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Title | Dihedral quintic fields with a power basis |
Creator |
Lavallee, Melisa Jean |
Publisher | University of British Columbia |
Date Created | 2008-11-14 |
Date Issued | 2008-11-14 |
Date | 2008 |
Description | Cryptography is defined to be the practice and studying of hiding information and is used in applications present today; examples include the security of ATM cards and computer passwords ([34]). In order to transform information to make it unreadable, one needs a series of algorithms. Many of these algorithms are based on elliptic curves because they require fewer bits. To use such algorithms, one must find the rational points on an elliptic curve. The study of Algebraic Number Theory, and in particular, rare objects known as power bases, help determine what these rational points are. With such broad applications, studying power bases is an interesting topic with many research opportunities, one of which is given below. There are many similarities between Cyclic and Dihedral fields of prime degree; more specifically, the structure of their field discriminants is comparable. Since the existence of power bases (i.e. monogenicity) is based upon finding solutions to the index form equation - an equation dependant on field discriminants - does this imply monogenic properties of such fields are also analogous? For instance, in [14], Marie-Nicole Gras has shown there is only one monogenic cyclic field of degree 5. Is there a similar result for dihedral fields of degree 5? The purpose of this thesis is to show that there exist infinitely many monogenic dihedral quintic fields and hence, not just one or finitely many. We do so by using a well- known family of quintic polynomials with Galois group D₅. Thus, the main theorem given in this thesis will confirm that monogenic properties between cyclic and dihedral quintic fields are not always correlative. |
Extent | 530363 bytes |
Subject |
Dihedral Quintic Fields Power Bases Algorithms Algebraic Number Theory |
Genre |
Thesis/Dissertation |
Type |
Text |
File Format | application/pdf |
Language | Eng |
Collection |
Electronic Theses and Dissertations (ETDs) 2008+ |
Date Available | 2008-11-14 |
DOI | 10.14288/1.0066793 |
Degree |
Master of Science - MSc |
Program |
Interdisciplinary Studies |
Affiliation |
Science, Faculty of |
Degree Grantor | University of British Columbia |
Graduation Date | 2008-11 |
Campus |
UBCO |
Scholarly Level | Graduate |
URI | http://hdl.handle.net/2429/2788 |
Aggregated Source Repository | DSpace |
Digital Resource Original Record | https://open.library.ubc.ca/collections/24/items/1.0066793/source |
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DIHEDRAL QUINTIC FIELDS WITH A POWER BASIS by Melisa Jean Lavallee B.Sc., The University of British Columbia, 2006 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in The College of Graduate Studies (Interdisciplinary Studies) THE UNIVERSITY OF BRITISH COLUMBIA (Okanagan) August 2008 c Melisa Jean Lavallee, 2008 ABSTRACT Cryptography is de ned to be the practice and studying of hiding information and is used in applications present today; examples include the security of ATM cards and computer passwords ([34]). In order to transform information to make it unreadable, one needs a series of algorithms. Many of these algorithms are based on elliptic curves because they require fewer bits. To use such algorithms, one must nd the rational points on an elliptic curve. The study of Algebraic Number Theory, and in particular, rare objects known as power bases, help determine what these rational points are. With such broad applications, studying power bases is an interesting topic with many research opportunities, one of which is given below. There are many similarities between Cyclic and Dihedral elds of prime degree; more speci cally, the structure of their eld discriminants is comparable. Since the existence of power bases (i.e. monogenicity) is based upon nding solutions to the index form equation - an equation dependant on eld discriminants - does this imply monogenic properties of such elds are also analogous? For instance, in [14], Marie-Nicole Gras has shown there is only one monogenic cyclic eld of degree 5. Is there a similar result for dihedral elds of degree 5? The purpose of this thesis is to show that there exist in nitely many monogenic dihedral quintic elds and hence, not just one or nitely many. We do so by using a well- known family of quintic polynomials with Galois group D5. Thus, the main theorem given in this thesis will con rm that monogenic properties between cyclic and dihedral quintic elds are not always correlative. ii Contents ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi Coauthorship Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xii Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1. Algebraic Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1. Classic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2. Algebraic Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2.1. Minimal Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2.2. Conjugates of an Algebraic Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.3. Algebraic Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3.1. Representation of Algebraic Number Fields . . . . . . . . . . . . . . . . . . . . . . . . 10 Chapter 2. Introduction to Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.1. Galois Groups of Polynomials - Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.2. Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.3. Quadratic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.4. Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 iii 2.4.1. Irreducible Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.4.2. Reducible Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.5. Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.5.1. Irreducible Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.5.2. Reducible Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.6. Quintic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.6.1. Irreducible Quintic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.6.2. Reducible Quintic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 2.7. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.8. Finding the Galois Group of an Algebraic Number Field . . . . . . . . . . . . . . 47 Chapter 3. Integral and Power Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.1. Integral Bases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.1.1. Minimal Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.1.2. Examples of Finding Integral Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.2. Field Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.3. Power Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.3.1. Index Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Chapter 4. Cyclic and Dihedral Fields of Prime Degree . . . . . . . . . . . . . . . . . . . . . . . 69 4.1. Field Discriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 4.2. Abelian and Cyclotomic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Chapter 5. Cyclic and Dihedral Quintic Fields with a Power Basis . . . . . . . . . . . . 76 5.1. Some Important Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 5.2. Proof of Main Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 5.3. Further Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Chapter 6. Number Fields without a Power Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 6.1. Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 6.2. Common Index Divisors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Chapter 7. Conclusion and Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 7.1. Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 iv 7.2. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 v List of Tables 2.1 Group Table for the Cyclic Group of Order 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.2 Group Table for the Cyclic Group of Order 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.3 Group Table for the Symmetric Group of Order 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.4 Summary of the Possible Galois Groups for Irreducible Polynomials . . . . . . . . . 47 vi List of Figures 2.1 Subgroups of S3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 vii List of Symbols Z Set of integers, 3 Q Set of rationals, 3 D(f) Discriminant of a polynomial f(x), 8 N() Norm of , 8 Tr() Trace of , 8 [K : Q] Degree of K over Q, 10 Aut(K) Collection of automorphisms of K, 11 AutQ(K) Collection of automorphisms of K which x Q, 11 Gal(K=Q) Galois group of K over Q, 11 C Set of complex numbers, 12 Gal(f(x)) Galois group of the splitting eld of f(x) 2 F [x] over F , 12 Sn Symmetric group of order n, 13 E L Sub elds, 14 jG : Hj Index of H in G, 45 Q( = 1; 2; : : : ; n) Normal closure of K = Q(), 47 OK Ring of integers in the number eld K, 50 d(K) Field discriminant of the eld K, 61 ind() Index of , 63 ind() = 1 Index form equation, 67 viii i(K) Field index of K, 67 m(K) Minimal Index of K, 67 &n Primitive nth root of unity, 74 Q(&n) Cyclotomic Field, 75 Ra;b(x) Kondo family of quintics, 76 ix Acknowledgements God has blessed me with an amazing network of family, friends and professors who have been consistently supportive and encouraging throughout my post-secondary experience; my thanks go to these wonderful people. Thanks to my committee of brilliant professors: Dr. Blair Spearman, Dr. Qiduan Yang, Dr. Shawn Wang, Dr. Heinz Bauschke and Dr. Yong Gao. Your guidance and constructive criticism proved invaluable and contributed greatly to the completion of this thesis. Furthermore, I wish to acknowledge and express my appreciation to Dr. Blair Spearman. Your ability to see the potential in others and the con dence you instill in them has contributed immensely to the furtherance of my education and as a result this thesis. The research for this paper was supported by a scholarship from the Natural Sciences and Engineering Research Council of Canada. x Dedication I would like to dedicate this paper to my mother. Your strength and perseverance never cease to amaze me. xi Coauthorship Statement The main result of this thesis is from the paper [18]. All authors of [18] con- tributed equally to this paper. Using MAPLETM, I was responsible for nding solutions to the index form equation of our problem. I also contributed in editing this paper. xii Introduction Although there are in nitely many integral bases of the ring of integers of an algebraic number eld, seldom are these power bases. When a power basis is found, its corresponding algebraic number eld is said to be monogenic. Even now, identifying monogenic elds is still an active area of research. Firstly, we know that all quadratic elds are monogenic. Also, much research has been conducted on cubic and quartic elds. Dedekind gave an example of a non- monogenic cubic eld (see [7]). However, there are in nitely many cyclic cubic elds as well as non-cyclic cubic elds that are monogenic (see [10] and [28], respectively). In regards to quartic elds, we now know that there are in nitely many pure quartic elds ([12]), bicyclic quartic elds ([22]) and dihedral quartic elds ([15]) that are monogenic. However, the cyclic quartic case is still open. On the other hand, higher degrees can make these problems quite challenging. Numerical methods are limited to lower degrees, usually at most 6. Thanks to Marie-Nicole Gras ([14]), we do know that there is only one monogenic cyclic quintic eld. The essence of this thesis is to show by way of comparison that there are an in nite number of monogenic dihedral quintic elds. However, a large portion is dedicated to establishing the foundations of the theory as well as important results needed to prove this focal point. The contents of this thesis are divided into seven chapters. Chapter 1 serves as an introduction to basic properties of Algebraic Number Theory. Chapter 2 gives some basic theory of Galois Theory. It also gives methods on nding the Galois group of a polynomial up to and including degree ve. Chapter 3 is devoted to the study of integral bases and in particular, power bases; it also discusses the Index Form and Index Form Equation and related properties. Chapter 4 compares eld discriminants of cyclic and dihedral elds of prime degree. In Chapter 5, the tools developed in earlier chapters are used to prove our main result (that there are an in nite amount 1 of monogenic dihedral quintic elds). It also discusses related results to this proof. Chapter 6 is concerned with ideals and common index divisors. Speci cally, it studies properties of elds that do not contain a power basis. Lastly, Chapter 7 gives numerous examples of future research in this particular area of study. 2 Chapter 1 Algebraic Preliminaries 1.1. Classic Theorems We begin this chapter with two classic theorems that will be quoted throughout this thesis. We begin with what is known as the Rational Root Theorem, Rational Zero Theorem or Rational Root Test (pg. 214-215, [20]). Theorem 1.1. Consider the polynomial equation anx n + an1xn1 + : : :+ a1x+ a0 = 0 with integer coe¢ cients. Let a0 and an be nonzero. Then each rational solution x of this equation can be written in the form x = p q where p and q satisfy the two properties: p is an integer factor of the constant term a0, and q is an integer factor of the leading coe¢ cient an: We will be using this theorem to determine if a given polynomial has any rational roots (and hence any linear factors). Now we state Gausss Lemma (Proposition 5, pg. 303, [9]). Theorem 1.2. If a polynomial with integer coe¢ cients is irreducible over Z (the set of integers) then it is also irreducible over Q (the set of rationals). We will be looking at problems where we need to determine if a given polynomial is irreducible over Q. We can then use Gausss Lemma to show a polynomial is irreducible over Q by showing it is irreducible over Z. 3 1.2. Algebraic Numbers De nition 1.1. is an algebraic number if it is a root of a non-constant monic polynomial f(x) with rational coe¢ cients (i.e. f(x) 2 Q [x]). De nition 1.2. If the coe¢ cients in the above polynomial are integers (i.e. f(x) 2 Z [x]), then is called an algebraic integer. The term "monic" means the coe¢ cient of the highest power of x is equal to 1. Example 1.1. Let f(x) = x2 1 2 . It has roots x = 1p 2 . Since f is monic and has coe¢ cients in Q, 1p 2 are both algebraic numbers. Example 1.2. Let f(x) = x2 2. It has roots x = p2. Since f is monic and has coe¢ cients in Z, p2 are both algebraic integers. 1.2.1. Minimal Polynomial De nition 1.3. The minimal polynomial of an algebraic number (or integer) is the monic polynomial A(x) 2 Q [x] (resp. Z [x]) of least degree which has as a root. We now introduce a simple property of the minimal polynomial. Lemma 1.1. If is an algebraic number then the minimal polynomial of divides every polynomial of which is a root. Proof. Let A(x) be the minimal polynomial of . Suppose p(x) 2 Q[x] such that p() = 0. By the division algorithm, there exists polynomials q(x) and r(x) in Q[x] such that p(x) = A(x) q(x) + r(x) where deg(r(x)) < deg(A(x)). Then 0 = p() = A() q() + r() = 0 + r()) r() = 0: If r(x) 6= 0 then we can nd a suitable constant multiplier such that r(x) is monic. Then deg(r(x)) < deg(A(x)) would contradict the minimality of A(x) (as r() = 0). Thus, r(x) = 0 and so p(x) = A(x) q(x)) A(x) j p(x). 4 We now give two examples on nding the minimal polynomial. Example 1.3. Let = p 2 + p 3. We want to nd the minimal polynomial of . Squaring we obtain 2 = 5 + 2 p 6 so that 2 5 = 2 p 6: Squaring 2 5 yields 4 102 + 25 = 24: Thus, is a root of the monic quartic polynomial f(x) = x4 10x2 + 1 2 Z[x]: This shows that is an algebraic integer. We will show that this polynomial f(x) is A(x) - the minimal polynomial of . First we show that this polynomial is irreducible over Z. Assume by way of contradiction that f(x) = x4 10x2 + 1 is reducible over Z. First of all, if f(x) were to have a linear factor, then there is a rational root, say x1 = p q where p; q 2 Z, such that f(x1) = 0. Then by the Rational Root Theorem, p j 1 and q j 1. So our choices for x1 are 1. Since f(1) = 8 6= 0, f(x) has no linear factors in Q[x] and hence in Z[x]. Thus, f(x) factors into two irreducible quadratic factors. So f(x) = x4 10x2 + 1 = (x2 + ax+ b) (x2 + cx+ d) where a; b; c; d 2 Z. Equating coe¢ cients yields the following equations: a+ c = 0 d+ b+ ac = 10 ad+ bc = 0 bd = 1. From the rst equation we have c = a and so the second equation becomes d+ba2 = 10. From the last equation we have (1) b = 1 and d = 1 (2) b = 1 and d = 1 5 So, b+d = 2. Thus, from the revised second equation, we have a2 = 12 or 8, which is a contradiction. Thus, f(x) is irreducible over Z and by GaussLemma, f(x) is irreducible over Q. By Lemma 1.1, the minimal polynomial of must divide f(x) and hence must be a factor of f(x) over Z. This implies that f(x) must be the minimal polynomial of = p 2 + p 3 since f(x) does not factor over Z. Thus, A(x) = x4 10x2 + 1. Example 1.4. Let = p 2 + p 2. We want to nd the minimal polynomial of . Squaring we obtain 2 = 2 + p 2 so that 2 2 = p 2: Squaring 2 2 yields 4 42 + 4 = 2: Thus, is a root of the monic quartic polynomial f(x) = x4 4x2 + 2 2 Z[x]: This shows that is an algebraic integer. By GaussLemma, we know that if f(x) is irreducible over Z then it is irreducible over Q. Thus, we will now show that f(x) is irreducible over Z. So we assume, by way of contradiction, that f(x) factors over Z. First of all, if f(x) were to have a linear factor, then there is a rational root, say x1 = p q where p; q 2 Z, such that f(x1) = 0. Then by the Rational Root Theorem, p j 2 and q j 1. So our choices for x1 are 2. Since f(2) = 2 6= 0, f(x) has no linear factors in Q[x] and hence in Z[x]. Since we assumed f(x) factors over Z, f(x) must factor as a product of two quadratic polynomials in Z[x]: x4 4x2 + 2 = (x2 + ax+ b) (x2 + cx+ d); 6 where a; b; c; d 2 Z. Equating coe¢ cients, we get a+ c = 0 d+ b+ ac = 4 ad+ bc = 0 bd = 2. From the rst equation we have c = a and so the second equation becomes d+ba2 = 4. From the last equation we have (1) b = 2 and d = 1 (2) b = 2 and d = 1 (3) b = 1 and d = 2 (4) b = 1 and d = 2. So, b + d = 3. Thus, from the revised second equation, we have a2 = 7 or 1. But a2 = 7 is impossible. If a2 = 1 then 0 = ad+ bc = ad ba = a(d b) = d b, which implies d = b, also impossible (see above possible values for b and d). Thus, f(x) is irreducible over Z and hence over Q (later on we shall see Eisensteins Crite- rion which can also be used to determine irreducibility). By the Lemma, the minimal polynomial of must divide f(x) and hence must be a factor of f(x) over Z. This implies that f(x) must be the minimal polynomial of = p 2 + p 2 since f(x) does not factor over Z. Note that in Example 1.4, is also a root of x5 x4 4x3 + 4x2 + 2x 2 but we are looking for the polynomial of least degree. Now the work done in the above example can be simpli ed immensely with the use of computer software programs like MAPLETM. In MAPLETM, we need only 7 use the MinimalPolynomial() command to determine the minimal polynomial of an algebraic element. The de nition of the minimal polynomial A(x) implies that A(x) must be irre- ducible. Otherwise, we would be able to factor A(x) and one of the factors would have as a root and hence contradict the minimality of A(x). While on the topic of polynomials, we will introduce a de nition which will be needed later on in this paper. De nition 1.4. The discriminant of a polynomial f(x) with roots 1 = ; 2; : : : ; n is de ned by D(f) = Y 1i<jn (i j)2: That is, the discriminant of a polynomial is equal to the product of the squares of the di¤erences of the polynomials roots. As a note, we sometimes use the symbol D to refer to the discriminant of f . With the above de nition, we are now able to de ne the discriminant of an alge- braic number. De nition 1.5. The discriminant of an algebraic number is equal to the dis- criminant of its minimal polynomial and is denoted as D(). 1.2.2. Conjugates of an Algebraic Number De nition 1.6. Let be an algebraic number and let A(x) be its minimal poly- nomial. Then the conjugates of are all the roots of A(x), including itself. Notation 1.1. We write 1 = ; 2; :::; n to denote the n conjugates of where n is equal to the degree of A(x). We say is an algebraic number of degree n. De nition 1.7. If A(x) is the minimal polynomial of an algebraic number of degree n then the norm of , denoted as N(), is the product of the conjugates of ; the trace of , denoted as Tr(), is the sum of the conjugates of . That is, N() = 1 2 n and Tr() = 1 + 2 + :::+ n. 8 Example 1.5. The minimal polynomial of = p 2 is x22. Thus, the conjugates of are itself and p2. So we have N() = p2 p2 = 2 and Tr() =p 2 + (p2) = 0. 1.3. Algebraic Number Fields De nition 1.8. An algebraic number eld is a sub eld of C of the formQ( 1; 2; : : : ; n) where 1; 2; : : : ; n are algebraic numbers. That is, an algebraic number eld is a eld K obtained from the eld of rational numbers Q by adjoining nitely many algebraic numbers 1; 2; : : : ; n. This eld is the smallest eld that contains Q and the algebraic numbers 1; 2; : : : ; n. Example 1.6. Q( p 2; p 3) is an algebraic number eld as both p 2 and p 3 are algebraic numbers. Now we will state a classic theorem known as the Primitive Element Theorem to simplify the representation of elds like Q( p 2; p 3). The proof can be found on page 595, [9]. Theorem 1.3. If K = Q( 1; 2; : : : ; n) is an algebraic number eld then there exists an algebraic number such that K = Q(). This means that a composite eld can be expressed in the form Q(), where is an algebraic number, and Q() is an algebraic number eld formed by constructing all possible numbers from and the set of rationals using the operators +;;;. Example 1.7. To demonstrate the Primitive Element Theorem, we will show that Q( p 2; p 3) = Q() where is the algebraic number p 2 + p 3. By Example, 1.3, is an algebraic number with minimal polynomial f(x) = x4 10x2 + 1 2 Q[x]. Now, Q() Q(p2;p3) as p2 +p3 2 Q(p2;p3) . Now, p 2 = 9 2 + 1 2 3 2 Q() and p3 = 11 2 1 2 3 2 Q() so that Q(p2;p3) Q(): Thus Q() = Q( p 2; p 3) and so the theorem holds. Remark 1.1. The expression K = Q() is not unique as Q() = Q() = Q(1 + ) = :::etc. 9 1.3.1. Representation of Algebraic Number Fields We can also view K = Q() as an n-dimensional vector space over Q as seen in the following result (Theorem 6.1.3, pg. 110, [2]). Theorem 1.4. Let K = Q() be an algebraic number eld, where is an algebraic number. Let A(x) be the minimal polynomial of with degree n. Then every element of K is uniquely expressible in the form a0 + a1 + : : : + an1 n1, where a0; : : : ; an1 2 Q. Thus, K = a0 + a1 + : : :+ an1 n1 j a0; : : : ; an1 2 Q . De nition 1.9. K = Q() is an n-dimensional vector space over Q with basis 1; ; : : : ; n1 and the degree of K over Q is n. We denote the degree of K over Q as n = [K : Q]. Algebraic number elds are classi ed according to their degree. So K is called a quadratic eld if n = 2, cubic if n = 3, quartic if n = 4, quintic if n = 5, etc. We can also term algebraic number elds as being pure or non-pure. De nition 1.10. K = Q() is pure if is a root of the polynomial xn a where a 2 Q and n is the degree of the minimal polynomial of . If K is not pure, then we say K is non-pure. Algebraic number elds can also be classi ed using Galois theory. We will study this in detail in the following chapter. 10 Chapter 2 Introduction to Galois Theory We begin this chapter with some basic de nitions that will be used to establish the foundation of Galois Theory. De nition 2.1. An isomorphism of a eld K with itself is called an automor- phism of K. The collection of automorphisms of K is denoted Aut(K). Notice that every eld will have at least one automorphism - the identity map. De nition 2.2. An automorphism 2 Aut(K) is said to x an element 2 K if () = . If F K then an automorphism is said to x F if it xes all the elements of F . That is, (a) = a, 8a 2 F . De nition 2.3. Let K be an algebraic number eld. Let AutQ(K) be the collec- tion of automorphisms of K which x Q. Since Q K, any automorphism of K xes Q. This is because xes 1, the identity, and Q can be built up from 1 using +;; and . Hence, Aut(K) = AutQ(K): IfK is not a eld extension of Q then AutQ(K) is a subgroup of the group Aut(K). De nition 2.4. An algebraic number eld K is separable if for each algebraic number 2 K, the minimal polynomial of over K has no repeated roots. De nition 2.5. An algebraic number eld K is normal if for each algebraic num- ber 2 K, the minimal polynomial of over K factors entirely in K[x]. That is, it completely factors into linear factors over K. De nition 2.6. An algebraic number eld is Galois if it is both separable and normal. The group AutQ(K) is called the Galois group of K and is denoted as Gal(K=Q). 11 2.1. Galois Groups of Polynomials - Overview De nition 2.7. Consider a polynomial f(x) 2 Q[x]. The splitting eld L of the polynomial f(x) over Q is the smallest sub eld of C (set of complex numbers) containing Q and all the roots of f(x). De nition 2.8. A polynomial f(x) over Q is called separable if it has no multiple roots (i.e. all its roots are distinct). We now state an important property of the splitting eld of a separable polynomial (Corollary 6, pg. 562, [9]). Corollary 2.1. If L is the splitting eld over Q of a separable polynomial f(x) then L is Galois. Because L is Galois, we are able to nd its Galois group. This leads us to the following de nition. De nition 2.9. If f(x) is a separable polynomial over Q, then the Galois group of f(x) over Q is the Galois group of the splitting eld of f(x) over Q and is denoted as Gal(f(x)). Now it is important to note that the above corollary implies the splitting eld of any polynomial over Q is Galois. Why? Clearly, the splitting eld of f(x) is the same as the splitting eld of the product of the irreducible factors of f(x) (i.e. the polynomial obtained by removing multiple factors). Now this latter polynomial is separable as it is the product of distinct irreducible polynomials (and so the factors never have zeros in common). Then using the above corollary on this polynomial, we get the result. From this, we can also conclude that every polynomial over Q has a Galois group. Now we consider Galois theory from a permutational perspective. Given a poly- nomial, we wish to nd its Galois group. First we introduce a short de nition. De nition 2.10. An algebraic equation is one of the form P = 0 where P is a polynomial with coe¢ cients in Q. 12 The main idea of Galois theory is to consider those permutations of the roots of our polynomial having the property that any algebraic equation satis ed by the roots is still satis ed after the roots have been permuted. These permutations form a group and is called the Galois group of the polynomial over Q. Thus, the elements of Gal(f(x)) are permutations that act on the roots of f(x). So, Gal(f(x)) is a subgroup of Sn - the group of all permutations of the set f1; 2; : : : ; ng. Now, we say subgroup because we may not have all permutations of the roots of f(x) as elements of Gal(f(x)). So jGal(f(x))j j jSnj: How does this correspond to the splitting eld of the polynomial f(x)? If f(x) is a separable polynomial over Q and if L denotes its splitting eld over Q, then we know that Gal(f(x)) = Gal(L=Q). But Gal(L=Q) consists of automorphisms from L to itself that x Q. Thus, any permutation of the roots which respects algebraic equations as described above gives rise to an automorphism of L over Q, and vice versa. That is, the automorphisms of L correspond to elements of Gal(f(x)): We will now do a small example to illustrate the above concept. This example parallels the "second example" in [33]. Example 2.1. Consider the polynomial x4 5x2 + 6 = (x2 3)(x2 2). We wish to describe the Galois group of this polynomial over the eld of rational numbers. The polynomial has four roots: (1) A = p 2 (2) B = p2 (3) C = p 3 (4) D = p3 There are 24 possible ways to permute these four roots, but not all of these permu- tations are members of the Galois group. The members of the Galois group must preserve any algebraic equation with rational coe¢ cients involving A, B, C and D. One such equation is: A+B = 0: 13 Therefore the permutation (A;B;C;D) ! (A;C;B;D) is not permitted because it transforms the valid equation A+B = 0 into the equation A+C = 0, p2+p3 = 0, which is invalid. Another equation that the roots satisfy is (A+B)2 + (C +D)2 = 0. This will exclude further permutations, such as (A;B;C;D)! (A;C;B;D): Continuing in this way, we nd that the only permutations satisfying both equations simultaneously are: (1) (A;B;C;D)! (A;B;C;D) (trivial permutation) (2) (A;B;C;D)! (B;A;C;D) (3) (A;B;C;D)! (A;B;D;C) (4) (A;B;C;D)! (B;A;D;C). Thus, the Gal(f(x)) = V = Z2 Z2 (the Klein 4-group). Now showing the above four permutations satisfy every possible algebraic equa- tion satis ed by A;B;C and D is less obvious and requires the theory of symmetric polynomials. Also, as we increase the number of roots of our given polynomial, this method becomes less e¢ cient. We can see that calculating the Galois group of a polynomial is a nontrivial task. However, with technology, the task no longer seems daunting. For example, MAPLETM will calculate the Galois group of a polynomial up to degree 9. We use the galois() command to do so. We now introduce an important property needed in order to examine the Galois group of polynomials, in particular, those of degree 3. For proof, see Theorem 14, pp. 523-524, [9]. Proposition 2.1. The multiplicative property of elds states that if E L F where E, L and F are elds and "" denotes "sub eld", then [F : E] = [F : L] [L : E]. Recall [F : E] denotes the degree of F over E. The next corollary is important when we examine the Galois group of polynomials. We refer the reader to pp. 51 of [3] for more details and the proof. 14 Corollary 2.2. Let K be a nite extension of the eld Q. If K is Galois then jAutQ(K)j = [K : Q]: As a result, we know that AutQ(L) = Gal(L=Q) and since L is Galois (Corollary 2.1), [L : Q] = jGal(L=Q)j . Lastly, we now present a proposition (Proposition 34, pp. 611, [9]) that will be encompassed in the theorems presented in these next sections. Proposition 2.2. The Galois group of f(x) 2 Q[x] is a subgroup of An if and only if the discriminant D(f) is a square in Q. 2.2. Irreducible Polynomials De nition 2.11. A polynomial p(x) in Q[x] is called irreducible over Q if it is non-constant and cannot be represented as the product of two or more non-constant polynomials from Q[x]. A useful test for irreducibility is Eisensteins Criterion. Proposition 2.3. Suppose we have the following polynomial with integer coe¢ - cients: f(x) = anx n + an1xn1 + : : :+ a1x+ a0. Assume that there exists a prime number p such that p divides each ai for i 6= n p does not divide an p2 does not divide a0. Then f(x) is irreducible over Q. Proof. Suppose f(x) = anxn + an1xn1 + : : : + a1x + a0 2 Z[x] satis es the above conditions for some prime p. By way of contradiction, assume f is reducible over Q: That is, f(x) = g(x) h(x) 15 where g(x) and h(x) are non-constant polynomials over Q. The reduction map mod p is given by ' : Z[x] ! Zp[x] and is a homomorphism. Thus, '(f) = '(g) '(h): But '(f) = '(anx n + an1xn1 + : : :+ a1x+ a0) = xn as p divides each ai for i 6= n and where an (mod p). So '(g) '(h) = xn and so '(g) = xk and '(h) = xnk for some 0 < k < n, where (mod p) and since Zp[x] is a unique factorization domain. This implies p divides the constant term c1 of g(x) and the constant term c2 of h(x). That is, ( c1 = pm c2 = pl where m; l 2 Z. Since f(x) = g(x) h(x), the constant term a0 of f(x) is a0 = c1 c2 = (pm)(pl) = p2(ml): Thus, p2 j a0 which is a contradiction. 16 Example 2.2. Consider g(x) = 3x4 + 15x2 + 10. Let p = 5. Clearly, 5 j ai for i = 0 : : : 3. Also, 5 - 3 and 52 - 10: Thus, g(x) is irreducible over Q by Eisensteins Criterion. We give another Theorem that may be helpful in trying to prove a polynomial is irreducible (see Proposition 12, pp. 309, [9]). Theorem 2.1. If f(x) is a nonconstant monic polynomial and f(x) factors over Z then f(x) factors over Zp for all primes p. Remark 2.1. The contrapositive of this theorem is very useful in showing a poly- nomial f(x) is irreducible. We need only nd one prime p for which f(x) does not factor mod p to show f(x) is irreducible. Example 2.3. Let f(x) = x5 5x+ 12: Then f(x) x5 + 2x+ 5 (mod 7) which does not factor over Z7. Hence, f(x) is irreducible. We now present one last theorem that may help in proving the irreducibility of a polynomial. Theorem 2.2. Let f(x) 2 Q[x]. f(x) is reducible if and only if f(x + a) is reducible for a 2 Z. Proof. For a 2 Z, f(x) is reducible , f(x) = g(x) h(x) , f(x+ a) = g(x+ a) h(x+ a) , f(x+ a) is reducible, where g(x) and h(x) are non-constant polynomials over Q. We illustrate this Theorem with the same polynomial used in Example 2.3. Example 2.4. Let f(x) = x5 5x + 12: As is, we cannot use Eisensteins Criterion on this polynomial. However, if we make the translation x = y 2 we 17 obtain f(y 2) = y5 10y4 + 40y3 80y2 + 75y 10: Now we can use Eisensteins Criterion with p = 5 to deduce f(y 2) is irreducible. Hence, by the contrapositive of Theorem 2.2, f(x) is also irreducible. The focus of the following sections is to present detailed methods for nding the Galois group of a polynomial of at most degree 5. For the most part, we are mainly concerned with nding the Galois group of an irreducible polynomial. Although we still present the cases of nding the Galois group of the reducible polynomial, we do so in less detail. 2.3. Quadratic Polynomials Recall that the Galois group of a polynomial f(x) 2 Q[x] is de ned as the group of automorphisms of the splitting eld of f that leave Q xed. To construct this splitting eld, we merely adjoin to the elements of Q the roots of f(x). So consider the general quadratic polynomial f(x) = x2 + bx+ c 2 Q[x], the roots of which are bpb2 4c 2 : It is easy to show that the discriminant of a quadratic polynomial of the form f(x) = x2 + bx+ c is D(f) = D = b2 4c. To form the splitting eld L of f(x), we need to adjoin the two roots above to Q. However, both roots are of the form m + n p D for some m;n 2 Q. Thus, Q[b+ p b24c 2 ; b p b24c 2 ] = Q[ p D]. So we need only adjoin the number p D to the set of rationals Q to get the splitting eld L of f(x). That is, L = Q[ p D]: Note that Q[ p D] = n m+ n p D j m;n 2 Q o (use the fact that p D is an algebraic number and Theorem 1.4) and we see that both roots of f(x) are in this eld. So we now consider Gal(f(x)) = Gal(L=Q). Now there are two possibilities for Gal(L=Q). (1) If D is the square of a rational number, say D = r2 where r 2 Q, thenp D = r implies L = Q. This implies that f(x) is reducible over Q (as its roots are in Q). 18 But there is only one group of automorphisms of Q that leave Q xed and satisfy the above - it is the identity mapping. So Gal(f(x)) = Gal(Q=Q) which consists of a single element - namely the identity automorphism. (2) If D is not the square of a rational number, then the splitting eld L is more complicated and is of the form Q[ p D] = n m+ n p D j m;n 2 Q o , for some non-square D. Since D is not a square, this implies that f(x) is irreducible over Q. Now Gal(L=Q) possesses one non-trivial automorphism given by (m+ n p D) = m n p D where m;n 2 Q: That is, is the mapping that maps one root of f(x) to the other (and visa versa). To show that is an automorphism, we note that it is one-to-one and onto and we have the relations ((m+ n p D) + (r + s p D)) = ((m+ r) + (n+ s) p D) = (m+ r) (n+ s)pD = (m npD) + (r spD) = (m+ n p D) + (r + s p D)) and ((m+ n p D) (r + spD)) = ((mr + nsD) + (ms+ nr)pD) = (mr + nsD) (ms+ nr)pD = (m npD) (r spD) = (m+ n p D) (r + spD)) where m;n; r; s 2 Q. Thus, is an automorphism that xes Q and hence in Gal(L=Q). We see that the composition of with itself yields the identity I. Thus, Gal(L=Q) consists of the two automorphisms I and and we have the fol- lowing group operation table (Table 2.1): 19 Table 2.1. Group Table for the Cyclic Group of Order 2 I I I I Thus, Gal(f(x)) = Z2. In conclusion, there is only one possible Galois group for irreducible quadratic polynomials: the group isomorphic to Z2 - also known as the cyclic group of order 2. 2.4. Cubic Polynomials Now we consider the general cubic polynomial f(x) = x3 + ax2 + bx + c 2 Q[x] with roots r1, r2 and r3. Then we have D = (r1 r2)2(r1 r3)2(r2 r3)2; where D is the discriminant of f(x). Now thanks to computer software programs like MAPLETM, we have a formula for D in terms of the coe¢ cients of f(x): D = a2b2 4b3 4a3c 27c2 + 18abc: It can be done by hand but the algebra can be challenging. 2.4.1. Irreducible Cubic Polynomials First we assume that f(x) is irreducible. Now in order to determine Gal(f(x)), we consider two cases for D. (1) D = n2 for some n 2 Q. That is, D is equal to a square number. Then, any permutation on the roots of f(x) must also satisfy (r1 r2)(r1 r3)(r2 r3) = n and in particular, (r1 r2)(r1 r3)(r2 r3) = n: 20 The following permutations do not satisfy the above algebraic equation: (r1; r2; r3)! (r2; r1; r3) (r1; r2; r3)! (r1; r3; r2) (r1; r2; r3)! (r3; r2; r1) So we nd that the only permutations satisfying both of the equations (r1 r2)(r1 r3)(r2 r3) = n simultaneously are: (r1; r2; r3)! (r1; r2; r3) (trivial permutation) (r1; r2; r3)! (r2; r3; r1) (r1; r2; r3)! (r3; r1; r2) If the above permutations correspond to I, 1 and 2, respectively, then we obtain the following group operation table (Table 2.2): Table 2.2. Group Table for the Cyclic Group of Order 3 I 1 2 I I 1 2 1 1 2 I 2 2 I 1 Thus, Gal(f(x)) = A3 = Z3 (the cyclic group/alternating group of order 3). As before, using an argument based on symmetry of polynomials will show the above three permutations satisfy every possible algebraic equation satis ed by r1, r2 and r3 and hence are indeed in Gal(f(x)). (2) D 6= n2 for any n 2 Q. That is, D is not equal to a square number. Now we consider the splitting eld of f(x) - the smallest eld containing Q and the roots of f(x). It is L = Q(r1; r2; r3). Since p D is the product of the di¤erences of the roots of f(x), that is,p D = (r1 r2)(r1 r3)(r2 r3), then p D 2 L = Q(r1; r2; r3): But p D =2 Q. Thus, Q(pD) = n m+ n p D j m;n 2 Q o and is a sub eld of L. So, [Q( p D) : Q] = 2 (since p D is not a square, the minimal polynomial of p D over Q is x2 D and then use Theorem 1.4). 21 Using the multiplicative property, we get [L : Q] = [L : Q( p D)] [Q( p D) : Q] = [L : Q( p D)] 2 and so 2 j [L : Q]. Since f(x) is irreducible, we can form the algebraic number eld K = Q(r1) which has the form fa+ br1 + cr21 j a; b; c 2 Qg with f1; r1; r21g as a basis (see Theorem 1.4). So [K : Q] = 3. Since r1 2 L, we must have K L = Q(r1; r2; r3): Now using the multiplicative property again yields [L : Q] = [L : K] [K : Q] = [L : K] 3 and so 3 j [L : Q]. Since L can have at most degree 6 (using Corollary 2.2 and Gal(L=Q) S3), we must have [L : Q] = 6 and by Corollary 2.2 again, Gal(L=Q) = S3. Note that we could have used any of the roots of f(x) to form the eld K and still yield the same result. Also, it is very important that f(x) is irreducible else [K : Q] may not be 3. 2.4.2. Reducible Cubic Polynomials Now if f(x) = x3 + ax2 + bx + c is reducible over Q, then it splits either into three linear factors or into a linear factor and an irreducible quadratic. (1) f(x) splits into three linear factors over Q. Then the three roots of f(x) are in Q and so adjoining the roots of f(x) to Q results in Q itself. This implies that Gal(f(x)) = Gal(Q=Q), the trivial Galois group. (2) f(x) splits into a linear factor and an irreducible quadratic over Q. Then adjoining the root of the linear factor to Q results in Q itself (as that root is in Q). So nding the splitting eld of f(x) is equivalent to nding the splitting eld of its quadratic factor. Since this factor is irreducible over 22 Q, its roots are not in Q and so the discriminant of this polynomial is not a square. Thus, Gal(f(x)) = Z2 (see quadratic polynomial section). Example 2.5. Consider the irreducible polynomial f(x) = x3 3x + 1. Using the formula for D, its discriminant is equal to 81, a square. Thus, Gal(f(x)) = A3. Example 2.6. Consider the irreducible polynomial f(x) = x3 2. Using the formula for D, its discriminant is equal to 108, not a square. Thus, Gal(f(x)) = S3. Example 2.7. Consider the polynomial f(x) = x32x+1. It has roots 1; p 5 2 1 2 and p 5 2 1 2 and factors into (x 1) (x2 + x 1) over Q. Thus, Gal(f(x)) = Z2. Check: If we take the discriminant of this quadratic polynomial, we obtain D(x2 + x 1) = (( p 5 2 1 2 ) ( p 5 2 1 2 ))2 = 5 which is not a square. Remark 2.2. Note that for the last example, we could have also used the formula for the discriminant of a quadratic polynomial to obtain D(x2 + x 1) = 5. In conclusion, there are only two possible Galois groups for irreducible cubic poly- nomials: the group isomorphic to Z3 (also known as the cyclic group C3 or the Alternating group A3 of order 3) and the group isomorphic to S3 (also known as the symmetric group of order 6). 2.5. Quartic Polynomials Now we consider the general quartic polynomial f(x) = x4+ ax3+ bx2+ cx+ d 2 Q[x] with roots r1, r2, r3 and r4. Also, let L be the splitting eld of f(x) over Q. 2.5.1. Irreducible Quartic Polynomials First we assume that f(x) is irreducible over Q. Recall that Gal(f(x)) is a subgroup of S4 - the symmetric group on 4 letters which has order 4! = 24. 23 De nition 2.12. The cubic resolvent of f(x) is the polynomial r(x) = x3 bx2 + (ac 4d)x (a2d 4bd+ c2) whose roots are t1 = r1r2 + r3r4 t2 = r1r3 + r2r4 t3 = r1r4 + r2r3. Denote the splitting eld of r(x) as E. Remark 2.3. The discriminant of f(x) and r(x) are the same. That is, D = Y i<j (ri rj)2 = Y i<j (ti tj)2 or equivalently, D = 4b3A+ b2B2 + 18dBAB3 27A2 where A = a2d+ c2 4bd and B = ac 4d. Theorem 2.3. Suppose f(x) is irreducible over Q, r(x) its resolvent with splitting eld E, and D the discriminant of f(x). Then (1) Gal(L=Q) = S4 if and only if r(x) is irreducible over Q and D 6= n2 for any n 2 Q. In this case, Gal(L=Q) is known as the symmetric group of order 24. (2) Gal(L=Q) = A4 if and only if r(x) is irreducible over Q and D = n2 for some n 2 Q. In this case, Gal(L=Q) is known as the alternating group of order 12. (3) Gal(L=Q) = V = Z2 Z2 if and only if r(x) splits into linear factors over Q. In this case, Gal(L=Q) is known as the Klein 4-group of order 4. (4) Gal(L=Q) = C4 if and only if r(x) has exactly one root t in Q and g(x) = (x2 tx + d) (x2 + ax + (b t)) splits over E. In this case, Gal(L=Q) is known as the cyclic group of order 4. (5) Gal(L=Q) = D4 if and only if r(x) has exactly one root t in Q and g(x) does not split over E. In this case, Gal(L=Q) is known as the dihedral group of order 8. 24 Remark 2.4. For further details and derivations of the Galois groups of quartic polynomials, we refer the reader to the paper written by Luise-Charlotte Kappe and Bette Warren [16]. Example 2.8. Let f1(x) = x4 4x3 + 4x2 + 6. In order to use the above theorem, we rst must determine if f1(x) is irreducible over Q. Using Eisensteins Criterion with p = 2, we conclude that indeed f1(x) is irreducible over Q. Now r(x) = x3 4x2 24x+ 0 = x (x2 4x 24). Thus, r(x) is reducible over Q and this eliminates cases 1 and 2 of the theorem. Since x2 4x 24 is irreducible over Q, r(x) does not split into three linear factors and hence eliminates case 3. Now the one and only rational root of r(x) is t = 0. So g(x) = (x2 tx+ d) (x2 + ax+ (b t)) = (x2 + 6) (x2 4x+ 4) = (x2 + 6) (x 2)2. We now need to nd the splitting eld E of r(x). The roots of r(x) are 0 and 22p7. Thus, E = Q(0; 2 2p7) = Q(p7). So E has the form E = a+ bp7 j a; b 2 Q : Now clearly g(x) does not split completely over E (as two roots are p6i =2 Q(p7)). Thus, Gal(f1(x)) = D4 (case 5). Example 2.9. Let f2(x) = x4 + 4x3 + 6x2 + 4x+ 2. Using p = 2, by Eisensteins Criterion, f2(x) is irreducible over Q. Now r(x) = x3 6x2 + 8x+ 0 = x (x2 6x+ 8) = x (x 4) (x 2). Since r(x) splits into three linear factors (case 3), Gal(f2(x)) = V . Example 2.10. Let f3(x) = x4 + x3 + x2 + x+ 1. Note that f3(x) is known as a cyclotomic polynomial of degree 4 and is irreducible over Q. We have r(x) = x3 x2 3x + 2 = (x 2) (x2 + x 1). So r(x) has exactly one root in Q and it is t = 2. The other roots of r(x) are 1 p 5 2 . Thus the splitting eld of r(x) is E = Q( p 5). Now g(x) = (x22x+1) (x2+x1) = (x1)2 (x2+x1). The roots of g(x) are 1 and 1 p 5 2 . Clearly, the roots of g(x) are in E = Q( p 5) and so g(x) splits over E (case 4). In fact, r(x) and g(x) both have the same splitting eld E = Q( p 5). Thus, Gal(f3(x)) = C4. 25 We now consider one more example that is more complicated than the above. Example 2.11. Let f4(x) = x4 + px+ p for each prime p. Using p = p, Eisensteins Criterion implies that f4(x) is irreducible over Q. Now r(x) = x3 4px p2. By GaussLemma, if r(x) is irreducible over Z then it is irreducible over Q. If r(x) factors over Z then it factors as a product of either three linear factors or a linear factor and an irreducible quadratic factor. Either case, there is a linear factor present. Thus, r(x) is reducible over Z if and only if there is at least one integer root. By the Rational Root Theorem, the possible roots are the divisors of p2. They are 1, p and p2. Now r(1) = 1 4p p2 r(1) = 1 + 4p p2 r(p) = p3 4p2 p2 = p2 (p 5) r(p) = p3 + 4p2 p2 = p2 (3 p) r(p2) = p6 4p3 p2 = p2(p4 4p 1) r(p2) = p6 + 4p3 p2 = p2(p4 + 4p 1). Thus, r(x) is reducible if and only if p = 3 or p = 5. So r(x) is irreducible if and only if p 6= 3 and p 6= 5. Now we have D(f4(x)) = 72p4 + 64p3 27p4 = p3 (99p 64) < 0. Thus, D(f4(x)) is not equal to a square in Q. So, Gal(f4(x)) = S4 if p 6= 3; 5 (case 1). If p = 3 then r(x) is reducible and r(x) = x3 12x 9 = (x+3) (x2 3x 3). Now r(x) has exactly one rational root t = 3. So g(x) = (x2 + 3x + 3) (x2 + 3) which has roots p3i and 3 p 3i 2 . Thus g(x) has no real roots. Since the splitting eld of r(x) is E = Q(3; 3 p 21 2 ) = Q( p 21), g(x) does not split over E (case 5). So, Gal(f4(x)) = D4 if p = 3. If p = 5 then r(x) is reducible and r(x) = x320x25 = (x5) (x2+5x+5). Now r(x) has exactly one rational root t = 5. So g(x) = (x2 5x+5) (x2 5) which has 26 roots p5 and 5 p 5 2 . Since the splitting eld of r(x) is E = Q(5; 5 p 5 2 ) = Q( p 5), the roots of g(x) are in E and so g(x) splits over E (case 4). So, Gal(f4(x)) = C4 if p = 5. In conclusion we have the following: Gal(f4(x)) = 8>><>>: D4 if p = 3 C4 if p = 5 S4 otherwise. Now Cases 4 and 5 in the above theorem can be di¢ cult to use. Determining whether or not a polynomial splits over a eld is not always an easy problem. Thus, we now introduce an alternate method in di¤erentiating between Cases 4 and 5 for the irreducible quartic. Theorem 2.4. Suppose f(x) is irreducible over Q with roots r1, r2, r3 and r4. Let K be the algebraic number eld formed by adjoining Q with any one of the roots of f(x), say r1. That is, K = Q(r1). Assume Gal(f(x)) = D4; C4 or V . Then we can nd integers a; b; c such that K = Q( p a+ b p c) and if b 6= 0 (1) Gal(f(x)) = V if and only if a2 b2c = k2 for some integer k. (2) Gal(f(x)) = C4 if and only if a2 b2c = ck2 for some integer k. (3) Gal(f(x)) = D4 if and only if a2 b2c 6= k2 or ck2 for any integer k. Remark 2.5. For further details and derivations regarding this alternate method for the above irreducible quartic cases, we refer the reader to the paper written by Luise-Charlotte Kappe and Bette Warren [16]. For more information regarding the eld K = Q( p a+ b p c) and the above example, we refer the reader to the paper written by James Huard, Blair Spearman and Kenneth Williams [15]. Remark 2.6. An obvious question arises from this theorem: what happens if b = 0? If Gal(f(x)) = D4; C4 or V , then we can nd integers a; b; c such thatK = Q( p c; p a+ b p c) (see [15]). This means if b 6= 0 then p a+ b p c 2 K implies a+ bpc 2 K. So, pc 2 K. Thus, K = Q( p c; p a+ b p c) = Q( p a+ b p c) if b 6= 0. 27 If b = 0 then K = Q( p c; p a+ b p c) = Q( p c; p a). Now this case is examined in the reducible quartic case when the polynomial reduces into two irreducible quadratic factors (see Theorem 2.5). We will see if b = 0 then Gal(f(x)) = Z2 if a c = square in Q and Gal(f(x)) = V if a c 6= square in Q. Now the obvious di¢ culty to this alternative method is trying to nd the integers a; b; c such that K = Q( p a+ b p c). However, with the aid of MAPLETM, this method is quite appealing. Example 2.12. Let f(x) = x4 4x3 + 4x+ 1. Using the solve() command in MAPLETM, we nd the roots of f(x): 1+ p 2 2 + p 10 + 4 p 2 2 ; 1+ p 2 2 p 10 + 4 p 2 2 ; 1 p 2 2 + p 10 + 4 p 2 2 ; 1 p 2 2 p 10 + 4 p 2 2 : We now form the algebraic number eld K = Q(1+ p 2 2 + p 10+4 p 2 2 ) = Q( p 10 + 4 p 2) and thus take a = 10 b = 4 c = 2: So a2 b2c = 68 6= k2 or 2k2 for any integer k. Thus, Gal(f(x)) = D4. 2.5.2. Reducible Quartic Polynomials Now lets take a look at the case when the general quartic polynomial f(x) = x4 + ax3 + bx2 + cx+ d with roots r1, r2, r3 and r4 is reducible over Q. (1) If f(x) splits into a product of a linear factor and an irreducible cubic factor, then Gal(f(x)) will be equal to the Galois group of the irreducible cubic factor which we determined how to nd in the previous section. There are two possibilities for Gal(f(x)) in this case: Z3 or S3. 28 (2) If f(x) splits into a product of four linear terms then Gal(f(x)) = Gal(Q=Q) and thus is the trivial Galois group. (3) We now consider the case when f(x) splits into a product of two irreducible quadratic factors whose discriminants are D1 and D2. In order to fully understand this case, we need the following Corollary (Corol- lary 22, pp. 593, [9]). Corollary 2.3. Let K1 and K2 be Galois elds both containing a eld F with K1 \K2 = F . Then Gal(K1K2=F ) = Gal(K1=F )Gal(K2=F ): We note that K1K2 stands for the composite eld of K1 and K2; it is the smallest eld containing Q, K1 and K2. This leads us to the following theorem regarding this case: Theorem 2.5. Let f(x) be a quartic polynomial with rational coe¢ cients and suppose that f(x) = f1(x) f2(x) over Q where f1(x) and f2(x) are irreducible quadratics. Let D1 and D2 denote the discriminants of f1(x) and f2(x), respectively. Then we have Gal(f(x)) = Z2 if D1D2 = a square in Q and Gal(f(x)) = Z2 Z2 = V if D1D2 6= a square in Q. Proof. Since both f1(x) and f2(x) are irreducible quadratics, their splitting elds are Q( p D1) and Q( p D2), respectively. The splitting eld of f(x) is then L = Q( p D1; p D2). If D1D2 = square in Q then D1D2 = n2 for some n 2 Q. So, p D1 D2 =p D1 p D2 = n , p D1 = np D2 . Thus, Q( p D1) = Q( p D2) and so L = Q( p D1). So, [L : Q] = 2 which implies jGal(L=Q)j = 2. But the only group of order 2 is Z2. Thus, Gal(f(x)) = Z2 if D1D2 = square in Q: 29 If D1D2 6= square in Q then Q( p D1) 6= Q( p D2). Let E = Q( p D1) \Q( p D2): By de nition of intersection, E Q(pD1) and E Q( p D2): However, Q( p D1) 6= Q( p D2) and neither is a proper subset of the other (as both have degree 2), so E < Q( p D1). The multiplicative property of elds implies [Q( p D1) : Q] = [Q( p D1) : E] [E : Q] or equivalently, 2 = [Q( p D1) : E] [E : Q]. Thus, [E : Q] = 1 or 2. But [E : Q] = 2 implies Q( p D1) = Q( p D2) which is a contradiction. Thus, [E : Q] = 1 and so E = Q( p D1) \Q( p D2) = Q. So by Corollary 2.3 (using F = Q), we get Gal(Q( p D1; p D2)=Q) = Gal(Q( p D1)=Q)Gal(Q( p D2)=Q): But Gal(Q( p D1)=Q) = Z2 and Gal(Q( p D2)=Q) = Z2 (as f1(x) and f2(x) are irreducible quadratics). Thus, Gal(f(x)) = Z2 Z2 = V if D1D2 6= square in Q: In conclusion, there are only ve possible Galois groups for irreducible quartic polynomials: the group isomorphic to S4, the group isomorphic to A4, the group isomorphic to V , the group isomorphic to C4 and the group isomorphic to D4. 2.6. Quintic Polynomials Now we consider the general quintic polynomial f(x) = x5+ px3+ qx2+ rx+ s 2 Q[x] with roots r1, r2, r3, r4 and r5. Also, let L be the splitting eld of f(x) over Q. Note that we simpli ed the general quintic polynomial g(x) = x5 +mx4 + nx3 + ux2 + vx+w 2 Q[x] to the polynomial f(x) given above by making the substitution y = x+ m 5 , x = y m 5 in g(x). Now Gal(g(x)) = Gal(f(x)) as translations over Q do not a¤ect the Galois group over Q. 30 2.6.1. Irreducible Quintic Polynomials First we assume that f(x) is irreducible over Q. De nition 2.13. The sextic resolvent of f(x) is the polynomial f20(x) = x 6 + 8rx5 + (2pq2 6p2r + 40r2 50qs)x4 + +(2q4 + 21pq2r 40p2r2 + 160r3 15p2qs 400qrs+ 125ps2)x3 +(p2q4 6p3q2r 8q4r + 9p4r2 + 76pq2r2 136p2r3 + 400r4 50pq3s+ 90p2qrs 1400qr2s+ 625q2s2 + 500prs2)x2 +(2pq6 + 19p2q4r 51p3q2r2 + 3q4r2 + 32p4r3 + 76pq2r3 256p2r4 +512r5 31p3q3s 58q5s+ 117p4qrs+ 105pq3rs+ 260p2qr2s 2400qr3s 108p5s2 325p2q2s2 + 525p3rs2 + 2750q2rs2 500pr2s2 + 625pqs3 3125s4)x +(q8 13pq6r + p5q2r2 + 65p2q4r2 4p6r3 128p3q2r3 + 17q4r3 +48p4r4 16pq2r4 192p2r5 + 256r6 4p5q3s 12p2q5s+ 18p6qrs +12p3q3rs 124q5rs+ 196p4qr2s+ 590pq3r2s 160p2qr3s 1600qr4s 27p7s2 150p4q2s2 125pq4s2 99p5rs2 725p2q2rs2 + 1200p3r2s2 +3250q2r2s2 2000pr3s2 1250pqrs3 + 3125p2s4 9375rs4). In the particular case when f(x) = x5 + ax+ b, the resolvent simpli es to: f20(x) = x 6 + 8ax5 + 40a2x4 + 160a3x3 + 400a4x2 +(512a5 3125b4)x+ (256a6 9375ab4). Recall that Gal(f(x)) is a subgroup of S5 - the symmetric group on 5 letters which has order 5! = 120. 31 Using MAPLETM we can nd the polynomial discriminant of f(x) as a function of its coe¢ cients. It is D = 560p2r2qs 72p4rsq 630prsq3 + 2250q2s2r 1600qsr3 + 16p4r3 128p2r4 + +256r5 + 3125s4 + 108p5s2 + 144pr3q2 900p3rs2 + 2000pr2s2 3750pqs3 + +825p2q2s2 + 16p3q3s 4p3q2r2 + 108q5s 27q4r2: We now break down the problem in nding Gal(f(x)) based on the existence of a rational root of the resolvent f20(x). Theorem 2.6. Suppose f(x) = x5 + px3 + qx2 + rx + s 2 Q[x] is irreducible over Q with resolvent f20(x) given by the above formula. Also, suppose f20(x) has no rational roots. Let D be the discriminant of f(x). Then (1) Gal(f(x)) = S5 if and only if D 6= n2 for any n 2 Q. In this case, Gal(f(x)) is known as the symmetric group of order 120. (2) Gal(f(x)) = A5 if and only if D = n2 for some n 2 Q. In this case, Gal(f(x)) is known as the alternating group of order 60. Example 2.13. Let f1(x) = x5 + 3x 3. Using p = 3, by Eisensteins Criterion, f1(x) is irreducible over Q. Now since f1(x) is of the form x5+ax+b, we can use the simpli ed resolvent formula. So, f20(x) = x 6 + 24x5 + 360x4 + 4320x3 + 32400x2 128709x 2091501. Now using MAPLETM and the factor() command, we deduce that f20(x) has no ra- tional roots. Equivalently, using the Rational Root Theorem, the possible rational roots are the divisors of 2091501. However, one can easily show that none of these divisors are roots of f20(x). Thus, Gal(f1(x)) = S5 or A5. To determine which is the correct group, we must now calculate D - the discriminant of f1(x). It is D = 315333: This can be done using the formula given for the discriminant of a quintic polynomial or by using the discrim() command in MAPLETM. 32 Now p D = p 315333 = 9 p 3893 and hence, D is not a square in Q. Thus, Gal(f1(x)) = S5. Example 2.14. Let f2(x) = x5 + 20x+ 16. In this example, we cannot use Eisensteins Criterion as the only prime p dividing all coe¢ cients of f2(x), other than the leading one, is p = 2. However, p2 cannot divide the constant term which does not hold. If we make the translation x = y 1 we obtain f2(y 1) = y5 5y4 + 10y3 10y2 + 25y 5: Now using Eisensteins Criterion with p = 5, we deduce f2(y1) is irreducible. Thus, by the contrapositive of Theorem 2.2, f2(x) is also irreducible. Note that we can use MAPLETM and its irreduc() or factor() command. The irreduc() command returns true if the polynomial is irreducible over Q and false if the polynomial is reducible over Q. We conclude that f2(x) is indeed irreducible. Now since f2(x) is of the form x5+ax+b, we can use the simpli ed resolvent formula. So, f20(x) = x 6 + 160x5 + 16000x4 + 1280000x3 + 64000000x2 +1433600000x+ 4096000000. Now using MAPLETM and the factor() command, we deduce that f20(x) has no ratio- nal roots. Equivalently, using the Rational Root Theorem, the possible rational roots are the divisors of 4096000000. However, one can show, with a little bit of work, that none of these divisors are roots of f20(x). Thus, Gal(f2(x)) = S5 or A5. To deter- mine which is the correct group, we must now calculate D - the discriminant of f2(x). It is D = 1024000000. This can be done using the formula given for the discriminant of a quintic polynomial or by using the discrim() command in MAPLETM. Now p D = p 1024000000 = 32000 2 Q and hence D is a square. Thus, Gal(f2(x)) = A5. Now in order to use this next theorem, we need to be able to calculate d(K) - the eld discriminant. We will de ne this value and examine this quantity in detail 33 in Chapter 3. For the examples proceeding this theorem, we will assume the given value of the eld discriminant is true. Theorem 2.7. Suppose f(x) = x5 + px3 + qx2 + rx + s 2 Q[x] is irreducible over Q with resolvent f20(x) given by the previous formula. Also, suppose f20(x) has exactly one rational root. Let D be the discriminant of f(x) and let d(K) be the eld discriminant of the eld Q(ri), where ri is any of the roots of f(x). Then (1) Gal(f(x)) = F20 if and only if D 6= n2 for any n 2 Q. In this case, Gal(f(x)) is known as the Frobenius or metacyclic group of order 20. (2) Gal(f(x)) = D5 if and only if D = n2 for some n 2 Q and d(K) is not equal to the fourth power of an odd integer. In this case, Gal(f(x)) is known as the dihedral group of order 10. (3) Gal(f(x)) = C5 if and only if D = n2 for some n 2 Q and d(K) is equal to the fourth power of an odd integer. In this case, Gal(f(x)) is known as the cyclic group of order 5. Remark 2.7. For further details and derivations regarding the irreducible quintic case, we refer the reader to the paper written by D. S. Dummit [8]. Remark 2.8. We will prove later that eld discriminants distinguish between the Galois groups D5 and C5. In fact, for cyclic quintic polynomials, its eld discriminant is always a fourth power, denoted as f 4 for some f 2 Z. We call f 4 the conductor of the eld. For dihedral quintic polynomials, its eld discriminant is of the form d2 f 4 where f 2 Z. In this case, d is the eld discriminant of the quadratic sub eld of the normal closure of K (normal closure will be de ned later). So, the eld discriminant of a dihedral polynomial is not a fourth power with the single exception of d2 = 24. This is the case when the quadratic sub eld is Q( p1). The topic of conductors and eld discriminants will be examined later in greater detail. Example 2.15. Let f3(x) = x5 + 15x+ 12. 34 Using p = 3, by Eisensteins Criterion, f3(x) is irreducible over Q. Now since f3(x) is of the form x5 + ax+ b, we can use the simpli ed resolvent formula. So, f20(x) = x 6 + 120x5 + 9000x4 + 540000x3 + 20250000x2 +324000000x. We can easily see that x = 0 is a rational root of f20(x). So we use Theorem 2.7 to deduce the Galois group of f3(x). We will nd the discriminant of f3(x) using the formula given in the beginning of this section. We get p D = p 259200000 = 7200 p 5. So D is not a square and thus Gal(f3(x)) = F20. Example 2.16. Let f4(x) = x5 5x+ 12. In this example, we cannot use Eisensteins Criterion as there is no prime p dividing all coe¢ cients (excluding the leading one) of f4(x). Using Example 2.3, we concluded that f4(x) was indeed irreducible over Q. Now since f4(x) is of the form x5+ ax+ b, we can use the simpli ed resolvent formula. So, f20(x) = x 6 40x5 + 1000x4 20000x3 + 250000x2 66400000x+ 976000000. Now using MAPLETM and the factor() command, we deduce that f20(x) factors into f20(x) = (x 40) (x5 + 1000x3 + 20000x2 + 1050000x 24400000) and thus has one rational root, x = 40. Once again, we use Theorem 2.7 to deduce the Galois group of f4(x). We will nd the discriminant of f4(x) using the formula given in the beginning of this section. We get p D = p 64000000 = 8000. So D is a square and we now need to nd d(K) - the eld discriminant of the eld formed by adjoining Q with one of the roots of f4(x). We will show this in detail in 35 the following chapter. In this example, d(K) = 1000000 = (2)6 (5)6. Since d(K) is not the fourth power of an odd integer, Gal(f4(x)) = D5. Example 2.17. Let f5(x) = x5 110x3 55x2 + 2310x+ 979. Using p = 11, by Eisensteins Criterion, f5(x) is irreducible over Q. Now since f5(x) is not of the form x5 + ax + b, we cannot use the simpli ed resolvent formula. So, we need to use the general formula to obtain f20(x) = x 6 + 18480x5 + 47764750x4 580262760000x3 1796651418959375x2 +2980357148316659375x 360260685644469671875. Now using MAPLETM and the factor() command, we deduce that f20(x) factors into f20(x) = (x+ 9955) (x5 + 8525x4 37101625x3 210916083125x2 +303018188550000x 36188918698590625) and thus has one rational root, x = 9955. Once again, we use Theorem 2.7 to deduce the Galois group of f5(x). We will nd the discriminant of f5(x) using the formula given in the beginning of this section. We get p D = p 1396274566650390625 = 1181640625. So D is a square and we now need to nd d(K) - the eld discriminant of the eld formed by adjoining Q with one of the roots of f5(x). We will show this in detail in the following chapter. In this example, d(K) = 14641 = (11)4. Since d(K) is the fourth power of an odd integer, Gal(f5(x)) = C5. Distinguishing between Galois groups D5 and C5 requires nding the eld discrim- inant; this is no easy task. Thus, we provide the reader with an alternate method 36 to make this distinction. The basis of this method is formed from the following Corollary (Corollary 41, pp. 641, [9]). Corollary 2.4. For any prime p not dividing the discriminant of f(x) 2 Z[x], each factorization of f(x) mod p exposes an element of the Galois group with that cycle structure. We now present the method for distinguishing between the Galois groups D5 and C5. Theorem 2.8. Let f(x) 2 Z[x] be a monic irreducible quintic polynomial. Let p be a prime and write f(x) (d1)n1 (d2)n2 (dr)nr (mod p) to denote the fact that f(x) factors modulo p into r distinct irreducible factors of degrees d1; : : : ; dr and multiplicities n1; : : : ; nr; respectively. The following procedure has been given to determine Gal(f(x)): (1) If there exists a prime p1 such that f(x) (2)(3) (mod p1) then Gal(f(x)) = S5. (2) If there exists a prime p2 such that f(x) (1)(1)(3) (mod p2) and case 1 does not hold then Gal(f(x)) = A5. (3) If there exists a prime p3 such that f(x) (1)(4) (mod p3) and cases 1 and 2 do not hold then Gal(f(x)) = F20. (4) If there exists a prime p4 such that f(x) (1)(2)(2) (mod p4) and cases 1; 2 and 3 do not hold then Gal(f(x)) = D5. (5) If for every prime q either f(x) (1)(1)(1)(1)(1) (mod q) or f(x) 5 (mod q) then Gal(f(x)) = C5. Remark 2.9. For further details and derivations regarding this alternate method for the irreducible quintic case, we refer the reader to the paper written by Blair Spearman and Kenneth Williams [27]. Firstly, the reader should notice that the above method (as well as its preceding corollary) is de ned for polynomials over Z. However, given an irreducible quintic 37 polynomial over Q, we can multiply f(x) by scalars to produce a function f(x) that is over Z. Since both f(x) and f(x) have the same roots, they have the same splitting eld and thus the same Galois group. So the theorem is applicable to our general quintic polynomial with rational coe¢ cients. Also, we should mention that we do not consider p to be those primes found in the prime factorization of the polynomial discriminant (as seen in the preceding corollary). This is because such primes produce unpredictable factorizations. We will now illustrate the use of this theorem. We will be using the Factor() mod p command in MAPLETM to determine whether the function factors modulo p. It should be noted that we are using this theorem to only distinguish between the dihedral and cyclic Galois groups of an irreducible quintic polynomial. So we still must calculate the sextic resolvent and determine the existence of rational roots. However, now we do not need to include the calculation of the eld discriminant (a not so easy procedure). Example 2.18. Let f1(x) = x5 5x+ 12. Recall from Example 2.16 that the sextic resolvent f20(x) had one rational root. This implies from Theorem 2.7 that the Galois group of f1(x) is either F20; D5 or C5. We also found that the polynomial discriminant was a square thus restricting the Galois group to either D5 or C5. So we can now use Theorem 2.8 to determine which one it is. First we nd the prime factorization of the polynomial discriminant of f1(x). It is D = (2)12 (5)6. This implies that we should not choose p to be 2 or 5. Thus, lets try p = 3. Using the notation given in the theorem, factoring f1(x) mod 3 yields f1(x) = (1)(2)(2). Thus, from the theorem, we have Gal(f1(x)) = D5: 38 We now illustrate an example on which you could determine the Galois group using only Theorem 2.8. Example 2.19. Let f2(x) = x5 21x+ 12. The prime factorization of its polynomial discriminant is D = (2)14 (3)4 (739): So we should avoid choosing p to be 2; 3 and 739. So lets try p = 5: We get f2(x) = (5) which does not help us. So lets try p = 7 to get f2(x) = (1)(4): This tells us that we could have cases 1; 2 or 3. Trying p = 11 yields f2(x) = (1)(2)(2): This implies we have cases 1; 2; 3 or 4 which yields no new information. Trying p = 13 yields f2(x) = (2)(3): This implies we have case 1. Thus Gal(f2(x)) = S5. Now Theorem 2.8 has its aws. It is not always practical to use this theorem to determine the Galois group of an irreducible quintic polynomial. The theorem only guarantees the existence of a prime p. Theoretically, we could use the theorem to nd the Galois group but we may be searching through a countless number of primes to nd it. Thus, using Theorems 2.6 and 2.7 may be more e¢ cient. However, with this said, the theorem is best used to di¤erentiate between the Galois groups D5 and C5, especially if the eld discriminant is not known. 39 2.6.2. Reducible Quintic Polynomials Now lets consider the case where the general quintic polynomial f(x) = x5 + px3 + qx2 + rx+ s 2 Q[x] with roots r1, r2, r3, r4 and r5 is reducible over Q. (1) If f(x) splits into a product of ve linear terms then Gal(f(x)) = Gal(Q=Q) and thus is the trivial Galois group (as its roots are all in Q). (2) If f(x) splits into a product of two irreducible quadratic factors and one linear factor, then Gal(f(x)) will be equal to the Galois group of the two irreducible quadratic factors which we determined how to nd in the quartic section. There are thus two possibilities for Gal(f(x)) in this reducible case: Z2 or V . (3) If f(x) splits into a product of three linear factors and one irreducible qua- dratic factor, then Gal(f(x)) will be equal to the Galois group of the irre- ducible quadratic factor which we determined how to nd in the quadratic section. There is thus only one possibility for Gal(f(x)) in this case: Z2. (4) If f(x) splits into a product of an irreducible quartic factor and one linear factor, then Gal(f(x)) will be equal to the Galois group of the irreducible quartic factor which we determined how to nd in the quartic section. There are thus ve possibilities for Gal(f(x)) in this reducible case: S4, A4, V , C4 or D4. (5) If f(x) splits into a product of two linear factors and one irreducible cubic factor, then Gal(f(x)) will be equal to the Galois group of the irreducible cubic factor which we determined how to nd in the cubic section. There are thus two possibilities for Gal(f(x)) in this case: Z3 or S3. (6) Now we consider the case when f(x) splits into a product of an irreducible quadratic factor and an irreducible cubic factor. Theorem 2.9. Let f(x) be a quintic polynomial with rational coe¢ cients and suppose that f(x) = f1(x) f2(x) over Q where f1(x) is an irreducible quadratic and f2(x) is an irreducible cubic. Let D1 and D2 denote the discriminants of f1(x) and f2(x), respectively. Then we have Gal(f(x)) = Z2 Z3 = Z6 if D2 = a square in Q, 40 Gal(f(x)) = S3 if D1 D2 = a square in Q, Gal(f(x)) = C2 S3 if both D2 and D1 D2 6= a square in Q. Proof. Firstly, since f1(x) is an irreducible quadratic, we know D1 is not a square in Q. The splitting eld of f1(x) is L1 = Q( p D1) and Gal(f1(x)) = Z2: If D2 = a square in Q then D2 = n2 for some n 2 Q. Then from the cubic section we know jGal(f2(x))j = 3 (i.e. Gal(f2(x)) = Z3). So from the multiplicative property of elds, any sub eld of the splitting eld of f2(x) must have degree dividing 3. Thus, this splitting eld, call it L2, has only the trivial sub elds: itself and Q. So, L1 \ L2 = Q. So using Corollary 2.3, Gal(f(x)) = Gal(f1(x))Gal(f2(x)) and so Gal(f(x)) = Z2 Z3 if D2 = a square in Q. If D1 D2 = a square in Q then D1 D2 = n2 for some n 2 Q. Equivalently, D2 = n2 D1 . But D1 6= a square in Q and so D2 6= a square in Q. Then from the cubic section we know jGal(f2(x))j = 6 (i.e. Gal(f2(x)) = S3). So from the multiplicative property of elds, any sub eld of the splitting eld of f2(x) must have degree dividing 6 which implies there is a quadratic sub eld (i.e. a sub eld of order 2). We will now use a result (from pp.345, [6]) that states the discriminant D2 of f2(x) can be written in the unique form dk2, d; k 2 Z, where d is the eld discriminant of the quadratic sub eld of the splitting eld of f2(x): This 41 means D1 D2 = n2 ) D1 dk2 = n2 ) D1 d = a square in Q ) p D1 2 Q( p d) ) Q( p D1) Q( p d) as p D1 generates Q( p D1) ) Q( p D1) = Q( p d) as both elds have degree 2 ) Q( p D1) splitting field of f2(x) as Q( p d) is the quadratic sub eld of the splitting eld of f2(x): So the splitting eld of f2(x) contains Q( p D1) - the splitting eld of f1(x). Therefore, the splitting eld of f(x) is the splitting eld of f2(x) (that is, the roots of f(x) are in the splitting eld of f2(x)). So the Galois group of f(x) is the Galois group of f2(x). Thus, Gal(f(x)) = S3 if D1 D2 = a square in Q: Assume D1 D2 6= a square in Q and D2 6= a square in Q: Then from the cubic section D2 6= a square in Q implies jGal(f2(x))j = 6 (i.e. Gal(f2(x)) = S3). Now let E = L1 \ L2 - the intersection of the splitting elds of f1(x) and f2(x); respectively. The only sub elds of L1 = Q( p D1) are itself and Q. Thus, E = Q( p D1) or Q. However, using the same result from [6] as before, D1 D2 6= a square in Q implies D1 dk2 6= n2 for any n 2 Q. That is, D1 d 6= a square in Q. So 42 we have D1 d 6= a square in Q ) p D1 p d =2 Q ) p D1 =2 Q( p d) ) Q( p D1) * Q( p d) as p D1 generates Q( p D1) ) Q( p D1) 6= Q( p d) ) Q( p D1) * splitting field of f2(x) (see remark at the end of proof). If Q( p D1) * L2 then E 6= Q( p D1). Thus, E = Q. So using Corollary 2.3, we get Gal(f(x)) = Z2 S3 if both D2 and D1 D2 6= a square in Q: 43 Remark 2.10. In the above proof, we have Gal(f2(x)) = S3. Now we want to show why the quadratic sub eld of f2(x) is unique. Firstly, we wish to nd the subgroup diagram of the group S3. Recall that S3 is the permutation group on the set f1; 2; 3g. We will rst list the permutations of this set and assign to each a Greek letter for a name. Let = 1 2 3 1 2 3 ! = 1 = 1 2 3 2 3 1 ! = (1 2 3) = 1 2 3 3 1 2 ! = (1 3 2) = 1 2 3 1 3 2 ! = (2 3) = 1 2 3 3 2 1 ! = (1 3) = 1 2 3 2 1 3 ! = (1 2): We will now show the multiplication table for S3 (Table 2.3). Table 2.3. Group Table for the Symmetric Group of Order 6 Note that = and = 2 (recall for permutations you read right to left). Now we want to nd all the possible subgroups of S3. Since the order of S3 is 6, any subgroup must have order dividing 6. Thus, the possible 44 orders for the subgroups of S3 are 1; 2; 3 and 6: The subgroup of order 1 is the trivial subgroup fg. Remembering that each subgroup has to contain the identity permutation and must be closed under function composition, we get the following subgroups of S3: hi = fg = 1 hi = f; ; g = hi hi = f; g , 2 = f; 2g h i = f; g , hi = f; g hi = f; g h; i = f; ; ; ; ; g = S3: Using these subgroups, we can make a subgroup diagram for S3 (see Figure 2.1). When using the Fundamental Theorem of Galois Theory, we invert the subgroup diagram so it corresponds to the sub eld diagram of a given eld. This is why Figure 2.1 is inverted. Figure 2.1. Subgroups of S3 Next we introduce a simple de nition. De nition 2.14. Let G be a nite group with subgroup H. Then the index of H in G is de ned to be jG : Hj = order of G order of H = jGj jHj : 45 Since both G and H are nite groups, jGj (or jHj) is the number of ele- ments in G (resp. H). Given a eld K that is a nite extension of Q, the Fundamental Theorem of Galois Theory states that there is a one-to-one correspondence between the intermediate elds of K (i.e. elds E satisfying Q E K) and subgroups of the Galois group G of K over Q. That is,8>>>>>>><>>>>>>>: sub elds E of K containing F K j E j F 9>>>>>>>=>>>>>>>; ! 8>>>>>>><>>>>>>>: subgroups H of Galois group G 1 j H j G 9>>>>>>>=>>>>>>>; This theorem implies that the dimension of the intermediate eld E over Q will equal the index of a subgroup of Gal(K=Q). Recall Gal(f2(x)) = S3. Thus, the quadratic sub eld of the splitting eld of f2(x) corresponds to the subgroup of S3 that has index 2 (and hence order 3). Since there is only one subgroup of S3 that has order 3 (see Figure 2.1), the quadratic sub eld of the splitting eld of f2(x) is unique. Hence, Q( p d) is the unique quadratic sub eld of the splitting eld of f2(x). Now Q( p D1) is a quadratic eld; since it is not equal to Q( p d), it is not a quadratic sub eld of the splitting eld of f2(x). This is the result we want. In conclusion, there are ve possible Galois groups of the irreducible quintic poly- nomial: the group isomorphic to C5, the group isomorphic toD5, the group isomorphic to F20, the group isomorphic to A5 and the group isomorphic to S5. The latter three groups prove challenging when determining whether the eld is monogenic or not. Numerical methods used in the cubic and quartic cases that took only minutes to perform on the fastest PCs could take several hours when solving one of these quintic cases. In this paper, we focus on the former cases - cyclic and dihedral quintic elds. 2.7. Summary The following is a summary of the possible Galois groups for irreducible polyno- mials of the speci ed degree (Table 2.4). 46 Table 2.4. Summary of the Possible Galois Groups for Irreducible Polynomials Quadratic Cubic Quartic Quintic C2 S3 S4 S5 C3 A4 A5 V = Z2 Z2 F20 C4 C5 D4 D5 2.8. Finding the Galois Group of an Algebraic Number Field Given an algebraic number eld K, we can always nd an algebraic number such that K = Q() (recall the Primitive Element Theorem). Now given any alge- braic number , we can always nd its minimal polynomial A(x). Performing these two calculations - nding and its minimal polynomial - is not always easy to do; nevertheless, it can be done. In the previous sections we looked at how to nd the Galois group of polynomials of degree ve or less. Using this knowledge, we can now apply it to the problem of classifying an algebraic number eld according to the Galois group of its normal closure (de ned below). De nition 2.15. The normal closure of K = Q() is Q( = 1; 2; : : : ; n) where = 1; 2; : : : ; n are the roots of the minimal polynomial of . That is, adjoin and its conjugates to Q to obtain the normal closure of K. Thus the normal closure of K is the smallest eld containing K and has the property that any element and its conjugates are also in this eld (the latter property is termed normal). Recall a eld is Galois if it is both separable and normal. We usually do not have the latter property. Thus, if given an algebraic number eld that is not normal, we then have to consider its normal closure. Because its normal closure is normal, we can nd its Galois group. On the other hand, if an algebraic number eld is normal, then it is equal to its normal closure and you can nd the Galois group of the eld directly. Now in order to classify elds according to their Galois group, we need to nd the Galois group of the normal closure of K over Q; in general, this is a hard problem. 47 We must adjoin and its conjugates to Q to obtain the normal closure of K. Now the roots of the minimal polynomial A(x) of are itself and its conjugates. Thus, the splitting eld of A(x) is Q( = 1; 2; : : : ; n). So Gal(A(x)) = Gal(Q( = 1; 2; : : : ; n)=Q). This means that the Galois group of the minimal polynomial of is the same as the Galois group of the normal closure of K (because the splitting eld of the minimal polynomial of is equal to the normal closure of K). So nding the Galois group of the normal closure of an algebraic number eld K = Q() is equivalent to nding the Galois group of the minimal polynomial of : As a note, the Galois group of an algebraic number eld refers to the Galois group of the elds normal closure. The reader should note this when we refer to the Galois group of an algebraic number eld as Gal(K=Q): The steps to nd the Galois group of an algebraic number eld are outlined below: 1: Find an algebraic number such that K = Q(): 2: Find the minimal polynomial A(x) of : 3: Find the Galois group of A(x): 4: Then Gal(Q( = 1; 2; : : : ; n)=Q) = Gal(A(x)): Like before, we will only be looking at algebraic number elds of degree ve or less and, in particular, of degree ve. We can now see why we focused on the cases where we were nding the Galois group of an irreducible polynomial f(x). If we are nding the Galois group of an algebraic number eld, then we are dealing with minimal polynomials which are always irreducible. Example 2.20. Find the Galois group of the algebraic number eld Q( p 2; p 3). That is, nd the Galois group of the normal closure of Q( p 2; p 3): We have seen this eld before and have shown Q( p 2; p 3) = Q( p 2 + p 3) (Example 1.7). Thus, = p 2 + p 3: Now we must nd the minimal polynomial of . From Example 1.3, we found that the minimal polynomial of was x4 10x2 + 1. Thus, A(x) = x4 10x2 + 1. 48 Now we must nd the galois group of A(x). Since A(x) is a quartic, we examine its cubic resolvent. We get r(x) = x3 + 10x2 4x 40 = (x+ 10) (x 2) (x+ 2). Thus, r(x) is reducible over Q and moreover, splits into linear factors over Q. So by Theorem 2.3 (case 3), Gal(A(x)) = V , the Klein 4-group, and so Gal(Q( p 2; p 3)=Q) = V: Example 2.21. Find the Galois group of the algebraic number eld Q( p 2 + p 2): In this example, our algebraic number eld is already of the form Q(), where =p 2 + p 2: We must nd the minimal polynomial of . We did so in Example 1.4. The minimal polynomial of is A(x) = x4 4x2 + 2: Now we must nd the Galois group of A(x). Since A(x) is a quartic, we examine its cubic resolvent. We get r(x) = x3 + 4x2 8x 32 = (x+ 4) (x2 8): Thus, r(x) is reducible over Q. This implies that Gal(A(x)) = C4; D4 or V . Because our algebraic number eld is of the form Q( p a+ b p c), we will use the alternate method (Theorem 2.4) to nd the Galois group. We calculate a2 b2c = 22 122 = 2 12 = c k2 where k = 1 2 Z: Thus, Gal(A(x)) = C4 and so Gal(Q( q 2 + p 2=Q) = C4: 49 Chapter 3 Integral and Power Bases 3.1. Integral Bases De nition 3.1. The set of all algebraic integers that lie in the algebraic number eld K is denoted by OK. In fact, OK forms a ring and thus is called the ring of integers of the algebraic number eld K. Remark 3.1. In fact, OK is an integral domain. So the ring of integers of an algebraic number eld K consists of those numbers in K whose monic minimal polynomials have integer coe¢ cients. De nition 3.2. An integral basis of an algebraic number eld K of degree n is a set f1; : : : ; ng of n algebraic integers in K such that every element in OK can be written uniquely in the form c11 + : : :+ cnn where ci 2 Z for i = 1; ::; n. Note that a basis of OK is an integral basis of K. Thus, if we are able to nd an integral basis of K then we know that OK consists of all integral linear combinations of the elements in the integral basis of K. Now we can always make a slight adjustment in an integral basis of K so that 1 = 1 (Theorem 13, pp. 36-38, [19]). Another result worth mentioning is that every algebraic number eld has an integral basis (Theorem 2.10, pp. 55, [23]). However, these bases are hard to nd. We can explicitly compute such a basis but it is now standard for computer algebra systems such as MAPLETM to have built-in programs to do this. We rst will present a theorem that will give us a basis of the ring of integers for a quadratic eld. Afterwards, we discuss how to nd integral bases for elds of higher dimension. 50 Theorem 3.1. Let K be the quadratic eld Q( p m) such that m is a squarefree integer not equal to 1. If t 2 Z then OK is given by OK = ( Z+ Z p m = fa+ bpm j a; b 2 Zg if m 6= 4t+ 1 Z+ Z(1+ p m 2 ) = fa+ b(1+ p m 2 ) j a; b 2 Zg if m = 4t+ 1: Proof. We prove this theorem using a dual-inclusion argument. First we show part of the dual-inclusion argument. Assume m 6= 4t + 1 for any t 2 Z: Let = a + bpm 2 Z + Zpm where a; b 2 Z. Then = a bpm: So the minimal polynomial of is f(x) = (x (a+ bpm)) (x (a bpm)) = x2 2ax+ (a2 mb2): Since a; b;m 2 Z, f(x) 2 Z[x]. Thus 2 OK . So Z+ Z p m OK : Now assume m = 4t+ 1 for some t 2 Z: Let = a+ b(1+ p m 2 ) 2 Z+ Z(1+ p m 2 ) where a; b 2 Z: Then = a+ b(1 p m 2 ). So the minimal polynomial of is g(x) = (x (a+ b(1 + p m 2 )) (x (a+ b(1 p m 2 )) = x2 + (2a b)x+ (a2 + ab b2t) as m = 4t+ 1: Since a; b; t 2 Z, g(x) 2 Z[x]. Thus 2 OK : So Z + Z(1+ p m 2 ) OK . So we have shown part of the dual-inclusion argument. Now we show the part of the dual-inclusion argument. Let 2 OK : We must show is in either Z+Z p m or Z+Z(1+ p m 2 ). Since OK K, 2 K = Q( p m). So = a+ b p m where a; b 2 Q. If b = 0 then is an algebraic integer if and only if = a 2 Z and the result holds. Assume b 6= 0: The minimal polynomial of is f(x) = (x (a+ bpm)) (x (a bpm)) = x2 2ax+ (a2 mb2) 51 where a; b 2 Q. Since 2 OK , this implies f(x) 2 Z[x]. So we conclude that( 2a 2 Z a2 mb2 2 Z: Thus, (2a)2 4(a2 mb2) 2 Z (as Z is closed under addition and multiplication). Simplifying yields (2b)2m 2 Z: Since m is squarefree (by assumption), 2b 2 Z. So we now have 2a 2 Z and 2b 2 Z. Thus = a+ b p m) = 1 2 (c+ d p m) where c; d 2 Z: Case #1: c and d are both even Then = 1 2 (2c1 + 2d1 p m) = c1 + d1 p m for some c1; d1 2 Z. Thus, 2 Z+ Zpm and the result holds. Case #2: c is even and d is odd Then = 1 2 (2c1 + (2d1 + 1) p m) = c1 + d1 p m+ p m 2 for some c1; d1 2 Z. Now the minimal polynomial of d1 p m is x2 d21m 2 Z[x]. So d1 p m 2 OK . But and c1 are both in OK and since OK is closed under addition, it follows that p m 2 2 OK : This implies m 4 2 OK . But m is squarefree and hence m4 =2 Z. But Q \ OK = Z. Since m4 2 Q and m4 2 OK this implies m4 2 Z; which contradicts the above. Therefore, this case is not allowable. Case #3: c is odd and d is even This case is also not allowable using a similar argument to Case #2. Case #4: c and d are both odd Then = 1 2 ((2c1+1)+(2d1+1) p m) = c1+d1 p m+ 1 2 + p m 2 for some c1; d1 2 Z. Now we have seen that d1 p m 2 OK and we know and c1 are both in OK : By closure, 1 2 + p m 2 2 OK . So the minimal polynomial of 12 + p m 2 must be in Z[x] and so h(x) = (x (1 2 + p m 2 )) (x (1 2 p m 2 )) = x2 x+ 1m 4 2 Z[x]: 52 So we require 1m 4 2 Z. That is, 1m 4 = n for some n 2 Z , m = 4t+ 1 for some t 2 Z: Now recall that = 1 2 ((2c1+1)+ (2d1+1) p m). We wish to show 2 Z+Z(1+ p m 2 ) (as m = 4t+ 1): So we set 1 2 ((2c1 + 1) + (2d1 + 1) p m) = u+ v( 1 + p m 2 ) and show u and v are in Z: The above equation yields , ( u = c1 d1 2 Z v = 2d1 + 1 2 Z: Thus, = (c1 d1) + (2d1 + 1)(1+ p m 2 ) 2 Z+ Z(1+ p m 2 ) where m = 4t+ 1 and so the result holds and the proof is complete. 3.1.1. Minimal Integers We rst need to go over some theory regarding minimal integers before we give an explicit method used to nd an integral basis of K. Further information regarding proofs of theorems and derivations can be found on pp. 160-170, [2]. First, let K be an algebraic number eld of degree n and let 2 OK such that K = Q(). So by Theorem 1.4, every 2 OK can be expressed in the form = a0 + a1 + a2 2 + : : :+ an1 n1 where a0; a1; : : : ; an1 2 Q and are uniquely determined by and . If k 2 f0; 1; 2; : : : ; n 1g such that ak 6= 0 and ak+1 = ak+2 = : : : = an1 = 0 (which implies = a0 + a1+ a2 2 + : : :+ ak k) then is called an integer of degree k in . A speci c case is when a1 = a2 = : : : = an1 = 0 (i.e. k = 0). This implies that = a0 and we say that is an integer of degree 0 in . 53 Now for k 2 f0; 1; 2; : : : ; n 1g, de ne the set Sk by Sk = ak 2 Q j a0 + a1 + : : :+ akk 2 OK for some a0; a1; : : : ; ak1 2 Q . So we have S0 = fa0 2 Q j a0 2 OKg = Z and Z Sk for k = 0; 1; 2; : : : ; n 1. Also, it can be shown that Sk has a least positive element, say ak. Since S0 = Z, we know a0 = 1. De nition 3.3. With the proceeding notation, any integer of K that is of the form a0 + a1 + a2 2 + : : :+ ak1 k1 + ak k, where a0; a1; : : : ; ak1 2 Q, is called a minimal integer of degree k in . Now the following theorem gives the form of the element ak. Theorem 3.2. For k 2 f0; 1; 2; : : : ; n 1g ak = 1 dk for some dk 2 N. Now there exists a property to the numbers dk and it is that each dk1 divides its successor dk (for k = 1; 2; : : : ; n 1). Theorem 3.3. For k = 1; 2; : : : ; n 1 dk1 j dk. We are now able to give the form of an integer of degree k in for k = 0; 1; 2; : : : ; n 1 and as a result, the form of a minimal integer of degree k in . Theorem 3.4. If is an integer of degree k in then there exists a0; a1; : : : ; ak 2 Z such that = a0 + a1 + a2 2 + : : :+ ak1 k1 + ak k dk . 54 In particular, if is a minimal integer of degree k in then there exists a0; a1; : : : ; ak1 2 Z such that = a0 + a1 + a2 2 + : : :+ ak1 k1 + k dk . Remark 3.2. The reader should note that the ais in the rst expression are generally not the same as the ones in the second expression. We have now reached the climax of this theory to obtain a method of nding an integral basis for an algebraic number eld of degree n. Theorem 3.5. Let K be an algebraic number eld of degree n. Let 2 OK such that K = Q(). For k = 0; 1; 2; : : : ; n 1 let k be a minimal integer in of degree k. Then f0; 1; : : : ; n1g is an integral basis for K. Remark 3.3. So in order to nd an integral basis for an algebraic number eld of degree n, we need only nd a minimal integer of each degree up to n 1. We now come to one last theorem which will also be useful in nding these integral bases. Theorem 3.6. Let K be an algebraic number eld of degree n. Let 2 OK such that K = Q(). For k = 0; 1; 2; : : : ; n 1 let k be a minimal integer in of degree k of the form k = ak0 + ak1 + ak2 2 + : : :+ akk1 k1 + k dk where ak0; ak1; : : : ; akk1 2 Z and 0 = d0 = 1. Then d0d1 dn1 = ind(), di j di+1 and d 2(ni) i j D() for i = 0; 1; : : : ; n 1. 55 Remark 3.4. The notation ind() means the index of . We shall see more of this in a later section. Currently, we are more interested in the second property. Also, recall that D() is the discriminant of the algebraic integer which is equal to the polynomial discriminant of the minimal polynomial of . There may be many values of di that yield an algebraic integer. However, since we are nding a minimal integer k, we choose the biggest di. That is, the minimal integer k is one whose denominator is largest yet still satis es the above theorem. 3.1.2. Examples of Finding Integral Bases As stated before, we know that every algebraic number eld has an integral basis. However, explicitly nding these bases are di¢ cult; with the help of computer software programs like MAPLETM, the problem becomes quite easy. Before we do an example of nding an integral basis, we need to be able to determine whether a given element of K is in OK . Now this means, we need to nd the minimal polynomial of . If this polynomial has integer coe¢ cients then 2 OK . For the integral basis calculations, given an algebraic number eld K = Q() where 2 OK , we know for any 2 K that = f() for some polynomial f 2 Q[x]. We illustrate how to nd the minimal polynomial of via an example. We note that these calculations were performed on MAPLETM as the algebra can prove very challenging. Example 3.1. Let f(x) = x5 5x + 12 and let K = Q() where is a root of f(x). That is, 2 K whose minimal polynomial is f(x). Given = 2 , determine whether or not is in OK : In order to do this, we must nd the minimal polynomial A(x) of . Now the degree of A(x) must be less than or equal to the degree of K and in fact, must divide the degree of K (see Remark 3.5): Thus, deg(A(x)) = 5. So let A(x) = x5 + ax4 + bx3 + cx2 + dx+ e 56 where a; b; c; d; e 2 Q. We know that is a root of its minimal polynomial A(x) and so A() = ( 2 )5 + a( 2 )4 + b( 2 )3 + c( 2 )2 + d( 2 ) + e = 0 or equivalently, A() = 1 32 5 + a 16 4 + b 8 3 + c 4 2 + d 2 + e = 0: Thus, A() is a polynomial of which is a root. So let A(x) = 1 32 x5 + a 16 x4 + b 8 x3 + c 4 x2 + d 2 x+ e: So by Lemma 1.1, f(x) j A(x). Or, equivalently, A(x) 0 mod f(x): So we now compute the remainder of A(x) divided by f(x) and set its coe¢ cients equal to zero. Using MAPLETM, we obtain the following: a = 0 b = 0 c = 0 d = 5 16 e = 3 8 : Thus the minimal polynomial of is A(x) = x5 5 16 x+ 3 8 =2 Z[x]: Thus, =2 OK. Remark 3.5. Since 2 K we adjoin to Q to get the following sub eld diagram: K j Q() j Q 57 Using multiplicative property of elds, we get [K : Q] = [K : Q()] [Q() : Q]: That is, [Q() : Q] j [K : Q] . But deg(A(x)) = [Q() : Q] and so the degree of A(x) divides the degree of K. We now illustrate via example how to nd an integral basis for an algebraic number eld K by using Theorems 3.5 and 3.6 (this method is also provided in detail with proof on pp. 36-38, [19]). For e¢ ciency, the search for algebraic integers in this example was done using MAPLETM. Example 3.2. Let K = Q() where is a root of f(x) = x5 5x+12. We want to nd an integral basis for K. Firstly, f(x) is irreducible. We can check this by using the factor() command in MAPLETM. We also showed f(x) was irreducible in Example 2.3. So K is a quintic eld (i.e. n = 5). In fact, we have seen in Example 2.16 that K is a quintic eld with the galois group of the normal closure of K equal to D5. The discriminant of f(x) is D = (2)12 (5)6. Thus, D() = (2)12 (5)6. By Theorems 3.5 and 3.6, an integral basis for K has the form 1; a10 + d1 ; a20 + a21 + 2 d2 ; a30 + a31 + a32 2 + 3 d3 ; a40 + a41 + a42 2 + a43 3 + 4 d4 where the positive integers d1; d2; d3; d4 satisfy d1 j d2 j d3 j d4, (d1) 8 j (2)12 (5)6, (d2) 6 j (2)12 (5)6, (d3) 4 j (2)12 (5)6, (d4) 2 j (2)12 (5)6. 58 Firstly, we have d1 = 1 or 2. Trying d1 = 2 yields 1 = a10 + 2 where a10 = 0 or 1 (see Remark 3.6). However, neither 2 nor +1 2 is an algebraic integer of K = Q() (method to determine this is seen in last example). Since each k 2 OK, we must have d1 = 1. So a10 = 0 and 1 = 2 OK : Secondly, we have d2 = 1; 2; 4; 5; 10 or 20. We rst try the prime possibilities for d2. Trying d2 = 5 yields 2 = a20 + a21 + 2 5 where 0 a20 4 and 0 a21 4. Like before, we loop through all the possible combinations of a20 and a21 to see if any yield an algebraic integer in K; we get none. Thus, we now know that d2 = 10 or 20 also yield no algebraic integers (see Remark 3.7). Trying d2 = 2 yields no algebraic integers and so d2 = 4 yields no algebraic integers. Thus, d2 = 1 (and so a20 = a21 = 0) which implies 2 = 2. Thirdly, d3 = 1; 2; 4; 5; 8; 10; 20 or 40. We rst try the prime possibilities for d3. Trying d3 = 5 yields 3 = a30 + a31 + a32 2 + 3 5 where 0 a30 4, 0 a31 4 and 0 a32 4. From these possible choices, we nd that there are no algebraic integers. Thus, we know that d3 = 10; 20 or 40 also yield no algebraic integers (similar argument as Remark 3.7). So now we try d3 = 2 which yields 3 = a30 + a31 + a32 2 + 3 2 where 0 a30 1, 0 a31 1 and 0 a32 1. Looping through these possible choices, we nd an algebraic integer 3+ 2 . Since d3 = 4 and 8 yield no algebraic integers, d3 = 2 and so 3 = 3 + 2 . 59 Now we consider d4. This is the hardest of the four cases as it yields the greatest number of possibilities to test. We have 27 non-trivial choices for d4. Since d3 = 2, by Theorem 3.3, d4 6= 5. Using a similar argument as Remark 3.7, we can exclude the possible choices of d4 that are divisible by 5. So we now try the other prime choice, that is, d4 = 2. We nd an algebraic integer 4+3+2+ 2 . However, d4 = 4 also yields an algebraic integer 4+3+2+ 4 . Since none of the other larger choices for d4 yield an algebraic integer, it follows that d4 = 4 and 4 = 4 + 3 + 2 + 4 . So an integral basis for K is 1; ; 2; 3 + 2 ; 4 + 3 + 2 + 4 : Remark 3.6. When choosing to examine the possible choices for each di, we ignore the trivial case di = 1 and examine the others rst, in particular, the prime possibilities of di. Also, we need only consider the values of ak0; ak1; : : : ; akk1 mod dk. For example, when trying the case d2 = 5, we consider only values mod 5 for a20 and a21. Since OK is a ring, we can subtract o¤ integers or integer multiples of and still remain in OK. Choosing these appropriately has the e¤ect of reducing the coe¢ cients modulo the denominator. Remark 3.7. We found that d2 = 5 yields no algebraic integers. That is, 2 = a20 + a21 + 2 5 =2 OK. By way of contradiction, assume that 2 = a20 + a21 + 2 10 2 OK. That is, assume d2 = 10: Since OK is closed under multiplication, 2 = 2 a20 + a21 + 2 10 = a20 + a21 + 2 5 2 OK which contradicts our earlier result. Thus, d2 6= 10. We can do a similar argument for d2 = 20. 60 Now this method yields itself nicely to computer software programs. Since there is only a nite number of possibilities, MAPLETM can easily loop through the choices for ak0; ak1; : : : ; akk1 to determine whether an algebraic integer is revealed. Smaller degrees are easy to do by hand but require many calculations. Thus, from now on, we use MAPLETM to determine an integral basis of an al- gebraic number eld. We do so by using the integral_basis() command. This command computes an integral basis for an algebraic number eld K (and hence a basis of OK). If we de ne K by Q() where 2 OK and has minimal polynomial A(x) then we pass in the polynomial A(x) in the integral_basis() command. Note that in order to use this command, one must rst open the with(numtheory) package. 3.2. Field Discriminant We rst describe the discriminant of a set of elements of an algebraic number eld K. De nition 3.4. Let K be an algebraic number eld of degree n. If f1; : : : ; ng is a subset of the eld K then we may de ne a symmetric integral matrix M whose (i; j)-entry is de ned to be Tr(i j). The discriminant of f1; : : : ; ng is then de ned as (1; : : : ; n) = det(M): The proof of the next proposition can be found on page 15, [5] (Proposition 2.3). Proposition 3.1. (1; : : : ; n) 6= 0 if and only if f1; : : : ; ng is a basis of K over Q. The proof of the next theorem can be found on page 127, [2] (Theorem 6.4.4). Theorem 3.7. Let K be an algebraic number eld of degree n. (1) If 1; : : : ; n 2 K then (1; : : : ; n) 2 Q. (2) If 1; : : : ; n 2 OK then (1; : : : ; n) 2 Z. De nition 3.5. Let K be an algebraic number eld of degree n. If f1; : : : ; ng is an integral basis of the eld K then the eld discriminant d(K) of K is de ned as d(K) = (1; : : : ; n): 61 Remark 3.8. It is important to note that the value of the discriminant is inde- pendent of the integral basis that you choose (see Proposition 2.4, pg 16, [5]). Since the eld discriminant d(K) = (1; : : : ; n) where f1; : : : ; ng is an inte- gral basis, it follows from Theorem 3.7, d(K) 2 Z. Now nding the eld discriminant is not an easy task. Using the above de nition, it would require nding the trace of the product of two algebraic numbers a nite number of times. However, in order to do this, we rst need to nd an integral basis of K which, as seen previously, is not very easy to do by hand. We rst will examine the eld discriminant of a quadratic eld then give a method to nd the eld discriminant of a eld K of higher dimension that does not depend on the trace function. Theorem 3.8. Let K be a quadratic eld Q( p m) such that m is a squarefree integer not equal to 1. Then the eld discriminant d(K) of K is given by d(K) = ( 4m if m 6= 4t+ 1 m if m = 4t+ 1 where t 2 Z. Proof. From Theorem 3.1, we have OK = ( Z+ Z p m = fa+ bpm j a; b 2 Zg if m 6= 4t+ 1 Z+ Z(1+ p m 2 ) = fa+ b(1+ p m 2 ) j a; b 2 Zg if m = 4t+ 1: That is, an integral basis of OK is f1; p mg if m 6= 4t+ 1 or f1; 1+ p m 2 g if m = 4t+ 1: Using the de nition of the eld discriminant, we wish to form the matrixM . Assume m 6= 4t + 1 and let 1 = 1 and 2 = p m. Then f1; 2g is a basis of OK . So to form the matrix M we wish to nd Tr(i j). Since K has degree 2, we have Tr(1 1) = Tr(1) = 1 + 1 = 2 Tr(1 2) = Tr( p m) = p m+ (pm) = 0 Tr(2 1) = Tr( p m) = p m+ (pm) = 0 Tr(2 2) = Tr(m) = m+m = 2m: 62 So M = " 2 0 0 2m # and so d(K) = det(M) = 2 2m 0 = 4m: Assume m = 4t+ 1 and let 1 = 1 and 2 = 1+ p m 2 . Then f1; 2g is a basis of OK . So to form the matrix M we wish to nd Tr(i j). So we have Tr(1 1) = Tr(1) = 1 + 1 = 2 Tr(1 2) = Tr( 1 + p m 2 ) = 1 + p m 2 + 1pm 2 = 1 Tr(2 1) = Tr( 1 + p m 2 ) = 1 + p m 2 + 1pm 2 = 1 Tr(2 2) = Tr( (m+ 1) + 2 p m 4 ) = (m+ 1) + 2 p m 4 + (m+ 1) 2pm 4 = m+ 1 2 : So M = " 2 1 1 m+1 2 # and so d(K) = det(M) = 2(m+1 2 ) 1 = m. This completes the proof. We now will look at the eld discriminant of elds of higher dimension. First we recall two di¤erent discriminants: the discriminant of a polynomial f(x) and the discriminant of an algebraic number eld K. The latter was described in De nition 3.5. The former was equal to the product of the squares of the di¤erences of the roots of f(x). Using these two discriminants, we introduce the following de nition. De nition 3.6. The index of an algebraic number is de ned by ind() = s D(f) d(K) where f is the minimal polynomial of and d(K) is the eld discriminant of the algebraic number eld K formed by adjoining Q with . The reader may recognize the notation ind(). We rst introduced this notion in Theorem 3.6. In fact, we will now make use of the rst property that was found in that theorem. Recall d0d1 dn1 = ind() 63 where d0; d1; ; dn1 are the denominators present in an integral basis for the alge- braic number eldK. Now we saw that nding an integral basis proved di¢ cult with- out the aid of a computer software program. However, with the use of MAPLETM, we can nd an integral basis using only the integral_basis() command. Also, nding a polynomial discriminant is quite straightforward. We have pro- vided the reader with a number of formulas to do so that vary according to the degree of the polynomial. Thus, using De nition 3.6, we can now calculate the eld discriminant d(K) of the algebraic number eld K. We have d(K) = D(f) (ind())2 . Now we provide the reader with two examples illustrating the calculation of the eld discriminants used in Examples 2.16 and 2.17, respectively. Example 3.3. Let f(x) = x5 5x+ 12: If is a root of f(x) then let K = Q(). We wish to nd the eld discriminant of K. First we calculate the discriminant of f(x) to get D = (2)12(5)6. Now we need to nd an integral basis for K. We did so in Example 3.2 and found 1; ; 2; 3 + 2 ; 4 + 3 + 2 + 4 to be an integral basis for K. Thus ind() = d0d1d2d3d4 = 1 1 1 2 4 = (2)3: So the eld discriminant for K is d(K) = (2)12(5)6 ((2)3)2 = (2)6(5)6. Example 3.4. Let f(x) = x5 110x3 55x2 + 2310x+ 979. If is a root of f(x) then let K = Q(). We wish to nd the eld discriminant of K. The discriminant of f(x) is D = (5)20(11)4. 64 Using MAPLETM, we nd 1; + 4 5 ; 2 + 3 + 21 25 ; 3 + 22 + 18 + 104 125 ; 4 + 3 + 162 + 86 + 521 625 to be an integral basis of K. Thus ind() = d0d1d2d3d4 = 1 5 25 125 625 = (5)10: So the eld discriminant of the eld K is d(K) = (5)20(11)4 ((5)10)2 = (11)4. 3.3. Power Bases As stated before, an integral basis for an algebraic number eld K is a basis for OK . Moreover, there is a special case in which the basis is known as a power basis. These speci c bases are very rare and also hard to nd. De nition 3.7. Let K be an algebraic number eld (not necessarily equal to Q()) of degree n. If there exists an element 2 OK such that 1; ; 2; : : : ; n1 is an integral basis for K then K is said to be monogenic and the integral basis 1; ; 2; : : : ; n1 is called a power basis for K. Now we can nd power bases based on the index of the element as seen in the following theorem. Theorem 3.9. Let K be an algebraic number eld of degree n. Let 2 OK such that K = Q(). Then 1; ; 2; : : : ; n1 is an integral basis for K if and only if ind() = 1. Thus, if ind() = 1 then 1; ; 2; : : : ; n1 is a power basis for K. We now introduce one more theorem that will help us in the search for power bases. Theorem 3.10. Let K be an algebraic number eld of degree n. Let 2 OK such that K = Q(). If D() is squarefree then 1; ; 2; : : : ; n1 is a power basis for K. 65 Remark 3.9. Theorem 3.10 is a consequence of Theorem 2.17, pp. 48, [30]. Note that in Theorems 3.9 and 3.10, is a generator of K. Remark 3.10. The proofs for Theorems 3.9 and 3.10 are found on pp. 146-147, [2]. We also remind the reader that a number is said to be squarefree if its prime decom- position contains no repeated factors. Thus, all primes are trivially squarefree. We now provide the reader with an example illustrating the use of Theorem 3.10. We shall use these theorems, in particular Theorem 3.9, in a later chapter. Example 3.5. Let f(x) = x3 + x+ 1. We can use the Rational Root Theorem to show f(x) is irreducible over Q. Let K be the algebraic number eld Q() where is a root of f(x). Thus, K has degree 3. Since f(x) 2 Z[x], 2 OK. Now we must calculate D(). We recall that the discriminant of an algebraic integer is equal to the polynomial discriminant of its minimal polynomial. Using MAPLETM, we get D() = D(f) = 31: Since D() is squarefree, by Theorem 3.10, 1; ; 2 is a power basis for K. Thus, K is monogenic. We must warn the reader when using Theorem 3.9. If ind() 6= 1 then this does not mean K does not admit a power basis. It just means that using the speci c algebraic integer , we do not have a power basis. Thus, in determining whether or not K is monogenic, we want to examine a typical element in OK and examine its index (rather than a speci c element). In order to use the theorem, we impose the restriction that the element in OK must be a generator of K. We will examine this method in detail in the following subsection. 66 3.3.1. Index Form Suppose K is an algebraic number eld with integral basis f1; 2; 3; : : : ; ng . Let be a typical element in OK : Then we can write = X1 +X22 +X33 + : : :+Xnn for some X1; X2; : : : ; Xn 2 Z. As a note, in order to use Theorem 3.9, we need K = Q(). That is, must be a generator of K. Now in order to compute ind(); rst we must nd the minimal polynomial A(x) of (see Example 3.1). Secondly, we nd its polynomial discriminant D(A). Lastly, we nd the eld discriminant d(K) of K to get ind() = s D(A) d(K) : The above equation will be a function of the n variables X1; X2; : : : ; Xn and is known as the index form. Because power bases are equivalent if they di¤er by an integer (pp. 80, [23]), we can actually drop the X1 term in = X1 +X22 +X33 + : : :+Xnn. This then would leave us an index form that is a function of n 1 variables. In order to determine monogenicity of K, we want to solve the equation (over Z) ind() = s D(A) d(K) = 1 for X2; : : : ; Xn. This equation is known as the index form equation. Solving this equation is best suited for a computer software program. We illustrate, in detail, this method in a later chapter. We now introduce two more de nitions and theorems to complete this chapter. De nition 3.8. The index of K is i(K) = gcd find() j is a generator of Kg : We sometimes refer to i(K) as the eld index of K. De nition 3.9. The minimal index of K is m(K) = min find() j is a generator of Kg : 67 Theorem 3.11. Let K be an algebraic number eld. Then m(K) = 1 if and only if K possesses a power basis. Theorem 3.12. Let K be an algebraic number eld such that K possesses a power basis. Then i(K) = 1. Remark 3.11. For proofs and further information regarding the above two theo- rems, see pp. 178-179, [2]. The reader should note that the converse of the latter theorem is not true. How- ever, this theorem does have its advantages. The contrapositive states that if i(K) 2 then K does not contain a power basis. 68 Chapter 4 Cyclic and Dihedral Fields of Prime Degree As stated before, we will mainly concentrate on quintic elds, in particular, dihe- dral quintic elds. However, there are many similarities between cyclic and dihedral quintic elds. Thus, we rst examine cyclic and dihedral elds of prime degree and then move on to the speci c case of degree 5. 4.1. Field Discriminants The question was asked, "How many cyclic or dihedral elds have the same eld discriminant?" This problem prompted researchers to examine the eld discriminant in detail, looking for any pattern or structure. In doing so, we have the following results ([21]). Theorem 4.1. Let p be an odd prime and K be a eld containing Q of degree p such that the galois group of K is Cp (i.e. the cyclic group of order p). Then the eld discriminant of K is of the form d(K) = fp1 where f 2 Z. We call f the conductor of K. Remark 4.1. In fact, if the galois group of the normal closure of K is Cp then the conductor f of K must be greater than 1 (see pp. 58-66, [23]). Now using this theorem, we see that for cyclic quintic elds (i.e. p = 5), its eld discriminant is of the form d(K) = f 4 where f 2 Z and f > 1. Now we have the veri cation of the rst part of Remark 2.8. 69 Once again, there is a parallel between cyclic and dihedral elds of prime degree. So it comes as no surprise that there is a similar theorem for the eld discriminant of a dihedral eld of prime degree. Theorem 4.2. Let p be an odd prime and K be a eld containing Q of degree p such that the galois group of the normal closure of K is Dp (i.e. the dihedral group of order 2p). Then the eld discriminant of K is of the form d(K) = fp1 d p12 where f 2 Z and d is the eld discriminant of the quadratic sub eld of the normal closure of K. We call f the conductor of K. Remark 4.2. The quadratic sub eld of K = Q() can be found using the method described on page 156 in [35]. Also, in [21] (pp. 835), the de nition of the conductor f of a eld K with the galois group of the normal closure of K being Dp implies f must be greater than zero. Now using this theorem, we see that for dihedral quintic elds (i.e. p = 5), its eld discriminant is of the form d(K) = f 4 d2 where f 2 Z and f > 0. Now we have the veri cation of the second part of Remark 2.8. We now provide an example illustrating both theorems. Example 4.1. Let K be the eld formed by adjoining one of the roots of f(x) = x5 5x+ 12 to Q. In Example 2.16, we found Gal(f(x)) = D5 and d(K) = 1000000 = (2)6 (5)6: Since K is a quintic eld with the dihedral group of order 10 as the galois group of its normal closure, we know that d(K) = f 4 d2: To determine the value of the conductor 70 f , we must rst calculate d. In order to nd d, we must work out the generator of the quadratic sub eld of K. Recall the method can be found in [35]. Omitting details, we give the reader an overview on how to use this method via MAPLETM. First we must de ne our polynomial f(x). Now substitute the value X + x for x in f(x). We get f(X + x) = (X + x)5 5X 5x+ 12: Using the resultant() command, compute the resultant of the two polynomials f(x) and f(X + x) with respect to the indeterminate x. Factor this result to obtain X5(X10+10X8+125X6+500X4+2500X2+4000)(X1010X875X6+1500X45500X2+16000): Based on this method, the quadratic sub eld is of the form Q( p(constant term)) where the constant term is the constant present in the factored resultant (note that there will always be a constant term present). We have two constants in our resultant and it does not matter which one you use as both yield the same result. That is, Quadratic sub eld of K = Q( p4000) = Q( p (2)5(5)3) = Q(20 p2 5) = Q( p10): Since 10 = 4(3) + 2 6= 4t + 1 for any t 2 Z, by Theorem 3.8, we have d = 40: Thus, f = 5 as d(K) = f 4 d2: Example 4.2. Let K be the eld formed by adjoining one of the roots of f(x) = x5 110x3 55x2 + 2310x+ 979 to Q. In Example 2.17, we found Gal(f(x)) = C5 and d(K) = 14641 = (11)4. Thus, the conductor of K is 11. 71 Remark 4.3. In the dihedral case, more work is needed to determine the value of f and d if there is more than one possible choice of f . Now consider the following scenario. Given a quintic eld K, we wish to nd the galois group of its normal closure. Using Theorem 2.7 and assuming the discriminant of the irreducible polynomial de ning K is a square, we deduce the group is either C5 or D5: We wish to know when f 4 = d2 f 4 where f is the conductor of the cyclic quintic eld and f is the conductor of the dihedral quintic eld. That is, when could we not distinguish between the cyclic and dihedral quintic cases using eld discriminants? We answer this question in the next theorem. Theorem 4.3. Let K be a eld with the galois group of the normal closure of K either C5 or D5: Then the eld discriminant of K will distinguish between the Galois groups C5 and D5: Proof. By way of contradiction, assume there exists f; f 2 Z such that f 4 = d2 f 4 where d is the eld discriminant of the quadratic sub eld of the normal closure of K. Thus, d2 = l4 where l 2 Z: Hence, d = n2 where n 2 Z: But by Theorem 3.8, d = ( 4m if m is not of the form 4t+ 1 m if m = 4t+ 1 where m is a squarefree integer. So d 6= m which implies d = 4m where m 6= 4t+ 1: Thus, 4m = n2 , m = (n 2 )2 , n = 2 as m is squarefree , m = 1 as m 6= 4t+ 1: So d = 4. 72 On pp. 831, [21], the author states that for cyclic elds of degree p we have f = pe q1 qt where p is an odd prime, e = 0 or e = 2, t 0 and the qi are pairwise distinct rational primes satisfying qi 1 (mod p) for i = 1:::t. We will now use this for p = 5 to get f = 5e q1 qt where e = 0 or e = 2, t 0 and the qi are pairwise distinct rational primes satisfying qi 1 (mod 5): Since d = 4, f 4 = d2 f 4 , f 4 = (4)2 f 4 , 42 j f 4 , 4 j f 2 , 2 j f: However, 2 2 (mod 5) and hence a contradiction. In regards to the question posed in the beginning of this section, we know that the number of cyclic elds of prime degree p that share the same conductor f = pe q1 qt (and hence eld discriminant) is equal to (p 1)t+w1 where w = 1 2 e. Thus, as the number of prime factors increase, the number of cyclic elds of prime degree p that share the same conductor also increases. 73 Mayer also gives an expression for the number of dihedral elds of prime degree p that share the same conductor in [21], but it is more complicated than the above and beyond the scope of this paper. Now recall that there is only one monogenic cyclic quintic eld ([14]). As seen, there is a parallel between the eld discriminants of cyclic and dihedral elds of prime degree. Since monogenicity is based on eld discriminants (i.e. the index form equation), this leads us to the following question. Question: If there is only one monogenic cyclic eld of degree 5, then can we deduce that there is only one or nitely many monogenic dihedral eld of degree 5? This question is the motivation for this thesis and will be answered in the next chapter 4.2. Abelian and Cyclotomic Fields The main focus of this section is to provide the reader with more insight into the structure of the conductor of a cyclic quintic eld. We begin with a few basic de nitions. De nition 4.1. An algebraic number eld K containing Q is called abelian if K is both normal and separable over Q and Gal(K=Q) is an abelian group. Recall that a group is abelian if its binary operation is commutative. Example 4.3. The only abelian Galois group for a quintic eld is the cyclic Galois group Z5. Thus, any algebraic eld K that is both normal and separable containing Q with Gal(K=Q) = Z5 is abelian. We now discuss a few basic principles from complex analysis. First, let n 2 N. Then, over C, there are n distinct roots to the polynomial xn 1; these roots are called the nth roots of unity. The collection of these roots form a cyclic group under multiplication over C (pp. 539, [9]). We call the generator of the group of all nth roots of unity a primitive nth root of unity. We denote such a generator as n. Geometrically, one can nd n = e 2i n = cos( 2 n ) + i sin( 2 n ): 74 De nition 4.2. The eld Q(n) is called the cyclotomic eld of nth roots of unity. This now leads us to a well-known theorem used in the analysis of abelian elds K containing Q. It is known as the Kronecker-Weber Theorem. Theorem 4.4. Let K be an abelian algebraic number eld containing Q. Then K Q(n) for some n 2 N. It is clear that K is then contained in in nitely many cyclotomic elds. Now in terms of quintic elds, we know that the only abelian algebraic number elds K containing Q must have Galois group isomorphic to Z5. That is, for quintic elds, we can only apply the Kronecker-Weber Theorem to cyclic quintic elds. So assuming K containing Q is a cyclic quintic eld, the least such n such that K Q(n) is called the conductor f of the cyclic quintic eld. Now we can see why the conductors of cyclic and dihedral elds with prime degree (in particular, degree 5) have di¤erent structure (see Remark 2.8). The conductor is based on the structure of the algebraic number eld. Cyclic elds are abelian whereas dihedral elds are not. Such study of abelian elds and their structure is known as class eld theory. 75 Chapter 5 Cyclic and Dihedral Quintic Fields with a Power Basis We have now reached the main focal point of this paper. It is in this chapter that we prove the main result of the abstract. We wish to show that there exist in nitely many dihedral quintic elds with a power basis. Now that we have discussed the theory needed to prove this statement, we see that this is no trivial problem. Firstly, we have already seen that integral bases are hard to nd. Also, power bases are very rare so to say that there are in nitely many particular quintic elds that have a power basis seems unlikely. This is one reason why this problem is so interesting. 5.1. Some Important Lemmas We now introduce an important parametric family of quintic polynomials which was studied by A. Brumer [4] and T. Kondo [17], independently. De nition 5.1. The parametric family of quintics studied by T. Kondo is the following: Ra;b(x) = x 5 + (a 3)x4 + (b a+ 3)x3 + (a2 a 1 2b)x2 + bx+ a where a; b 2 Z. Remark 5.1. For our purposes, we set a = 1; in the above expression, to get Fb(x) = x 5 2x4 + (b+ 2)x3 (2b+ 1)x2 + bx+ 1 where b 2 Z. We begin our study on the irreducibility of Fb(x): Lemma 5.1. Fb(x) is irreducible over Q for all b 2 Z. Proof. Firstly, if Fb(x) is irreducible over Zp, where p is a prime, then Fb(x) is irreducible over Z (recall contrapositive of Theorem 2.1). If we let p = 2 then, using 76 MAPLETM, we obtain the following: Fb(x) x5 + bx3 + x2 + bx+ 1 mod 2. We can break down Fb(x) mod 2 into cases (b is even or odd) to get Fb(x) = ( x5 + x2 + 1 (mod 2) if b 0 mod 2 x5 + x3 + x2 + x+ 1 (mod 2) if b 1 mod 2. Finally, using MAPLETM and the Factor() mod 2 command, we see that both x5 + x2 + 1 and x5 + x3 + x2 + x + 1 are irreducible over Z2. Thus, Fb(x) is irreducible over Z for all b 2 Z. So by GaussLemma, Fb(x) is irreducible over Q. Now we look at the polynomial discriminant of Fb(x). Lemma 5.2. The polynomial discriminant of Fb(x) is D(Fb(x)) = (4b 3 + 28b2 + 24b+ 47)2: Proof. Using MAPLETM and the discrim() command, the result follows. Remark 5.2. We note that the cubic polynomial 4b3+28b2+24b+47 is irreducible over Q. We now wish to determine the Galois group of Fb(x): We rst narrow down the choices of Gal(Fb(x)) to two possibilities. Lemma 5.3. Gal(Fb(x)) = Z5 or D5. Proof. In order to use the theory of irreducible quintics, we rst must rewrite Fb(x) of the form x5 + px3 + qx2 + rx+ s by making the substitution Fb(x+ 25). We then must nd the Galois group of Fb(x+ 25). Since translations do not a¤ect the Galois group of a polynomial, the result will give us the Galois group of Fb(x). 77 We rst determine if the resolvent sextic for Fb(x+ 25) has any rational roots. The resolvent sextic is f20(x) = 1 3814697265625 (125x 579 + 230b)(648144531250bx3 + 4191965234375bx2 +8455564403125bx+ 30517578125x5 + 299873046875x3 + 2930874765625x2 +12425749097500x 30637508969600b+ 86669921875x4 40098357698500b2 85449218750bx4 + 413085937500b2x3 + 61035156250b3x3 29180050973750b3 15391108715625b4 + 446085937500b2x2 363164062500b3x2 151367187500b4x2 5671380700000b5 1077349609375b6 65917968750b7 + 3417867046875b2x +1665324531250b3x+ 1521822656250b4x+ 532031250000b5x +30517578125b6x 12808965183149). Thus, the resolvent sextic of Fb(x+ 25) has one rational root 579 125 46b 25 . So we now use Theorem 2.7 to deduce that Gal(Fb(x+ 25)) = F20;Z5 orD5. So, Gal(Fb(x)) = F20;Z5 or D5. But by Lemma 5.2, D(Fb(x)) = (4b 3 + 28b2 + 24b+ 47)2 = square in Z: However, D(Fb(x)) = D(Fb(x+ 25)) and so Gal(Fb(x+ 2 5 )) F20. Thus, Gal(Fb(x)) F20 and the result follows. Based on the polynomial discriminant of Fb(x), we can deduce the exact Galois group of Fb(x). Lemma 5.4. If 4b3 + 28b2 + 24b+ 47 is squarefree then Gal(Fb(x)) = D5. Proof. Assume 4b3 + 28b2 + 24b+ 47 is squarefree. By way of contradiction, assumeGal(Fb(x)) D5. Then by Lemma 5.3, Gal(Fb(x)) = Z5: Let 2 C be a root of Fb(x) and setK = Q(). Since Fb(x) is irreducible (Lemma 5.1) and Gal(Fb(x)) = Z5, K is a quintic eld with the galois group of its normal closure being Z5. Hence, the eld discriminant d(K) of K is of the form f 4 for some 78 f > 1; f 2 Z (see Remark 4.1). So (4b3 + 28b2 + 24b+ 47)2 = D(Fb(x)) (by Lemma 5.2) = g2 d(K) (by the index form and where g 2 Z) = g2 f 4 (by above explanation). So we have 4b3 + 28b2 + 24b+ 47 = g f 2: But by assumption, 4b3 + 28b2 + 24b + 47 is squarefree and so f = 1. This is a contradiction as f > 1. Therefore, Gal(Fb(x)) = D5: We now introduce another Lemma before we prove the abstract. Lemma 5.5. If 4b3 + 28b2 + 24b+ 47 is squarefree then the quadratic sub eld of the splitting eld of Fb(x) is Q( p 4b3 28b2 24b 47): Proof. Using Lemma 5.4, the proof is deduced from [29] (pp. 4748). We now state an important theorem, also known as ErdösTheorem ([11]). Theorem 5.1. Let f(x) 2 Z[x] a polynomial whose coe¢ cients have highest com- mon factor 1 (f(x) is thus called primitive). Let l 3 denote the degree of f(x). Also assume that the leading coe¢ cient in f(x) is positive and that f(x) has no fac- tors repeated (l 1) times. Lastly, assume the following condition holds: if l is a power of 2 then there exists some n such that f(n) 6= 0 (mod 2l1): Then there are in nitely many positive integers n for which f(n) is (l 1)-th power free. In this paper, we are interested in the case when l = 3. Lemma 5.6. There are in nitely many integers b such that 4b3+28b2+24b+47 is squarefree. 79 Proof. Let f(x) = 4x3 + 28x2 + 24x + 47 2 Z[x]. Then f(x) is primitive as gcd(4; 28; 24; 47) = 1 (as 47 is prime) and all coe¢ cients are positive (and hence the leading one is positive). Since f(x) is irreducible over Q, f(x) has no factors repeated 2 times. Let l = 3. Since l is not a power of 2 we do not have to nd an n such that f(n) 6= 0 (mod 22). So by ErdösTheorem, there are in nitely many positive integers n for which f(n) is squarefree and the result follows. We now have the tools to prove the central theorem of this paper. 5.2. Proof of Main Result Theorem 5.2. There are in nitely many integers b such that the dihedral quintic elds Q(), where 5 24 + (b+ 2)3 (2b+ 1)2 + b + 1 = 0, are distinct and monogenic. This theorem then implies that there are in nitely many dihedral quintic elds with a power basis. Proof. Recall Fb(x) = x5 2x4 + (b + 2)x3 (2b + 1)x2 + bx + 1 and is irreducible (Lemma 5.1). By Lemma 5.2, D(Fb(x)) = (4b3 + 28b2 + 24b + 47)2. By Lemma 5.6, there are in nitely many integers b such that 4b3 + 28b2 + 24b+ 47 is squarefree. Thus, by Lemma 5.4, Gal(Fb(x)) = D5 for in nitely many integers b. Now for each such b, let b be a root of Fb(x) and set Kb = Q(b): Then each Kb is a dihedral quintic eld (as Gal(Fb(x)) = D5 and Fb(x) is irreducible). Since each Kb is a dihedral quintic eld, the eld discriminant of Kb is of the form d(Kb) = f 4 b d2b where db is the eld discriminant of the quadratic sub eld of the splitting eld of Fb(x) and fb (2 N) is the conductor of Kb. By Lemma 5.5, the quadratic sub eld is Q( p 4b3 28b2 24b 47): 80 Now we will use Theorem 3.8 to determine the eld discriminant db of this quadratic sub eld. Firstly, m = 4b3 28b2 24b 47 = 4(b3 7b2 6b 12) + 1: Since m is squarefree by assumption (and hence m is squarefree), we have d(Q( p 4b3 28b2 24b 47)) = 4b3 28b2 24b 47 and so db = 4b3 28b2 24b 47: Thus, d(Kb) = (4b 3 + 28b2 + 24b+ 47)2 f 4b : Now by Lemma 5.2, D(Fb(x)) = (4b3 + 28b2 + 24b+ 47)2 so that d(Kb) = D(Fb(x)) f 4b : Now recall the index form: ind(b) = s D(Fb(x)) d(Kb) , (ind(b))2 = D(Fb(x)) d(Kb) : Since (ind(b))2 2 Z+, d(Kb) divides D(Fb(x)) over Z. This implies f 4b = 1 and so fb = 1 (as fb 2 N). Thus, d(Kb) = (4b 3 + 28b2 + 24b+ 47)2 = D(Fb(x)): Consequently, ind(b) = 1: So by Theorem 3.9, f1; b; 2b ; 3b ; 4bg is a power basis for the eld Kb. Hence, each eld Kb is monogenic. So we have shown that there are in nitely many integers b such that the dihedral quintic elds Q(), where 5 24 + (b+ 2)3 (2b+ 1)2 + b + 1 = 0, are monogenic. We now need to show that these elds are distinct. Recall that two elds are equal if they have the same eld discriminant. Since d(Kb) = (4b 3 + 28b2 + 24b + 47)2; we wish to know how many possible values of b 81 satisfy the given equation for xed k 2 Z: (4k3 + 28k2 + 24k + 47)2 = (4b3 + 28b2 + 24b+ 47)2; or equivalently, 4b3 + 28b2 + 24b+ 47 = (4k3 + 28k2 + 24k + 47): Firstly, since k is xed and 4b3+28b2+24b+47 is a degree 3 polynomial in terms of b, there can be at most three solutions for the equation 4b3 + 28b2 + 24b+ 47 (4k3 + 28k2 + 24k + 47) = 0 and three solutions for the equation 4b3 + 28b2 + 24b+ 47 + (4k3 + 28k2 + 24k + 47) = 0: Thus, there are at most six solutions for a given integer k. Recall that we have an in nite sequence of the integers b such that 4b3 + 28b2 + 24b + 47 is squarefree. So we can choose an in nite subsequence of this sequence such that the elds Kb have di¤erent discriminants (as we can have at most six duplications). Thus, we will have an in nite number of distinct elds Kb such that 4b3 + 28b2 + 24b+ 47 is squarefree. To conclude, if 4b3 + 28b2 + 24b + 47 is squarefree then the dihedral quintic eld Kb has the power basis f1; ; 2; 3; 4g where we have written for b. This concludes our proof. Remark 5.3. In comparison to the result found in [14], this theorem implies that there is not just one or nitely many monogenic dihedral elds of degree 5. In fact, there are in nitely many. 5.3. Further Results We will now show that the eld Kb possess more than one power basis. Claim 5.1. The eld Kb possesses at least two more power bases. 82 Proof. We will show that Kb has the power bases f1; 'i; '2i ; '3i ; '4i g for i = 1; 2 where '1 = b (b+ 1)2 + 3 4 and '2 = (2b+ 1) (b+ 2)2 + 23 4: Using MAPLETM and the argument of Example 3.1, we nd the minimal polynomials of '1 and '2 to be f1(x) = x 5 + x4 + (b+ 3)x3 + (b+ 4)x2 + 3x+ 1 and f2(x) = x 5 4bx4 + (6b2 2b 1)x3 + (4b3 + 6b2 + 4b+ 2)x2 +(b4 6b3 5b2 4b 2)x+ (2b4 + 2b3 + 2b2 + 2b+ 1), respectively. We can also use MAPLETM to show the discriminant of these polyno- mials is (4b3 + 28b2 + 24b+ 47)2. Since f1(x) and f2(x) are in Z[x] (as b 2 Z), '1; '2 2 OK and so D('1) = D(f1(x)) = (4b 3 + 28b2 + 24b+ 47)2 and D('2) = D(f2(x)) = (4b 3 + 28b2 + 24b+ 47)2: Thus D(f1(x)) = D(Fb(x)) and D(f2(x)) = D(Fb(x)) and hence are equal to d(Kb). We know that the minimal polynomials of '1 and '2 both have degree 5. Thus [Q('1) : Q] = 5 and [Q('2) : Q] = 5: Since Q('1) and Q('2) are both sub elds of Q() and [Q() : Q] = 5, we must have Q('1) = Q() = Q('2). Therefore, both '1 and '2 are generators of Kb. Thus, because '1 and '2 are generators of Kb; we have ind('1) = s D(f1(x)) d(Kb) = 1 83 and ind('2) = s D(f2(x)) d(Kb) = 1: So by Theorem 3.9, f1; 'i; '2i ; '3i ; '4i g for i = 1; 2 are power bases of Kb. Remark 5.4. We found '1 and '2 by using a similar method seen in Claim 5.2 (see below) with various values for b. By inspection, we were then able to build up the formulas for '1 and '2: Claim 5.2. The eld Kb possesses an additional eight power bases if b = 0: Proof. Now consider the special case when b = 0. Let 0 be a root of the polynomial F0(x) = x 5 2x4 + 2x3 x2 + 1: We know that f1; 0; 20; 30; 40g is a power basis of the eld K0 = Q(0) (as 4b3 + 28b2+24b+47 = 47 is squarefree). In order to nd additional power bases, we must set up and solve the index form equation. This is best suited for a computer software program so we just list the results. Firstly, we take a typical element in OK , say , and nd its minimal polynomial A(x). Since f1; 0; 20; 30; 40g is a power basis, we have = a+ b0 + c 2 0 + d 3 0 + e 4 0 where a; b; c; d; e 2 Z. As stated before, we could drop the "a" term in : If we choose not to, this term will disappear from the expression D(A(x)) yielding the same result. Using the argument of Example 3.1, we nd the minimal polynomial of . We then nd and factor the discriminant of A(x), denoted as D(A(x)) using the discrim() command. We know D(A(x)) = (d(K0)) (ind())2: From before, d(Kb) = (4b 3 + 28b2 + 24b+ 47)2 84 and so d(K0) = (47) 2: Thus, ind() = r D(A(x)) 472 which is an equation in the four variables b; c; d and e (as D(A(x)) no longer contains the variable a). Now we wish to nd solutions to the above index form equation by looping through small values of b; c and d and seeing if any yield integral values for e. We chose to loop through the integers in the interval [10; 10] and we got the following results: ind() = 1) 8>><>>: e = 2 e = 0 e = 2 and ind() = 1) 8>><>>: e = 1 e = 0 e = 1: Now using each value of e, we substitute e back into the corresponding index form equation (i.e. ind() = 1 or ind() = 1). We then loop through small values of b and c looking for integral values of d. Once again, we chose the interval [10; 10] to get: ind() = 1) 8>>>>>>><>>>>>>>: e = 2) d = 5 e = 0) 8>><>>: d = 1 d = 0 d = 1 e = 2) d = 5 85 and ind() = 1 =) 8>>>>>>>>>>>>><>>>>>>>>>>>>>: e = 1 =) ( d = 1 d = 2 e = 0 =) 8>><>>: d = 1 d = 0 d = 1 e = 1 =) ( d = 1 d = 2: Using the same procedure, we nd values for both b and c: We summarize these results as follows: ind() = 1) 8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>: e = 2) d = 5) c = 7) b = 6 e = 0) 8>>>>>>>>>><>>>>>>>>>>: d = 1) c = 1) ( b = 1 b = 0 d = 0) ( c = 1) b = 1 c = 1) b = 1 d = 1) c = 1) ( b = 0 b = 1 e = 2) d = 5) c = 7) b = 6 86 and ind() = 1 =) 8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>: e = 1 =) 8>>>>><>>>>>: d = 1) ( c = 1) b = 0 c = 0) b = 0 d = 2) c = 2) ( b = 1 b = 2 e = 0 =) 8>>>>>>>>>><>>>>>>>>>>: d = 1) ( c = 0) b = 1 c = 2) b = 1 d = 0) c = 0) ( b = 1 b = 1 d = 1) ( c = 2) b = 1 c = 0) b = 1 e = 1 =) 8>>>>><>>>>>: d = 1) ( c = 0) b = 0 c = 1) b = 0 d = 2:) c = 2) ( b = 2 b = 1: 87 Now we substitute each solution into (remembering that we can drop the "a" term). Disregarding those solutions that are the negation of another solution and the solutions '1 and '2 we obtain the following values of (letting = 0). '1 = 3 4; '2 = 2 22 + 23 4; '3 = + 3; '4 = 22 + 3; '5 = 6 72 + 53 24; '6 = 2 3; '7 = 2 + 3; '8 = 2: Hence, we have the additional eight power bases f1; 'i; '2i ; '3i ; '4i g for 1 i 8. Now the reader might wonder how we knew the Kondo family with a = 1 would yield monogenic elds. Why did we choose that speci c a? Is there a way of knowing a parametric eld will yield a power basis without going through the long process of solving the index form equation? To answer these questions, we illustrate using our family - the Kondo family. The rst step is to determine the Galois group. Kondo claimed his family of polynomi- als was dihedral under certain restrictions and we con rmed this already. Using MAPLETM, we now calculate and factor the polynomial discriminant of the Kondo family. We obtain a2(4b3 b2 + 30b2a+ 14ba 4a b2a2 + 34ba2 + 91a2 24a3b 40a3 4a4 + 4a5)2: Recall for a dihedral eld K the eld discriminant is of the form d(K) = f 4 d2 where d is the eld discriminant of the quadratic sub eld of the normal closure of K and f is called the conductor. Now in order for our eld to have a power basis, we 88 need a2(4b3b2+30b2a+14ba4ab2a2+34ba2+91a224a3b40a34a4+4a5)2 = f 4 d2 (that is, the index of a generator of K is equal to 1): Now from the above equation we see that we should let a = 1 in order to get rid of the a2 term out in front. Substituting a = 1 into this equation yields (4b3 + 28b2 + 24b+ 47)2 = f 4 d2: We now want f = 1 so that d = (4b3 + 28b2 + 24b + 47) (which is indeed true by [29]). But d is the eld discriminant of a quadratic eld Q( p m) where m is squarefree and hence must be one of the two forms: d = ( 4m if m is not of the form 4t+ 1 m if m = 4t+ 1: Now looking at d = (4b3+28b2+24b+47) we see that dmust be odd (as it is the sum of an even number and an odd number) and hence cannot be of the form 4m. Thus, d is of the form m and so d must be squarefree. We thus require 4b3+28b2+24b+47 to be squarefree. Thanks to Erdös, we know that there are in nitely many integers b such that 4b3+28b2+24b+47 is squarefree. Hence, we know we will have an in nite number of monogenic elds. As a note, a similar theorem as Erdösholds true for quadratics. However, for quartics, a parallel theorem has yet to be proved (although most suspect it to be true). We have discussed both cyclic and dihedral elds and the forms of the their eld discriminants. What happens if we nd a parametric family of polynomials and the Galois group is neither of these two? How do we then compare our polynomial discriminant to our eld discriminant if we do not know the structure of the eld discriminant? These cases can be done but require more experimentation. First determine the Galois group and polynomial discriminant as before and try to sim- plify the discriminant as much as possible by eliminating one of the parameters (for instance, if given the previous example, choosing a = 1 would help). Then using the 89 integral_basis() command in MAPLETM and trying values for the remaining parame- ter, see which ones yield a power basis. Hopefully, there will be in nitely many values that do. As before, this experimentation helps con rm your suspicions that you do have a parametric family yielding monogenic elds but is in no way a substitution of a formal proof. 90 Chapter 6 Number Fields without a Power Basis Because power bases are so rare, there are plenty of algebraic number elds that do not have a power basis (i.e. not monogenic). Now we have seen that an algebraic number eld K is not monogenic if the index form of all generators of K cannot equal 1. But this can be a di¢ cult condition to use for elds of degree six or more as the number of variables increase as the degree of the eld increases. Thus, we wish to avoid solving this equation and rather nd certain properties of the eld that allow us to conclude the eld is not monogenic. 6.1. Ideals We begin with a few basic de nitions and theorems. De nition 6.1. Let K be an algebraic number eld and OK its ring of integers. An ideal I of OK is a subring of OK with the property 8a 2 OK, aI I: An ideal I of OK is called a proper ideal of OK if f0g I OK. So we say an ideal I of OK is closed under external multiplication by elements of OK : De nition 6.2. A proper ideal I of OK is called a prime ideal if a; b 2 OK and ab 2 I ) a 2 I or b 2 I: In OK , every ideal I of OK is nitely generated. Hence I = h1; : : : ; ki 91 for some 1; : : : ; k 2 OK . In fact, for ideals of OK ; every ideal can be generated by two of its elements (this is known as the two-generator property). Ideals that can be generated by only one element are known as principal ideals. Now we know that any integer can be written as a unique factorization into primes (Fundamental Theorem of Arithmetic). However, this property does not always hold when working over OK - the ring of integers. In truth, it is rare for this property to hold over OK for some algebraic number eld K. Here is an example. Example 6.1. Let K = Q( p 5i): Clearly, 1p5i 2 K. Then 1p5i 2 OK as their minimal polynomial is x22x+6 2 Z[x]. We will show that 1 + p 5i is an irreducible over OK. By way of contradiction, assume 1 + p 5i is reducible over OK. Since K is degree 2, N(1 + p 5i) = (1 + p 5i)(1 p 5i) = 6. If 1+ p 5i is reducible then its nontrivial factors would have norms either 2 or 3. We will show there are no such elements of OK with such norms. By Theorem 3.1, OK = Z+ Z p5 = fa+ bp5 j a; b 2 Zg: Let = a+ b p5 be a typical element of OK (where a; b 2 Z). Then N() = (a+ b p5)(a bp5) = a2 + 5b2: If a2 + 5b2 = 2 then b = 0 and so a2 = 2, a contradiction as a 2 Z. If a2 + 5b2 = 3 then b = 0 and so a2 = 3, a contradiction as a 2 Z. Thus 1 +p5i is irreducible over OK. A similar argument can be used to show 1 p 5i is irreducible over OK : Now 6 = 2 3 = (1 + p 5i) (1 p 5i): So there exists two factorizations of 6 2 OK into irreducibles. The above equation implies 2 j (1+p5i) (1p5i). We now show that 2 is not prime by a contradiction argument. By way of contradiction, assume 2 j (1 +p5i). Then taking the norm of each side yields N(2) j N(1 + p 5i), 4 j 6; 92 a contradiction. Similarly, 2 - (1p5i). Thus, 2 is not prime. A similar argument can be made for 3. Thus, these factorizations are not into primes but irreducibles. When working with OK , we would like to have a unique factorization property that holds for any element in OK . This is the reason for working with ideals. This leads us to a very important theorem. Theorem 6.1. Let K be an algebraic number eld. Then every proper ideal of OK can be expressed uniquely up to order as a product of prime ideals. Remark 6.1. The proof of this theorem is found on pp. 200-202, [2]. Factoring ideals into prime ideals proves a di¢ cult problem. In [1], the authors give a method to nd the explicit decomposition of the principal ideal hpi in a cubic eld, where p is a prime. Rather than go into the details of this method, we list two major consequences that the reader should be aware of. (1) The prime numbers p which divide the eld discriminant of K are precisely those primes for which the prime ideal factorization of hpi contains at least one prime ideal divisor to an exponent greater than 1. (2) Certain prime ideals give rise to common index divisors. The latter consequence is seen in the following section. 6.2. Common Index Divisors One reason for an algebraic number eld K not to have a power basis is because the index of every element in OK is divisible by some xed prime p. Recall the index form equation ind() = s D(A) d(K) = 1 where is a typical element in OK . If there exists a xed prime p j ind() for all 2 OK then ind() 6= 1 (as p - 1). This implies that K has no power bases. Such a prime p is called a common index divisor. The common index divisors of an algebraic number eld K are precisely the factors of the eld index i(K). It is a well known result that p < n where n is the degree of the algebraic number eld K (see pp. 186, [2]). Thus, to be a common index divisor of a cubic eld, p must be 2. 93 We will do an example illustrating a method to nd a common index divisor when there exists one. Example 6.2. Let f(x) = x3 x2 2x 8 and let be a root of f(x). Let K = Q(). Using MAPLETM, we see that K is a cubic eld with Galois group S3. Next we nd an integral basis for K: Using MAPLETM, we get 1; ; 2 + 2 : Now we let be a typical element in OK: = a+ b + c( 2 + 2 ) where a; b; c 2 Z. We wish to nd the minimal polynomial A(x) of and nd its polynomial discriminant (see Example 3.1). Its discriminant (in factored form) is D(A) = 503(2c3 + 3bc2 + 5b2c+ 2b3)2: Next we need to nd the eld discriminant of K using the element : Using a similar argument as in Example 3.3 we get ind() = 2 and D(f) = 2012 where f(x) is the minimal polynomial of . This implies d(K) = D(f) ind()2 = 2012 22 = 503. Thus, ind() = s D(A) d(K) = 2c3 + 3bc2 + 5b2c+ 2b3 : But 2c3 + 3bc2 + 5b2c+ 2b3 bc2 + b2c 0 mod 2: Thus, 2 j ind() and so 2 is a common index divisor. Thus, K does not possess a power basis. 94 Remark 6.2. Firstly, the reader must not get the algebraic integers and confused. is a speci c element of OK such that K = Q() and represents a typical element of OK. Secondly, arriving to the conclusion bc2 + b2c 0 mod 2 is not so obvious. We see that bc2+ b2c = bc(c+ b). Now an argument by cases will su¢ ce. If either b or c is even, then bc2 + b2c 0 mod 2: So we need only consider the case when b and c are both odd. Then b + c is even and so bc(c + b) 0 mod 2 and the result follows. From the above example, we can nd that 2c3 + 3bc2 + 5b2c+ 2b3 = 2 for (b; c) = (1; 0) so that m(K) = min 2c3 + 3bc2 + 5b2c+ 2b3 : 2c3 + 3bc2 + 5b2c+ 2b3 6= 0 and b; c 2 Z = 2 and i(K) = gcd 2c3 + 3bc2 + 5b2c+ 2b3 : 2c3 + 3bc2 + 5b2c+ 2b3 6= 0 and b; c 2 Z = 2: As i(K) > 1, K does not possess a power basis. We also see that the prime factor of i(K) was indeed the common index divisor of K. 95 Chapter 7 Conclusion and Future Work 7.1. Future Research We rst introduce the following two de nitions. De nition 7.1. A polynomial is called homogeneous if all its terms have the same degree. Example 7.1. Let f(x) = x6 + 2x3y3 + 9x2y4. Then f(x) is a homogeneous polynomial of degree 6, in 2 variables; the sum of the exponents in each term is always 6. De nition 7.2. Let F 2 Z[X; Y ] (i.e. F has two variables X and Y ): Also, let F be an irreducible homogeneous polynomial of degree n 3. Let m be a non-zero integer. Then the following equation is known as a Thue equation: F (X; Y ) = m in X; Y 2 Z. Remark 7.1. We can assume the leading coe¢ cient of F is 1; otherwise, if the leading coe¢ cient of F is "a", we multiply the equation by an1, replace either (aX; Y ) or (X; aY ) (depending on which variable is the leading term) with (X; Y ) and get the required condition. For example, consider the Thue equation 3x6 + 2x3y3 + 9x2y4 = 7 then multiply both sides by 35 to get 36x6 + 2 35x3y3 + 359x2y4 = 357: Now we replace (3x; y) with (X;Y ) to get X6 + 2 32X3Y 3 + 339X2Y 4 = 357; an equivalent Thue equation with leading coe¢ cient equal to 1. 96 In 1909, Axel Thue rst proved that this equation has only nitely many integer solutions ([31]). However, Thues method did not enable one to nd the solutions. In 1989, Nikos Tzanakis and Benne de Weger gave a practical method for solving a Thue equation by bounding the size of the solution ([32]). The main tool of their method is the lattice reduction algorithm (LLL-algorithm). Other algorithms are also available to help one solve a Thue equations (see [24] and [25]). In order to determine the monogenicity of a given eld, one must rst solve an index form equation. Oftentimes, in cubic and quartic elds, these equations can be reduced to simpler types of Diophantine equations, in particular, Thue equations. If the eld given is indeed monogenic, with some di¢ culty, we are then able to nd a complete list of the power bases of this eld using those algorithms mentioned above. However, one must keep in mind, that although these algorithms exist to solve Thue equations, it does not mean they are "good", in terms of practicality and e¢ ciency (see [26]). For quintic elds and some elds of higher degree, methods exist for resolving the index form equation into other Diophantine equations called unit equations ([13]). Depending on the Galois group, such methods can be very time consuming. However, they allow us to nd complete solutions to the index form equation. In [13], the author states that the number of power bases of a given eld is small. We suspect that the list of power bases given in Claim 5.2 is complete. For future research, one might try to prove that this is indeed true. In order to do so, one would have to use those resolution methods discussed above to reduce the index form equation into simpler types of Diophantine equations. Once reduced, solving these equations, in principle, is possible. In this paper, we have shown an in nite family of dihedral quintic elds are monogenic. For future research, one might continue exhibiting in nite families of monogenic elds of given Galois group and degree. As mentioned in the introduction, it is not known if there are in nitely many monogenic cyclic quartic elds. The research potential is great in this area of mathematics and the study of power bases is a long term project with considerable scope for sub-projects. 97 7.2. Conclusion After building the necessary groundwork, we showed that there are in nitely many dihedral quintic elds that are monogenic and thus, not just one or nitely many; this result was based on a parametric family of quintic polynomials studied by A. Brumer and T. Kondo, independently. Much work was needed to prove this result, however, experimentation with MAPLETM con rmed our conjecture. We left the reader with a challenging problem to nd a complete list of power bases for the case b = 0. As an introduction, we have discussed various elds of given degree and Galois group that also possess power bases; our result is a continuum of these studies. There is much research that can be done in this area of mathematics. Finding parametric families that yield monogenic elds are hard to come by and constructing your own family proves even more di¢ cult. However, there is an abundant number of elds of higher degree and given Galois group unexplored and so the research potential is great. For instance, our focus was on quintic elds of which there are only ve possible Galois groups. For octic elds (degree 8) there are fty possible Galois groups, much of which have not been analyzed; with the increase in technology, the ability to study higher degrees becomes possible and seems less daunting. May the reader be encouraged to study power bases for themselves and delve into this great subject. 98 References [1] S. Alaca, B. K. Spearman and K. S. Williams. Explicit Decomposition of a Ratio- nal Prime in a Cubic Field. International J. of Math. and Math. Sciences (2006), Vol. 2006, Article ID 17641, pp. 1-11. [2] S. Alaca and K. S. Williams. Introductory Algebraic Number Theory. United Kingdom: Cambridge Univ. Press, 2004. [3] A. Baker. Notes for 4H Galois Theory 2003-2004. Home page. U. of Glasgow. 19 June. 2007 <http://www.maths.gla.ac.uk/~ajb>. [4] A. Brumer, preprint. [5] R. Chapman. Algebraic Number Theory - Summary of Notes. Home page. U. of Exeter. 10 Sept. 2007 <http://www.secamlocal.ex.ac.uk/people/sta¤/rjchapman/rjc.html>. [6] H. Cohen. A Course in Computational Algebraic Number Theory. Germany: SpringerVerlag, 1993. [7] R. Dedekind. Über den Zusammenhang zwischen der Theorie der Ideale und der Theorie der höheren Kongruenzen. Abh. Kgl. Ges. Wiss. Göttingen (1878), Vol. 23, pp. 1-23. [8] D. S. Dummit. Solving Solvable Quintics. Math. of Computation (Jul., 1991), Vol. 57, No. 195, pp. 387-401. [9] D. S. Dummit and R. M. Foote. Abstract Algebra. 3rd ed., John Wiley & Sons Inc., 2004. [10] D. S. Dummit and H. Kisilevsky. Indices in Cyclic Cubic Fields. Number Theory and Algebra, Academic Press (1977), pp. 29-42. [11] P. Erdös. Arithmetic Properties of Polynomials. J. London Math. Soc. (1953) Vol. 28, pp. 416-425. 99 [12] T. Funakura. On Integral Bases of Pure Quartic Fields. Math. J. of Okayama Univ. (1984), Vol. 26, pp. 27-41. [13] I. Gaál. Diophantine Equations and Power Integral Bases. Boston: Birkhäuser, 2002. [14] M.-N. Gras. Non Monogénéité de Lanneau des Entiers des Extensions Cycliques de Q de Degré Premier l 5. J. Number Theory (1986), Vol. 23, pp. 347-353. [15] J. G. Huard, B. K. Spearman and K. S. Williams. Integral Bases for Quartic Fields with Quadratic Sub elds. J. Number Theory (1995), Vol. 51, pp. 87-102. [16] L.-C. Kappe and B. Warren. An Elementary Test for the Galois Group of a Quartic Polynomial. American Math. Monthly (Feb., 1989), Vol. 96, No. 2, pp. 133-137. [17] T. Kondo. Some Examples of Unrami ed Extensions over Quadratic Fields. Sci. Rep. Tokyo Womans Christian Univ. (1997), No. 120-121, pp. 1399-1410 [18] M. J. Lavallee, B. K. Spearman, K. S. Williams and Q. Yang. Dihedral Quintic elds with a power basis. Math. J. of Okayama Univ. (2005), Vol. 47, pp. 7579. [19] D. A. Marcus. Number Fields. 3rd ed., Springer-Verlag, 1977. [20] A. R. Marshall. Precalculus: Functions and Graphs. 1st ed., Addison-Wesley Pub. Co., 1990. [21] D. C. Mayer. Multiplicities of Dihedral Discriminants. Math. of Computation (April, 1992), Vol. 58, No. 198, pp. 831-847. [22] T. Nakahara. On the Indices and Integral Bases of Non-cyclic but Abelian Bi- quadratic Fields. Arch. Math (1983), Vol. 41, pp. 504-508. [23] W. Narkiewicz. Elementary and Analytic Theory of Algebraic Numbers. 3rd ed., New York: SpringerVerlag, 2004. [24] A. Pethö. Algebraische Algorithmen. Vieweg Verlag (1999), Braunschweig- Wiesbaden. [25] N. P. Smart. The Algorithmic Resolution of Diophantine Equations. London Math. Soc. (1998), Student Texts 41, Cambridge Univ. Press. [26] N. P. Smart.How Di¢ cult is it to Solve a Thue Equation? Lecture Notes in Com- puter Science - Algorithmic Number Theory (1996), Berlin/Heidelberg: Springer, Vol. 1122, pp. 363-373. 100 [27] B. K. Spearman and K. S. Williams. On a Procedure for Finding the Galois Group of a Quintic Polynomial. Sci. Math. Jpn. 53 (2001), No. 2, pp. 323-333. [28] B. K. Spearman and K. S. Williams. Cubic Fields with a Power Basis. Rocky Mountain J. (2001), Vol. 31, pp. 1103-1109. [29] B. K. Spearman, K. S. Williams and Q. Yang. The 2-Power Degree Sub elds of the Splitting Fields of Polynomials with Frobenius Galois Groups. Comm. Algebra (2003), Vol. 31, pp. 4745-4763. [30] I. Stewart and D. Tall. Algebraic Number Theory and Fermats Last Theorem. 3rd ed., Massachusetts: A K Peters, 2002. [31] A. Thue. Über Annäherungswerte algebraischer Zahlen. J. Reine Angew Math (1909), 135, pp. 284-305. [32] N. Tzanakis and B. de Weger. On the Practical Solution of the Thue Equation. J. Number Theory (1989), Vol. 31, pp. 99-132. [33] -. Galois Theory. Wikipedia. 19 June. 2007 <http://en.wikipedia.org/wiki/Galois_theory>. [34] -. Cryptography. Wikipedia. 29 August. 2008 <http://en.wikipedia.org/wiki/Cryptography>. [35] C. J. Williamson. Odd Degree Polynomials with Dihedral Galois Groups. Diss. U of California, Berkeley, 1989. 101
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