Maximal Unramified Extensions of Cyclic Cubic Fields

Brigham Young University BYU ScholarsArchive All Theses and Dissertations 011-07-05 Maximal Unramified Extensions of Cyclic Cubic Fields Ka Lun Wong Brigham Young University - Provo Follow this and additional works at: https://scholarsarchive.byu.edu/etd Part of the Mathematics Commons BYU ScholarsArchive Citation Wong, Ka Lun, "Maximal Unramified Extensions of Cyclic Cubic Fields" (011). All Theses and Dissertations. 781. https://scholarsarchive.byu.edu/etd/781 This Thesis is brought to you for free and open access by BYU ScholarsArchive. It has been accepted for inclusion in All Theses and Dissertations by an authorized administrator of BYU ScholarsArchive. For more information, please contact scholarsarchive@byu.edu, ellen_amatangelo@byu.edu.

Maximal Unramified Extensions of Cyclic Cubic Fields Ka Lun Wong A thesis submitted to the faculty of Brigham Young University in partial fulfillment of the requirements for the degree of Master of Science Darrin Doud, Chair David Cardon Pace Nielsen Department of Mathematics Brigham Young University August 011 Copyright c 011 Ka Lun Wong All Rights Reserved

Abstract Maximal Unramified Extensions of Cyclic Cubic Fields Ka Lun Wong Department of Mathematics Master of Science Maximal unramified extensions of quadratic number fields have been well studied. This thesis focuses on maximal unramified extensions of cyclic cubic fields. We use the unconditional discriminant bounds of Moreno to determine cyclic cubic fields having no non-solvable unramified extensions. We also use a theorem of Roquette, developed from the method of Golod-Shafarevich, and some results by Cohen to construct cyclic cubic fields in which the unramified extension is of infinite degree.

Acknowledgments First, I want to thank my advisor Darrin Doud who introduced this project to me. He has been kind and enthusiastic to help me. I am grateful for his patience with my lack of knowledge. His wide knowledge has been of great value to me. He has made available his support in a number of ways. I have enjoyed the time and study with him. Besides Professor Doud, I would also like to thank my research committee, including Professor David Cardon and Professor Pace Nielsen, for their time and comments on my thesis. I also want to thank my wife and my family for their sacrifice and support. Without their support, it would have been impossible to complete this project.

Contents 1 Introduction 1 1.1 Preliminaries I................................... 1. Preliminaries II - Class Field Theory...................... 6 Cyclic Cubic Fields 9.1 General Parametric Description of Cyclic Cubic Fields............ 9. Specific Parametric Description of Cyclic Cubic Fields............. 17 Discriminant Bounds 9 4 Maximal Unramified Extensions of Cyclic Cubic Fields 4.1 Maximal Unramified Solvable Extensions.................... 4. Unramified Non-Solvable Extensions...................... 45 4. Unramified Extensions of Infinite Order..................... 49 iv

Chapter 1. Introduction Maximal unramified extensions of quadratic number fields have been well studied. Yamamura [11] studied maximal unramified extensions of imaginary quadratic number fields of small conductors 40 (and 1000 under the Generalized Riemann Hypothesis (GRH)) and determined the structures of the Galois groups Gal(K ur /K) of the maximal unramified extensions K ur of these imaginary quadratic number fields K. The main idea in Yamamura s paper is to use the discriminant bounds to show that certain fields with discriminant less than the bounds have no non-solvable unramified extension. Then, the maximal unramified extension will coincide with the top of the class field tower of K which is also the maximal unramified solvable extension. In his paper, he gives the explicit structure of K ur and Gal(K ur /K) in the following table. 1

In the above table, K 1 is the (first) Hilbert class field of K and K is the second Hilbert class field (Hilbert class field of K 1 ) of K. The constant l is the length of the class field tower, i.e., the smallest number l such that K l = K l+1, provided that such l exists. He also gives examples of unramified non-solvable extensions of K. The following is the first example. Proposition 1.1. The field Q( 1507) is the first imaginary quadratic number field having an unramified A 5 -extension which is normal over Q in the sense that none of Q( d) of discriminant d with 0 > d > 1507 has such an extension. Moreover, such an extension of K = Q( 1507) is given by the composite field of K with the splitting field of the quintic polynomial x 5 5x + 5x + 4x + 4, which is an A 5 -extension of Q. Our research focuses on maximal unramified extensions of cyclic cubic fields. We follow his ideas but use the unconditional discriminant bounds of Moreno [8] to determine cyclic cubic fields having no non-solvable unramified extensions. That will imply the maximal unramified solvable extensions, which are also the top of the class field tower, are the maximal unramified extensions. We give several examples of cyclic cubic fields with non-solvable unramified extensions as well. We also use a theorem of Roquette (see [5]), developed from the method of Golod-Shafarevich, to construct cyclic cubic fields in which the unramified extension is of infinite degree. Some results by Cohen [] on cyclic cubic fields are also very useful and important to our construction of certain examples. 1.1 Preliminaries I All of the results in this section can be found in most introductory books on algebraic number theory, such as Marcus [6]. Let K be a number field over Q and let O K be the ring of integers in K. Let L be a number field containing K and let O L be the ring of integers in L. Let P be a prime in O K. Then

the ideal P O L in O L factors uniquely into prime ideals in O L, i.e. P O L = Q 1 e 1 Q e...q g e g. The exponent e i of Q i is called the ramification index of Q i over P, denoted by e(q i P ). Definition 1.. A prime P is ramified if e i > 1 for some i. The prime P splits if g > 1. The prime P is inert (remains prime) if g = 1 and e 1 = 1. Here we also give some definitions from Cox [] about infinite primes. Definition 1.. Prime ideals of O K are often called finite primes to distinguish them from the infinite primes, which are determined by the embeddings of K into C. A real infinite prime is an embedding σ : K R, while a complex infinite prime is a pair of complex conjugate embeddings σ, σ : K C, σ σ. Given an extension L/K, an infinite prime σ of K ramifies in L provided that σ is real but it has an extension to L which is complex. For example, the infinite prime of Q is unramified in Q( ) but ramified in Q( ). We say L is an unramified extension of K if L is unramified at all primes, finite or infinite. Definition 1.4. Let P be a prime of O K and Q be a prime of O L. The fields O K /P and O L /Q are called the residue fields associated with P and Q, respectively. The degree f of O L /Q over O K /P is called the inertial degree of Q over P, denoted by f(q P ). Proposition 1.5. If U P Q are primes in three number rings O F O K O L, then e(q U) = e(q P )e(p U) and f(q U) = f(q P )f(p U). Proof. Let the factorization of U in O K be P e(p U) P e P...P e Pn n and the factorization of P in O L be Q e(q P ) Q e Q...Q e Qr r. Since every prime of O L lies over a unique prime of O K and every prime of O K lies over a unique prime of O F, the factorization of U in O L looks like [Q e(q P ) Q e Q...Q e Qr r ] e(p U) [...] e P...[...] e Pn. Thus, e(q U) = e(q P )e(p U). And f(q U) = [O L /Q : O F /U] = [O L /Q : O K /P ][O K /P : O F /U] = f(q P )f(p U). The following proposition can be found in Marcus [6]. It relates an important relation between the degree of L over K and the ramification indices and inertial degrees.

Proposition 1.6. Let n be the degree of L over K and let Q 1,..., Q g be the primes of O L over a prime P of O K. Denote by e 1,..., e g and f 1,..., f g the corresponding ramification indices g and inertial degrees. Then e i f i = n. i=1 Proposition 1.7. If L is normal over K and Q and Q are two primes lying over P, then e(q P ) = e(q P ) = e and f(q P ) = f(q P ) = f. Moreover, efg = n. Proof. Let P = Q e(q P ) Q e(q P ) Q e P...Q e Pn n. Let G = Gal(O L /O K ). We know that G permutes the primes lying over P and σ(q) = Q for some σ G. Thus, P = σ(p ) = Q e(q P ) σ(q ) e(q P ) σ(q ) e P...σ(Q n ) e Pn and e(q P ) = e(q P ). We define σ : O L /Q O L /Q by σ(x + Q) = σ(x) + Q where σ is the specific element in G such that σ(q) = Q as above. This function is a well-defined function. To show it is one-to-one, we suppose σ(x 1 + Q) = σ(x + Q). Then σ(x 1 ) + Q = σ(x ) + Q and σ(x 1 ) σ(x ) = σ(x 1 x ) Q. Thus, x 1 x Q and so x 1 + Q = x + Q. This shows σ is one-to-one. Also, for any y + Q O L /Q, σ 1 (y) + Q O L /Q where σ 1 G such that σ(σ 1 (y) + Q) = σ(σ 1 (y)) + Q = y + Q. This shows σ is onto. It is clearly a field homomorphism as it preserves addition and multiplication. Therefore, O L /Q = O L /Q and f(q P ) = f(q P ) = f. So if we have a cyclic cubic field K over Q and if p Z is ramified in K, then p = P for some prime P in K and e(p p) =. Moreover, if L is an unramified extension over K and Q is a prime in L over P, e(q P ) = 1 and e(q p) = e(p p) =. Definition 1.8. Let K be a number field. Let α 1,..., α n be an integral basis for K and let σ 1,..., σ n be the n embeddings of K in C. The discriminant of K is defined to be disc(k) = disc(α 1,..., α n ) = σ i (α j ) which is an invariant of K. 4

Definition 1.9. The root discriminant of K is defined to be rd K = disc(k) 1/[K:Q]. The following proposition can be found in Marcus [6]. From it, we know what primes in Z are ramified in K if we know the discriminant of K. Proposition 1.10. Let p be a prime in Z. Then p is ramified in O K if and only if p disc(k). Here we quote a proposition from Neukirch [9] which we will use. Proposition 1.11. Let F K L. Let denote the relative discriminant and N denote the relative norm. We have L/F = N K/F ( L/K ) [L:K] K/F. Corollary 1.1. If L is an unramified extension of K, then L/Q = [L:K] K/Q discriminant rd L of L is the same as the root discriminant rd K of K. and the root Proof. Let m = [K : Q] and n = [L : K]. By Proposition 1.11, L/Q = N K/Q ( L/K ) [L:K] K/Q. Since L over K is unramified, no prime in K divides L/K and L/K is the unit ideal. Then N K/Q ( L/K ) = 1. Thus, rd L = 1 mn L/Q = ( n K/Q ) 1 mn = 1 m K/Q = rd K. Thus, we see that if L is an unramified extension over K, then p Z divides disc(l) if and only if p divides disc(k). Now, we will look at a proposition of Marcus [6] which tells us necessary and sufficient conditions for when a prime splits completely in a subfield of cyclotomic field. Let Q(ζ) be a cyclotomic field with ζ = e πi/p. We know that the Galois group G of Q(ζ) is cyclic of order p 1, hence there is a unique subfield F d Q(ζ) having degree d over Q, for each divisor d of p 1. Proposition 1.1. Let p be an odd prime, and let q be any prime p. Fix a divisor d of p 1. Then q is a d-th power mod p if and only if q splits completely in F d. 5

1. Preliminaries II - Class Field Theory We now give some definitions and theorems from class field theory, which can be found in Cox []. Definition 1.14. A fractional ideal a of K is a nonzero finitely generated O K -submodule of K. O K. We list some properties of fractional ideals found in Cox []. (1) Any fractional ideal can be written in the form αi where α K and I is an ideal of () Any fractional ideal a is invertible, i.e., there is a fractional ideal b such that ab = O K. r () Any fractional ideal a can be written uniquely as a product a = p a i i, where the p i s are distinct prime ideals of O K, and the a i are integers. (4) Let I K denote the set of all fractional ideals of K. The set I K is closed under multiplication and I K is an Abelian group under this operation. The subset P K (i.e., those of all principal fractional ideals of the form αo K for some α K ) forms a subgroup. The quotient I K /P K is the ideal class group and is denoted by C K. The order of the ideal class group is called the class number and is denoted by h K (or h(k)). We will now introduce the Hilbert class field of K and the relation between the Hilbert class field and the ideal class group. The following propositions are found in Cox []. i=1 Proposition 1.15. Given a number field K, there is a finite Galois extension L of K such that: (i) The field L is an unramified Abelian extension of K. (ii) Any unramified Abelian extension of K lies in L. The field L of Proposition 1.15 is called the Hilbert class field of K. It is the maximal unramified Abelian extension of K and is unique. To see the relation between the Hilbert 6

class field and the ideal class group, we introduce a map, called the Artin map. We first define the Artin symbol. Lemma 1.16. Let L/K be a Galois extension, and let p be a prime of O K which is unramified in L. If P is a prime of O L containing p, then there is a unique element σ Gal(L/K) such that for all α in O L, σ(α) α N(p) (mod P), where N(p) = O K /p is the norm of p. The unique element σ of Lemma 1.16 is called the Artin symbol and is denoted ((L/K)/P) since it depends on the prime P of L. When L is an Abelian extension of K, the Artin symbol ((L/K)/P) depends only on the underlying prime p because of the following property: If σ Gal(L/K), then ( ) (L/K) = σ σ(p) ( (L/K) P ) σ 1. So the Artin symbol can be written as ((L/K)/p). Now we can extend the definition of Artin symbol. When L is an unramified Abelian extension, ((L/K)/p) is defined for all primes p of O K. If I K is the set of all fractional ideals, r then for any a = I K we can define the Artin symbol ((L/K)/a) to be the product i=1 p a i i ( ) (L/K) = a r ( ) ai (L/K). i=1 p i The Artin symbol thus defines a homomorphism, called the Artin map, ( ) (L/K) : I K Gal(L/K). The Artin reciprocity theorem for the Hilbert class field relates the Hilbert class field to 7

the ideal class group C K as follows: Theorem 1.17. If L is the Hilbert class field of a number field K, then the Artin map ( ) (L/K) : I K Gal(L/K) is surjective, and its kernel is exactly the subgroup P K of principal fractional ideals. Thus the Artin map induces an isomorphism C K Gal(L/K). If we apply Galois theory to Proposition 1.15 and Theorem 1.17, we get the following classification of unramified Abelian extensions of K. Corollary 1.18. Given a number field K, there is a one-to-one correspondence between unramified Abelian extensions M of K and subgroups H of the ideal class group C K. Furthermore, if the extension M corresponds to the subgroup H, then the Artin map induces an isomorphism C K /H Gal(M/K). All these results are the special case and consequences of the Existence Theorem and Isomorphy Theorem from class field theory which will not be stated here. 8

Chapter. Cyclic Cubic Fields This whole chapter is mostly a reproduction of results from Cohen [] with many of the details reproduced here for completeness. Let K be a number field of degree over Q, i.e. a cubic field. If K is Galois over Q, its Galois group must be isomorphic to the cyclic group Z/Z. Hence, we say that K is a cyclic cubic field. The Galois group has an identity element and two other elements which are inverses of each other. We denote them by σ and σ 1 = σ. Proposition.1. Let K = Q(θ) be a cubic field, where θ is an algebraic integer whose minimal polynomial will be denoted P (X). Then K is a cyclic cubic field if and only if the discriminant of P is a square. Proof. Since Gal(K/Q) is a subgroup of S, Gal(K/Q) = A (= Z ) or S. By a proposition of Cohen [], we know that Gal(P ) A n if and only if disc(p ) is a square, where P is the minimal polynomial of θ for K = Q(θ). Thus, K is a cyclic cubic field if and only if the discriminant of P is a square. Let K be a cyclic cubic field. Let θ be an algebraic integer such that K = Q(θ), and let P (X) = X SX + T X N be the minimal polynomial of θ, with integer coefficients S, T and N. Since any cubic field has at least one real embedding and since K is Galois, all the roots of P must be in K. Hence, they must all be real, so a cyclic cubic field must be totally real..1 General Parametric Description of Cyclic Cubic Fields From Cohen [], we know we can describe cyclic cubic fields parametrically. First, we set ζ = e πi/, i.e. a primitive cube root of unity. Since K is totally real, ζ K, hence the extension field K(ζ) is a degree six field over Q. The field K(ζ) is still Galois over Q 9

because it is the composite of two Galois extensions over Q. The Galois group is generated by commuting elements σ and τ, where σ acts on K by permuting the roots of P (X) transitively and trivially on ζ, and τ denotes complex conjugation. Then, the first result we need is as follows. Lemma.. Set γ = θ + ζ σ(θ) + ζσ (θ) K(ζ), and β = γ /τ(γ). Then β Q(ζ) and we have P (X) = X SX + S e X S Se + eu, 7 where e = βτ(β) and u = β + τ(β) (i.e. e and u are the norm and trace of β considered as an element of Q(ζ)). Proof. We have τ(γ) = τ(θ) + τ(ζ )τ(σ(θ)) + τ(ζ)τ(σ (θ)) = θ + ζσ(θ) + ζ σ (θ). One sees immediately that σ(γ) = σ(θ) + σ(ζ )σ(σ(θ)) + σ(ζ)σ(σ (θ)) = σ(θ) + ζ σ (θ) + ζθ = ζ(θ + ζ σ(θ) + ζσ (θ)) = ζγ. Hence, σ(β) = σ(γ ) σ(τ(γ)) = ζ γ ζ τ(γ) = γ τ(γ) = β. Thus, β is invariant under the action of σ, so by Galois theory β must belong to the quadratic subfield Q(ζ) of K(ζ). In particular, e and u as defined above are in Q. Also, ( ) e = βτ(β) = γ τ(γ) [τ(γ)] γ = γτ(γ), and eu = γτ(γ) τ (γ) τ(γ) + [τ(γ)] = γ + [τ(γ)]. γ Now we have the matrix equations: S θ + σ(θ) + σ (θ) 1 1 1 θ γ = θ + ζ σ(θ) + ζσ (θ) = 1 ζ ζ σ(θ). τ(γ) θ + ζσ(θ) + ζ σ (θ) 1 ζ ζ σ (θ) Hence, θ 1 1 1 σ(θ) = 1 ζ ζ σ (θ) 1 ζ ζ S γ = 1 1 1 1 S 1 ζ ζ γ. τ(γ) 1 ζ ζ τ(γ) 1 10

Thus, θ = 1 (S + γ + τ(γ)), σ(θ) = 1 (S + ζγ + ζ τ(γ)), and σ (θ) = 1 (S + ζ γ + ζτ(γ)). We compute that T =θσ(θ) + θσ (θ) + σ(θ)σ (θ) = 1 9 [(S + γ + τ(γ))(s + ζγ + ζ τ(γ)) + (S + γ + τ(γ))(s + ζ γ + ζτ(γ)) + (S + ζγ + ζ τ(γ))(s + ζ γ + ζτ(γ))] = 1 9 [S + ζsγ + ζ Sτ(γ) + Sγ + ζγ + ζ γτ(γ) + Sτ(γ) + ζγτ(γ) + ζ [τ(γ)] + S + ζ Sγ + ζsτ(γ) + Sγ + ζ γ + ζγτ(γ) + Sτ(γ) + ζ γτ(γ) + ζ[τ(γ)] + S + ζ Sγ + ζsτ(γ) + ζsγ + γ + ζ γτ(γ) + ζ Sτ(γ) + ζγτ(γ) + [τ(γ)] = 1 9 [S + Sγ(1 + ζ + ζ ) + Sτ(γ)(1 + ζ + ζ ) + γ (1 + ζ + ζ ) + [τ(γ)] (1 + ζ + ζ ) + γτ(γ)(ζ + ζ )] = 1 9 (S e) = S e. Next we compute N = θσ(θ)σ (θ) = 1 7 (S + γ + τ(γ))(s + ζγ + ζ τ(γ))(s + ζ γ + ζτ(γ)) = 1 7 [(S + ζsγ + ζ Sτ(γ) + Sγ + ζγ + ζ γτ(γ) + Sτ(γ) + ζγτ(γ) + ζ [τ(γ)] )(S + ζ γ + ζτ(γ))] 11

= 1 7 [S + ζs γ + ζ S τ(γ) + S γ + ζsγ + ζ Sγτ(γ) + S τ(γ) + ζsγτ(γ) + ζ S[τ(γ)] + ζ S γ + Sγ + ζsγτ(γ) + ζ Sγ + γ + ζγ τ(γ) + ζ Sγτ(γ) + γ τ(γ) + ζγ[τ(γ)] + ζs τ(γ) + ζ Sγτ(γ) + S[τ(γ)] + ζsγτ(γ) + ζ γ τ(γ) + γ[τ(γ)] + ζs[τ(γ)] + ζ γ[τ(γ)] + [τ(γ)] ] = 1 7 [S + (S γ + Sγ + S τ(γ) + S[τ(γ)] + γ τ(γ) + γ[τ(γ)] )(1 + ζ + ζ ) + Sγτ(γ)(ζ + ζ ) + γ + [τ(γ)] ] = S Se + eu. 7 This completes the proof. We will modify θ (hence its minimal polynomial P (X)) so as to obtain a unique defining polynomial for each cyclic cubic field. But before that, we need to prove a lemma. Lemma.. If Q(θ) is a cyclic cubic field, then Q(θ) = Q(bθ + cσ(θ)) for any b, c Q, where b and c are not both zero. Proof. First we note that 1 and θ are linearly independent over Q, otherwise θ Q. Now suppose 1, θ, σ(θ) are linearly dependent over Q. Then A + Bθ + Cσ(θ) = 0 for some A, B, C Q, but not all zero. If C = 0, then A + Bθ = 0 and A = B = 0. Thus C 0. Then, Bθ + Cσ(θ) = A B C θ + σ(θ) = A C σ(θ) = A C + B C θ. Let x = A C and y = B. We compute C 1

σ(θ) = x + yθ, σ (θ) = x + y(x + yθ) = (x + xy) + y θ, θ = σ (θ) = x + xy + y (x + yθ) = x(1 + y + y ) + y θ. Thus, y = 1, x = 0 because 1 and θ are linearly independent over Q. However, σ(θ) = θ is a contradiction. Thus, 1, θ, σ(θ) are linearly independent over Q. Then bθ + cσ(θ) Q if b, c Q and b, c are not both zero. Thus, Q Q(bθ + cσ(θ)) Q(θ) and so Q(θ) = Q(bθ + cσ(θ)). Now, we modify θ. First note that replacing γ by (b + cζ)γ is equivalent to changing θ (b + cζ) into bθ + cσ(θ), and β is changed into β b + cζ. To see this, set γ = (b + cζ)γ. Then we compute γ = (b + cζ)(θ + ζ σ(θ) + ζσ (θ)) = bθ + cζθ + bζ σ(θ) + cσ(θ) + bζσ (θ) + cζ σ (θ) = bθ + cσ(θ) + ζ [bσ(θ) + cσ (θ)] + ζ[cθ + bσ (θ)] = bθ + cσ(θ) + ζ σ(bθ + cσ(θ)) + ζ(bθ + cσ(θ)). Let θ = bθ + cσ(θ). Then γ = θ + ζ σ(θ ) + ζσ (θ ). Also, β = γ τ(γ ) = (b + cζ) γ τ((b + cζ)γ) = (b + cζ) γ τ(γ)(b + cζ ) = β (b + cζ) (b + cζ ). Now, let {p k } be the set of primes which split in Q(ζ) (they are the primes whose factorization looks like p k = π k π k ). By Proposition 1.1, if we choose p =, q = p k and d =, then p k splits completely in F = Q(ζ) if and only if p k is a square mod, i.e. p k 1 (mod ). Let {q k } be the set of inert primes, i.e. primes such that q k (mod ). Note that is the only prime ramified in Q(ζ) because the discriminant of Q(ζ) is. Let ρ = 1 + ζ = 1

denote the prime above. Then, we can write b + cζ = ( ζ) g ρ f π k e k πk f k qk g k. Hence, since b + cζ = b + cζ, we have (b + cζ) b + cζ = ( ζ)g ρ f e πk k f πk k g qk k ( ζ ) g ( ρ) f e π k k πk f k qk g k = ( 1) g+f ρ f π k e k f k πk f k e k qk g k. If the decomposition of β is ( ζ) n ρ m π k l k πk m k qk n k, then choose g k = n k and f = m. Thus, (b + cζ) β (b + cζ ) = ( 1)g+f+n ζ n l π k +e k f k m k πk k +f k e k. Furthermore, for each k consider the quantity m k + l k. mk l k + 1 Case(1): If m k + l k 0 or 1 (mod ), choose e k = and f k = l k + e k. Then, π l k+e k f k k π k m k +f k e k = π k m k +l k +e k = π k m k +l k + mk l k +1 mk l k + 1 Let m k +l k = M or M+1. Then m k l k +1 = M+1 or M. So, = M + 1 mk l k + 1 = M or M = M. Then, m k + l k + = 0 or 1. Thus, π l k+e k f k k π k m k +f k e k = 1 or π k. Case(): If m k + l k (mod ), then l k + m k 1 (mod ), and we choose f k =. 14

lk m k + 1 and e k = m k + f k. Then, π l k+e k f k k π k m k +f k e k = π k l k +m k +f k = π k l k +m k + lk m k +1. Letting l k + m k = M + 1, then l k m k = M 1, l k m k + 1 = M. So lk m k + 1 lk m k + 1 = M. Hence, l k + m k + = M + 1 M = 1. Thus, π l k+e k f k k π k m k +f k e k = π k. With this choice of exponents, β Z[ζ] because π k and π k are in Z[ζ]. Also, β is not divisible by any inert or ramified prime, and is divisible by split primes only to the first power. Also, at most one of π k or π k divides β. In other words, if e = β τ(β ) is the new value of the norm of β, then e is equal to a product of distinct primes congruent to 1 modulo. Now, K = Q(θ) = Q(θ ) and θ is a root of the polynomial F (X) = X S X +T X N and by Lemma., F (X) = X S X + S e ) Consider Q(X) = F (X + S ) ) = (X + S S (X + S + S e X S S e + e u. 7 ) (X + S S S e + e u 7 = X +S X + S X + S 7 S X S X S 9 + S X e X + S 9 S e 9 S 7 + S e 9 e u 7 = X e X e u 7 and let θ = θ S be a root of Q(X). ) ) Consider T (X) = 7Q ( X ( X = 7 7 e 9 X e u 7 = X e X e u. Since β = u + v and β is not divisible by the ramified prime ρ, u cannot be divisible by. Otherwise, ρ divides u and v and thus divides β. Then, by suitably choosing the exponent g above (which amounts to changing β into β ), we may assume u (mod ). In our process, because we want e to be equal to a product of primes congruent to 1 modulo, b and c are chosen uniquely and thus β is unique, hence so are e and u. We 15

restate this in the following proposition. Proposition.4. For any cyclic cubic field K, there exists a unique pair of integers e and u such that e is equal to a product of distinct primes congruent to 1 modulo, u (mod ) and such that K = Q(θ ) where θ is a root of the polynomial Q(X) = X e X eu 7, or equivalently K = Q(θ) where θ is a root of P (X) = 7Q(X/) = X ex eu. Example.5. Consider the cyclic cubic field represented by P (x) = x + 7x 1911x 0697. By Lemma., we find that β = 1140 7 ζ 118, e = 68796 and u = 096 7 7. When we factor β, we get β = ( ) ( + ζ) 1 (1 ζ) ( ζ)(). The first factor is the prime above, the second and third factors are primes above 7. The fourth factor is a prime above 1. The last factor is an inert prime. So we have n = 4, m =, n 1 = 1, l 1 = 1, m 1 =, l = 1, m = 0. So we choose g 1 = 1, f =, e 1 = 0, f 1 = 1, e = 0, f = 0. Then we have b + cζ = ( ζ) g ( ) (1 ζ) 1 and β (c + bζ) = β (b + cζ ) = ( ( 1) g+1 ( ζ) 4 (1 ζ)( ζ) = ( 1) g+1 8 + 10i ). We then choose g = 1. We get β = 8 + 10i. So e = 91 = (7)(1) and u = 8. The canonical defining polynomial is then P (X) = X (91)X (91)(8) = X 7X 78. Example.6. Consider the cyclic cubic field represented by P (x) = x + 551x 6777958x 5771714577. By Lemma., we find that β = 4505665 1091 ζ + 95576950, e = 801475 and u = 1091 1955959565. When we factor β, we get β = ( ζ) 4 (4 + ζ)(1 + 6ζ)(7 + 9ζ) 1 ( + 9ζ) (1 + 1091 9ζ)( 14ζ) 1 ( 11 14ζ) (5). The first factor is a unit. The second factor is a prime above 1. The third factor is a prime above 1. The fourth and fifth factors are primes 16

above 67. The sixth factor is a prime above 7. The seventh and eighth factors are primes above 16. The last factor is an inert prime. So we have n = 4, m = 0, n 1 = 1, l 1 = 1, m 1 = 0, l = 1, m = 0, l = 1, m =, l 4 = 1, m 4 = 0, l 5 = 1, m 5 =. So we choose g 1 = 1, f = 0, e 1 = 0, f 1 = 0, e = 0, f = 0, e = 0, f = 1, e 4 = 0, f 4 = 0, e 5 = 0, f 5 = 1. Then we have b + cζ = ( ζ) g ( + 9ζ) 1 ( 11 14ζ) 1 (5) 1 and β (c + bζ) = β (b + cζ ) = ( ( 1) g ζ 4 (4 + ζ)(1 + 6ζ)(1 + 9ζ) = ( 1) g 4 i ). We then choose g = 1. We get β = 4 + i. So e = (1)(1)(7) and u = 4. The canonical defining polynomial is then P (X) = X (1)(1)(7)X (1)(1)(7)( 4) = X 8857X + 10090717. In the next section, we will prove the converse of this proposition and show examples of how to find cyclic cubic fields with a given discriminant.. Specific Parametric Description of Cyclic Cubic Fields From the work of Cohen [], we know that depending on whether is ramified or not in K, the canonical minimal representing polynomial has different forms. We can see the following theorem due to Cohen. Theorem.7. All cyclic cubic fields K are given exactly once (up to isomorphism) in the following way: (1) If the prime is ramified in K, then K = Q(θ) where θ is a root of the equation P (X) = X e X eu 7 Z[X], where e = u + 7v, u 6 (mod 9), v, u v (mod ), 4 v > 0 and e is equal to the product of distinct primes congruent to 1 modulo. 9 () If the prime is unramified in K, then K = Q(θ) where θ is a root of the equation P (X) = X X + 1 e 1 e + eu X Z[X], where e = u + 7v, u (mod ), 7 4 u v (mod ), v > 0 and e is equal to the product of distinct primes congruent to 1 modulo. 17

In both cases, the discriminant of P is equal to e v and the discriminant of the number field K is equal to e. () Conversely, if e is equal to 9 times the product of t 1 distinct primes congruent to 1 modulo, (respectively is equal to the product of t distinct primes congruent to 1 modulo ), then there exist up to isomorphism exactly t 1 cyclic cubic fields of discriminant e defined by the polynomials P (X) given in (1) (respectively ()). To prove this theorem, we will need in particular to compute explicit integral bases and discriminants of cyclic cubic fields. So, let K be a cyclic cubic field. By Proposition.4, we have K = Q(θ) where θ is a root of the equation P (X) = X ex eu, where e = u + v, 4 u (mod ) and e is equal to a product of distinct primes congruent to 1 modulo. We first quote some definitions, a proposition and a few lemmas from Cohen []. Definition.8. An order R in field K is a subring of K which as a Z-module is finitely generated and of maximal rank n = deg(k). Definition.9. Let O be an order in a number field K and let p be a prime number. We say O is p-maximal if [O K : O] is not divisible by p. The following proposition, called Dedekind s criterion, gives the conditions for Z[θ] to be p-maximal. We only quote the second part that we will use. Proposition.10. Let K = Q(θ) be a number field, let T Z[X] be the minimal polynomial of θ and let p be a prime number. Denote by reduction modulo p (in Z, Z[X] or Z[θ]). Let k k T (X) = t i (X) e i be the factorization of T (X) modulo p in F p [X], and set g(x) = t i (X) i=1 where the t i Z[X] are arbitrary monic lifts of t i. () Let h(x) Z[X] be a monic lift of T (X)/g(X) and set f(x) = (g(x)h(x) T (X))/p Z[X]. Then Z[θ] is p-maximal if and only if (f, g, h) = 1 in F p [X]. Lemma.11. Let p e. Then the order Z[θ], where θ is a root of X ex + eu (as in Proposition.4), is p-maximal. i=1 18

Proof. We apply Dedekind s criterion. Since p e, P (X) = X, therefore, t 1 (X) = X, g(x) = X, h(x) = X g(x)h(x) P (X) and f(x) = = e p p X + eu. Since p e, we cannot p have p u, otherwise p v, hence p e which was assumed not to be true. Therefore, p eu p so (f, g, h) = 1, showing that Z[θ] is p-maximal. Corollary.1. The discriminant of P (X) is equal to 81e v. The discriminant of the number field K is divisible by e. Proof. The discriminant of X + ax + b is equal to 4a 7b, hence the discriminant of P is equal to 4( e) 7(eu) = 7e (4e u ) = 7e (v ) = 81e v, thus proving the first statement. For the second statement, we know that the discriminant of the field K is a square divisor of 81e v. We also know disc(p ) = disc(k)f where f = [O K : Z[θ]]. By the preceding lemma, Z[θ] is p-maximal for all primes dividing e, and e is coprime to 81v. That means if p e, p f and thus p disc(k) and e disc(k). Since, as we will see, the prime divisors of v other than are irrelevant, what remains is to look at behavior of the prime. Lemma.1. Assume that v. Then Z[θ] is -maximal. Proof. Again we use Dedekind s criterion. Since eu (mod ), we have P = X eu = X + 1 = (X + 1) in F [X], hence t 1 = X + 1, g(x) = X + 1, h(x) = (X + 1) and f(x) = (X + 1) (X ex eu) = X + (e + 1)X + 1 + eu eu + 1 e = (X + 1)(X + e) +. ( ) Hence (f, g, h) = (X +1, f) = X + 1,. Now we check that r = = (u + v )(u ) + 4 1 = (u ) (u + 1) + v (u ) eu + 1 e = (u (u ) + v (u ) + 4 1 eu + 1 e = (u u + 4) + v (u ) 1. Since u (mod ), 4r v (u ) (mod 9) and since 1 v, r (mod ) so (f, g, h) = 1, which proves the lemma. Lemma.14. With the above notation, let θ be a root of P (X) = X ex eu, where e = u + v and u (mod ). The conjugates of θ are given by the formulas σ(θ) = 4 e u + v v v θ + 1 v θ, σ (θ) = e v + u v v θ 1 v θ. 19

Proof. Let θ = σ(θ) and θ = σ (θ). The discriminant of P (X) = f where f = (θ θ )(θ θ )(θ θ) and f = ±9ev by Corollary.1. If necessary, by exchanging θ and θ, 9ev we may assume that θ θ = (θ θ )(θ θ ) = 9ev P (θ) = 9ev. Using the extended (θ e) Euclidean algorithm with A(X) = X ex eu and B(X) = X e, we find the inverse of B modulo A. Thus, A(X) = [B(X)]X ex eu 1 e A(X) = X e B(X) + X + u X + u = 1 e A(X) + X e B(X). And, So, ( B(X) = X + u ) ( X u ) + u 4e ( 4 u 4e = B(X) 1 4 e A(X) + X ) ( e B(X) X u ) u 4e = 1 ( X u ) [ A(X) + 1 X ( X u ) ] B(X) 4 e e 1 = 4 ( X u ) A(X) + e(u 4e) ( 4 u 4e A(X)r(X) + B(X)s(X) = 1, ) [ 1 X ( X u ) ] B(X). e where r(x) = 4 ( X u ) e(u 4e) ( 4 and s(x) = u 4e [ 4 X = 4e u e 1 = 4e u ) [ 1 X ( e ( X u ) X u ] 1 ] [ X (X u) 4 e = 1 [X(X u) 4e] v e = X ux 4e. v e ) ] 0

Hence, B(X)s(X) 1 (mod A(X)) and s(x) is the inverse of B(X) modulo A(X). Also, s(θ) = 1 B(θ). Thus, θ 9ev θ = (θ e) = 9ev B(θ) = 9ev ( ) ( ) 9ev θ s(θ) = uθ 4e = v e 1 v (θ uθ 4e). Then, since θ + θ + θ = 0, θ θ θ = θ. Hence, θ = ( ) 1 1 v (θ uθ 4e) θ = e v u + v v θ + 1 v θ. And we obtain θ = θ θ = e v + u v v θ 1 v θ. Now we will prove a theorem that implies the first two statements of Theorem.7. Theorem.15. Let K = Q(θ) be a cyclic cubic field where θ is a root of X ex eu and where, as above, e = u + v is equal to a product of distinct primes congruent to 1 4 modulo. (1) Assume that v. Then (1, θ, σ(θ)) (where σ(θ) is given by the above formula) is an integral basis of K and the discriminant of K is equal to (9e). () Assume that v. Then if θ = θ + 1, (1, θ, σ(θ )) is an integral basis of K and the discriminant of K is equal to e. Proof. (1) Since θ = vσ(θ)+ u + v θ +e, the Z-module O generated by (1, θ, σ(θ)) contains Z[θ]. We also see Z[θ] = 1, θ, vσ(θ). Thus [O : Z[θ]] = v. That means 81e v = disc(o)[o : Z[θ]] and hence disc(o) is equal to 81e. We know that Z[θ] is -maximal and p-maximal for every prime dividing e. Hence, [O K : Z[θ]] is not divisible by or p. Therefore, and p do not divide [O K : O], so O is -maximal and p-maximal for every prime p dividing e. Thus, [O K : O] = 1 and disc(k) = 81e and it follows that O is the maximal order and (1, θ, σ(θ)) is an integral basis of K. () We now consider the case where v. The field K can then be defined by the polynomial Q(X) = P (X 1)/7 = X X + 1 e 1 e + eu. Since e 1 (mod ), 7 u (mod ) and v, we know 1 e 1 e + eu Z and Z. To see the second one, 7 we let u = n+ and v = m for some n, m Z. Then, e = u + v = (n + ) + (m) 4 4 1

for some n, m Z. Also, 7 (1 e + eu) 7 (4 1e + 4eu) and 4 1e + 4eu =4 [(n + ) + (m) ] + [(n + ) + (m) ](n + ) =4 (9n + 1n + 4 + 7m ) + [9n + 1n + 4 + 7m ](n + ) =4 7n 6n 1 81m + 7n + 6n + 1n + 81m n + 18n + 4n + 8 + 54m =7n 7m + 7n + 81m n. 1 e + eu Thus, 7 (4 1e + 4eu) and 7 (1 e + eu) and Z. So Q(X) Z[X]. 7 ( θi + 1 Furthermore, the discriminant of Q(X) is θ ) j + 1 ( θi = θ ) j = 1 i<j 1 i<j 1 (θ 6 i θ j ) = 1 disc(p (X)) = e v 6. Set θ = θ + 1, which is a root of Q(X), and let 1 i<j ( ) θ + 1 e O be the Z-module generated by (1, θ, σ(θ )). So σ(θ u+v v ) = σ = θ + 1 v v θ + 1 = v e u + v v 6v θ + 1 v θ. Since θ = θ 1, σ(θ ) = v e v = v e v = v e = u + v 6v (θ 1) + 1 v (θ 1) u + v v θ + u + v + 1 6v v (9θ 6θ + 1) u + v v θ + u + v + 1 6v v (9θ 6θ + 1) v + u + v 4e 6v 4 + u + v θ + v v θ. Thus, θ + u + v 4e = + 4 + u + v θ + v 18 6 σ(θ ). Since 4e = u + v, u v (mod ) and thus u v (mod ). And since we let u = n + and v = m, we see that n m (mod ). So we have,

+ u + v 4e = + n + + (m) (n + ) (m) = 4 + n + 9m 9n 1n 4 7m = 9m 9n 9n 7m = 9(m n) 9(n m ) 18m. Hence, 18 ( + u + v 4e). Also, 4 + u + v = 4 + n + + m = 6 + (n + m), so 6 (4 + u + v). Thus, O Z[θ ] and since Z[θ ] = 1, θ, v σ(θ ), [O : Z[θ ]] = v. Therefore, the discriminant of O is equal to e. By Corollary.1, disc(k) must be divisible by e and so disc(k) = e and (1, θ, σ(θ )) is an integral basis of K. Now we will prove Theorem.7. Proof. First, we note that the polynomials given in Theorem.7 are irreducible in Q[X]. We use Eisenstein s Criterion for Z[X] for the first polynomial and the second one is obtained from the first one by changing X to X 1 and dividing by 7. Thus, the irreducibility follows. (1) From Theorem.15, one sees immediately that is ramified in K if and only if v. Hence Proposition.4 tells us that K is given by an equation P (X) = X ex eu. If we set u = u, v = v and e = 9e, we have e = u + 7v, u 6 (mod 9), v, and 4 P (X) = X e X e u 7. () Assume now that is not ramified, i.e. that v. From the proof of the second part of Theorem.15, we know that K can be defined by the polynomial X X + 1 e X 1 e + eu Z[X] and this time we set e = e, v = v 7 and u = u, it is clear that the second statement of Theorem.7 follows.

Now we prove that any two fields defined by different polynomials P (X) given in (1) or any two fields defined by different polynomials P (X) given in () are not isomorphic, i.e. the pair (e, u) determines the isomorphism class. This follows immediately from the uniqueness statement of Proposition.4. (Note that the e and u in Proposition.4 are either equal to the e and u of the theorem (in case()), or to e/9 and u/ (in case (1)).) Let us prove (). Assume that e is equal to a product of t distinct primes congruent to 1 modulo. Let A = Z[(1 + )/] be the ring of algebraic integers of Q( ). If α A with N (α), there exists a unique α associate to α (i.e. generating the same principal ideal) such that α = (u + v )/, u (mod ). To see this, we look at the following: Let α = a + b where a b (mod ). Suppose N (α) = a + b. Then, N (α) (a + b ) a a. 4 Let ζ = 1 +, ζ = 1 + α = a + b αζ = α = a b αζ = a b + (a + b) 4, ζ = 1, ζ 4 = 1 a+b + a b = αζ 4 = αζ = αζ = a b + (a b) 4 αζ 5 = αζ = a+b + a+b a b + a+b = a b + a b, ζ 5 = 1. Then, Consider the following cases: (i) If b and a (mod ), then take α = α. (ii) If b and a 1 (mod ), then take α = α. 4

(iii) If b 1 (mod ) and a 1 (mod ), then let b = t + 1 and a = k + 1. We know that t k (mod ). Then a + b ( ) ( ) t + k + t + k a + b = = + 1, so. ( ) a b a b = t k, so. We would take α = αζ or αζ 5. Now look at ( ) a b k 1 9t k + t = = t 1 (mod ). Thus we take α = αζ 5. (iv) If b 1 (mod ) and a (mod ), then let b = t + 1 and a = k +. We know that t + 1 k (mod ). Then a + b ( ) ( ) t + k + t + k + 1 a + b = =, so. ( ) a b a b = t k + 1, so. We would take α = αζ or αζ 4. Now look at a + b k + 9t + 9t + k + 1 (t k + 1) = = = (mod ). Thus, we will take α = αζ 4. (v) If b (mod ) and a 1 (mod ), we will take α = αζ. (vi) If b (mod ) and a (mod ), we will take α = αζ. Again, by Proposition 1.1, if p i is a prime congruent to 1 modulo, then p i splits in A and p i = α i α i for a unique α i = (u i + v i )/ with ui (mod ) and v i > 0 because N (α i ) = p i. Hence, if e = = 1 i t p i, then e = ( ) ( ) u + v u v 1 i t p i = 1 i t α i α i = u + 7v 4 if and only if u + v ( ) u + v = N = β i where β i = α i or β i = α i and this gives t solutions to the equation e = u + 7v. (We choose this specific 4 associate in order to get u (mod ) and v.) But, we have seen above that the isomorphism class of a cyclic cubic field is determined uniquely by the pair (e, u) satisfying appropriate conditions. Since e = u + 7( v) gives 4 1 i t 5

the same field as e = u + 7v, this shows, as claimed, that there exactly t 1 distinct value 4 of u, hence t 1 non-isomorphic fields of discriminant e. This finishes the proof of one case. Assume e is equal to 9 times the product of t 1 distinct primes congruent to 1 modulo. If α A with N (α), there exist unique α associates to α (i.e. generating the same principal ideal) such that α = (a + b )/, a (mod ) and b. Hence, if e = 9 p i, then e = 9 p i = 9 ( ) α iα i = u + 7v u + v = N 4 1 i t 1 1 i t 1 1 i t 1 ( ) ( ) u + v u v = if and only if u + v = β i where β i = α i 1 i t 1 or β i = α i. Now, we want to count how many unique solutions we can get from this equation. First, we note that any p i = α i α i, where we can set α i to be the unique form in the previous part and α i ζ and α i ζ 4 to be the unique associates that we want. Then u + v always has the form α 1 β i ζ k where β i = α i or β i = α i and k = 0, i t 1 k = or k = 4 (The case with α 1 β i ζ k just give the same u and v). All possible i t 1 solutions will be generated and we just need to count the wanted unique solutions. (The case where k = 0 does not give any solutions.) We see that the number of unique solutions t ( ) t is = (1 + 1) t = t 1. We have seen above that the isomorphism class of a i i=0 cyclic cubic field is determined uniquely by the pair (e, u) satisfying appropriate conditions. Hence there are t 1 non-isomorphic fields of discriminant e. This finishes the proof of the second case. Example.16. We can continue Example.5. We have the cyclic cubic field K = Q(θ) where θ is a root of x 7x 78 with e = 7 1, u = 8 and v = 10 under the notation of Proposition.4. By Theorem.15, disc(k) = (9e). Using notation of Theorem.7(1), we get new values of e and u, namely e = 9 7 1 and u = 8 (with v = 10). We nevertheless obtain the same defining polynomial P (x) = x e x eu 7 = x 7x 78. In addition, we are now able to find all cyclic cubic fields ramified at, 7 and 1. According to Theorem 6

.7(), there exist up to( isomorphism exactly 4 cyclic cubic fields of discriminant e. To find 1 + i ) ( 1 i ) ( 5 i ) ( 5 + i ) them, we see that 7 = and 1 =. Then, u + vi = or = or = or = ( ( ( ( 1 + i ) ( ) ( 1 + i 1 + i 1 + i ) ( ) ( 5 i ) ( ) ( 5 i 5 + i 5 + i 1 + i ) = 57 + i, 1 + i ) 4 = 4 0i, ) ( 1 + i ) = i, ) ( 1 + i ) 4 = 51 + 15i. Thus, the 4 cyclic cubic fields are determined uniquely by the pairs (e, 57), (e, 4), (e, ) and (e, 51), with e = 7 1. Example.17. We can continue Example.6. We have the cyclic cubic field K = Q(θ) where θ is a root of x 8857x + 10090717 with e = 1 1 7, u = 4 and v = under the notation of Proposition.4. By Theorem.15, disc(k) = e. Using notation of Theorem.7(), we have the same e and u, but with a new v, namely v = 1. Thus, we have the defining polynomial P (x) = x x + 1 e 1 e + eu x = x x 9806x+76999. Also, as in the 7 previous example, we can find all cyclic cubic fields of discriminant e. According to Theorem.7(), there exist up to( isomorphism exactly 4 cyclic cubic fields of discriminant e. To 5 i ) ( 5 + i ) ( 4 + 6i ) ( 4 6i ) find them, we see that 1 =, 1 = ( 7 + 9i ) ( 7 9i ) and 7 =. Then, 7

u + vi = or = or = or = ( ( ( ( 5 i ) ( ) ( 5 i 5 i 5 i ) ( ) ( 4 + 6i ) ( ) ( 4 6i 4 + 6i 4 6i ) ( ) ( 7 + 9i ) ) 7 + 9i 7 9i 7 9i ) ) = 4 + i, = 51 15i, = 4 150i, = 8 + 198i. Thus, the 4 cyclic cubic fields are determined uniquely by the pairs (e, 4), (e, 51), (e, 4) and (e, 8), with e = 1 1 7. Example.18. In this example, we want to find all cyclic cubic fields that ramify at 7, 1, and 19. According to Theorem.7(), there exist up to isomorphism ( exactly 4 cyclic cubic fields of discriminant e 1 + i ) ( 1 i ) where e = 7 1 19. We have 7 =, ( 5 i ) ( 5 + i ) ( 7 i ) ( 7 + i ) 1 = and 19 =. Then, u + vi = or = or = or = ( ( ( ( 1 + i ) ( ) ( 1 + i 1 + i 1 + i ) ( ) ( 5 i ) ( ) ( 5 i 5 + i 5 + i ) ( ) ( 7 i ) ) 7 + i 7 i 7 + i ) ) = 48i, = 79 15i, = 8 + i, 9 45i. Thus, the 4 cyclic cubic fields are determined uniquely by the pairs (e, ), (e, 79), (e, 8) and (e, 9), with e = 7 1 19. 8

Chapter. Discriminant Bounds Discriminant bounds have been a popular topic in number theory. Minkowski gave the first proof that d K > 1 for any number field K using the geometry of numbers, based on the ( π ) ( ) r n n lower bound d K (see [6]). Discriminant bounds have been improved over 4 n! the years. Odlyzko [10] utilized the zeros of the Dedekind zeta function to get better lower bounds. Also, if Generalized Riemann Hypothesis (GRH) is assumed, much better bounds can be obtained. Serre suggested using explicit formulae to achieve greater flexibility in the choice of parameters. The unconditional bound that we use from C.J. Moreno [8] is also derived by using Weil s explicit formulas and the bound is given by: rd K (60.8) r 1 n (.) r n e 8.6 n /. (.1) If K is a totally real field, the inequality becomes: rd K (60.8)e 8.6 n / (.) Now we look at the following proposition from Yamamura [11] and see how we use the lower bound to determine that a field has no non-solvable unramified extension. Proposition.1. Let B(n K, r 1, r ) be the lower bound for the root discriminant of K of degree n K with signature (r 1, r ). Suppose that K has an unramified normal extension L of degree m. If h(l) = 1 and rd K < B(60mn K, 60mr 1, 60mr ), then K ur = L. Proof. Suppose that L/K is normal, that h(l) = 1 and that rd K < B(60mn k, 60mr 1, 60mr ). Given any normal unramified extension F of K, we see that LF is a normal unramified extension of L. Set h = [LF : L] and set G = Gal(LF/L). Clearly, G must be non-solvable, since otherwise it would have an Abelian quotient (i.e. G/G 1 is Abelian where G 1 is a normal 9

subgroup of G) which would yield an Abelian unramified extension of L. Such an extension can not exist, because L has class number 1. h 60 LF 5 5555555 L F m 4 4444444 K n K Q Since G is non-solvable, we have h 60 where 60 is the order of A 5, the smallest non-solvable group. In addition, since LF/K is unramified, we have rd LF = rd K < B(60mn k, 60mr 1, 60mr ) B(hmn k, hmr 1, hmr ). This is a contradiction, since the degree of LF is hmn k, and it has hmr 1 real places and hmr pairs of complex places (no real places turns to complex places as LF is unramified over L). Thus, all unramified extensions of K are contained in L and K ur = L. Let K be a cyclic cubic field and L is the top of the class field tower of K. We will make a table of unconditional lower bounds for totally real fields F with degree n over Q using inequality (.). [L : K] n lower bounds for rd F 1 180 46.4 540 5.40 4 70 54.6 7 160 56.47 8 1440 56.8 Table.1 0

The conditional (under GRH) lower bounds for totally real fields F are found in the unpublished tables due to A.M. Odlyzko, which are copied in Martinet s expository paper [7]. We only list part of the tables here. n lower bounds for rd F 180 7.55 60 87.64 480 9.555 70 101.488 1000 107.548 100 110.78 Table. 1

Chapter 4. Maximal Unramified Extensions of Cyclic Cubic Fields With the information above, we proceed to look at cyclic cubic fields and their maximal unramified extensions. 4.1 Maximal Unramified Solvable Extensions Since we know the canonical representing polynomial of cyclic cubic fields, we can generate a full list of cyclic cubic fields of root discriminant less than a certain number. We first look at the cyclic cubic fields of root discriminant less than 7.55 (the lower bound for totally real fields of degree 180 under GRH in Table.), ramified at only one prime and having h(k) = 1. The first column gives the canonical representing polynomials. P (x) disc(k) rd K h(k) x x x + 1 7.66 1 x x 1 4 4. 1 x x 4x 1 1 5.5 1 x x 6x + 7 19 7.1 1 x x 10x + 8 1 9.87 1 x x 1x 11 7 11.1 1 x x 14x 8 4 1.7 1 x x 0x + 9 61 15.5 1 x x x 5 67 16.5 1 x x 4x + 7 7 17.47 1 x x 6x 41 79 18.41 1 x x x + 79 97 1.11 1 x x 4x + 61 10 1.97 1 x x 6x + 4 109.8 1 P (x) disc(k) rd K h(k) x x 4x 80 17 5.7 1 x x 46x 10 19 6.8 1 x x 50x + 1 151 8.6 1 x x 5x 64 157 9.1 1 x x 60x + 67 181 1 x x 64x 14 19.4 1 x x 66x 59 199 4.09 1 x x 70x + 15 11 5.44 1 x x 74x + 56 6.77 1 x x 76x + 1 9 7.4 1 x x 80x 15 41 8.7 1 x x 90x 61 71 41.88 1 x x 94x 04 8 4.1 1 x x 10x + 16 07 45.51 1

(Continued) P (x) disc(k) rd K h(k) x x 110x + 49 1 47.85 1 x x 11x 5 7 48.4 1 x x 1x 45 67 51.6 1 x x 14x + 1 7 51.8 1 x x 16x 65 79 5.7 1 x x 16x + 515 409 55.1 1 x x 140x + 4 41 56.17 1 x x 144x + 16 4 57. 1 x x 146x + 504 49 57.76 1 P (x) disc(k) rd K h(k) x x 154x 4 46 59.85 1 x x 16x + 505 487 61.9 1 x x 166x 56 499 6.91 1 x x 174x + 891 5 64.91 1 x x 180x 51 541 66.9 1 x x 190x + 719 571 68.8 1 x x 19x 171 577 69.1 1 x x 00x 51 601 71. 1 x x 04x 999 61 7.16 1 x x 15x + 0 457 59. 1 Table 4.1 The following table shows cyclic cubic fields of root discriminant less than 7.55, ramified at only one prime but having h(k) 1. We will explain the first three lines in this table in Example 4., 4., 4.4. The others are proven similarly. P (x) disc(k) rd K h(k) h(k 1 ) h(k ) x x 54x + 169 16 9.84 4 1 1 x x 9x 6 77 4.49 4 1 x x 104x 71 1 46.1 7 1 1 x x 116x + 517 49 49.57 4 1 1 x x 1x + 544 97 54.0 4 1 1 x x 18x + 81 547 66.88 4 1 1 x x 0x + 1169 607 71.69 4 1 Table 4.

The following table shows cyclic cubic fields of root discriminant less than 7.55, ramified at two primes. The first pair of cyclic cubic fields in the first two lines of this table will be explained in Example 4.5. The others are proven similarly. P (x) disc(k) rd K h(k) h(k 1 ) x 1x 5 4 7 15.8 1 x 1x 8 4 7 15.8 1 x x 0x 7 7 1 0. 1 x x 0x + 64 7 1 0. 1 x 9x 6 4 1.9 1 x 9x 91 4 1.9 1 x x 44x + 64 7 19 6.06 1 x x 44x 69 7 19 6.06 1 x 57x 15 4 19 0.81 1 x 57x 19 4 19 0.81 1 x x 7x 09 7 1 6.11 1 x x 7x + 5 7 1 6.11 1 x x 8x + 64 1 19 9.7 1 x x 8x + 11 1 19 9.7 1 x x 86x 48 7 7 40.6 1 x x 86x + 11 7 7 40.6 1 x 9x 41 4 1 4.7 1 x 9x 17 4 1 4.7 1 x x 100x + 79 7 4 44.91 1 x x 100x 7 4 44.91 1 x 111x 70 4 7 48.04 1 x 111x 7 4 7 48.04 1 4

(Continued) P (x) disc(k) rd K h(k) h(k 1 ) x 19x 15 4 4 5.11 1 x 19x 559 4 4 5.11 1 x x 14x 09 1 1 54.56 1 x x 14x + 597 1 1 54.56 1 x x 14x 601 7 61 56.7 1 x x 14x + 680 7 61 56.7 1 x x 156x + 799 7 67 60.6 1 x x 156x 608 7 67 60.6 1 x x 160x 677 1 7 61.9 1 x x 160x 196 1 7 61.9 1 x x 170x 776 7 7 6.9 1 x x 170x + 757 7 7 6.9 1 x 18x 854 4 61 67.05 1 x 18x 79 4 61 67.05 1 x x 184x 41 7 79 67.7 1 x x 184x + 51 7 79 67.7 1 x x 186x 911 1 4 67.86 1 x x 186 + 07 1 4 67.86 1 x x 196x 89 19 1 70.7 1 x x 196x + 49 19 1 70.7 1 x 01x 77 4 67 71.7 1 x 01x 107 4 67 71.7 1 Table 4. 5

According to Table 4.1, the first 8 cyclic cubic fields with class number 1 have root discriminant less than 46.4 which is the unconditional lower bound for totally real fields of degree 180 (from Table.1), and the first 47 cyclic cubic fields with class number 1 have root discriminant less than 7.55 which is the conditional lower bound for totally real fields of degree 180 (from Table.). Therefore, they do not have non-solvable unramified extension. Applying Proposition.1, we can conclude as follows: Theorem 4.1. The first 8 (47 under GRH) cyclic cubic fields with class number 1 have trivial maximal unramified extension. Now we examine each of the cyclic cubic fields with h(k) 1 in the tables above and demonstrate how we determine the maximal unramified extensions. Example 4.. Let K be the cyclic cubic field defined by x x 54x+169 with d K = 16 and rd K = 9.84. It is the first cyclic cubic field with h(k) = 4 (from Table 4.). The class field K 1 of K is the splitting field L (A 4 -extension over Q) of the sextic field L represented by x 6 x 5 11x 4 + 7x x 11x + 1 with field discriminant 16 4, found from John Jones website Number Fields. To confirm L is K 1, we see that L has degree 1 and the prime 16 is unramified in L /K. Also, we know that Gal(L /K) is Abelian (Z Z, normal subgroup of A 4 ). We then use PARI to compute the class number of K 1. Since h(k 1 ) = 1 and rd K = 9.84 < 54.6 which is the lower bound for totally real fields of degree 70, we have K ur = K 1 by Proposition.1. In this case, we have Gal(K ur /K) = Z Z and Gal(K ur /Q) = A 4. K 1 4 L 6 6666666 K Q 6

Example 4.. Let K be the cyclic cubic field defined by x x 9x 6 with d K = 77 and rd K = 4.49. It is the second cyclic cubic field with h(k) = 4 (from Table 4.). The class field K 1 of K is the splitting field of the sextic field L represented by x 6 x 5 19x 4 + 4x + 47x 69x + 16 with field discriminant 77 4. This sextic field is obtained from John Jones website Number Fields. We use PARI to find that h(k 1 ) =. And K turns out to be the splitting field of the octic field F represented by x 8 x 7 11x 6 + 1x 5 + x 4 41x x + x 1 with field discriminant 77 4. We use PARI again to find that h(k ) = 1. Since rd K = 4.49 < 56.8 which is the lower bound for totally real fields of degree 1440, we have K ur = K by Proposition.1. In this case, we have Gal(K ur /K) = Q 8 and Gal(K ur /Q) = SL(, ). K K 1 F 4 L F 4 K Q 7