 20 √3 25 , 3 q ³ ´ (ii) √3 28 √3 27 = 1 √3 98 √3 28 1 . 3 q ³ ´ In one of his notebooks [1], chapter 22, entry 23, we can nd the theorem this question must have been based upon.

Theorem 99. If m, n are arbitrary, then

m 3 4(m 2n) + n√3 4m + n q p = 1 3 (4m + n)2 + 3 4(m 2n)(4m + n) 3 2(m 2n)2 . (3.1) 3 ³p p p ´ Remark: In the examples in Ramanujan’s notebook the theorem is only applied for integers m and n. We will consider rationals m, n, as we study denestability over the eld Q. 3.2. DENESTING CONDITION 59

It is easy to verify that equation (3.1) holds for real numbers m and n by squaring both sides, but it is much harder to understand why it holds. However, for integers and with √3 + √3 contained in Q(1) it is not clear at all whether or not there exist integers m and n as given in the theorem. p In section 3.3 we show that radicals of the form √3 + √3 , for rational numbers and with / / Q3, are contained in Q(1) if and only if the polynomial ∈ p F = t4 + 4t3 + 8 t 4 / has a rational root. Given and such that F/ has a rational root we express √3 + √3 in terms of this root. At the beginning of section 3.4 we prove that F has a rational root if and only if there exist integers m, n such that p/ (4m + n)n3 = . 4(m 2n)m3 Note that for rational numbers and with / a cube in Q the nested radical √3 + √3 is always contained in Q(1): there exist c Q with c3 = / and d C ∈ ∈ 6 3 √3 √ pwith d = such that we have √ + = d 1 + c. Theorem 98 therefore shows the generality of Ramanujan’s formula. p Example 100. If the quotient / is a cube, then the statements (2) and (3) in theorem 98 are still equivalent, but the statements (1) and (2) are not equivalent. As 3 remarked above for / Q statement (1) holds. However, if we have / = 1, 4 ∈ 3 2 2 then the polynomial t + 4t + 8 t 4 = (t + 2)(t + 4t 2) has no rational roots. 3 There also exist and with / Q for which statement (2) does hold. For example if ∈ 73 (4 2 + 7)73 = = 29 4( 2 2 7)( 2)3 the polynomial t4 + 4t3 + 8 t 4 has two rational roots: 7/4 and 7/2. At the end of section 3.4 we work out some examples and show, for integers and , that it is not possible in general to nd integers m and n such that

= 4(m 2n)m3 and = (4m + n)n3 although there do exist m and n such that / equals (4m + n)n3/(4(m 2n)m3). 3.2 Denesting condition

In this section we show that a nested radical of the form √ for some in Q(√3 ) 3 (1) with Q Q is contained in Q if and only if there exist f Q and e Q(√3 ) such that∈ \equals f e2. ∈ ∈ 60 CHAPTER 3. RAMANUJAN’S NESTED RADICALS

We use the following notation. If K/Q is a eld extension then K norm denotes the normal closure of K over Q. If G is a group then G0 denotes the commutator subgroup of G and G00 denotes the group (G0)0 and, as in chapter 2, we denote a primitive n-th root of unity by n.

Lemma 101. Let be an element of Q Q3, let be an element of Q(√3 ) Q and norm \ norm \ let K be the eld Q() = Q(, 3). Let G be the group Gal(Q(√ ) /Q) and let 1 = , 2, 3 be the conjugates of in K/Q. Then the following implication holds

2 G00 = 1 = K . { } ⇒ 2 3 ∈

Proof. Suppose that 2 3 is not a square in K. Then K(√2 3 ) is an extension of degree 2 over K. As , and are conjugates, also and are non- 1 2 3 1 2 1 3 squares in K. It follows that √1 2 is not contained in K(√2 3 ) as otherwise, by corollary 18, we have K 2. 1 3 ∈ Dene L as the eld K(√2 3, √1 2 ). We saw above that [L : K] equals 4 and we get the following chain of eld extensions

L

V4

K

S3 Q.

This gives a contradiction with G00 = 1 as we will see below. { } norm Suppose that G00 equals 1 . First we remark that L is contained in Q(√ ) . { } Therefore we have Gal(L/Q)00 G00 = 1 . There is an exact sequence { } 0 Gal(L/K) Gal(L/Q) Gal(K/Q) 0 .

As Gal(L/K) V4 is abelian, this gives an action of Gal(K/Q) S3 on Gal(L/K) dened by the'map '

Gal(L/K) Gal(K/Q) Gal(L/K) → 1 sending (v, ) to v Gal(L/K) where is an element of Gal(L/Q) with = . ∈ |K We remark that Gal(L/K) is the group 1, 1, 2, 12 , where 1 and 2 are dened by { }

: : 1 2 3 7→ 2 3 2 2 3 7→ 2 3 p p p p 1 2 7→ 1 2 1 2 7→ 1 2 p p , p p . 1 3 7→ 1 3 1 3 7→ 1 3 p p p p 3.2. DENESTING CONDITION 61

The isomorphism from Gal(K/Q) to S3 is given by the action of Gal(K/Q) on the set of points 1, 2, 3. Denote by the map from Gal(L/Q) to Gal(K/Q) in the exact sequence above. When a, b and c in Gal(L/Q) are lifts under of the maps corresponding to the elements (12), (23) and (13) in S3 then we have 1 1 a2a 2 = 1 1 1 b(12)b (12) = 2 1 1 c1c 1 = 12 and thus Gal(L/K) is contained in Gal(L/Q)0. As (123) equals (13)(23)(13)(23) 1 1 we see that the element d = cbc b Gal(L/Q)0 is a lift under of (123). As ∈ 1 1 Gal(L/Q)0 is abelian and Gal(L/K) is contained in Gal(L/Q)0 we have dvd v = 1 1 for all v Gal(L/K). This gives a contradiction as d1d (√2 3 ) equals √2 3. We conclude∈ that is a square in K. 2 3 Theorem 102. Let be an element of Q Q3 and let be an element of Q(√3 ). \ Denote by G the Galois group of the normal closure of Q(√ ) over Q. Then the following are equivalent: (1) The element √ is contained in Q(1).

(2) The group G00 is 1 . { } 2 (3) There exist f Q and e Q() such that = f e . ∈ ∈ Proof. Implication (3) (1) is trivial. ⇒ (1) If √ is an element of Q then there exist 1, . . . , k C and a positive integer n with n Q for all i such that √ is contained in Q( , ∈, . . . , ). Let L denote the i ∈ n 1 k eld Q(n, 1, . . . , k). As both L/Q(n) and Q(n)/Q are abelian extensions, the second derived subgroup Gal(L/Q)00 of Gal(L/Q) is trivial. Now the normal closure of Q(√ ) over Q is contained in L and hence also the second derived subgroup of its Galois group is trivial. If is a rational number then implication (2) (3) is trivial. Assume that is ⇒ not contained in Q and let K be Q(, 3). Assume that G00 is 1 . Let 1 = , 2, 3 { } 2 be the conjugates of in K. In the previous lemma we showed that 2 3 K . Now let be an element of K with 2 = . ∈ 2 3 Suppose that is not contained in Q(). As 2 3 is an element of Q() and K 2 is Q(, √ 3 ) we see by corollary 18 that 2 3 equals 3 for some in Q(). It follows N Q()()2 = N Q()( ) N Q()( ) = 27 N Q()()2. Q Q 2 Q 3 Q As we have 27 / Q2 this gives a contradiction, so is an element of Q(). We have ∈ 1 2 3 Q() 2 = = N () Q 2 3 Q() 1 and take f = NQ () and e = . 62 CHAPTER 3. RAMANUJAN’S NESTED RADICALS 3.3 A special polynomial

3 3 Given the nested radical √ + √ we dene a polynomial F/ depending on and . p

Denition 103. Let , be elements of Q. The polynomial F/ Q[t] is dened by ∈ F = t4 + 4t3 + 8 t 4 . /

In this section we show that √3 + √3 is contained in Q(1) if and only if the polynomial F has a rational root. In the previous section we saw that the radical / p

√3 + 3 = √3 1 + 3 / q p q q p (1) 3 is contained in Q if and only if there exist f Q and e Q( / ) such that ∈ ∈ 2 1 + 3 / = f e2. In theorem 104 we write e on the basis 1, 3 /, 3 / of p Q( 3 / ) as a vector space over Q and compare the coecients in the resulting p p p equation to obtain F . At the end of the section we provide some examples. p /

Theorem 104. Let , be elements of Q such that / is not a cube in Q. The nested radical √3 + 3 q (1) p is contained in Q if and only if the polynomial F/ has a root in Q.

Proof. The radical √3 + √3 is contained in Q(1) if and only if 1 + 3 / is (1) contained in Q . Byp theorem 102, this holds if and only if there existq f pQ and x, y, z Q with ∈ ∈ 2 1 + 3 / = f x + y 3 / + z 3 (/)2 . µ ¶ p p q The elements 1, 3 / and 3 (/)2 are linearly independent over Q. Therefore coecients of like powers on both sides must be equal. This leads to the following p p equations:

1 = x2 + 2yz , (3.2) f 0 = y2 + 2xz, (3.3) 1 = z2 + 2xy. (3.4) f 3.3. A SPECIAL POLYNOMIAL 63

We may assume that z = 0 (since z = 0 implies y = 0 and this gives a contradiction in (3.4)). Substitution of6 x 1 y 2 = , z 2 z ³ ´ in (3.2) and (3.4) yields y 4 y 4 + 8 = , z z f z2 ³ ´ y 3 1 = . z f z2 ³ ´ The combination of the above two equalities leads to the following quartic equation in y/z y 4 y 3 y + 4 + 8 4 = 0. z z z ³ ´ ³ ´ ³ ´ 3 3 (1) So, if √ + √ is contained in Q then F/ has a rational root. For the other implication we assume that a rational root s of F is given. We p / use the relations y x 1 y 2 y 3 1 s = , = and = z z 2 z z f z2 ³ ´ ³ ´ that we derived above and nd

2 3 √3 + 3 = z2 f √3 1 y + y 3 / + (/)2 2 z z q µ ¶ p q ¡ ¢ p q = z2 f √3 1 s2 + s 3 / + 3 (/)2 2 q µ q ¶ 1 3 p = 1 s2 √2 + s 3 + 3 2 . (3.5) s3 2 ³ p p ´ 3 3 p (1) So √ + √ is an element of Q if and only if F/ has a rational root.

Corollaryp 105. Let , be elements of Q, such that / is not a cube in Q. Then the nested radical √3 + 3 q is contained in Q(1) if and only if there existspan element s Q such that ∈

1 3 √3 + 3 = 1 s2 √2 + s 3 + 3 2 , √t 2 q p ³ p p ´ where t = s3. We give some examples. 64 CHAPTER 3. RAMANUJAN’S NESTED RADICALS

Example 106. We compute the rational roots of F 4/5 and F 27/28 to derive the equalities (i) and (ii) from section 3.1. As we have

F 4/5( 2) = F 27/28( 3) = 0, we nd the equalities

1 √3 5 √3 4 = 2√3 25 + 2√3 20 + 2√3 2 √36 q ³ ´ = 1 √3 25 + √3 20 + √3 2 3 ³ ´ and

1 3 3 √3 28 √3 27 = 9 √282 3√3 27 28 + √272 √272 2 q ³ ´ = 1 √3 98 + √3 28 + 1 . 3 ³ ´ Example 107. The nested radicals do not have to be real. For example, when we take = 27 and = 7 we get √3 + 3 = 4√3 2 + √3 7. q q For these and we have the polynomialp

4 3 56 28 F 7/27 = t + 4t t + , 27 27 with rational root 1/2. Substituting this in equation (3.5) gives

1 3 4√3 2 + √3 7 = 1 √214 1 3 7 27 + √3 49 √7 16 8 2 q ³ p ´ = i 1 2√3 4 + 2√3 14 + √3 49 . 3 ³ ´ Example 108. Until now, we only considered nested radicals for which both and were integers. Now let = 2 and = 1/2, then / = 1/4. The rational root of F 1/4 is 1. This leads to the equality 1 √3 2 3 1/2 = 1 √3 4 √3 1 + 3 1/4 , 3/2 2 q p ³ p ´ which we simplify to p

√2 √3 2 3 1/2 = 1 + 1 √3 2 1 √3 4 . √3 2 2 q p ³ ´ 3.4. COMPARISON TO RAMANUJAN’S METHOD 65 3.4 Comparison to Ramanujan’s method

In this section we derive equation (3.1) from equation (3.5). We show in theorem 109 that integers m and n that satisfy

(4m + n)n3 = 4(m 2n)m3 give a rational root n/m of F/. Conversely, if s is a rational root of F/ and s = r/t for integers r and t, then we can take n = r and m = t and the equality above holds. Theorem 98 follows directly from it. Moreover we show that there are , Q with √3 + √3 Q(1) for which there do not exist rational numbers m and n∈ with = 4(m 2n)m∈3 and = (4m + n)n3. p

Theorem 109. Let , be elements of Q such that / is not a cube in Q. Let s be a rational number. Then s is a root of the polynomial F/ if and only if there exist integers m, n such that s = n/m with

(4m + n)n3 = . 4(m 2n)m3

Proof. Let s be a rational root of F/. Then we have

s4 + 4s3 + 8 s 4 = 0, so s3(s + 4) equals 4(1 2s)/. Remark that s = 1/2 as F (1/2) = 9/16. We 6 / see that s is a rational root of F/ if and only if the equality

s3(s + 4) = 4(1 2s) holds. Writing s = n/m for integers n, m gives

(4m + n)n3 = 4(m 2n)m3 which proves the theorem.

Suppose that there exist integers m and n with F/(n/m) = 0 and

(4m + n)n3 = . 4(m 2n)m3 66 CHAPTER 3. RAMANUJAN’S NESTED RADICALS

3 3 Then there exists some c Q with = c 4(m 2n) m and = c (4m + n) n . We use equation (3.5) to ∈derive equation (3.1); by theorem 109 we have s = n/m .

1 3 √3 + 3 = 1 s2 √2 + s 3 + 3 2 s3 2 q ³ ´ p 2 p p 1 1 n 2 3 2 2 = p 2 m (4(m 2n)) c √9 c n4 2 m ³ p + n nm 3 4(m 2n)(4m + n) c2 + n2 3 (4m + n)2 c2 m √3 c2 p p ´ = 3 2(m 2n)2 + 3 4(m 2n)(4m + n) 3√c ³ p p + 3 (4m + n)2 (3.6) ´ If we take c = 1 in equationp 3.6, this gives equation 3.1 again. In the case that both and are integers the question arises if there always exist integers m, n such that = (4m + n)n3 and = 4(m 2n)m3. Unfortunately, this is not the case as the examples 111 and 112 show.

Lemma 110. Let and be elements of Q and let m and n be integers with (4m + n)n3 = . 4(m 2n)m3 Let a and b be coprime integers with / = b/a. Then we have both

n 3 m 3 4b and a. gcd(m, n) gcd(m, n) µ ¶ ¯ µ ¶ ¯ ¯ ¯ Proof. Let a and b be coprime ¯integers with / = b/a. If we¯take d = gcd(m, n) then also the quotient of (4m/d+n/d)(n/d)3 and 4(m/d 2n/d)(m/d)3 equals /. Now replace m by m/d and replace n by n/d. Then gcd(m, n) = 1 and we have (4m + n)n3 a = 4(m 2n)m3 b, where also gcd(a, b) = 1. If p is a prime diving m, then p does not divide (4m+n)n3. Similarly, if p is a prime dividing n, then p does not divide (m 2n)m3. Hence m3 is a divisor of a and n3 is a divisor of 4b. Example 111. Let us look at the nested radical √3 5 √3 4 from example 106 again. When we take = 4 and = 5, we see that for m = n = 1 we have p = 4(m 2n)m3 and = (4m + n)n3. So, again we nd the equality √3 5 √3 4 = 1 √3 25 √3 20 √3 2 . 3 q ³ ´ 3.4. COMPARISON TO RAMANUJAN’S METHOD 67

However, when we take = 5 and = 4, which is a more obvious thing to do, there do not exist integers, or even rationalnumbers, m and n with = 4(m 2n)m3 and = (4m + n)n3, although we have ( 4 + 2)23 = . 4( 1 4)( 1)3 Example 112. Now look at 5√3 7 8. For this nested radical, for each of theq choices = 7 53 and = 83, we are not able to nd integers m, n such that = (4m + n)n3 and = 4(m 2n)m3, although we have 7 53 (16 + 5)53 83 ( 20 + 8)83 = and = . 83 4(4 10)43 7 53 4( 5 16)( 5)3 To give 5√3 7 8 as an element of Q(1), we take = 83 and = 7 53 and use the equality p 7 53 (16 + 5)53 = = . 83 4(4 10)43 As the quotient of and 4(4 10)43 is 3, we see that √3 3 8 + 5√3 7 = 4 3 4(4 10) + 5√3 16 + 5. q q 1 q p Using equation (3.6) with c = 3 we nd

3 √3 3 8 5√3 7 = √212 + 2√3 63 + 2√3 9 . 3√3 q ³ ´ Finally, one could wonder, as Ramanujan’s formula (3.1) lacks obvious symmetry, what happens if the roles of and interchange. The answer is that we nd the same simplication, putting p = n/√2 and q = m√2. (This explains why in example 111 we were not able to nd integers m and n such that = 4(m 2n)m3 and = (4m + n)n3 for the choice = 5 and = 4.) If we dene p = n/√2 and q = m√2, it holds that = 4(m 2n)m3 = (4p + q)q3 = (4m + n)n3 = 4(p 2q)p3. Since we have 2 (4m + n)2 = 2√2q √2p = 2(p 2q)2 2(m 2n)2 = 2³ 1 q2 + 4pq +´8p2 = (4p + q)2 2 4(m 2n)(4m + n) = 4 2q2 + 7pq 4p2 = 4(p 2q)(4p + q), ¡ ¢ ¡ ¢ 68 CHAPTER 3. RAMANUJAN’S NESTED RADICALS we may replace m and n simply by p and q in the right hand side of Ramanujan’s formula:

3 + √3 = 1 3 (4p + q)2 3 4(p 2q)(4p + q) + 3 2(p 2q)2 . 3 qp ³ p p p ´ Chapter 4

Nested radicals of depth one

Borodin, Fagin, Hopcroft and Tompa [5] gave conditions for the nested radical a + b√r F (2) to be an element of F (1), for a real eld F and a, b, r elements ∈ 3 √3 pof F . In the previous chapter we gave conditions for √ + , with and in Q, to be an element of Q(1) and more generally we proved theorem 102. We would like p to have a general condition for elements of K(2) to be contained in K(1), where K is a eld of characteristic 0. Richard Zippel [40] formulated the following conjecture for denesting radicals of nesting depth 2 that consist of one term.

Denition 113. Let K be a eld and let L/K be a radical extension. We say that L/K is a simple radical extension if there is an element in L with minimal polynomial xr r for some r Z such that L = K(). ∈ >0 Conjecture 114 (Zippel). Let K be a eld of characteristic 0 and let K be a xed algebraic closure of K. Let 1, . . . , k in K be such that there exist positive integers d1 dk d1, . . . , dk with 1 , . . . , k K. Dene L as the eld K(1, . . . , k). Let F be a composite of ∈simple radical extensions over K. Let be an element of L and let n be a positive integer. If √n is an element of the composite LF , then there exist integers s1, . . . , sk and an element w of K such that the product s1 sk n w is an element of L . 1 k In example 117, section 4.1, we show that the conjecture is false. In proving denesting conditions, most of the diculties are caused by adjoining roots of unity in a eld extension. In example 117 we consider a eld extension generated by a root of unity in which we nd a nested radical of depth 2 over Q that violates conjecture 114. So, it is not sucient that our eld extension is a composite of simple radical extensions; we need something stronger. In this chapter we denote for positive integers n a primitive n-th root of unity by n. Let K be a eld of characteristic 0

69 70 CHAPTER 4. NESTED RADICALS OF DEPTH ONE and let K be such that t K for some integer t. If the minimal polynomial ∈ ∈n n of over K is not of the form x for some n Z>0, then K() K contains roots of unity. In the reformulationof conjecture 114∈we therefore give a\condition on the roots of unity in the extension F/K instead of demanding that F is a composite of simple radical extensions. If for some odd prime p the root of unity is contained in K, then for all r Z p ∈ >0 the extension K(pr )/K is cyclic and the minimal polynomial of pr is of the form n r x a for some n Z>0 and some a K. So, adjoining a p -th root of unity to a eldK with K∈ gives a simple radical∈ extension. The same holds for adjoining p ∈ 2r for some integer r if i is contained in K. So, if a radical extension is simple, it is pure (denition 61) as well. For a composite of radical extensions this is not necessarily true. In section 4.2 we prove Zippel’s conjecture for pure extensions.

Theorem 115. Let K, L and 1, . . . , k be dened as in conjecture 114. Assume that L/K is pure. Let n be a positive integer, let be an element of L and let F be a pure radical extension of K with F K (1). If we have F with n ∈ = , then there exist elements w K and e L and integers s1, . . . , sk such s ∈ ∈ that is equal to w en i . i i Another way to makeQsure that the minimal polynomials of the generating rad- icals are of the form xn n for some positive integer n is to adjoin all roots of unity to the ground eld. We dene K = K((K )) for a xed algebraic closure K of K. ∞ Inspired by Zippel’s idea we formulate for a eld K a condition for elements of K(2) to be contained in K(1).

Theorem 116. Let K, L and 1, . . . , k be dened as in conjecture 114. Let be an element of L and let n be a positive integer. If there exists a eld F K(1) and an element F such that n = , then there exist elements ∈ n si w K L and e L and integers s1, . . . , sk such that is equal to w e i . ∈ ∞ ∩ ∈ i This theorem we will prove in section 4.3. Q

4.1 Counterexample to Zippel’s conjecture

In the introduction of this chapter we gave a conjecture due to Richard Zippel. He gives a condition for the existence of a special kind of denesting of radical expressions n (1) of the form √ for some n Z>0 and some in K . This denesting is required to be an element of some comp∈osite of simple radical extensions. To construct a counterexample to conjecture 114 we rst forget about the con- dition on the eld F . A well known radical equation is

1 + √ 1 = √8 4. q 4.2. DENESTING WITHOUT ROOTS OF UNITY 71

If we take, in conjecture 114, ground eld Q, L = Q(i), = 1 + i and n = 2 then we see that there exists a denesting √8 4 in Q(i) Q(√8 4 ). But there do not exist s Z and w Q such that ∈ ∈ w is (1 + i) is a square in Q(√ 1 ); the norm of w is (1 + i) is 2 w2 and this is not a square in Q. So, without the extra condition on F the conjecture does not hold. However, we can embed Q(√8 4 ) in a composite of simple radical extensions as we will see in example 117. This directly gives a counterexample to the conjecture itself.

Example 117. Let K = Q, L = Q(i), = 1+i, n = 2 and dene F = Q(√8 2, √8 2 ). As both Q(√8 2 ) and Q(√8 2 ) are simple radical extensions, F is of the form required in conjecture 114. But similarly as before, there do not exist s Z and w Q such that w is (1 + i) is a square in Q(√ 1 ). ∈ ∈ In my opinion, the eld F in this example clearly is not of the form that Zippel had in mind for the conjecture. Probably he wanted his conjecture to give a condition for denesting nested radicals of depth two consisting of one term without using roots of unity in the denesting. That is, without using any other roots of unity than those already contained in the ground eld.

4.2 Denesting without roots of unity

In this section we give some condition for denesting without roots of unity. First we have to dene what we mean by this. Assume that K is a eld of characteristic zero and let L be a subeld of K(1) with [L : K] nite. Let be some element of L. A denesting of √n is a denesting without roots of unity if we can nd an element in some pure subextension F of K(1)/K with n = . Below we use theorem 58 to prove a statement for non-Kummer extensions similar to corollary 18. The only extra condition is that the extension is pure.

Proposition 118. Let K be a eld of characteristic 0 and let K be a xed alge- braic closure of K. Let , 1, . . . k K be such that there exist natural numbers d d ∈ n, d , . . . , d with n, 1 , . . . k K. Let L be the eld K(, , . . . ). If L/K 1 k 1 k ∈ 1 k is a pure extension then is an element of K 1, . . . , k if and only if there exist b K and l , . . . l N such that can be written in the form ∈ 1 k ∈ ¡ ¢ k = b li . i i=1 Y

Proof. Take groups C = K, (i I) and C = K, , (i I) . Then we 1 h i ∈ i 2 h i ∈ i have L = K(C1) = K(C2). As the extension L/K is pure, both C1 and C2 satisfy the conditions R1 and R2 from theorem 58. We conclude that the index (C1 : K) 72 CHAPTER 4. NESTED RADICALS OF DEPTH ONE

equals the index (C2 : K). As C1 is contained in C2 the groups are equal. The element therefore is contained in C1, hence we have k = b li , i i=1 Y for an element b of K and integers l1, . . . , lk. If is of the above form, then clearly is an element of K(1, . . . , k). We apply this proposition and the theory of cogalois extensions described in section 2.1 to prove theorem 115. Proof. Let be an element of F with n = . We consider the following diagram F

L F ∩

K. As F/K is a pure extension, both the extensions F/(L F ) and (L F )/K are pure (1)∩ ∩ as well. As F is a radical extension contained in K there are radicals 1, . . . , t ni in F and integers n1, . . . , nt with i K for all i in 1, . . . , t , such that the i generate F over L F . We apply prop∈osition 118 with{ground}eld L F and ∩ ∩ equal to . This gives us an element b (L F ) and integers l , . . . , l with ∈ ∩ 1 t t = b li . i i=1 Y So, there exists some in F with m K for some positive integer m, such that equals b . Raising both sides of the ∈equation to the power n gives = bn n. As both and b are elements of L F , also n is an element of L F . As a subextension of the pure extension L = K(∩, . . . , )/K, by proposition 62,∩ also (L F )/K is a 1 k ∩ radical extension generated by monomials in 1, . . . , k. We apply proposition 118 n again, this time with ground eld K and equal to . We nd an element c in K and integers s1, . . . , sk with k n = c si . i i=1 Y Hence we have k = bn n = bn c si , i i=1 Y for some b in (L F ) L, some c K and integers s , . . . , s . ∩ ∈ 1 k 4.3. DENESTING ALLOWING ROOTS OF UNITY 73 4.3 Denesting allowing roots of unity

Let K be a eld of characteristic 0. Let n be a positive integer and let be an element of K(1). In theorem 116 we give a necessary condition for the nested radical √n to be of depth at most one. We rst prove this theorem, next we derive some results in special cases, and nally we give an example that shows that the condition in this theorem in general is not sucient. Proof of theorem 116. Let be an element of L and let n be a positive integer. Assume that there exist some eld F K(1) and some in F with n = . Then, (1) there exists a eld L L0 LK of nite degree over L with in L0 that is ∞ ∞ generated by radicals of depth 1 over K; that is L0 equals L (1, . . . , l) for elements ∞ mi i K for which there exist positive integers m1, . . . , ml with i K for all i. Applying∈ corollary 18 with ground eld L and equal to gives ∈ ∞ L , ∈ ∞ m for an element of K satisfying K for some m in Z>0. We now study the action of elemen∈ ts of Gal L ( )/L on . A basis of L ∞ ∞ over K is given by a subset c1, . . . , cs of the basis of the extension L/K. There are ∞ ¡ ¢ unique a1, . . . , as in K such that we have ∞ = (a c + + a c ). 1 1 s s Let be an element of Gal(L ()/L), then we have ∞ () = (a )c + (a )c . () 1 1 s s Because () equals and () is for roots of unity and there exists some root of unity with (a )c + + (a )c = (a c + + a c ). 1 1 s s 1 1 s s

As the ci form a basis of L over K we have (ai) = ai for all i 1, 2, . . . , s . ∞ ∞ ∈ { } It follows that the quotient ai/a1 is contained in L for all i. Therefore we have a L, ∈ m where a is contained in K and is an element of K with K for some ∞ ∈ m Z>0. ∈ m Raising to the powers m respectively n shows that ( a) is an element of K n ∞ and that ( a) is contained in L. We apply corollary 18 again, this time with n m ground eld K and equal to ( a) . As is an element of K we see that ∞ ∞ = w si , i i Y 74 CHAPTER 4. NESTED RADICALS OF DEPTH ONE

holds for certain w K and s1, . . . sk N. This gives the following equation ∈ ∞ ∈ = w en si , i i Y where we have e L, w K and si N for all i. Because all the other terms are elements of L it follo∈ ws that∈ ∞w is contained∈ in L as well. We now have the following result similar to theorem 102. Corollary 119. Let K be a eld of characteristic 0 and let K be a xed algebraic closure of K. Let , . . . , K be such that there exist d in Z with di K for 1 k ∈ i >0 i ∈ all i and let L be K (1, , k ). Let be an element of L and let n be a positive (1) integer with gcd(n, di) = 1 for all i 1, . . . , k . If there exists a eld F K and an element in F with n = , then∈ {there exist} w K L and e L such that equals w en. ∈ ∞ ∩ ∈ Proof. Let be an element of L and let n be a positive integer. Assume that there exist some eld F K(1) and an element of F with n = . From the theorem 116 it follows = w en li , i i Y for e L, w K and li N for all i. As we have gcd(n, di) = 1 for all i, there are ∈ ∈ ∞ ∈ l0 N with nl0 = l mod d . We conclude that there is some w0 w K with i ∈ i i i ∈ n 0 li = w0 e . i à i ! Y As is an element of L and also 1, . . . , k and e are elements of L, we see that w0 is contained in L as well. We therefore have w0 K L. ∈ ∞ ∩ Let K be a real eld of characteristic 0 and let a, b, r be elements of K with √r / K such that the nested radical a + b√r is real. Borodin et al [5] showed that∈the following are equivalent. p The expression a + b√r is an element of K(1). We have √4 r√spa + b√r K(√r ) or √s a + b√r K(√r ) for some s in K. ∈ ∈ p p Either √a2 b2r or r(b2r a2) is an element of K. Example 120. The radicalp 12 + 5√6 is contained in Q(1) because the number 6(52 6 122) = 36 is a square in Q. To determine w in Q and e in Q(√6 ) such l p ∞ that 12 + 5√6 = w e2 √6 for some l 0, 1 we remark ∈ { } 12 + 5√6 = √6(5 + 2√6 ). 4.3. DENESTING ALLOWING ROOTS OF UNITY 75

If there exist a, b, c in Q with c(a + b√6 )2 = 5 + 2√6 then a and b satisfy 10ab = 2(a2 + 6b2). For a = 2 and b = 1 this equality holds and indeed we have

1 12 + 5√6 = √6 (2 + √6 )2. 2

This gives

12 + 5√6 = √4 6 (√2 + √3 ), q where √4 6 is the positive real fourth root of 6.

We see that both in this case and in theorem 102 it holds that not only we have w K L but we even have w K. However, we will not be able to prove that w is con∈ tained∞ ∩ in K in general as w∈e saw in example 117. In some special cases theorem 116 gives a both necessary and sucient condition for a nested radical to be of depth one as the following corollary shows.

Lemma 121. Let , . . . , be elements of R with di Q for odd positive integers 1 k i ∈ d1, . . . , dk and dene L = Q (1, , k ). Then the intersection L Q equals Q. ∩ ∞ Proof. When we apply the theory of chapter 1 we see

Q L Q(√a : a Q). ∞ ∩ ∈ From proposition 84 we derive that 2 does not divide [L : Q] and thus we have L Q = Q. ∩ ∞

Corollary 122. Let , . . . , be elements of R with di Q for odd positive 1 k i ∈ integers d1, . . . , dk. We dene L = Q (1, , k ). Let be an element of L and let n be a positive integer, then √n is containe d in Q(1) if and only if there exist w Q, e L and l , . . . , l N such that equals w en li . ∈ ∈ 1 k ∈ i i That the condition in theorem 116 is not always sucienQt we see in the following example.

Example 123. With notations as in theorem 116, we take K = Q, L = Q(√2 ), = 1+√2 and n = 3. Let be an element of R with 3 = 1+√2. We show that is not contained in Q(1), although we have 3 = w en with w = 1+√2 Q(√2 ) Q and e = 1. ∈ ∩ ∞ First we remark that we cannot use the same strategy we followed in theorem 102. As 1+√2 is the fundamental unit in the ring of integers of Q(√2 ) we see that Q()/Q is an extension of degree 6. We have ( 1/)3 = 1 √2 and the normal closure of 76 CHAPTER 4. NESTED RADICALS OF DEPTH ONE

Q() over Q is Q(, 3). This gives the following chain of eld extensions

Q(, 3)

C3

Q(√2, 3)

V4

Q.

It immediately follows that Gal(Q(, 3)/Q)00 is 1 . Suppose that is contained in Q(1). Applying{ } corollary 18 with ground eld Q = Q(√2 ) gives ∞ ∞ Q , ∈ ∞ for some Q with m Q for some positive integer m and d / Q for all divisors d of m. We assume∈ without∈ loss of generality that R and that∈b = m > 0. First we show that m equals 3 or 6. Because ∈is not an element of Q and 3 is contained in Q we have / Q and 3 Q . By lemma 68 we have∞[Q () : Q ] = 3 and therefore∞ [Q() :∈Q(∞3)] = 3. As∈ 3∞is contained in Q the extension∞ Q(∞3)/Q is abelian and as 3 R the degree [Q(3) : Q] is 1 or 2. As∞b is positive, it follows from lemma 68 that xm∈ b is the minimal polynomial of over Q. Hence m equals 3 or 6. We saw above that there exists some w Q with w3 = (1+√2)/3. As Q(w)/Q is an abelian extension, the polynomial x∈3 ∞w3 is reducible over K = Q(√2, 3). So, there exists some y in K such that y3 equals w3. For this element y3 the norm in the extension K/Q is a third power in Q. We have

3 K 3 K 1 + √2 Q( ) K 1 + √2 NQ y = NQ 3 = NQ NQ(3) 3 Ã ! Ã Ã !! ¡ ¢ Q 1 1 NQ b2 = b2 if m = 3 Q(√2 ) 1+√2 1 3 √ = NQ ¡ ¢ 3 = b if m = 6 and Q( ) = Q( 2) 3  Q( ) 1 1 3 N ³ =´ 2 if m = 6 and Q( ) = Q(√2). Q b b 6  ¡ ¢ We conclude b Q3, but this is a contradiction with the denition of b, so 3 1 + √2 is not an elemen∈t of Q(1). p Chapter 5

Rogers-Ramanujan continued fraction

This chapter consists of a paper co-authored with Alice Gee. It was accepted by the Ramanujan Journal in May 2001 and can also be found in [10]. The paper, with some minor corrections, follows after a brief introduction. In this chapter we determine the class elds generated by singular values of the famous Rogers-Ramanujan continued fraction and give a method for writing these values as nested radicals. Where, in chapters 3 and 4, our goal was to construct a radical expression of minimal nesting depth for a given nested radical, in this chapter we are satised if we can give a radical expression. The elements that we consider are given by values of analytic functions and we rst have to prove that these elements are nested radicals. It is well known that, for a perfect eld K, the element is a nested radical over K if the Galois group of K()/K is solvable. In section 5.1 we introduce the Rogers-Ramanujan continued fraction, R, which is a function on the complex upper half plane. Moreover, we give some history in constructing nested radicals for special values of this function. In the second section we give two formulas due to Watson [37] that relate R(z), for some z in the upper half plane, to the Dedekind -function

∞ (z) = e2iz/24 1 e2izn . m=1 Y ¡ ¢ This relation we use to prove that R is a modular function of level 5. Moreover, we show that R generates the eld of modular functions of level 5 over Q(5). Let be an element of the upper half plane with [1, ] a Z-basis of some imaginary quadratic order. In section 5.3 we explain, for a modular function h of level N Z>0, how to use the structure of the Galois groups in the following chain of elds ∈ Q K = Q() H H(h()), 77 78 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION with H the Hilbert class eld of K, to determine a nested radical for h() R. In this computation we again use the Dedekind -function. As all elements ∈that we compute have minimal polynomial in Z[x] we compute suciently accurate real approximations of its coecients and round these. In section 5.4 we restrict to modular functions of level 5 again. We introduce functions v, w and w˜ with

1 w(z) n 1 w(z) = R(z), w˜(z) = and v(z) = w˜(z) + w˜(z) , R(z) √5 5 µ ¶ n where 5 denotes the Jacobi symbol (denition 43). Another way to represent a nested radical is to give its minimal polynomial. It is clear from¡ ¢ the relations above that given a representation for v(z) a representation for R(z) can easily be derived. As the representations for v(z) are more compact we provide in section 5.4 a way to calculate the minimal polynomials for the algebraic integers w˜(n) and v(n) where n is dened by

√ n if n 3 mod 4 n = 5+√ n 6 if n 3 mod 4. ( 2

In the last section we compute nested radicals for the elements v(n) for 1 n 16. In section 5.3, example 128, we give an extensive explanation how we useLagrange resolvents and the Galois action in each of the eld extensions in the chain of elds

Q Q() H H(h()) to calculate these expressions. 5.1. INTRODUCTION 79 Singular values of the Rogers-Ramanujan continued fraction

Abstract Let z C be imaginary quadratic in the upper half plane. Then the Rogers-Ramanujan∈ continued fraction evaluated at q = e2iz is contained in a class eld of Q(z). Ramanujan showed that for certain values of z, one can write these continued fractions as nested radicals. We use the Shimura reciprocity law to obtain such nested radicals whenever z is imaginary quadratic.

5.1 Introduction

The Rogers-Ramanujan continued fraction is a holomorphic function on the complex upper half plane H, given by

n 1 ∞ n ( ) 2iz R(z) = q 5 1 q 5 , with q = e and z H. (5.1) ∈ n=1 Y ¡ ¢ n Here 5 denotes the Legendre symbol. The function R owes part of its name to the expansion ¡ ¢ 1 q 5 R(z) = (5.2) q 1 + q2 1 + 1 + q3 . . . as a continued fraction. In their rst correspondence of 1913, Ramanujan astonished Hardy with the assertion

2 e 5 5 + √5 √5 + 1 2 = . (5.3) e s 2 2 1 + 4 e 1 + 6 1 + e . . . Hardy was unaware of the product expansion 5.1 that Ramanujan had used to com- pute identity 5.3, which is none other than the evaluation of R at i. In the same correspondence, Ramanujan expressed the equality

5 + i 5 √5 √5 1 R = (5.4) 2 s 2 2 µ ¶ with a similar dramatic air. The radical symbol in 5.3 and 5.4 should be interpreted as the real positive root on R. Ramanujan communicated radical expressions for 80 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

5+√ 5 R √ 5 and R 2 in his second letter to Hardy, and several other values of R atimaginaryquadratic arguments are recorded in his notebooks. The other name connected¡ ¢ to the ¡function¢R is that of L.J. Rogers, who proved the equality of 5.1 and 5.2 in 1894. This was discovered by Ramanujan after his arrival in England. In this paper, we evaluate singular values of the Rogers-Ramanujan continued fraction. These are the function values of R taken at imaginary quadratic H. As R is a modular function of level 5 —a classical fact for which we furnish a∈proof— these values generate abelian extensions of Q(). Exploiting the Galois action given by the Shimura reciprocity law, we give a method for constructing a nested radical for R() that works whenever is imaginary quadratic. Our systematic approach extends the results of [2], [6], [13] and [27], which only apply to individual examples. By way of example, we provide nested radicals for R √ n for the integers n = 1, 2, . . . , 16 with n 3 mod 4. Writing down nested radicals for R() becomes increasingly unwieldy as6the discriminant of grows, so in¡ the case¢ n 3 mod 4, 5+√ n 5+√ n where Q and R( 2 ) generate a subeld of Q(R(√ n )), we evaluate R( 2 ) 5+z instead of R(√ n ). In the classical literature, the notation S(z) = R( 2 ) is frequently used.

5.2 The modular function eld of level 5

A modular function of level N is a meromorphic function on the extended complex upper half plane H P1(Q) that is invariant under the natural action of the modular ∪ group (N) = Ker[SL2(Z) SL2(Z/NZ)] of level N. As such functions are invariant → 1 2iz under z z + N, they admit a Fourier expansion in the variable q N = e N . The 7→ 1 modular functions of level N with Fourier expansion in Q(N )((q N )) form a eld FN , the function eld of the modular curve X(N) over Q(N ). The extension F is Galois over F with group GL (Z/NZ)/ 1 . For a proof, N 1 2 { } see [23], page 66, Theorem 3. One can describe the action of GL2(Z/NZ) on FN explicitly. The group (Z/NZ) acts as a group of automorphisms of FN over F1, 1 by restricting its natural cyclotomic action on Q(N )((q N )). The natural action of (1) = SL2(Z) on H induces a right action of (1)/(N) = SL2(Z/NZ) on FN which leaves F1 invariant. The homomorphisms

(Z/NZ) Gal (F /F ) and SL (Z/NZ) Gal (F /F ) → N 1 2 → N 1 can be combined into an action of the semi-direct product

(Z/NZ) n SL (Z/NZ) GL (Z/NZ) 2 ' 2 1 0 on FN . For this isomorphism, we identify d (Z/NZ) with the element 0 d GL (Z/NZ). The resulting sequence ∈ ∈ 2 ¡ ¢ 1 1 GL (Z/NZ) Gal (F /F ) 1 (5.5) → { } → 2 → N 1 → 5.2. THE MODULAR FUNCTION FIELD OF LEVEL 5 81 is exact. The modular invariant j generates F1 over Q, and induces the isomorphism X(1) P1(Q). In a similar fashion, the curve X(5) has genus 0, thus its function ' eld F5 can be generated by a single function over Q(5). The Rogers-Ramanujan continued fraction R is such a generator. There are several ways to prove this classical fact. Our proof is based up the following two formulas of Ramanujan proofs of which were given by Watson [37] 1 (z/5) R(z) 1 = , (5.6) R(z) (5z)

1 (z) 6 R5(z) 11 = , (5.7) R5(z) (5z) ³ ´ which relate R to Dedekind’s -function

∞ (z) = q1/24 1 qn , q1/24 = e2iz/24. m=1 Y ¡ ¢ The above formulas 5.6 and 5.7 will prove useful in section 5.4, where we evaluate singular values of R. We dene functions

1 0 5 0 0 5 0 1 h0 = and h5 = √5 , ¡ ¢ ¡ ¢ so that equations 5.6 and 5.7 become 1 h R 1 = √5 0 , (5.8) R h5

3 1 5 5 5 R 11 = 6 . (5.9) R h5 We will derive the classical fact that R is modular from the modularity of the func- 6 tions appearing on the right hand side of 5.8 and 5.9. This is well known for h5 ([26], page 619), but for lack of a reference in the case of 5.8, we provide a proof that works in both cases. In order to compute the action of SL2(Z) on h0 and h5, we begin by observing 0 1 1 1 that the generating matrices S = 1 0 and T = 0 1 of SL2(Z) act on the Dedekind -function as ¡ ¢ ¡ ¢ S (z) = √ iz (z) and T (z) = (z) . (5.10) 24 The radical sign stands for the holomorphic branch of the square root on iH that is positive on the real axis. The observation 1 0 S = S 5 0 gives h S = h . 0 5 0 1 0 5 ¡ ¢ ¡ ¢ 82 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

with the help of 5.10. Let 5 denote the set of 2 2 matrices with coecients in Z that have determinant 5. The matrices

1 i 5 0 Mi = 0 5 , i = 0, 1, . . . , 4 and M5 = 0 1 form a set of representativ¡ es ¢for 5. For A SL2(Z) and ¡i ¢0, 1, . . . , 5 , we can nd B SL (Z) and j 0, 1, . . .\, 5 such that∈ M A = B M∈ {holds. We }put ∈ 2 ∈ { } i j M M h = √5 5 and h = i for i = 0, 1, . . . , 4. (5.11) 5 i Using 5.10 one computes

1 h0 h5 h0 24 h1 3 1 h1 24 h4 h1 24 h2        1  h2 h3 h2 24 h3 S = and T = 1 . (5.12) h3  h2  h3 24 h4    3      h4  24h1  h4  h0         4  h5  h  h5  h     0     24 5          6 Lemma 124. The functions h5 and h0/h5 are modular of level 5. 6 Proof. We will show that each of the functions hi and hi/hj with 0 i, j 5 are invariant under the action of (5). From [17] we know that (5) is the normal 5 closure of T in SL2(Z). This means that (5) is generated by matrices of the 5 h 1 i form AT A with A SL (Z). From 5.10 we observe ∈ 2 5 1 h T = h for i = 0, 1, . . . , 5 . i 6 i

For A SL2(Z) and j 0, 1, . . . , 5 , the equations 5.12 show that all hj A are of the form∈ h A = h ∈for{ some i }0, 1, . . . , 5 and some root of unity .Similarly j i ∈ { } 5 1 h AT = h A j 6 j 6 5 1 holds for every A SL (Z). Thus h is invariant under AT A for all A SL (Z), ∈ 2 i ∈ 2 as well as every quotient hi/hj . 6 6 It is easy to check that the Fourier expansions of the functions h0, h5, h0/h5 and 1/5 1/5 h5/h0 are in Q(5)((q )). Those of the other functions are in Q(20)((q )). Lemma 125. The Rogers-Ramanujan continued fraction R is modular of level 5. Proof. From 5.1 we know that R is holomorphic on H and that its q-expansion is an 1/5 5 1 element of Q(5)((q )). Therefore it suces to show that R AT A = R for all A SL (Z). From formula 5.8 one derives ∈ 2 1 h X R X + = X2 + √5 0 + 1 X 1. (5.13) R h5 ³ ´³ ´ ³ ´ 5.2. THE MODULAR FUNCTION FIELD OF LEVEL 5 83

5 1 As AT A acts trivially on √5 h0/h5, it maps R to either R or 1/R. Suppose the latter to be true. Then R AT 5 = 1/(R A) holds. As the translation T 5 xes the cusp i , we have ∞ 1 R A(i ) = R AT 5(i ) = , ∞ ∞ R A(i ) ∞ which implies R A(i ) = i. Then formula 5.9 yields ∞ 3 5 1 5 6 = ( i) 11 = 2i 11. (5.14) h A(i ) ( i)5 5 ∞ On the other hand,¡ we can ev¢ aluate h A(i ) by considering the product expansion 5 ∞ for h5 A at q = 0. By 5.12, one has h5 A = hj for some root of unity and some j 0, 1, . . . , 5 . For j = 0, 1, . . . , 4, we compute ∈ { } 2i iN+j e ( 5 ) hj (i ) = lim = 0 . ∞ N e2i(iN) →∞ A similar calculation shows that h has a pole at i . Contradiction with 5.14. 5 ∞ 5 Lemma 126. The minimum polynomial of R over F1 = Q(j) is

P (X) = X12 + 1 + (j 684)(X11 X) + (55j + 157434)(X10 + X2) + (1205j 12527460)(X9 X3) + (13090j + 77460495)(X8 + X4) + (69585j 130689144)(X7 X5) + (134761j 33211924)X6. The minimum polynomial of R over Q(j) is P (X 5), with P as above.

2 6 3 Proof. Weber shows, [38] page 256, that h0 is a zero of X + 10X 2X + 5, with a cube root of j. Another zero is h2 = (h S)2 because S xes . We obtain 2 5 0 2 3 h12 + 10h6 + 5 j = 5 5 . (5.15) h6 ¡ 5 ¢ Rewriting 5.9 gives the identity

53 R5 h6 = . 5 R10 11R5 + 1 6 Substituting the above relation for h5 into 5.15, we have

3 1 + 228R5 + 494R10 228R15 + R20 j = 5 , (5.16) R + 11R6 + R11 ¡ ¢ ¡ ¢ 84 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION which readily yields P (R5) = 0, with P as in lemma 126. To see that P is irreducible in Z[X, j], compose the evaluation map Z[X, j] Z[X] dened by j 1 with reduction modulo 2. We obtain a homomorphism→Z[X, j] F [X] that sends7→ P to → 2 the cyclotomic polynomial 13 F2[X], which is irreducible because 2 is a primitive root modulo 13. As P is a monic∈ polynomial in X, we conclude that it is the minimum polynomial of R5 over Q(j). In order to see that Q(R) has degree 5 Q(R5) : Q(j) = 60 over Q(j), it suces to observe that 1 1 SL (Z), which acts as R(z) R(z + 1) = R(z) induces 0 1 2 £ ¤ 5 an automorphism of order∈ ve of Q(R) over Q(R5). 7→Thus P (X5) is the minimum polynomial of R¡over¢Q(j).

Theorem 127. The Rogers-Ramanujan continued fraction R generates F5 over Q(5).

Proof. As R has rational Fourier coecients, the subelds Q(R) = F1(R) and F1(5) of F5 are linearly disjoint extensions of F1 having degree 60 and 4, respectively. Their composite, which has degree 240 = # GL2(Z/5Z)/ 1 over F1 is therefore equal to F . { } 5 ¡ ¢ The rational function on the right hand side of 5.16 appears in Klein’s study of 1 the nite subgroups of Aut(P (C )). His icosahedral group A5 is isomorphic to 1 SL2(Z/5Z)/ 1 , and the natural map from P (C ) to the orbit space of the icosa- hedral group{ramies} above 3 points. The relation 5.16 denes a generator, [15] page 61 and 65, for the eld of functions invariant under the icosahedral group. In our situation the natural map X(5) X(1) ramies over 3 points and the Galois group of C (R) over C (j) is SL (Z/5Z→)/ 1 . 2 { } The subgroups of GL2(Z/5Z)/ 1 that stabilise the functions appearing in the { } 6 equations 5.8 and 5.9 are given in the diagram below. The stabilisers of h5 and √5 h /h in SL (Z/5Z)/ 1 can be determined using 5.12. 0 5 2 { } 5.3. GALOIS THEORY 85

F5 = Q(R, 5)

( 1 0 ) h 0 2 i 4

Q(R) ( 2 0 ) h 0 3 i 2 1 1 ( ) h0 0 1 Q(√5 ) SL2(Z/5Z)/ 1 h i 5 h5 60 { }

Q(R5)

6 Q(h5)

Q(j, 5) 6

4 ( 1 0 ) h 0 2 i

Q(j)

5.3 Galois theory for singular values of modular functions

Let be an imaginary quadratic order having Z-basis [, 1]. Dene HN = HN, to be theO eld generated over K = Q() by the function values h(), where h rangesO over the modular functions in FN that are pole-free at . The rst main theorem of complex multiplication [23] states that HN is an abelian extension of K. For N = 1, the eld H is the ring class eld for . If is a maximal quadratic order with 1 O O eld of fractions K, then HN is the ray class eld of conductor N over K, and H1 is the Hilbert class eld of K. For ray class elds of non-maximal orders, see for example [34]. Before we can describe the explicit action of Gal (HN /K) on elements of HN , we rst look at Gal (HN /H1), which ts in a short exact sequence

A 1 ( /N ) Gal (H /H ) 1. → O → O O → N 1 → 86 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

In order to describe the Artin map A, we write the elements of /N as row vectors with respect to the Z/NZ-basis [, 1]. If has minimum polynomialO OX 2 +BX +C Z[X], dene the homomorphism ∈

g : ( /N ) GL2(Z/NZ) O O → t Bs Cs s + t . (5.17) 7→ s t ¡ ¢ The matrix g (x) represents multiplication by x on /N with respect to the Z/NZ- basis [, 1]. For h F , the Shimura reciprocityOlaw O[30] gives the action of x ∈ N ∈ ( /N ) on h() as O O x1 h() = hg (x)(). (5.18) ¡ ¢ Here g (x) GL2(Z/NZ) acts on h FN as described in 5.5. Moreover, if h FN is a function∈for which Q(h) F is∈Galois, then K h() is the xed eld of∈ N g (x) ¡ ¢ x ( /N ) h = h ( /N ). (5.19) { ∈ O O | } O O For any h F , we aim to compute the conjugates of h() with respect to the ∈ N full group Gal (HN /K). In the case N = 1, the Galois group of H1 = K j() over K is isomorphic to the ideal class group C( ) of . The elements of C( ) can be represented as primitive quadratic forms [a,Ob, c] ofOdiscriminant D = b2O¡ 4ac¢, b+√D where D is the discriminant of . The Z-module having basis a, 2 is an -ideal in the class of [a, b, c], andO the class of [a, b, c] acts on j() as O £ ¤

[a, b,c] b + √D j() = j . (5.20) 2a µ ¶

In the general case for N > 1, we need the elements of Gal (HN /K) that lift (3.5) for each representative [a, b, c] in C( ). The formula [9], Theorem 20 produces one O element Gal (HN /K) along with a matrix MN = MN (a, b, c) in GL2(Z/NZ) such that for∈ all h F , the relation ∈ N b + √D h() = hMN (5.21) 2a µ ¶ holds. The automorphism clearly lifts the action in 5.20 to Gal (HN /K) because M GL (Z/NZ) acts trivially on j F . As every automorphism in Gal (H /K) N ∈ 2 ∈ 1 N is obtained by composing elements of Gal (HN /H1) with one of the coset repre- sentatives for Gal (H1/K) in 5.21, we can determine the conjugates of h() under Gal (H /K) for any h F . N ∈ N Given this explicit action on HN over K we can compute representations for singular values of modular functions by minimal polynomials as well as radical ex- pressions over Q. 5.3. GALOIS THEORY 87

The natural way to describe an algebraic number is its minimum polynomial over Q. Let h FN be a function for which h() is an algebraic integer. The conjugates of h(∈) over K can be approximated numerically when the Fourier ex- pansion for h is known. One expresses each conjugate in the form hM (), with M GL (Z/NZ) and K, and then writes M = 1 0 A with x = det(M) ∈ 2 ∈ 0 x and A SL2(Z/NZ). After modifying the Fourier coecients of h with respect to ∈x ˜ ¡ ¢ ˜ N N , one evaluates the new expansion at A(z), where A SL2(Z) is a lift of A7→. We calculate the minimum polynomial f of h() over Q ∈by approximating the conjugates of h() over K. Adjoining complex conjugates gives a full set of conjugates over Q. In order to determine the polynomial f Z[X], we need only to approximate its coecients accurate to the nearest integer.∈ Because H5 is abelian over K, one can also express h() as a nested radical over Q in the spirit of Ramanujan’s evaluations 5.3 and 5.4. Unlike the minimum polynomial f, which is unique, many dierent nested radicals over Q exist that all represent h(). Given any abelian extension H/K of degree greater than 1 and any w H, the following standard procedure expresses w as a radical expression over a ∈ eld H0 with the property [H0 : K] < [H : K]. Applying the procedure recursively produces a nested radical for w over K. We choose an automorphism Gal (H/K) of order m > 1 and set H 0 = ∈ H (m), where H denotes the xed eld of . Then H 0/K is an abelian extension of degree h i

[H0 : K] ϕ(m) [H : K] < m [H : K] = [H : H ][H : K] = [H : K] . We write 1 w = h0 + h1 + h2 + + hm 1 , (5.22) m where ¡ ¢ m k h = ik w( ) , i = 0, 1, . . . , m 1 i m kX=1 are the Lagrange resolvents for w with respect to . Note that h0 = TrH/H (w) is an element of H0. Every Gal (H(m)/H0) acts trivially on m and as some a on H. For i = 1, 2, . . .∈, m 1, we have ∈ h i n ik (k+a) ia h = w = h , i m m i kX=1 m m m m m which means h1 , h2 , . . . , hm 1 H0. As hi = hi for the appropriate choice ∈ of the m-th root, equation 5.22 represents w as a radical expression over H 0. The m m m p recursion step is applied to h0, h1 , h2 , . . . , hm 1 H0. Suppose h F such that h() is an algebraic ∈ integer. In order to apply the ∈ N recursive procedure above to h(), one needs not only the action of Gal (HN /K), 88 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

but also that of Gal (HN (d)/K) for various numbers d > 1. This is obtained by restricting the action of Gal (HdN /K) to HN (d). The elements computed in the nal recursion step are in , which is a discrete subgroup of C . An approximation OK of their coordinates with respect to a Z-basis for K , that is accurate to the nearest integer, produces a nested radical for h() over QO. The methods above can be extended to arbitrary imaginary quadratic numbers H that are not necessarily algebraic integers. In order to compute the conjugates ∈ a b of h() over K = Q() we take an integral basis [, 1] for K and write = d + d a b with a, b, d Z. One evaluates h 0 d FadN at , which is contained in HadN . Again, 5.18∈and 5.21 allow us to calculate ∈the conjugates of h() over K. ¡ ¢ Example 128. Dene = √ 6, we will explain how to compute a nested radical 6 for v(6) where h h v = 0 + 5 . h5 h0

Let K denote Q(√ 24 ) and let H be the Hilbert class eld of K. As v F5 we nd in [9] that the Galois action of H(v( ))/H is given by the matrix A = ∈1 4 6 1 1 ∈ GL2(Z/5Z)/ 1 . The Galois group of the extension H/K is given by the class group of K. {The}tower of eld extensions in a diagram: ¡ ¢

H v(6)

¡2 ¢ h i H

2 h i K = Q(√ 24 ) 2

Q

2 If we have Gal(H(v(6))/H) = then v(6)+v(6) and v(6) v(6) H. We determine the conjugates of theseh i elements by the action of Gal(H/K) =∈ . The elements of the class group of K are the classes of the quadratic¡ forms¢[1, 0h, 6]i 1 0 2 0 and [2, 0, 3]. They induce ([9], Theorem 20) matrices I = 0 1 and M = 0 1 . Formula 5.21 tells us how to determine v( ) + v( ) . We nd 6 6 ¡ ¢ ¡ ¢ ¡ ¢ 0 + √ 24 v( ) + v( ) = (v + vA)( )[2,0,3] = (vM + vAM ) . 6 6 6 4 ¡ ¢ ³ ´ To determine vM and vAM we rst write each of the matrices as a product of a matrix 1 0 of the form 0 d , where d denotes the determinant, and a matrix in SL2(Z/5Z). ¡ ¢ 5.3. GALOIS THEORY 89

This gives

M = 2 0 = 1 0 2 0 and A M = 2 4 = 1 0 2 4 . 0 1 0 2 0 3 2 1 0 4 3 4

Then we determine¡ ¢ matrices¡ ¢ of¡ determinan¢ t 1 in GL2(¡Z) that¢ are¡ represen¢ ¡ tativ¢ es for 2 5 2 1 the matrices in SL2(Z/5Z). They are 5 13 and 3 1 . Then we get 0 + √ 24 2 ¡ 1/2¢ √ 6¡+ 5 ¢ 4 √ 6 1 (vM + vAM ) = v + v , 4 5 5/2 √ 6 + 13 5 3/2 √ 6 1 ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ where ( ) denotes the Jacobi symbol. Similar computations yield all conjugates of the elements v(6) + v(6) and 2 v( ) v( ) . Dene v , v , v , v by 6 6 1 2 3 4

¡ ¢ 6+5 v1 = v(6) v3 = v 5/2 +13 6 11 6+29 ³6 1 ´ v2 = v 3 +8 v4 = v 3/2 1 , 6 6 ³ ´ ³ ´ then these conjugates are v + v , v + v , v2 v2 and v2 v2 in H. 1 2 3 4 1 2 3 4 To compute a nested radical for v(6) = v1 we determine an accurate approxi- mation for v + v + v + v R = Z, rounding gives that 1 2 3 4 ∈ OK ∩

v1 + v2 + v3 + v4 = 8.

By approximating we also nd

((v + v ) (v + v ))2 = 200 and (v + v ) (v + v ) > 0, 1 2 3 4 1 2 3 4 (v v )2 + (v v )2 = 60, 1 2 3 4 ((v v )2 (v v )2)2 = 3200 and (v v )2 (v v )2 > 0. 1 2 3 4 1 2 3 4 Thus we have the equalities

1/2 (8 + √200 ) = v + v 1 2 and 1/2 (60 + √3200 ) = (v v )2. 1 2 From which follows, as v v > 0, 1 2 v( ) = v = 1/4 8 + √200 + 1/2 30 + 1/2 √3200 6 1 q = 1/2 ¡ 4 + 5√2 +¢ 10 3 + 2√2 µ q ¶ = 1/2 4 + 5√2 + √10 + 2√5 ¡ ¢ 90 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

5.4 The ray class eld H5 We turn our attention back to the functions of level 5 from section 2. In this section, we compute some singular values R() of the Rogers-Ramanujan continued fraction. As the singular values of j are known to be algebraic integers, the same holds true for R because the polynomial P of lemma 126 has coecients in Z[j]. We x = [, 1] to be an order in some imaginary quadratic number eld K. We state a fewOproperties of R() before computing some examples.

Lemma 129. The class eld H5 = H5, is generated by R() over K. O

Proof. As we have F5 = Q(R, 5) by Theorem 127, the extension F5/Q(R) is Galois and we are in the situation for which 5.19 applies. As Q(R) is the subeld of F5 xed by 1 0 d (Z/5Z) GL (Z/5Z) , 0 d | ∈ 2 the class eld K R() ©¡is the¢ subeld of H5ªxed by

¡ ¢ 1 0 G = x ( /5 ) g (x) = for some d (Z/5Z) ( /5 ) . { ∈ O O | 0 d ∈ } O O From formula 5.17 we see that the ¡only ¢diagonal matrices appearing in the image g [( /5 )] are scalar. We conclude G = 1 and K R() = H . O O { } 5 z ¡ ¢ Let w(z) = ( 5 )/(5z) denote the function that appears on the right hand side of equation 5.6. Thus we have 1 R(z) 1 = w(z). (5.23) R(z)

Lemma 130. The singular values R() and 1/R() are conjugate over the eld K w() . Furthermore, H5 is generated over K by 5 together with w(). Pr¡oof. The¢ polynomial

X2 + w() + 1 X 1 K w() [X] (5.24) ∈ derived from 5.23 has zeroes¡ R() and¢ 1/R(). To¡show ¢that 5.24 is irreducible in K w() [X] we consider the homomorphism g : ( /5 ) GL (Z/5Z) in 5.17. O O → 2 By 5.18, the group Gal (H /H ) contains the automorphism R() Rg (2)(). In ¡ ¢ 5 1 order to determine the action action of 7→

g (2) = 2 0 = 1 0 2 0 GL (Z/5Z) 0 2 0 4 0 3 ∈ 2 ¡ ¢ ¡ ¢¡ ¢ on F5, we recall that R and w have rational Fourier coecients and thus are xed by 1 0 . Using 5.12 one checks that w is stabilised by 2 0 SL (Z/5Z). Theo- 0 4 0 3 ∈ 2 rem 127 together with equation 5.23 tells us that this matrix 2 0 sends R to 1/R, ¡ ¢ ¡ 0¢ 3 ¡ ¢ 5.4. THE RAY CLASS FIELD H5 91

so R() and 1/R() are conjugates over K. As K R() = H5 contains 5, the situation R() = 1/R() = i is impossible. We conclude that 5.24 is irreducible in K w() [X]. ¡ ¢ We have [H : K w() ] = 2. In fact, K w() is the subeld of H xed ¡ ¢ 5 5 by the subgroup of ( /5 ) generated by 2 and the image of . By 5.24 the O¡ O¢ 1¡ ¢ O action of 2 ( /5 ) on H is . We conclude K w() and ∈ O O 5 ∈ 5 5 7→ 5 5 6∈ H = K w(), . 5 5 ¡ ¢ To determine¡ ¢the minimum polynomial of R() over Q, it is convenient, although certainly not necessary, to rst compute the polynomial for w() and then recover R() with 5.23. As both values R() and 1/R() are algebraic integers, it follows that w() is an algebraic integer too. In particular, the method of section 5.4 for w() computing fQ works here. Working with values of w is easier than working with R directly as there are only half as many conjugates over K to compute. More importantly, the Dedekind -function is implemented in several software packages that quickly compute (z) to a high degree of accuracy. These routines make use of SL2(Z)-transformations to ensure that the imaginary part of z is suciently large to guarantee rapid convergence of the Fourier expansion of (z). For the expansion 5.1 of the function R(z) such a standard implementation does not appear to be available. w() One obtains the minimal polynomial of R() over Q from fQ by writing

1 R R2 w = R using 5.6. Then R() is a zero of the monic polynomial

1 X X2 Xdeg f w() Q[X] Q X ∈ ³ ´ w() with deg = deg(fQ ). According to lemma 130, the resulting polynomial is irre- ducible because its degree is 2 deg. An inspection of product expansion 5.1 shows R(z) R whenever the real part of z H is an integer multiple of 5 . For the singular argumen∈ ts ∈ 2 √ n if n 3 mod 4 n = 5+√ n 6 if n 3 mod 4 ( 2 the value R(n) is a real, and its minimum polynomial over K is contained in Z[X] because complex conjugation acts as

R(n) R(n) R(n) fK = fK = fK . 92 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

R() 2d i Lemma 130 implies that fQ = i=0 ciX is the minimum polynomial for both i R() and 1/R(), thus the coecients satisfy ci = ( 1) c2d i. For this reason we P only list the rst half of the coecients c2d, c2d 1, . . . , cd in table 1, where we give the minimum polynomials for R( ) with 1 n 16. n Table 1. The minimum polynomials of R(n) over Q

n degree rst half of coecients c2d, c2d 1 . . . cd 1 4 1, 2, 6 2 12 1, 6, 1, 0, 50, 14, 16 3 4 1, 3, 1 4 8 1, 14 , 22, 22, 30 5 20 1, 10, 90, 280, 730, 1022, 2410, 2540, 3330, 1730, 2006 6 16 1, 28, 140 , 60, 365 , 264, 482, 340, 2035 7 12 1, 4, 1, 25, 25, 14, 31 8 24 1, 32 , 96,268, 51 , 328 , 1446, 5112, 996, 3972, 10594, 4208, 6924 9 16 1, 38, 240, 300, 235, 726, 92, 1840, 675 10 20 1, 60, 360 , 120 , 120, 1728 , 3540, 840 , 4320, 7620, 1006 11 8 1, 6, 13, 28, 5 12 24 1, 82 , 329 , 282, 74, 3672, 3846, 4238, 13521, 9028, 7844, 2408, 43651 13 24 1, 82, 996, 968, 1051, 1422, 96, 24912, 7896, 16722, 28844, 13658, 114024 14 32 1, 116, 614 , 3040, 25230, 17988, 103372, 184292, 207725, 409400, 323390, 129140, 2879690 , 3515800, 5057000, 4838560, 7624315 15 20 1, 15, 60, 270, 720, 1353, 2115, 2610, 2970, 1095, 3119 16 16 1, 148 , 670, 240, 1570, 2616, 302,1180, 1610

A nice way of generating H5 = K(w(n), 5) comes from lemma 130. The subeld K(w(n)) of H5 is the xed eld for the subgroup generated by 2 and in ( /5 ). √ 2 0 √ O O O Because 5 is invariant under gn (2) = 0 2 , we conclude 5 K(w(n)). Thus for the function ∈ w¡ ¢h w˜ = = 0 √5 h5 we have w˜( ) K(w( )) and H = K(w˜( ), ). n n 5 n 5 Lemma 131. The value w˜(n) is an algebraic integer. If 5 - n then w˜(n) is a unit in H , the ray class eld of conductor 5 over = [ , 1]. 5 O n Proof. Hasse and Deuring, [8] page 43, determine exactly the ideals generated by singular values of the lattice functions z M 1 ϕM (z) = z , ¡ 1¡ ¢¢ ¡ ¢ 5.4. THE RAY CLASS FIELD H5 93

with M a 2 2 matrix having coecients in Z. Our functions h0 and h5 were dened in 5.11 as M M h = 0 , h = √5 5 with M = 1 0 , M = 5 0 . 0 5 0 0 5 5 0 1 ¡ ¢ ¡ ¢ Thus we have z z ϕ = h (z)24 and ϕ = h (z)24 , M0 1 0 M5 1 5 µ ¶ µ ¶ and n 24 ϕM0 1 w˜(n) = . ϕ n M5 ¡ 1 ¢ If n is not divisible by 5, then ¡ ¢

M n = [ , 5] and M n = [5 , 1] 0 1 n 5 1 n µ ¶ µ ¶ are both proper ideals of = [n, 1]. Deuring’s theorem, [8] page 42, shows that n n O ϕM0 1 and ϕM5 1 are associate elements in the ring of integral algebraic numbers; one writes ¡ ¢ ¡ ¢ ϕ n ϕ n . M0 1 M5 1 µ ¶ µ ¶ 24 It follows that the quotient w˜(n) is a unit. If 5 n but 25 - n, then ϕ n is again a proper -ideal. However, the | M0 1 O multiplicator ring for M n is not , but [5 , 1]. Deuring’s formulas, [8] page 43, 5 1 ¡ ¢ n yield O ¡ ¢ ϕ n 56 in and ϕ n 56/5 in [5 , 1]. M0 1 O M1 1 n µ ¶ µ ¶ We nd w˜( ) 51/5. n When n is divisible by 25, the multiplicator rings for M n and M n are 0 1 5 1 O and [1, n/5] respectively. In this case, the formulas, [8] page 43, show that w˜(n) is again associated to a positive rational power of 5. ¡ ¢ ¡ ¢

When n Z is not divisible by 5, the Galois action 5.5 for the matrix gn (n) of ∈ n 1 5.17 sends w˜ to w˜ . We dene 5 ¡ ¢ n 1 v( ) = w˜( ) + w˜( ) . n n 5 n µ ¶ 94 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

Clearly we have w˜(n) = v(n) when n is divisible by 5. However, if n > 1 with 5 - n then w˜(n) has degree 2 over v(n). In these cases the minimum polynomial for w˜(n) satises 2 n w˜(n) deg v(n) X + 5 fQ = X fQ X ¡ ¢ v(n) ³ ´ with deg = deg(fQ ). Table 2 lists the minimum polynomials over Q for v(n) for 1 n 16. Table 2. The minimum polynomials over Q for v(n)

n degree coecients 1 1 1, 2 2 3 1, 2, 3, 4 3 1 1, 1 4 2 1, 6, 4 5 10 1, 10, 25, 30, 25, 10, 25, 0, 25, 0, 25 6 4 1, 8, 16,28, 31 7 3 1, 2, 3,9 8 6 1, 16, 20, 100, 25, 156, 124 9 4 1, 22, 54, 62 , 59 10 10 1, 20, 75, 60, 75, 20, 25, 0, 25, 0, 25 11 2 1, 4, 1 12 6 1, 34, 5, 150, 75, 144, 99 13 6 1, 46, 210 , 290,905, 456, 576 14 8 1, 44, 238, 88, 520, 2508, 4978, 176, 2711 15 10 1, 5, 0, 15, 0, 5, 25, 0,25, 0, 25 16 4 1, 68, 14, 328, 284

5.5 Nested radicals

In order to obtain nested radicals for R(n) over Q it suces to have radicals for w˜(n). On the imaginary axis, R() and w() and w˜() take positive real values, 5 and when Re() = 2 , each of the values R() and w() and w˜() are real negative numbers. As the conjugate of R() over K(w) is 1/R(), equation 5.13 gives 2 1+w() + 1+w() + 1 if n 3 mod 4 R() = 2 2 6 2  1+w() q¡ 1+w() ¢ + 1 if n 3 mod 4  2 2 q where √ is always thepositive square¡ root of¢a positive real number. When n > 1 and 5 - n, the algebraic number w˜(n) has degree 2 over Q(v(n)). As the absolute value of w˜( ) satises w˜( ) > 2 when n > 1, one recovers w˜( ) n | n | n 5.5. NESTED RADICALS 95

from v(n) as

2 n v(n) + v(n) 4 5 if n 3 mod 4 2w˜(n) = 6 v( ) qv( )2 4¡n¢ if n 3 mod 4 .  n n 5 q ¡ ¢  n Note that the radicands above are positive real numbers. This is obvious for 5 = 1. When n = 1, we have v( ) = w˜( ) + 1/w˜( ) > 2. One easily recovers 5 | n | | n n | ¡ ¢ R( ) from √5 w˜( ) = w( ). In the case n is divisible by 5, one simply has n ¡ ¢ n n v(n) = w˜(n).

Below, we give nested radicals for v(n) with 1 n 16, computed as in example 128. In many cases the radicals below have undergone some cosmetic mod- ications made by factorising elements in real quadratic orders of class number one. Every root appearing in our examples should be interpreted as the real positive root of its real positive argument. Our computation v(1) = 2 for example, leads to w˜(√ 1) = 1 and w(√ 1) = √5, which gives Ramanujan’s formula 5.3. The value v(3) also gives a trivial extension of Q.

v(1) = 2

v( ) = 1 3

For n = 4, 6, 9, 11, 14, 16 the degree [Q(v(n)) : Q] is a power of 2. In these cases we opt for a tower of quadratic extensions in solving for v(n).

v(4) = 3 + √5 1 √ √ √ v(6) = 2 4 + 5 2 + 10 + 2 5 1 √ √ √ v(9) = 2 ¡11 + 5 3 + 3 5 + 3 15¢

v(11) = ¡2 √5 ¢ v( ) = ( 1+√2 )2 6 + √2 + 5 2 + 4√2 14 2 µ q + 2 5 21 10√2 + (15 2√2) 11 + 8√2 r ³ q ´ ¶ 1 √ √ √ v(16) = 2 34 + 25 2 + 11 10 + 14 5 . ¡ ¢ For n 2 mod 5, the group ( /5 ) is cyclic of order 24. If the discriminant D of =[, 1] satises D < 4,OthenO v( ) generates a degree 3 extension over the O n n ring class eld H . In the examples below, we choose the eld tower H (v(n)) O O 96 CHAPTER 5. ROGERS-RAMANUJAN CONTINUED FRACTION

H Q(√ n) to solve for v(n). O

3 3 v( ) = 1 2 + 35 + 15√6 35 + 15√6 2 3 q q v( ) = 1 ¡ 4 + 3 20( 41 + 9√21 ) 3 20(41¢ + 9√21 ) 7 6 ³ q q ´ 1 √ 3 5 √ √ v(8) = 3 8 + 5 2 + 2 782 + 565 2 + 3 6(7771 + 5490 2 ) r µ ³ q ´ 3 5 √ √ + 2 782 + 565 2 3 6(7771 + 5490 2 ) r ³ q ´ ¶ 1 √ 3 √ 3 √ v(12) = 3 17 + 10 3 + 4 260 + 150 3 + 23975 + 13875 3 q q 1 ¡ √ 3 5 √ ¢ √ v(13) = 3 23 + 5 13 + 2 14123 + 3905 13 + 9 274434 + 76110 13 r µ ³ q ´ 3 5 √ √ + 2 14123 + 3905 13 9 274434 + 76110 13 r ³ q ´ ¶ When n is divisible by 5, the value of v at n generates a eld extension of degree 5 over the ring class eld for = [ , 1]. In applying the algorithm of section 3, our O n rst step solves for v(n) over H . O

v( ) = 1 + 1 √5 a + √5 a + √5 a + √5 a , where 5 √5 1 2 3 4 ¡ ¢ a , a = 10 55 + 25√5 5050 + 2258√5 1 2 ³ q ´ a , a = 5 55 + 25√5 50 + 22√5 3 4 2 ³ q ´ v( ) = 1 5 + 2√5 + √5 a + √5 a + √5 a + √5 a , where 10 √5 1 2 3 4 ³ ´ 1 + √5 6 a , a = 20 5(3 + √5 )(16 + √5 )(9 + 4√5 ) 51 2(5 + 2√5 ) 1 2 2 q ³ 1 + √5 12 1 + √5³ 6 ´ ´ a , a = 5 5 (22 3√5 ) 3 2(5 + 2√5 ) 3 4 2 2 ³ ³ ´ ³ ´ q ´ v( ) = 1 1 (5 + 5√5) + √5 a + √5 a + √5 a + √5 a , where 15 5 2 1 2 3 4 ³ ´ a , a = 125 5(25 + 13√5 ) 14 15 (25 + 11√5 ) 1 2 4 2 ³ q ´ a , a = 125 5(15 + 7√5 ) 2 15 (25 + 11√5 ) 3 4 4 2 ³ q ´ Bibliography

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Wiskundigen gebruiken computeralgebrapakketten om lastig rekenwerk voor hen op te knappen. Kenmerkend voor deze pakketten is dat ze symbolisch kunnen rekenen. Een getal als √2 wordt niet afgerond op een aantal decimalen, maar wordt weergegeven als een symbool waarvan de computer weet dat het in het kwadraat gelijk aan 2 is. Dit heeft het grote voordeel dat je exacte berekeningen kunt doen. Helaas kleven hier bij berekeningen met wortels ook wel nadelen aan. Er zijn bij- voorbeeld twee getallen die in het kwadraat 2 opleveren: √2 en √2. Algebrasch gezien zijn ze ononderscheidbaar, beide hebben minimumpolynoom x2 2. Wil je toch speciek voor een van deze getallen je berekening doen, dan zul je naast de sym- bolische weergave ook een numerieke benadering bij moeten houden. Verder kan het rekenen met wortels nodeloos ingewikkelde termen opleveren. Neem nu bijvoorbeeld het getal 3 3 A = √5 + 2 √5 2. We bedoelen hier, zoals gebruikelijk,q met √5 deqpositieve reele wortel uit het getal 5; ook de twee derdemachtswortels geven hier de unieke reele derdemachtswortel uit de getallen √5 2 aan. Ook in de andere voorbeelden in deze samenvatting zullen we alleen de positiev e reele wortels uit positieve reele getallen beschouwen. Als je A numeriek benadert lijkt het wel heel veel op het getal 1:

3 3 √5 + 2 √5 2 = 1.618033988 . . . 0.618033988 . . . . Met behulp vqan symboliscqh rekenen kun je laten zien dat A een nulpunt is van het polynoom x3 + 3x 4, dat gelijk is aan (x 1)(x2 + x + 4). Het enige reele nulpunt van dit polynoom is 1, dus geldt A = 1. Graag zou je hebben dat de computer voor je controleert of de uitdrukking een- voudiger geschreven kan worden. De eerste vraag die hierbij beantwoord moet worden is wat je onder een zo eenvoudig mogelijke uitdrukking verstaat. In het verleden hebben verschillende wiskundigen voorbeelden gegeven van gelijkheden van worteluitdrukkingen ([1], [5], [20], [21], [28], [40]). Zo is de volgende gelijkheid van Richard Zippel ([40]) afkomstig

6 7√3 20 19 = 3 5/3 3 2/3. q p p 101 102

In de uitdrukking aan de rechterkant van het gelijkteken staan minder worteltekens onder elkaar dan in de uitdrukking aan de linkerkant van het gelijkteken; we zullen een worteluitdrukking eenvoudig noemen als er weinig worteltekens onder elkaar staan. We denieren de worteldiepte van een getal als het minimum van dit aantal onder elkaar staande worteltekens voor alle worteluitdrukkingen voor . Zo is dus de worteldiepte van het getal 6 7√3 20 19 gelijk aan 1: omdat √3 20 niet rationaal is hebben we minstens 1 worteltekennodig en hierboven zagen we al dat p je dit getal in een uitdrukking zonder geneste wortels weer kunt geven. Susan Landau ([20]) gaf, gebruikmakend van Galoistheorie, een methode om een worteluitdrukking W te berekenen voor een getal van worteldiepte r N, zodat in W ten hoogste r + 1 worteltekens onder elkaar staan. ∈ In specieke gevallen kun je ook eenvoudige algoritmen geven die een uitdrukking van minimale worteldiepte opleveren. Dit is wat we in het derde en vierde hoofdstuk van dit proefschrift doen voor uitdrukkingen van worteldiepte 2 die uit een enkele term bestaan. Laat en gehele getallen zijn en denieer B = √3 + √3 . In hoofdstuk 3 laten we, met behulp van de Galoisgroep van de normale afsluiting van Q(B) over Q, p zien dat B te vereenvoudigen is dan en slechts dan als √3 + √3 gelijk is aan het produkt van een kwadraat in Q(√3 , √3 ) en een eenvoudige factor. Daarmee bewij- zen we dat B te vereenvoudigen is dan en slechts dan als of / een derdemacht is of er gehele getallen m en n zijn met (4m + n)n3 = . 4(m 2n)m3 Als zulke m en n bestaan geven we een formule die ons de vereenvoudiging geeft. Zo vinden we voor m = n = 1 bijvoorbeeld

√3 5 √3 4 = 1 (√3 2 + √3 20 √3 25 ). 3 q Algemener denieren we voor een lichaam K van karakteristiek 0 het lichaam van alle getallen die met worteldiepte 1 over K geschreven kunnen worden: K(1) = K( K met n K zekere n Z ). { ∈ ∈ ∈ >0} Laat L een deellichaam van K(1) zijn. We beschouwen elementen van de vorm √n , met in L K. We laten zien dat ook in dit geval √n K(1) impliceert dat een n-de macht\in L is vermenigvuldigd met een eenvoudige factor.∈ Zo laten we ook zien dat 3 1 + √2 geen element is van K(1): het element 1 + √2 is niet te schrijven als een kwadraat in Q(√2 ) maal een eenvoudige factor. p Ook de gelijkheid van Zippel geldt omdat 7√3 20 19 ‘bijna een zesde macht is in het lichaam Q(√3 20 )’; er geldt 1 6 7√3 20 19 = √3 20 2 144 ³ ´ SAMENVATTING 103 en we hebben

1 6 √3 20 2 = 3 1/12 √3 20 √3 8 = 3 5/3 3 2/3. r144 ³ ´ p ³ ´ p p Dit ‘bijna-n-de macht zijn van een worteluitdrukking’ is een voorwaarde voor vereenvoudiging die intutief voor de hand ligt. We leiden deze af met behulp van Galoistheorie voor radicaaluitbreidingen. In de eerste twee hoofdstukken van dit proefschrift bestuderen we Galoisgroepen van dit soort uitbreidingen. Laat L over K een lichaamsuitbreiding zijn voortgebracht door n-de machtswortels uit elementen van K voor een zeker positief geheel getal n. Als het grondlichaam K een primitieve n-de eenheidswortel bevat, dan heet de uitbreiding een Kummeruitbreiding. Zo is bijvoorbeeld Q(√3, √5 ) een Kummeruitbreiding voor n = 2 omdat Q de primitieve 2-de eenheidswortel 1 bevat. Als we nu W denieren als de multiplicatieve groep Q, √3, √5 , dan wordt elk tussenlichaam van Q(√3, √5 ) voortgebracht door een h i ondergroep W 0 van W . De tussenlichamen van Q(√3, √5 ) zijn dus

Q, Q(√3 ), Q(√5 ), Q(√15 ) and Q(√3, √5 ).

In het algemene geval hebben we een grondlichaam K, met K voor een n ∈ primitieve n-de eenheidswortel n. Het lichaam L is van de vorm K(1, 2, . . .) voor n n i K met i K en W is K, 1, 2, . . . . We leiden voor K af dat √ in L is b∈evat dan en ∈slechts dan alsh een elemenit van W is. ∈ We bepalen ook de Galoisgroep van uitbreidingen van de vorm

L = K( K met n K voor zekere n I ), { ∈ ∈ ∈ } waarbij I een deelverzameling is van Z zodat K voor alle n I. >0 n ∈ ∈ Verder bestuderen we de situatie waar het grondlichaam de vereiste eenheids- wortels uit Kummertheorie niet bevat. In dit geval schrijven we de Galoisgroep als een semidirekt produkt van twee groepen, een ‘Kummerachtige’ groep en de Galois- groep van de uitbreiding voortgebracht door de eenheidswortels uit de verzameling n K : K voor zekere n I . { ∈ ∈ ∈ } Wanneer het grondlichaam de vereiste eenheidswortels niet bevat blijk je ook tussenlichamen te kunnen krijgen die niet voortgebracht worden door een onder- groep van de voortbrengende radicalen W . Laat bijvoorbeeld 7 een primitieve 7-de eenheidswortel zijn. Het lichaam Q(7) heeft twee niet-triviale tussenlichamen, die geen van beide door een macht van 7 voortgebracht worden. Toch zijn er ook uit- breidingen, zoals Q()/Q met 4 = 3, waarbij zo’n eenheidswortel niet in het grondlichaam zit en toch alle deellichamen van de verwachte vorm zijn. In het tweede hoofdstuk bestuderen we het eenvoudigste geval dat op kan treden: we nemen K gelijk aan Q en beschouwen een uitbreiding voortgebracht door een radicaal . We karakteriseren de radicalen waarvoor alle tussenlichamen van Q() voortgebracht worden door een ondergroep van Q, . h i 104

Tot nog toe hebben we het vooral gehad over het vereenvoudigen van worteluit- drukkingen. Echter, het is ook goed mogelijk dat je een getal op een andere wijze gegeven hebt en dat je bijvoorbeeld uit een stelling weet dat er een worteluitdrukking voor gegeven kan worden. Een voorbeeld is dat je het getal gegeven hebt als nulpunt van een oplosbare veelterm. In hoofdstuk 5 kijken we naar zo’n situatie. We bepalen worteluitdrukkingen voor singuliere waarden van de Rogers-Ramanujan kettingbreuk

n 1 ∞ n ( ) R(z) = q 5 1 q 5 ; n=1 Y ¡ ¢ 2iz n hierbij is z een element uit het complexe bovenhalfvlak, q is e en 5 is een Legendre symbool. We bewijzen dat voor een voortbrenger van een orde in een ima- ginair kwadratisch lichaam geldt dat R() met een worteluitdrukking weergegev¡ ¢ en kan worden. We beschrijven een algemene methode om zo’n worteluitdrukking te bepalen door weer de bijbehorende lichaamsuitbreiding en Galoisgroep uit te reke- nen en deze te gebruiken om een worteluitdrukking te construeren. We geven zo’n uitdrukking voor R(n), met

√ n indien n 3 mod 4 n = 5+√ n 6 indien n 3 mod 4 ( 2 voor 1 n 16. Curriculum Vitae

Mascha Honsbeek is geboren op 25 oktober 1971 in Haarlem. Al op jonge leeftijd bleek haar interesse voor rekenen. Een van haar favoriete spelletjes was het cijfers- deel uit het tv-programma ‘Cijfers & Letters’. Tijdens de vwo-opleiding aan het Liemers College te Zevenaar ging haar voorkeur uit naar de betavakken. Ze rondde de opleiding in 1990 cum laude af. Geheel naar verwachting ging ze daarna wiskunde studeren aan de Katholieke Universiteit Nijmegen. Naast studeren besteedde ze veel tijd aan sport en aan aller- lei activiteiten van de studievereniging Desda. Verder was ze studentassistent en studentlid van de opleidingscommissie wiskunde. Haar scriptie ‘Het vereenvoudi- gen van worteluitdrukkingen’, geschreven onder leiding van Frans Keune, was een opstapje voor het onderzoek in dit proefschrift. Ze studeerde af in augustus 1996. Aansluitend startte ze als aio met haar promotieonderzoek. Toen deze aanstel- ling na vier jaar ophield, werd ze coordinator van het door Frans Keune opgestarte project Ratio, waar wiskunde-onderwijs voor het vwo ontwikkeld wordt, en werkte ze als voorlichter van de wiskundestudie aan de kun. Dit deed ze tot september 2004. In de vrije uurtjes rondde ze het proefschrift af. Nu is ze een geheel nieuwe carriere begonnen als docente wiskunde aan het Bernard Nieuwentijt College, locatie Damstede, een havo-vwo school in Amsterdam Noord.

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