Minimal Indices of Pure Cubic FieldsbyJeewon YooB.Sc., The University of British Columbia, 2014A THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFMaster of ScienceinTHE COLLEGE OF GRADUATE STUDIES(Mathematics)The University of British Columbia(Okanagan)May 2016c© Jeewon Yoo, 2016 The undersigned certify that they have read, and recommend to the College of Graduate Studies for acceptance, a thesis entitled: Minimal Indices of Pure Cubic Fields submitted by Jeewon Yoo in partial fulfilment of the requirements of the degree of Master of Science in Mathematics. Blair Spearman, Faculty of Mathematics, UBC Supervisor, Professor (please print name and faculty/school above the line) Qiduan Yang, Faculty of Mathematics, UBC Supervisory Committee Member, Professor (please print name and faculty/school in the line above) Edward Butz, Faculty of Mathematics, UBC Supervisory Committee Member, Professor (please print name and faculty/school in the line above) Jim Robinson, Faculty of Philosophy, UBC University Examiner, Professor (please print name and faculty/school in the line above) Bernard Bauer, Faculty of Earth and Environmental Sciences, UBC External Examiner, Professor (please print name and university in the line above) May 26, 2016 (Date Submitted to Grad Studies) ii AbstractDetermining whether a number field admits a power integral basis is a classicalproblem in algebraic number theory. It is well known that every quadratic numberfield is monogenic, that is they admit power bases. However, when we talk aboutcubic or higher degree number fields we may discover fields without power integralbases. In 1878, Dedekind gave the first example of a cubic field without a powerintegral basis. It is known that a number field is monogenic if and only if theminimal index is one. In 1937, Hall proved that the minimal index of pure cubicfields can be arbitrarily large. We extend this result by showing that the minimalindex of a family of infinitely many pure cubic fields have an element of index nbut no element of index less than n for a positive integer n.iiiPrefaceThe main result in this thesis is written in the paper [16] accepted on August 30,2015. All authors of [16] contributed equally.ivContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivContents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiList of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ixAcknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x1 Introduction and Historical Remarks . . . . . . . . . . . . . . . . . 12 Number Theory Preliminaries . . . . . . . . . . . . . . . . . . . . . 32.1 Elementary Number Theory . . . . . . . . . . . . . . . . . . . . 32.2 Abstract Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.1 Group Theory . . . . . . . . . . . . . . . . . . . . . . . . 82.2.2 Ring Theory and Fields . . . . . . . . . . . . . . . . . . . 153 Algebraic Number Theory . . . . . . . . . . . . . . . . . . . . . . . 173.1 Algebraic Number Fields . . . . . . . . . . . . . . . . . . . . . . 173.2 The Set OK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Discriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22v3.4 Integral Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.5 Index Forms and Minimal Indices . . . . . . . . . . . . . . . . . 314 Galois Groups and Chebotarev Density Theorem . . . . . . . . . . . 364.1 Galois Groups of Cubics . . . . . . . . . . . . . . . . . . . . . . 364.2 The Chebotarev Density Theorem . . . . . . . . . . . . . . . . . 455 Minimal Indices in Pure Cubic Fields . . . . . . . . . . . . . . . . . 525.1 Pure Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 525.2 Computing Index Forms . . . . . . . . . . . . . . . . . . . . . . 555.3 Hall’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 585.4 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 Conclusion and Future Work . . . . . . . . . . . . . . . . . . . . . . 666.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 666.2 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70viList of TablesTable 2.1 The values of φ(n) for 1≤ n≤ 12 . . . . . . . . . . . . . . . . 5Table 2.2 The cubic residues mod 11 . . . . . . . . . . . . . . . . . . . 6Table 2.3 The cubic residues mod 13 . . . . . . . . . . . . . . . . . . . 6Table 2.4 C2 : a set of cubic non-residues mod 13 . . . . . . . . . . . . . 7Table 2.5 C3 : a set of cubic non-residues mod 13 . . . . . . . . . . . . . 7Table 2.6 Group table for Z3 under addition . . . . . . . . . . . . . . . . 8Table 2.7 Elements of S3 . . . . . . . . . . . . . . . . . . . . . . . . . . 11Table 2.8 Group table for S3 . . . . . . . . . . . . . . . . . . . . . . . . 12Table 3.1 Computing Integral Bases . . . . . . . . . . . . . . . . . . . . 26Table 3.2 Examining α∗ ∈ OK . . . . . . . . . . . . . . . . . . . . . . . 28Table 3.3 Examining θ ∗ ∈ OK . . . . . . . . . . . . . . . . . . . . . . . 29Table 3.4 Examining θ ∗ ∈ OK . . . . . . . . . . . . . . . . . . . . . . . 30Table 4.1 The elements in the Galois group of S3 . . . . . . . . . . . . . 43Table 4.2 Cycle types in S3 . . . . . . . . . . . . . . . . . . . . . . . . . 46Table 4.3 Polynomial factorization over Fp . . . . . . . . . . . . . . . . 46Table 4.4 Cycle types in D4 . . . . . . . . . . . . . . . . . . . . . . . . 48Table 4.5 Polynomial factorization over Qp . . . . . . . . . . . . . . . . 48viiList of FiguresFigure 2.1 The lattice of subgroups diagram of S3 . . . . . . . . . . . . . 13Figure 4.1 The lattice of subfields diagram of Q(α,ω) :Q . . . . . . . . 44Figure 4.2 The lattice of subgroups diagram of Gal(Q(α,ω) :Q) . . . . 44viiiList of SymbolsN Set of natural numbersZ Set of integersQ Field of rational numbersR Field of real numbersC Field of complex numbersφ(n) Euler φ functionSn Symmetric group of order nZ+Zi Set of Gaussian integersZ[x] Set of polynomials with integer coefficientsQ[x] Set of polynomials with rational coefficients〈 f (x)〉 Ideal generated by the polynomial f (x)[K :Q] Degree of K over Qdisc( f (x)) Discriminant of the polynomial f (x)D(α) Discriminant of the element α in an algebraic number fieldd(K) Field discriminantN(α) Norm of the element α in an algebraic number field KTr(α) Trace of the element α in an algebraic number field Kindα The index of an element αi(K) Index of an algebraic number field Km(K) Minimal index of an algebraic number field KGal(K :Q) Galois group of K over QGal( f (x)) Galois group of the polynomial f (x)ixAcknowledgmentsI am a blessed one to meet my supervisor Dr. Blair Spearman. Without his support,I would not be able to accomplish this work and complete my graduate studies as aMaster’s student. Thanks for replying to my email and arranged a meeting with mefour years ago when I was afraid of taking first step towards a mathematics student.My life changed gradually since our very first meeting.Thanks to Dr. Qiduan Yang for teaching me the core courses that are indispens-able to this achievement. I really appreciate your valuable comments and advicesthroughout my studies.Thanks to my colleagues Chad Davies, Lindsey Reinholz, and Paul Lee forhelping me throughout my undergraduate and graduate life. Your advices and ex-periences became a valuable model to follow.Thanks to Cindy Bourne, the manger in the Math and Science Centre, for giv-ing me opportunities to work as a tutor and as a SL leader. I learned the respon-sibility as a member of an organization and as a leader of students. Thanks forproofreading this paper.Also, I would like to thank Mr. and Mrs. Mitsouras for treating me as yourreal grandchild although I am from totally different nation. Your life experiencesand advices helped me a lot when I was left all alone in Canada since when I wasa young kid.xLastly, I would like to give a special thanks to my parents and my sister for theendless support and love.xiChapter 1Introduction and HistoricalRemarksThree positive integers a,b, and c do not satisfy the equationan+bn = cn for any integer value of n greater than two.— Pierre de Fermat (1637)Algebraic number theory has been widely studied since 500 BC when the Pythagoreantheorem was first introduced. It was developed in two different ways. One for theFermat equations, and the other for class field theory. In either way, we have thesame purpose: solving Diophantine equations. A Diophantine equation (named af-ter Diophantus of Alexandria) is a polynomial equation in two or more unknowns,such that only the integer solutions are studied. One of the easiest Diophantineequations we have isX2+Y 2 = Z2which is related to the Pythagorean theorem. Infinitely many integral solutionshave been found for this equation such as(3,4,5),(5,12,13),(8,15,17), ...which we call them Pythagorean triples. One of the most famous and interesting1Diophantine equations in the history of mathematics isxn+ yn = znwhere n is a positive integer. Pierre de Fermat claimed that there are no integralsolutions to the Diophantine equation above when n ≥ 3. This is called the “Fer-mat’s Last Theorem.” This theorem was first conjectured in 1637. Fermat claimedthat he had a proof, but he did not show it to the public. This problem was left un-solved for more than 350 years until the first successful proof was released in 1994by Andrew Wiles. Hence Fermat’s Last Theorem shows that solving a Diophan-tine equation can be extremely difficult. The best possible situation for solvingDiophantine equations is when we work over a unique factorization domain. Thecomplexity of calculation is simpler than the equations without unique factoriza-tion domains. However, this only means that the calculation is simpler than theother; it still can be extremely difficult.We say an algebraic number field is a monogenic field if it posesses a powerintegral basis, denoted by {1,θ ,θ 2, ...,θ n−1}, where θ is a root of a minimal poly-nomial of degree n. It is well known that every number field has an integral basis.The next question then might be “how can we decide whether a field is mono-genic?” It is known that a field is monogenic if and only if the absolute value of theindex |I| = 1 is solvable, where I denotes the index form of the field. Using this,one can get a measure of how far away from being monogenic the number field is.Also, it is proven that the values of |I| can be arbitrarily large. In [1], Hall provedthat the minimal index of pure cubic fields can be arbitrarily large. The main aimof this thesis is to develop Hall’s result so that we may construct an infinite familyof pure cubic fields with an element of index equal to a particular positive integern and we show the impossibility of the index being equal to any positive integersless than n.2Chapter 2Number Theory PreliminariesIn this chapter, we review some of the basic definitions and theorems in elementarynumber theory and abstract algebra courses. In particular, we focus on primes, con-gruences, and cubic residues that are building blocks to the main result in chapter5.2.1 Elementary Number TheoryDefinition 2.1.1. If a and b are integers with a 6= 0, we say that a divides b if thereis an integer c such that b = ac, and we write a | b. If a does not divide b, we writea - b.Definition 2.1.2. (Greatest Common Divisor) The greatest common divisor of twointegers a and b, which are not both 0, is the largest positive integer that dividesboth a and b. It is written as gcd(a,b).Definition 2.1.3. (Relatively Prime) The integers a and b, with a,b 6= 0, are rela-tively prime if gcd(a,b) = 1.Theorem 2.1.1. (Euclid) There are infinitely many primes.Theorem 2.1.2. (Dirichlet’s Theorem on Primes in Arithmetic Progressions) Sup-pose that a and b are relatively prime positive integers. Then the arithmetic pro-gression an+b, n = 1,2,3....., contains infinitely many positive primes.3Definition 2.1.4. Let m be a positive integer. If a and b are integers, we say that ais congruent to b modulo m, denoted a ≡ b (mod m) if m | (a−b).Example 2.1.1. We have 5 ≡ 2 (mod 3), since 3 | (5− 2). However, 8 6≡ 2(mod 5) since 5 - (8−2).Theorem 2.1.3. If a,b,c, and m are integers, with m> 0, such that a ≡ b (mod m),then(i) a+ c ≡ b+ c (mod m),(ii) a− c ≡ b− c (mod m),(iii) ac ≡ bc (mod m).Theorem 2.1.4. If a ≡ b (mod m1),a ≡ b (mod m2), ...,a ≡ b (mod mk),where a,b,m1,m2, ...,mk are integers with m1,m2, ...,mk positive, thena ≡ b (mod [m1,m2, ...,mk]),where [m1,m2, ...mk] denotes the least common multiple of m1,m2, ...,mk.Theorem 2.1.5. (The Chinese Remainder Theorem) Let m1,m2, ...,mr be pairwiserelative prime positive integers. Then the system of congruencesx ≡ a1 (mod m1)x ≡ a2 (mod m2)...x ≡ ar (mod mr)has a unique solution modulo M = m1m2 · · ·mr.Example 2.1.2. We solve the following system of congruences.x ≡ 1 (mod 3)x ≡ 2 (mod 5)x ≡ 3 (mod 7)4Let M = 3 ·5 ·7= 105, and M1 =M/3= 35,M2 =M/5= 21, and M3 =M/7= 15.Now, we find y1,y2, and y3 such thatx≡ y1M1+2y2M2+3y3M3 (mod 3 ·5 ·7).To do this, we set up a congruence like following. M1y1 = 35y1 ≡ 1 (mod 3).Then we get 2y1 ≡ 1 (mod 3) by the replacement principle. Then, it is easy to seethat y1 ≡ 2 (mod 3). Similarly, y2 ≡ 1 (mod 5), and y3 ≡ 1 (mod 7). Hence,x ≡ 1 ·35 ·2+2 ·21 ·1+3 ·15 ·1 ≡ 157 ≡ 52 (mod 3 ·5 ·7).Lastly, we check if 52 satisfies the desired congruences. Indeed, 52≡ 1 (mod 3),52≡2 (mod 5), and 52 ≡ 3 (mod 7). Therefore, we found the desired solution to thesystem of congruences above.Definition 2.1.5. (Euler φ function) Let n be a positive integer. The Euler phi-function φ(n) is defined to be the number of positive integers not exceeding n thatare relatively prime to n.Example 2.1.3. Consider φ(n) for 1≤ n≤ 12.Table 2.1: The values of φ(n) for 1≤ n≤ 12n 1 2 3 4 5 6 7 8 9 10 11 12φ(n) 1 1 2 2 4 2 6 4 6 4 10 4Theorem 2.1.6. (Euler’s Theorem) If m is a positive integer and a is an integerwith (a,m) = 1, thenaφ(m) ≡ 1 (mod m).We proceed to the notion of cubic residues since this concept will be needed toprove Hall’s results [1].Definition 2.1.6. Let p be a prime and let q be an integer not divisible by p. Ifthere is an integer x such that x3 ≡ q (mod p), then q is said to be a cubic residuemodulo p. If not, q is said to be a cubic non residue modulo p.5Definition 2.1.7. Two integers a,b are equivalent mod p ifax3 ≡ b (mod p)⇐⇒ y3 ≡ a2b (mod p)is solvable.Theorem 2.1.7. Let p be a prime with p ≡ 2 (mod 3). Then every integer is acube modulo p.Proof. Suppose p ≡ 2 (mod 3). Set p = 3e+2 for some e ∈ Z. Clearly, p ≡ 0(mod p). Consider the integers x, where x ∈ {1, ..., p−1}.By Theorem 2.1.6,xp−1 = x3e+1 ≡ 1 (mod p).Thus,x = 1 · x ≡ x3e+1x3e+2 ≡ x6e+3 ≡ (x2e+1)3 (mod p)This shows that every integer is a cube mod p.The following example illustrates Theorem 2.1.7.Example 2.1.4. Consider p = 11 ≡ 2 (mod 3).Table 2.2: The cubic residues mod 11x 1 2 3 4 5 6 7 8 9 10x3 (mod 11) 1 8 5 9 4 7 2 6 3 10But, choosing p ≡ 1 (mod 3) gives totally different result.Example 2.1.5. Consider p = 13 ≡ 1 (mod 3).Table 2.3: The cubic residues mod 13x 1 2 3 4 5 6 7 8 9 10 11 12x3 (mod 13) 1 8 1 12 8 8 5 5 1 12 5 126We only obtained four integers mod 13 that are cubes mod 13. Let C1 be theset of integers mod 13 that are cubes mod 13. In this case,C1 = {1,5,8,12}Define the set of integers cubic non residues mod 13 byCNR = {2,3,4,6,7,8,10,11}.CNR can be further divided into two sets C2 and C3. To do this, pick any elementin CNR, say 2. Then, calculate 2x3 (mod 13).Table 2.4: C2 : a set of cubic non-residues mod 13x 1 2 3 4 5 6 7 8 9 10 11 122x3 (mod 13) 2 3 2 11 3 3 10 10 2 11 10 11Then setC2 = {2,3,10,11}We define C3, pick any element in CNR excluding C1 and C2, say 4. Then,calculate 4x3 (mod 13).Table 2.5: C3 : a set of cubic non-residues mod 13x 1 2 3 4 5 6 7 8 9 10 11 124x3 (mod 13) 4 6 4 9 6 6 7 7 4 9 7 9By Table 2.5, we find thatC3 = {4,6,7,9}Observe that C1∪C2∪C3 gives every integer mod 13. By Definition 2.1.7, if we7pick elements a,b where a ∈Ci and b ∈C j where i 6= j, then the modular equationax3 ≡ b (mod p)is insolvable. Hall uses this argument in his main result, which will be discussedin Chapter 6.2.2 Abstract AlgebraIn this section, we introduce basic definitions and theorems from abstract algebra.2.2.1 Group TheoryDefinition 2.2.1. (Group) A group 〈G,∗〉 is a set G, closed under a binary opera-tion ∗, such that the following axioms are satisfied:For all a,b,c ∈ G, we have1. (a∗b)∗ c = a∗ (b∗ c) associativity of ∗2. there exists e ∈ G, such that for all x ∈ G, e∗ x = x∗ e = x identity element e3. For each a ∈ G, there exists a′ ∈ G such that a∗a′ = a′ ∗a = e inverse a′Example 2.2.1. The integers modulo 3, denoted Z3 form a group under addition,which is readily verified. We can describe all the elements in this group via agroup table as follows. Later, we see that more complicated groups admit moreTable 2.6: Group table for Z3 under addition+ 0 1 20 0 1 21 1 2 02 2 0 1complicated group tables.Definition 2.2.2. (Subgroup) A subset H of a group G, (denote H ≤ G) is a sub-group of G if and only if1. H is closed under the binary operation of G,82. the identity element e of G is in H.3. for all a ∈ H it is true that a−1 ∈ H also.Definition 2.2.3. A group (G,∗) is called abelian (or commutative) if a∗b = b∗afor all a,b ∈ G.Definition 2.2.4. Let G be a group. The order of G is its cardinality. That is, thenumber of elements in G. The order of an element a ∈ G, is the smallest positiveinteger m such that am = e, where e denotes the identity element in G. If no such mexists, then a is said to have infinite order.Definition 2.2.5. A permutation of a set X is a function φ : X → X that is both oneto one and onto.Definition 2.2.6. Let X be a set. SX := { f : X 7→X | f is a bijection}. Multiplicationis compositionSX ×SX 7→ SX( f ,g) 7→ g◦ f .In case X = {1,2, ...,n} for some n∈N,write Sn for SX (called a symmetric group).If G≤ Sn for some n ∈ N, G is a permutation group of degree n.Note that the order of Sn is n! = n · (n−1) · · ·2 ·1.Definition 2.2.7. A given binary relation ∼ on a set X is said to be an equivalencerelation if and only if it is reflexive, symmetric and transitive. That is, for all a,band c in X ,• a∼ b⇒ b∼ a. (Reflexivity)• a∼ b. (Symmetry)• If a∼ b and b∼ c then a∼ c. (Transitivity)Lemma 2.2.1. Let σ be a permutation of a set A. For a,b ∈ A, we have a ∼ b ifand only if b= σn(a) for some integer n. Here, ∼ denotes an equivalence relation.Definition 2.2.8. Let σ be a permutation of a set A. The equivalence classes in Adetermined by the equivalence relation in Definition 2.2.7 are the orbits of σ .9Definition 2.2.9. A permutation σ ∈ Sn is a cycle if it has at most one orbit con-taining more than one element. The length of a cycle is the number of elements inits largest orbit.Theorem 2.2.2. Every permutation σ of a finite set is a product of disjoing cycles.Definition 2.2.10. A cycle of length 2 is a transposition.Example 2.2.2. Consider the following permutation in S8.σ =(1 2 3 4 5 6 7 83 8 6 7 4 1 5 2)First, we find the orbit containing 1. Applying σ repeatedly, we see that1→ 3→ 6→ 1→ 3→ 6→ 1→ ··· .This shows the orbit containing 1 is {1,3,6}. We now choose an integer from 1to 8 not in the set {1,3,6}, say 2, and similarly find that the orbit containing 2is {2,8}. Finally, we find that the orbit containing 4 is {4,7,5}. Since these threeorbits include all integers from 1 to 8, we see that the complete list of orbits of σ is{1,3,6}, {2,8}, {4,5,7}.Notice that the permutations corresponding to these three sets are cycles sincethey contain more than one element and we write these cycles as (1 3 6),(2 8) and(4 5 7) respectively. Since σ is a permutation in the finite set S8, Theorem 2.2.1shows that it can be written as a product of disjoint cycles:σ = (1 3 6)(2 8)(4 7 5).Clearly, {1,3,6}∩{2,8}∩{4,7,5}= /0. Moreover, (2 8) is a transposition by Def-inition 2.2.10.Note that the sign of a permutation σ can be defined its decomposition into theproduct of transpositions assgn(σ) = (−1)m,10where m is the number of transpositions in the decomposition. If sgn(σ) = +1,then σ is even. If sgn(σ) =−1, then σ is odd.Example 2.2.3. Consider the symmetric group on 3 letters, S3. Note that Sn has n!elements. That means S3 = 3! = 6 elements. Table 2.7 shows all the elements inS3. Note that the abbreviation “CCW” stands for counterclockwise.Table 2.7: Elements of S3Elements Standard Notation Mappingρ0(1 2 31 2 3)Identityρ1(1 2 32 3 1)Rotation by 120◦ CCWρ2(1 2 33 1 2)Rotation by 240◦ CCWµ1(1 2 31 3 2)Reflection by fixing 1µ2(1 2 33 2 1)Reflection by fixing 2µ3(1 2 32 1 3)Reflection by fixing 3Now, we calculate a few compositions of permutations in S3.µ1 ◦ρ1 =(1 2 31 3 2)(1 2 32 3 1)= µ2ρ1 ◦µ1 =(1 2 32 3 1)(1 2 31 3 2)= µ3This reveals that S3 is not abelian. In other words, the elements in S3 do notcommute except for one special case, which will be discussed later. Continuingwith the similar calculations for all other compositions in the group, we obtain thegroup table for S3 given below.11Table 2.8: Group table for S3◦ ρ0 ρ1 ρ2 µ1 µ2 µ3ρ0 ρ0 ρ1 ρ2 µ1 µ2 µ3ρ1 ρ1 ρ2 ρ0 µ3 µ1 µ2ρ2 ρ2 ρ0 ρ1 µ2 µ3 µ1µ1 µ1 µ2 µ3 ρ0 ρ1 ρ2µ2 µ2 µ3 µ1 ρ2 ρ0 ρ1µ3 µ3 µ1 µ2 ρ1 ρ2 ρ0Theorem 2.2.3. If n≥ 2, then the collection of all even permutations of {1,2,3, ...,n}forms a subgroup of order n!/2 of the symmetric group Sn.Definition 2.2.11. The subgroup of Sn consisting of the even permutations of nletters is the alternating group An on n letters.By Theorem 2.2.3, the order of An is n!/2, and we say An has index 2 in Snbecause intuitively “half” of the elements of Sn lie in An.Example 2.2.4. Going back to Table 2.7,ρ0 = (1)(2)(3), Evenρ1 = (1 2 3), Evenρ2 = (1 3 2), Evenµ1 = (1)(2 3), Oddµ2 = (1 3)(2), Oddµ3 = (1 2)(3), OddBy Example 2.2.3, we showed that S3 is not abelian except in one special case.That is, the elements ρ0,ρ1,ρ2 are commutative and these elements have an evennumber of transpositions. Thus, by Definition 2.2.11, A3 = {ρ0,ρ1,ρ2}. The sub-group diagram for S3 is constructed below in Figure 2.1. The number representingeach lines joining each subgroups is the order of the corresponding subgroups. Forexample, the order of (1 2 3) is 3, where as the index of A3 in S3 is 2..12Figure 2.1: The lattice of subgroups diagram of S3S3〈2 3〉A3 =〈1 2 3〉 〈1 2〉 〈1 3〉{1}32 3 33 2 2 2Definition 2.2.12. In a group G, two elements g and h are called conjugate ifh = xgx−1for some x ∈ G.Definition 2.2.13. For an element g of a group G, its conjugacy class is the set ofelementsCg = {xgx−1 | x ∈ G}.We know thatS3 = {(1),(1 2),(1 3),(2 3),(1 2 3),(1 3 2)}Example 2.2.5. It is easy to see that g = (1) results in its own conjugacy class.Thus, we set the first trivial conjugacy class asC1 = {1}.13σ σ (1 2)σ−1 Cycle length(1) (1 2) 2(1 2) (1 2) 2(1 3) (2 3) 2(2 3) (1 3) 2(1 2 3) (2 3) 2(1 3 2) (1 3) 2Thus, the conjugates of (1 2) are (1 2),(1 3), and (2 3). We define the secondconjugacy class of S3 asC2 = {(1 2),(1 3),(2 3)}.Similarly,σ σ (1 2 3)σ−1 Cycle length(1) (1 2 3) 3(1 2) (1 3 2) 3(1 3) (1 3 2) 3(2 3) (1 3 2) 3(1 2 3) (1 2 3) 3(1 3 2) (1 2 3) 3and we define the final conjugacy class of C3 asC3 = {(1 2 3),(1 3 2)}.Definition 2.2.14. (Cosets) Let H be a subgroup of a group G written multiplica-tively. The subset aH = {ah | h ∈H} of G is the left coset of H containing a, whilethe subset Ha = {ha | h ∈ H} is the right coset of H containing a.Theorem 2.2.4. (Theorem of Lagrange) Let H be a subgroup of a finite group G.Then the order of H is a divisior of the order of G.Definition 2.2.15. (Homomorphism) A map φ : G→ G′, where (G,∗) and (G′,∗′)are groups is called a group homomorphism ifφ(a∗b) = φ(a)∗′ φ(b)14for all a,b ∈ G.Definition 2.2.16. (Normal Subgroup) A subgroup H of a group G is normal ifgH = Hg for all g ∈ G.Theorem 2.2.5. (Factor Group) Let φ : G→ G′ be a group homomorphism withkernel H. Then the cosets of H form a group called the factor group and denotedG/H, where the group operation ∗ is given by (aH)∗ (bH) = (ab)H.2.2.2 Ring Theory and FieldsDefinition 2.2.17. (Rings) A ring 〈R,+, ·〉 is a set R together with two binary op-erations + and ·, which we call addition and multiplication, defined on R such thatthe following axioms are satisfied:1. 〈R,+〉 is an abelian group. That is, a+b = b+a for all a,b ∈ R.2. For all a,b,c ∈ R, (a ·b) · c = a · (b · c).3. For all a,b,c ∈ R, a · (b+c) = (a ·b)+(a ·c) and (a+b) ·c= (a ·c)+(b ·c)hold.If a ·b = b ·a for all a,b ∈ R, we say R is commutative.From here, we consider commutative rings with unity. That is, we only look atcommutative rings with multiplicative identity, 1R.Example 2.2.6. Z,nZ,Zn,Z+Z√d, where d is squarefree integer, and Z+Zi areexamples of rings.Definition 2.2.18. (Zero Divisors and Units) Let R be a ring.1. A element a 6= 0 in R is called a zero divisor in R, if there exists b 6= 0 in R,such that ab = 0.2. Let R be a ring with the multiplicative identity 1. An element u ∈ R is calleda unit in R, if there exists v ∈ R such that uv = 1 = vu.Definition 2.2.19. (Integral Domain) An integral domain is a commutative ringthat has a multiplicative identity 1, and has no divisors of zero.15Example 2.2.7. Z,Q,R,Z+ Zi,Z+ Zω, where ω = −1+√−32 , and Z+ Z√m,where m is a square free integer are examples of integral domain.Definition 2.2.20. (Irreducible) A nonzero, nonunit element a of an integral do-main D is called an irreducible, or said to be irreducible, if a= bc, where b,c ∈D,implies that either b or c is a unit.Definition 2.2.21. (Prime) A nonzero, nonunit element p of an integral domain Dis called a prime if p | ab, where a,b ∈ D, implies that p | a or p | b.Definition 2.2.22. (Fields) Let R be a ring with unity 1 6= 0. If every nonzero el-ement of R is a unit, then R is a division ring. A field is a commutative divisionring.Example 2.2.8. Q,R,C and Zp, p is a prime are examples of fields.Example 2.2.9. The ring Zn is of characteristic n, while Z,Q,R, and C all havecharacteristic 0.Definition 2.2.23. (Unique Factorization Domain) Let D be a factorization do-main. Suppose that every nonzero, nonunit element a of D has a unique factoriza-tion as a product of irreducible elements of D. Then D is called a unique factor-ization domain.16Chapter 3Algebraic Number TheoryIn this chapter, we start by introducing algebraic elements and algebraic integers.Then we define the ring of algebraic integers in an algebraic number field. InSection 3.2 and 3.4, we study integral bases of cubic fields. Further, we find ex-pressions for the index forms of cubic fields in Section 3.5. Using the index formof a number field, we are able to determine whether the field is monogenic or not.3.1 Algebraic Number FieldsDefinition 3.1.1. (Algebraic Numbers) An element α ∈ C is an algebraic numberif f (α) = 0 for some polynomial f (x) with rational coefficients.Definition 3.1.2. (Algebraic Integers) An element α ∈ C is an algebraic integerif f (α) = 0 for some monic polynomial f (x) with integer coefficients. A monicpolynomial is a polynomial in which the leading coefficient (the nonzero coefficientof highest degree) is equal to 1.Consider the following example.Example 3.1.1. α =√3−√5 is an algebraic number as α is a root of the poly-nomial f (x) = x4−6x2+4 ∈Q[x]. Moreover, α is also an algebraic integer sincef (x) ∈ Z[x] is a monic polynomial.Example 3.1.2. Let f (x) = x3− p2x−2p2, where p is an odd prime. We show that17if θ is a root of f (x), θ2p is an algebraic integer. Since θ is a root of f (x),f (θ) = θ 3− p2θ −2p2 = 0,θ 3 = p2θ +2p2. (3.1)Let α = θ2p . We show thatα3+Aα2+Bα+C = 0,where A,B,C ∈ Z. By (3.1), we can simplify the following expressions for α3,α2.α3 =(θ 2p)3=θ 6p3=θ 3θ 3p3=(p2θ +2p2)(p2θ +2p2)p3= p(θ 2+4θ +4),andα2 =(θ 2p)2=θ 3p2=θθ 3p2=θ(p2θ +2p2)p2=p2(θ 2+2θ)p2= θ 2+2θ .Thus,α3+Aα2+Bα+C = 0is equivalent top(θ 2+4θ +4)+A(θ 2+2θ)+B(θ 2p)+C = 0.Rearranging terms,(A+Bp+ p)θ 2+(2A+4p)θ +(4p+C) = 0.Since {1,θ ,θ 2} is linearly independent overQ, each coefficient must vanish. ThusA+ Bp + p = 02A+4p = 04p+C = 018Three equations give C = −4p ∈ Z,A = −2p ∈ Z,B = p2 ∈ Z. Thus, we have apolynomial g(x) = x3−2px2+ p2x−4p such that g(α) = 0. Since g(x) is a monicpolynomial with integer coefficients, Definition 3.1.2 tells us that α is an algebraicinteger.We claim that C is a field. It is enough to say that for all z ∈ C, there existsz−1 = z/|z|2 ∈ C such that zz−1 = 1 = z−1z ∈ C. We now define field extensionsand algebraic number fields that are stated on pages 98 and 109, [2].Definition 3.1.3. (Field Extension) Let K be a subfield of C and let α ∈ C. LetK(α) =⋂Fα∈FK⊆F⊆CF,where the intersection is taken over all subfields F of C, which contain both Kand α. The intersection is nonempty as C is such a field. Since the intersection ofsubfields of C is again a subfield of C,K(α) is the smallest field containing both Kand α. We say that K(α) is formed from K by adjoining a single element α. K(α)is called a simple extension of K. If α1, ...,αk ∈ C for k ≥ 2, K(α1, ...,αk) is thesmallest subfield of C that contains both K and α1, ...,αk, and is called a multipleextension of K.Definition 3.1.4. (Algebraic Number Field) An algebraic number field is a subfieldof C of the form Q(α1, ...,αn), where α1, ...,αn are algebraic numbers.Definition 3.1.5. (Irreducible Polynomials) Let K be a subfield of C. A non-constant polynomial f (x) ∈ K[x] is irreducible over K if it cannot be factored intothe product of two nonconstant polynomials g(x) ∈ K[x] and h(x) ∈ K[x], wheredeg(g(x)),deg(h(x))< deg( f (x)).Theorem 3.1.1. (Eisenstein’s Irreducibility Criterion) Letf (x) = anxn+an−1xn−1+ · · ·+a1x+a0 ∈ Z[x].If there exists a prime number p such that1. p | ai, i = 0,1, ...,n−1192. p - an,3. p2 - a0then f (x) is irreducible over the rational numbers.Example 3.1.3. Continue with Example 3.1.2. We show that the polynomial f (x)=x3− p2x− 2p2 is irreducible over Q. We begin by finding the reverse polynomialg(x) of f (x). Considerf(1x)=(1x)3− p2(1x)−2p2.Then, multiply both sides by x3. We getx3 f(1x)= x3(1x)3− p2(1x)x3−2p2x3.Then, the reverse polynomial of f (x) isg(x) = 1− p2x2−2p2x3.Now, set y = xp. Then,g(yp)= −2p2(yp)3− p2( yp)2+1=−2y3p− y2+1.Finally, multiplying both sides by p givesh(y) =−2y3− py2+ p.Since p does not divide −2, and p2 - p by Theorem 3.1.1, h(y) is irreducibleover Q. Because h(y) is related to g(x) by a linear change of variables, h(y) isirreducible over Q implies that g(x) is also irreducible over Q. Note that if f (x)has a nonzero constant coefficient, then f (x) is irreducible if and only if its reversepolynomial is irreducible. This statement shows that f (x) is irreducible over Q.20Theorem 3.1.2. (Minimal polynomial of θ over K) Let θ be an algebraic integer.Then there exists a monic polynomial f (x) ∈ Z[x] of least degree such that f (θ) =0. f (x) is called the minimal polynomial of θ with the following properties:1. If g(x) ∈Q[x], then g(θ) = 0 if and only if f (x) divides g(x).2. f (x) is the unique monic irreducible polynomial such that f (θ) = 0.Note that the degree of K is equal to the degree of f (x), and denoted by[K :Q] = deg( f (x)).Definition 3.1.6. (Conjugates of α over K) Let α ∈C be algebraic over a subfieldK of C. The conjugates of α over K are the roots in C of the minimal polynomialof α over K.3.2 The Set OKDefinition 3.2.1. (The set OK) The set of all algebraic integers that lie in an alge-braic number field K is denoted by OK , that is,OK =Ω∩Kwhere Ω is the set of all algebraic integers in C. OK is called the ring of integersof the algebraic number field K.Theorem 3.2.1. Let K be an algebraic number field. Then OK is an integral do-main.Theorem 3.2.2. Let K be a quadratic field. Then there exists a unique squarefreeinteger m such that K =Q(√m).Theorem 3.2.3. Let K be a quadratic field. Let m be the unique squarefree integersuch that K =Q(√m). Then the set OK of algebraic integers is given byOK =Z+Z√m, if m 6≡ 1 (mod 4)Z+Z(1+√m2), if m ≡ 1 (mod 4).21Example 3.2.1. Consider α = 3√5 ∈ C. Let K = Q( 3√5). Then K is an algebraicnumber field. Note that α is a root of the irreducible polynomial x3− 5, hence, itis an algebraic integer. The set OK for K contains elements of the form:a+bα+ cα2,where a,b,c ∈ Z. We will show this in Section 3.4.3.3 DiscriminantsFrom high school, we know that the discriminant of a quadratic equation ax2 +bx+ c is b2− 4ac. The discriminant of a polynomial of degree n is defined in thefollowing way.Definition 3.3.1. (Discriminant of a polynomial)Let f (x) = anxn + an−1xn−1 + · · ·+ a1x+ a0 ∈ C[x], where n ∈ N and an 6= 0. Letθ1, ...,θn ∈ C be the roots of f (x). The discriminant of f (x) is the quantitydisc( f (x)) = a2n−2n ∏1≤i< j≤n(θi−θ j)2.Example 3.3.1. Consider f (x) = x2 + bx+ c, where b,c ∈ Z. Let θ1,θ2 ∈ C bethe roots of f (x). Then, f (x) = (x− θ1)(x− θ2) = x2 − (θ1 + θ2)x+ θ1θ2. Setb=−(θ1+θ2), and c= θ1θ2. By Definition 3.3.1, (θ1−θ2)2 = θ 21 −2θ1θ2+θ 22 =θ 21 +2θ1θ2+θ 22 −4θ1θ2 = (θ1+θ2)2−4θ1θ2 = b2−4c, which confirms the dis-criminant of a quadratic equation.Definition 3.3.2. Let K be an algebraic number field of degree n. Let ω1, ...,ωn ben elements of K. Let σk (k = 1,2, ...,n) denote the n distinct monomorphisms ofK in C. For i = 1, ...,n, letω(1)i = σ1(ωi) = ωi,ω(2)i = σ2(ωi), · · · ,ω(n)i = σn(ωi)22denote the conjugates of ωi. Then the discriminant of {ω1, ...,ωn} isD(ω1, ...,ωn) =∣∣∣∣∣∣∣∣∣∣ω(1)1 ω(1)2 · · · ω(1)nω(2)1 ω(2)2 · · · ω(2)n...... · · · ...ω(n)1 ω(n)2 · · · ω(n)n∣∣∣∣∣∣∣∣∣∣2.It is known that if ω1, ...,ωn are all algebraic integers then D(ω1, ...,ωn) is arational integer.Definition 3.3.3. (Discriminant of an element α) Let K be an algebraic numberfield of degree n. Let α ∈ K. Then we define the discriminant D(α) of α byD(α) = ∏1≤i< j≤n(α(i)−α( j))2,where α(1) = α,α(2), ...,α(n) are the conjugates of α . Note that if α is an alge-braic integer in K, then the discriminant of α is the discriminant of the minimalpolynomial of α over K.Example 3.3.2. Let K = Q(√2). The minimal polynomial of√2 over Q is givenbyf (x) = x2−2.It is easy to check that the discriminant of f (x) is 8. We may also confirm this usingDefinition 3.3.3. Since ±√2 are the roots of f (x),D(√2) = (√2+√2)2= (2√2)2= 8.3.4 Integral BasisIn section 3.2, we fully generalized the elements in the set OK for quadratic fields.Recall Theorem 3.2.3, if K is a quadratic field, and m is a squarefree integer suchthat K =Q(√m), then the set OK is described by23OK =Z+Z√m, if m 6≡ 1 (mod 4)Z+Z(1+√m2), if m ≡ 1 (mod 4).That is, every element in OK is an integer linear combination of the elementsin the following two sets.{1,√m}, if m 6≡ 1 (mod 4){1,(1+√m2)}, if m ≡ 1 (mod 4).For K = Q(√2), since 2 6≡ 1 (mod 4), the above observation implies that OKcan be generated by integral linear combinations of {1,√2}. For L = Q( 2√5) inExample 3.2.1, we claimed that the elements in OL can be expressed as integerlinear combinations of {1, 3√5,( 3√5)2}. This phenomenon leads us to the notion ofan integral basis.Definition 3.4.1. (Integral Basis) Let K be an algebraic number field of degree nwith ring of integers OK . An integral basis for OK is a set of n elements {η1,η2, ...,ηn}of OK such that for any algebraic integer α ∈ OK ,α =C1η1+C2η2+ · · ·+Cnηnwhere C1,C2, ...,Cn ∈ Z.Theorem 3.4.1. Every number field K has an integral basis.Theorem 3.4.2. Suppose {α1,α2, ...,αn} is a Q−basis for a number field K. IfD(α1, ...,αn) is squarefree, then {α1, ...,αn} is an integral basis.Definition 3.4.2. Let K be a number field, with degree n and the ring of integersOK . If OK =Z[α] for some α ∈OK , the set {1,α, ...,αn−1} is called a power basis.Theorem 3.4.2 provides a sufficient condition for a set to be an integral basis.But, the problem is this situation does not come up very often. It is also known thatevery integral basis has the same discriminant. Maple will calculate an integral24basis for a number field, but we will present an algorithm one can carry out byhand that will give an integral basis. First, we need to define the norm and trace ofan element.Definition 3.4.3. Let K be a number field of degree n. The norm of an elementα ∈ K is given byN(α) =n∏i=1α(i),and the trace is given byTr(α) =n∑i=1α(i),where α = α(1),α(2), ...,α(n) are the conjugates of α. Note that N(α),Tr(α) ∈ Zfor any algebraic integer α.Theorem 3.4.3. Let K be an algebraic number field of degree n. Let α,β ∈ K.ThenTr(α+β ) = Tr(α)+Tr(β ) and N(αβ ) = N(α)N(β ).Theorem 3.4.4. Suppose that {α1, ...,αn} is a Q−basis for K consisting of alge-braic integers, and let p be a prime such that p2 divides D(α1, ...,αn). Then thereis an algebraic integer of the form1p(λ1α1+ · · ·+λnαn),where 0≤ λi ≤ p−1, λi ∈ Z.Now, we adapt the algorithm from Cook [18] to compute integral bases.Let K =Q(α) be an algebraic number field.25Table 3.1: Computing Integral BasesSteps Description Mathematical Description1 Set up a Q−basis for K.{α1,α2, ...,αn}.2 Calculate the discriminant of the set inStep 1. If it is squarefree, then by Theo-rem 3.4.2, process complete.D(α1,α2, ...,αn).3 Find all primes p such that p2 divides thediscriminant in Step 2. p2 | D(α1,α2, ...,αn).4 Select one of the primes found in Step 3.Define a new algebraic integer in the formdescribed in Theorem 3.4.4. Then com-pute the trace of the element.α∗=1p(λ1α1+ · · ·+λnαn), Find Tr(α∗).5 Find the norm of α∗.N(α∗).6 Using the fact that the norm of an alge-braic integer is a rational integer, check allthe cases of λi to see if α∗ is the algebraicinteger for the given prime.See the examples below.7 If all λis do not satisfy the conditions inStep 6, then α∗ is not an algebraic integer.Return to Step 3 and repeat the same pro-cess by selecting the other primes. Other-wise, proceed to Step 8.See the examples below.8 At this stage, α∗ is tested to be an alge-braic integer in K. Adjoin the new alge-braic integer to the basis in step 1. {α1,α2, ...,αn,α∗}→ {β1,β2, ...,βn}.9 Calculate the discriminant of the new ba-sis in Step 8. If it is not squarefree, repeatthe same process from Step 3 until no newalgebraic integers are found.D(β1,β2, ...,βn).26Example 3.4.1. Let K =Q( 3√5).Step 1. A natural guess for a Q−basis for K. That is, {1, 3√5,( 3√5)2}.Step 2. Calculate the discriminant of the set in Step 1.D(1, 3√5,( 3√5)2) =−3352.Step 3. Find all primes p such that p2 divides the discriminant in Step 2. Hence,p = 3,5.Step 4. Select one of the primes found in Step 3. Define a new algebraic integerin the form described in Theorem 3.4.4. Then compute the trace of theelement. Asα∗ =15(λ1+λ23√5+λ3(3√5)2),Tr(α∗) =(15) 3∑i=1α(i) =3λ15,where 0≤ λi ≤ 4, and since 3λ15 ∈ Z, it follows that λ1 = 0.Step 5. Find the norm of α∗.N(α∗) =3∏i=1α(i) =λ 32 +5λ3325.Step 6. Using the fact that the norm of an algebraic integer is a rational integer,check all the cases of λi to see if α∗ is the algebraic integer for the givenprime. See Table 3.2.Step 7. The algebraic integer α∗ in Step 6 is not in OK . Return to Step 3.Step 8. Choose the other prime, which is p = 3. Repeating Steps 4 - 6, there isno algebraic integers to be added in the basis described in Step 1. Theintegral basis is{1, 3√5,( 3√5)2}.27Table 3.2: Examining α∗ ∈ OKλ2 λ3 Divisible by 25? YES/NO0 1 5 NO0 2 40 NO0 3 135 NO0 4 320 NO1 0 1 NO1 1 6 NO1 2 41 NO1 3 136 NO1 4 321 NO2 0 8 NO2 1 13 NO2 2 48 NO2 3 143 NO2 4 328 NO3 0 27 NO3 1 32 NO3 2 67 NO3 3 162 NO3 4 347 NO4 0 64 NO4 1 69 NO4 2 104 NO4 3 199 NO4 4 384 NOExample 3.4.1 is one where our natural guess of the integral basis turns outto be the correct one. We now give an example due to Dedekind which is not assimple as the previous one.Example 3.4.2. Let K = Q(θ) where θ is a root of the polynomial f (x) = x3−x2−2x−8 ∈Q[x]. We find the integral basis for K.Step 1. A natural guess for a Q−basis for K. That is, {1,θ ,θ 2}.Step 2. Calculate the discriminant of the set in Step 1.D(1,θ ,θ 2) =−(2)2(503).28Step 3. Find all primes p such that p2 divides the discriminant in Step 2. Hence,p = 2.Step 4. Select one of the primes found in Step 3. Define a new algebraic integerin the form described in Theorem 3.4.4. Then compute the trace of theelement. Asθ ∗ =12(λ1+λ2θ +λ3θ 2),Tr(θ ∗) =(12) 3∑i=1α(i) =3λ1+λ2+5λ32,where 0≤ λi ≤ 1.Table 3.3: Examining θ ∗ ∈ OKλ1 λ2 λ3 3λ1 +λ2 +5λ3 Divisible by 2?0 0 1 5 NO0 1 0 1 NO0 1 1 6 YES1 0 0 3 NO1 0 1 8 YES1 1 0 4 YES1 1 1 9 NOThe table above shows that we only need to test(λ1,λ2,λ3) = (0,1,1),(1,0,1) and (1,1,0).Step 5. Find the norm of θ ∗N(θ ∗) =18∏3i=1α(i)=λ 31 +8λ32 +64λ33 −2λ1λ 22 −12λ1λ 23 −16λ2λ 23 +λ 21 λ2+5λ 21 λ3+8λ 22 λ3−26λ1λ2λ38.Step 6. Check all the cases of λi to see if θ ∗ is the algebraic integer for the givenprime.29Table 3.4: Examining θ ∗ ∈ OKλ1 λ2 λ3 λ 31 +8λ32 +64λ33 −2λ1λ 22 −12λ1λ 23 −16λ2λ 23 +λ 21 λ2 +5λ 21 λ3 +8λ 22 λ3−26λ1λ2λ3 Divisible by 8?0 1 1 64 YES1 0 1 58 NO1 1 0 8 YESStep 7. Two algebraic elements are survived.1+θ2,θ +θ 22.The minimal polynomial of each elements are 4x3−8x2+3x−4 and x3−3x2− 10x− 8 respectively. Since the first one is not a monic polynomial,(1+θ)/2 is not an integer in K.Step 8. Adjoin any new algebraic integers to the original basis in Step 1.{1,θ ,θ 2,θ +θ 22}.Observe thatθ 2 = 2(θ +θ 22)−θ ,hence the Z−linearly independent basis is{1,θ ,θ +θ 22}.Step 9. Calculate the discriminant of the new basis.D(1,θ ,θ +θ 22)= 503.Since the discriminant D(1,θ ,(θ +θ 2)/2)is square free, we have found an inte-gral basis for K. Therefore, we terminate this algorithm. The integral basis for Kis {1,θ ,θ +θ 22}.30In Section 3.3, we discussed various kinds of discriminants. However, therewas one type of discriminant we didn’t mention, which we now define.Definition 3.4.4. (Field Discriminant) Let K be an algebraic number field of de-gree n. Let {η1, ...,ηn} be an integral basis for K. Then D(η1, ...,ηn) is called thediscriminant of K and is denoted by d(K).The field discriminant is used to derive index forms. To compute the fielddiscriminant d(K) for a number field K, we obtain a integral basis for K first.Then, use the determinant equation expressed in Definition 3.3.2.The number field in Example 3.4.1 possesses a power basis, whereas the onein Example 3.4.2 does not. Determining whether a number field admits a powerintegral basis is a classical problem in algebraic number theory. Example 3.4.2was the first example given of an algebraic number field without a power integralbasis. If number fields admit power bases, we call them monogenic fields. Numbertheorists discovered a tool to measure how far a number field can be away frombeing monogenic. We use indices of number fields to determine whether they aremonogenic or not. These will be discussed in the next section.3.5 Index Forms and Minimal IndicesThe minimal index of a number field measures how close it is to being monogenic.If the minimal index is 1, then the corresponding number field is monogenic. Wewill discuss how to find minimal indices in this section, but first we need a defini-tion.Definition 3.5.1. (Index of α) Let K be an algebraic number field. Let α ∈ OKbe such that K = Q(α). Then the index of α, written indα , is the positive integergiven byD(α) = (indα)2d(K),where D(α) is the discriminant of α defined in Definition 3.3.3, and d(K) is thefield discriminant defined in Definition 3.4.4. Equivalently, we may use the follow-ing expression for the index of an element in K.31indα =√D(α)d(K).Now we will calculate indices in some number fields, beginning with an ar-bitrary quadratic field. In Theorem 3.2.3, we stated that the general elements inthe ring of integers of a quadratic field can be described by Z+Z√m if m 6≡ 1(mod 4), or Z+Z(1+√m)/2 if m ≡ 1 (mod 4). By Alaca [2], we state anotherimportant result about the field discriminant of a quadratic field.Theorem 3.5.1. Let K be a quadratic field. Let m be the unique squarefree integersuch that K =Q(√m). Then the field discriminant d(K) of K is given byd(K) =4m, if m 6≡ 1 (mod 4),m, if m ≡ 1 (mod 4).Using this result, we are ready to compute the index of α ∈ OK for somequadratic field K.Example 3.5.1. Let K be a quadratic field. Then by Theorem 3.2.2, there exists aunique squarefree integer m such that K =Q(√m). First, we assume that m ≡ 1(mod 4). Then, the corresponding integral basis is{1, 1+√m2}, with d(K) = m.Pick α ∈ OK . Thenα = a+b(1+√m2), a,b ∈ Z.Now, by Definition 3.3.2, we getD(α) =∣∣∣∣∣∣1 a+b(1+√m2)1 a+b(1−√m2)∣∣∣∣∣∣2= (−b√m)2 = b2m.Thus,indα =√D(α)d(K)=√b2mm= |b|.Now, if m 6≡ 1 (mod 4), then the integral basis is {1,√m} and d(K) = 4m. Fol-lowing a similar process as the first case, we pick α ∈ OK . Then α = a+ b√m,32where a,b ∈ Z. Then,D(α) =∣∣∣∣∣1 a+b√m1 a−b√m∣∣∣∣∣2= (−2b√m)2 = 4b2m.Thus,indα =√D(α)d(K)=√4b2m4m= |b|.Here, the equation I(a,b) = indα = |b| is called the index form of the quadraticfield K in the previous example. This is the simplest index form we get for analgebraic number field. The results in Example 3.5.1 show that the index formof any quadratic field can be described by the positive integer |b|. Therefore, theminimum of ind(α) is 1 for some α ∈ OK . Now, we consider finding the indexform of a cubic polynomial.Example 3.5.2. Consider K = Q(θ), where θ is a root of f (x) = x3−3x+9. Wecompute the index form, field index, and the minimal index of the field K.An integral basis for K is{1,θ , θ23}. Then,OK ={a+bθ + c(θ 23)| a,b,c ∈ Z}.Let α ∈ OK . Then α = a+ bθ + c(θ 23), where a,b,c ∈ Z. The conjugates of αover K are described below:α ′ = a+bθ ′+ c(θ ′23),α ′′ = a+bθ ′′+ c(θ ′′23).Obeserve that f (x) = x3−3x+9 = (x−θ)(x−θ ′)(x−θ ′′), where, θ ′ and θ ′′ arethe other roots of f (x). Expanding and simplifying gives x3− (θ + θ ′+ θ ′′)x2 +(θθ ′+θθ ′′+θ ′θ ′′)x−θθ ′θ ′′ Since the coefficient of x2 is zero, we have θ +θ ′+θ ′′ = 0, and so we obtain α−α ′, α−α ′′, and α ′−α ′′ as follws:α−α ′ = (θ −θ ′)(b+c3(θ +θ ′))= (θ −θ ′)(b− c3θ ′′),33α−α ′′ = (θ −θ ′)(b+c3(θ +θ ′′))= (θ −θ ′′)(b− c3θ ′),α ′−α ′′ = (θ ′−θ ′′)(b+c3(θ ′+θ ′′))= (θ ′−θ ′′)(b− c3θ).Hence, by Definition 3.3.3,D(α) = (α−α ′)2(α−α ′′)2(α ′−α ′′)2= (θ −θ ′)2(θ −θ ′′)2(θ ′−θ ′′)2 (b− c3θ)2 (b− c3θ ′)2 (b− c3θ ′′)2= D(θ){( c3)3f(3bc)}2=−33 ·7 ·11(b3− bc23+c33)2=−3 ·7 ·11(3b3−bc2+ c3)2.Therefore, the index form of the field K isindα =√D(α)d(K)=√−3 ·7 ·11(3b3−bc2+ c3)2−3 ·7 ·11 = |3b3−bc2+ c3|.Now, we are ready to give two new definitions about the index forms.Let K be an algebraic number field of degree n.Definition 3.5.2. (Index of a field) The index of the field K isi(K) = gcd{indα | α ∈ OK}.Definition 3.5.3. (Minimal index of a field) The minimal index of K ism(K) = min{indα | α ∈ OK}.It is well known that the field index of a cubic field is either 1 or 2 (see [5]).Referring back to Example 3.5.2, the field index of K is 1 as we have indα =|3b3−bc2+c3|= 1 if (b,c) = (1,−1). Hence, m(K) = 1. Similarly, we found thatthe index form of any quadratic field can be expressed by |b|,b ∈ Z\{0}. Thus,the field and minimal index of any quadratic field is equal to 1. There is a closerelationship between the minimal index and the integral basis.34Theorem 3.5.2. Let K be an algebraic number field, Then m(K) = 1 if and only ifK possesses a power basis.Proof. (⇒) Suppose m(K) = 1. Then there exists a generator α of K such thatind α = 1.Hence, D(1,α, ...,αn−1)=D(α)= (ind α)2d(K)= d(K) so that {1,α, ...,αn−1}is an integral basis for K. Hence, K possesses a power basis.(⇐) Conversely, suppose K possesses a power basis, say {1,α, ...,αn−1}. Then{1,α, ...,αn−1} is an integral basis for K and so D(1,α, ...,αn−1) = d(K). But,D(1,α, ...,αn−1) = D(α) = (ind α)2d(K),which forces ind α = 1 and hence m(K) = 1.35Chapter 4Galois Groups and ChebotarevDensity TheoremFor our discussion on Galois theory, we will restrict our attention to cubic polyno-mials.4.1 Galois Groups of CubicsIn Theorem 3.1.1, we introduced Eisenstein’s Irreducibility Criterion to test if apolynomial is irreducible. There are more techniques for testing irreducibility ofpolynomials. We begin this section by providing a few other techniques for irre-ducibility.Theorem 4.1.1. (Rational Root Theorem) Letf (x) = anxn+an−1xn−1+ · · ·+a1x+a0 = 0 ∈ Z[x],where an,a0 6= 0. Then, each rational solution x, when written as a fraction x= p/qwith gcd(p,q) = 1, satisfies• p is an integer factor of the constant term a0, and• q is an integer factor of the leading coefficient an.36Theorem 4.1.2. (Gauss) Let f be a polynomial over Z which is irreducible overZ. Then f , considered as a polynomial over Q, is also irreducible over Q.We demonstrate Gauss’s Lemma in the following example.Example 4.1.1. Considerf (x) = x2+ x+1 ∈ Z[x].Since f (±1) 6= 0, f (x) is irreducible by the Rational Root Theorem. Hence, byTheorem 4.1.2, f (x) is irreducible over Q.Another type of irreducibility test is called the modulo p test. Let p be a primeand suppose that f (x) = anxn+an−1xn−1+ · · ·+a1x+a0 ∈ Z[x] with the degree off (x), n≥ 1, and p - an. Let f (x) be the polynomial in Zp[x] obtained by reducingthe coefficients of f modulo p.Theorem 4.1.3. If f is irreducible overZp and the degree of f (x) equals the degreeof f (x), then f (x) is irreducible over Q.Proof. By way of contradiction, suppose f (x) = g(x)h(x), where g,h ∈ Z[x], with1≤ deg(g)≤ deg( f ) and 1≤ deg(h)≤ deg( f ). DefineΦp : Z[x]→ Zp[x]byf (x) =n∑k=0akxk 7−→n∑k=0akxk = f (x),where ak = ak + pZ. Note that Φp is a ring homomorphism. Apply Φp to bothsides of the equation f (x) = g(x)h(x). Then we obtainf (x) = g(x)h(x)By the assumption, deg( f ) = deg( f ) which means p - an. Letg(x) = brxr + (lowest terms)h(x) = csxs+ (lowest terms)37Then, an = brcs. Since p is a prime such that p - brcs, by Euclid’s Lemma, p - brand p - cs. This means deg(g) = deg(g), and deg(h) = deg(h). Hence, f (x) isreducible over Zp. This is a contradiction to the assumption on f irreducible overZp. The irreducibility over Zp implies irreducibility over Z. By Gauss’ Lemma,f (x) is irreducible over Q.Symmetries of roots of polynomials is the main idea behind Galois theory. Weconsider the following general cubic polynomial:f (x) = x3+a2x2+a1x+a0.Suppose that θ1,θ2, and θ3 are the roots of f (x). Then,f (x) = (x−θ1)(x−θ2)(x−θ3)= x3− (θ1+θ2+θ3)x2+(θ1θ2+θ2θ3+θ1θ3)x−θ1θ2θ3.For i = 1,2,3, define si(x1,x2,x3) as follows:s1 = x1+ x2+ x3s2 = x1x2+ x1x3+ x2x3s3 = x1x2x3.Then we see that evaluating each si at each of the roots of f gives us exactly thethree coefficients of f . Consider the symmetric group on 3 lettersS3 = {1,(1,2),(1,3),(2,3),(1,2,3),(1,3,2)}.Let σ ∈ S3. Then, we can easily compute that σ(si) = si for all i = 1,2,3. Forexample, if σ = (1 2 3),σ(s1) = s1(x3,x1,x2) = x3+ x1+ x2 = s1σ(s2) = s2(x3,x1,x2) = x3x1+ x3x2+ x1x2 = s2σ(s3) = s3(x3,x1,x2) = x3x1x2 = s3Thus, σ(si) = si for all i = 1,2,3. Repeating the similar process for all σ ∈ S3,we find that σ(si) = si for i = 1,2,3.38Definition 4.1.1. h(x1,x2, ..,xn)∈K[x1,x2, ...,xn] is called a symmetric polynomialfor x1,x2, ..,xn if for any σ in the symmetric group Sn,σ(h) = h.Definition 4.1.2. The elementary symmetric polynomials in n variables x1, ...,xn,written ek(x1, ...,xn) for k = 0, ...,n, are defined bye0(x1,x2, ...,xn) = 1,e1(x1,x2, ...,xn) = ∑1≤ j≤n x j,e2(x1,x2, ...,xn) = ∑1≤ j n.Theorem 4.1.4. (Fundamental Theorem of Symmetric Polynomials) Every sym-metric polynomial f (x1,x2, ...,xn)∈K[x1, ...,xn] is a polynomial of elementary sym-metric polynomials.In section 5.1, we will use a symmetric polynomial to derive the discriminantof a cubic polynomial.Next, we state a series of definitions and theorems regarding field extensions,automorphisms, splitting fields, separability, and normality.In Theorem 3.1.2, we mentioned that the degree of the field K over the groundfield Q can be expressed as [K :Q] . In general, if K is a subfield of a field L, thedegree of L over K can be written as [L : K] .Theorem 4.1.5. (Tower Law) Let K ≤ L≤M be field extensions. Then[M : K] = [M : L][L : K].Definition 4.1.3. L : K is called finite extension, if [L : K]< ∞.Definition 4.1.4. L : K is called an algebraic extension if for all λ ∈ L,λ is a rootof some nonzero polynomial in K[x].39Definition 4.1.5. An isomorphism σ is both one-to-one and onto map of a field Kwith itself is called an automorphism of K. The collection of automorphisms of Kis denoted byAut(K) = {σ : K→ K | σ is an automorphism}.Definition 4.1.6. An automorphism σ ∈ Aut(K) is said to fix an element α ∈ Kif σ(α) = α. If F ⊆ K then an automorphism σ is said to fix F if it fixes all theelements in F. That is, σ(α) = α for all α ∈ F.Observe that Aut(L) is a group under composition. It is easy to see that ifσ ,τ ∈ Aut(L),τ ◦σ ∈ Aut(L). Since σ is both one-to-one and onto, its inverse σ−1is in Aut(L).Definition 4.1.7. If K is a field and f is a polynomial over K, then f splits in K ifit can be expressed as a product of linear factorsf (x) = k(x−α1) · · ·(x−αn),where k,α1,α2, ...,αn ∈ K.Definition 4.1.8. (Splitting Fields) The field Σ is a splitting field for the polynomialf over the field K if K ⊆ Σ and1. f splits over Σ,2. Σ= K(α1,α2, ...,αn), where α1,α2, ...,αn are the roots of f .Definition 4.1.9. (Normality) An algebraic extension L : K is normal if every irre-ducible polynomial f over K which has at least one zero in L splits in L.Definition 4.1.10. (Separability) An irreducible polynomial f over a field K isseparable over K if it has no multiple zeros in a splitting field for K.Definition 4.1.11. Let K be a field and let f (x),g(x) ∈ K[x]. A common divisor off (x) and g(x) is a polynomial c(x) ∈ K[x] such that c(x) | f (x) and c(x) | g(x). Thegreatest common divisor (gcd) is a monic common divisor of the highest degree.40Corollary 4.1.6. If L is the splitting field over Q of a separable polynomial f (x),then L is normal.Note that a Galois extension is an algebraic field extension K ≤ L that is normaland separable. If K ≤ L is a Galois extension, then Aut(L/K) is called the Galoisgroup of L over K, and write Gal(L : K).Let K ≤ L be a finite Galois extension with Galois group G, which consists ofall K-automorphisms of L. LetF be the set of intermediate fields, that is, subfieldsM such that K ⊆ M ⊆ L, and let G be the set of all subgroups H of G. We havedefined two maps:∗ : F → G‡ : G →Fas follows: if M ∈F , then M∗ is the group of all M-automorphisms of L. If H ∈ G ,then H‡ is the fixed field of H. We have observed that the maps ∗ and ‡ reverseinclusions, that is, M ⊆M∗‡ and H ⊆ H‡∗. Now, we are ready to state the Funda-mental Theorem of Galois Theory from [6].Theorem 4.1.7. (Fundamental Theorem of Galois Theory) If L : K is a finite nor-mal field extension inside C, with Galois group G, and if F ,G ,∗,‡ are defined asabove, then:1. The Galois group G has order [L : K].2. The maps ∗ and ‡ are mutual inverses, and set up an order-reversing one-to-one correspondence betweenF and G .3. If M is an intermediate field, then[L : M] = |M∗| [M : K] = |G|/|M∗|4. An intermediate field M is a normal extension of K if and only if M∗ is anormal subgroup of G.5. If an intermediate field M is a normal extension of K, then the Galois groupof M : K is isomorphic to the quotient group G/M∗.41Theorem 4.1.8. Let K ≤ L be a field extension, where L is the splitting field of aseparable polynomial f ∈ K[x]. Then Gal(L : K) has order [L : K].It is well-known that the Galois group Gal(L : Q) is a subgroup of the sym-metric group Sn where n is the degree of an irreducible polynomial over Q. By theLagrange Theorem,|Gal(L :Q)| | |Sn|We now calculate the Galois group of some algebraic extensions. We considertwo examples in this section.Example 4.1.2. Let α = 3√2. The minimal polynomial m(x) of α ism(x) = x3−2 ∈Q[x].The roots of m(x) are α,ωα,ω2α, where ω = cos 2pi3 + isin2pi3 , and ω2 = cos 4pi3 +isin 4pi3 . The splitting field of m(x) is Q(α,ω). Now, we investigate the number ofelements in the Galois group Gal(Q(α,ω) : Q). Observe that the field Q(α,ω) isnormal by Corollary 4.1.6. Also, since m(x) is an irreducible polynomial over Q,we have [Q(α) :Q] = 3. Similarly, since ω is a root of the irreducible polynomialx2+ x+1 over Q(α), we have [Q(α,ω) :Q(α)] = 2. By the Tower law describedin Theorem 4.1.5, we obtain the following[Q(α,ω) :Q] = [Q(α,ω) :Q(α)] · [Q(α) :Q] = 6.Since the fieldQ(α,ω) is the splitting field of a separable polynomial m(x)∈Q[x],Q(α,ω) is normal. Thus,|Gal(Q(α,ω) :Q)|= 6Hence, Gal(Q(α,ω) :Q)' S3. Now, we explore the elements in Gal(Q(α,ω) :Q).42Defineσ : Q(α,ω) −→ Q(α,ω)α 7−→ ωαω 7−→ ωτ : Q(α,ω) −→ Q(α,ω)ωα 7−→ ω2αα 7−→ αNow, consider the following table.Table 4.1: The elements in the Galois group of S3Elements α ωα ω2α order1 α ωα ω2α 1σ ωα ω2α α 3σ2 ω2α α ωα 3τ α ω2α ωα 2στ ωα α ω2α 2σ2τ ω2α ωα α 2Hence, Gal(Q(α,ω) : Q) = {1,σ ,σ2,τ,στ,σ2τ}. Lastly, we relate the sub-group diagram in Figure 2.2 as follows:43Figure 4.1: The lattice of subfields diagram of Q(α,ω) :QQ(α,ω)Q(α)Q(ω) Q(ωα) Q(ω2α)Q23 2 22 3 3 3Figure 4.2: The lattice of subgroups diagram of Gal(Q(α,ω) :Q){1}〈τ〉〈σ〉 〈σ2τ〉 〈στ〉G ' S323 2 22 3 3 3Example 4.1.3. Consider the following polynomialf (x) = x3−6717x−203749.By Theorem 3.1.1, f (x) irreducible over Q as f (x) is 2239-Eisenstein. Let α1 be a44root of f (x). The discriminant of f (x) is∆[1,α,α2] = (3)6(5)2(2239)2 > 0,which means all the roots α1,α2, and α3 of f (x) are real and distinct. Then, thesplitting field K =Q(α1,α2,α3) is normal. By Theorem 4.1.8,|Gal(K :Q)|= [K :Q] = 3.This implies that the Galois group is isomorphic to Z3.4.2 The Chebotarev Density TheoremLet f (x) be monic and irreducible polynomial over Q. Let p be a prime such thatp does not divide the discriminant of f (x). Using Galois theory, we investigate thefactorization of polynomials in finite fields Fp that is related to the elements of theGalois group. Again, we restrict ourselves to the cubic and quartic polynomials inthis section. Consider the following example.Example 4.2.1. Let f (x) = x3−5. Note that f (x) is irreducible over Q because itis 5-Eisenstein. Let θ = 3√5 be a root of f (x). The discriminant∆[1,θ ,θ 2] =−675 =−(3)3(5)2.The splitting field of f (x) isL =Q(θ ,ω),where ω is a root of the polynomial x2 + x+ 1 = 0. We know that Gal(L : Q) isisomorphic to a subgroup of S3. In fact,Gal(L :Q)' S3as [L :Q] = 6. We categorize the subgroups of Gal(L :Q) as follows.45Table 4.2: Cycle types in S3Cycle Type # ElementsType 1 (1 2 3), (1 3 2)Type 2 (1 2)(3), (1 3)(2), (1)(2 3)Type 3 (1)(2)(3)Now, choose primes p such that p - ∆[1,θ ,θ 2] = −(3)3(5)2. We choose p =2,7,11,13,17,19, and check the factorization over the finite fields Fp with Maple.Table 4.3: Polynomial factorization over Fpp Factorization over Fp Cycle Type Type #2 (x+1)(x2+ x+1) (1)(2 3) 27 x3+2 (1 2 3) 111 (x2+3x+9)(x+8) (1 2)(3) 213 (x+2)(x+5)(x+6) (1)(2)(3) 317 (x2+11x+2)(x+6) (1 2)(3) 219 x3+14 (1 2 3) 1We notice that there are three types of factorizations of x3 − 5 over Fp forthe first 6 primes that do not divide the discriminant of f . If we associate eachtype of factorization to a certain cycle type in S3, we see that the number of eachfactorization type is equal to the number of each cycle type in S3.Maybe this observation was merely a fluke. Let’s see what happens when wetry another polynomial, which gives Galois group of D4. The dihedral group Dn isthe symmetry group of an n-sided regular polygon for n > 1. The group order ofDn is 2n. Dihedral groups Dn are non-abelian permutation groups for n > 2. Theelements in D4 can be expressed as follows.D4 = {1,σ ,σ2,σ3,τ,στ,σ2τ,σ3τ}, |D4|= 8,46satisfying the following property:σ4 = τ2 = 1,τσ = σ−1τ.Example 4.2.2. Let f (x) = x4−5. Clearly, f (x) is irreducible overQ. Let θ = 4√5be a root of f (x). The discriminant∆[1,θ ,θ 2,θ 3] =−32000 =−(2)8(5)3.The roots of f (x) are θ = 4√5,− 4√5, i 4√5,−i 4√5. The splitting field of f (x) is L =Q(θ , i). By Theorem 4.1.8, L is Galois, and so|Gal(L :Q)|= [L :Q] = 8.Defineσ : Q(θ , i) −→ Q(θ , i)θ 7−→ iθi 7−→ iτ : Q(θ , i) −→ Q(θ , i)iθ 7−→ −iθθ 7−→ θIt is easy to verify that σ3τ = τσ ,τσ2 = σ2τ, and τσ3 = στ. Hence,Gal(L :Q)' D4.We obtain the following cycle notations that are categorized in four differenttypes shown in Table 4.4.47Table 4.4: Cycle types in D4Cycle Type # Elements1 (1 2 3 4), (1 4 3 2)2 (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)3 (1)(2 4)(3), (1 3)(2)(4)4 (1)(2)(3)(4)Now, choose primes that do not divide both 2 and 5. Say,p = 3,7,11,13,17,19,23, and 29.Table 4.5: Polynomial factorization over Qpp Factorization over Qp Cycle Type Type #3 (x2+2x+2)(x2+ x+2) (1 2)(3 4) 27 (x2+6x+4)(x2+ x+4) (1 2)(3 4) 211 (x+2)(x2+4)(x+9) (1)(2 4)(3) 313 x4+8 (1 2 3 4) 117 x4+12 (1 2 3 4) 119 (x+3)(x2+9)(x+16) (1)(2 4)(3) 323 (x2+4x+8)(x2+19x+8) (1 2)(3 4) 229 (x2+18)(x2+11) (1 2)(3 4) 2We obtain 2/|D4| chances of the type 1, 4/|D4| chances of the type 2, 2/|D4|chances of the type 3, and no chance of the type 4. To verify this phenomenon, weused MAPLE to calculate the factorizations for a large number of primes p and weachieved the following results.48Cycle Pattern # of occurrences(1 2 3 4) ≈ 2/8(1 2)(3 4) ≈ 3/8(1)(2 4)(3) ≈ 2/8(1)(2)(3)(4) ≈ 1/8Thus, there seems to be a close relationship between the factorization of poly-nomials in finite fields and elements in their corresponding Galois groups. In [9],Lenstra computes this behaviour for polynomials f1(X) = X4−X2−1 ∈ Z[x], andf2(X) = X4−X −1 ∈ Z[x]. He finds 1000 primes that do not divide the discrimi-nant of each polynomials f1 and f2. The details of this is shown below.fi(X) p0 ∆( fi) # of p≤ p0 s.t. p - ∆( fi)X4−X2−1 7933 −24 ·52 1000X2−X−1 7927 −283 1000Then, he finds how often each factorization occurs among all primes up to p0.fi(X) 4 1,3 2,2 1,1,2 1,1,1,1X4−X2−1 254 0 379 251 116X4−X−1 258 337 117 253 35The numbers in the header of the table above represent the factorization pattern.For example, there are 254 primes p up to 7933 such that f = X4−X2−1 remainsirreducible modulo p. The pattern 1, 1, 2 represents that f splits into two linear andone quadratic factors for 251 primes.Note that Gal( f1(X))' D4 and Gal( f2(X))' S4. Thus, the probability of get-ting each factorization matches with our results in the previous two examples. Itturns out that the elements in each type of cycle notation are related to one anothervia conjugation in Sn. In Example 2.2.5, we showed that the elements in S3 can becategorized in different conjugacy classes. We will make this result more preciseby considering the conjugacy classes of S3 and show that these exactly match withthe cycle types from Example 4.2.1.Before we provide the Chebotarev Density Theorem, we summarize the previ-ous result. Let f be a monic polynomial with integer coefficients with degree n.49Write K =Q(α1,α2, ...,αn), where α1, ...,αn are the roots of f . The Galois groupG of f is the group of field automorphisms of K. We know that G is a subgroupof Sn. Writing an element σ ∈ G as a product of disjoint cycles (including cyclesof length 1), we obtain the cycle pattern of σ , which is a partition n1,n2, ...,nt ofn. If p is a prime number not dividing ∆( f ), then we can write f modulo p asa product of distinct irreducible factors over Fp. The degrees of these irreduciblefactors form the decomposition type of f modulo p. The following theorem tellsus that the number of primes with a given decomposition type is proportional tothe number of σ ∈ G with the same cycle pattern.Theorem 4.2.1. The density of the set of primes p for which f has a given decom-position type n1,n2, ...,nt exists, and it is equal to 1/#G times the number of σ ∈Gwith cycle pattern n1,n2, ...,nt .We verify Theorem 4.2.1 by looking at the following example.Example 4.2.3. Recall Example 4.2.1. We showed that the Galois group G of thepolynomial f (x) = x3− 5 is isomorphic to S3. The order of G is 6. Each of thefollowing subsetsC1 = {(1)(2)(3)}C2 = {(1 2)(3), (1 3)(2), (1)(2 3)}C3 = {(1 2 3), (1 3 2)}form a different conjugacy classes. First, consider the cycle pattern with ni = 1for i = 1,2,3. Only the identity permutation has this cycle pattern, which is inC1. Hence, by Theorem 4.2.1, the set of primes p for which f modulo p splitscompletely into three distinct linear factors has density 1/#G = 1/6. Similarly, thecycle pattern with n1 = 2,n2 = 1 or n1 = 1,n2 = 2 corresponds to the elements inC2. Again, using Theorem 4.2.1, the set of primes p for which f modulo p splitsinto linear-quadratic factors has density (1/#G)× #C2 = 3/6. Similarly the setof primes p for which f remains irreducible modulo p has density 2/6. Thesestatistics are also confirmed with our experiment in Table 4.3.Theorem 4.2.2. (Chebotarev Density Theorem) Let K be an algebraic number fieldof degree n over Q. Let C ⊂G = Gal(K :Q) be a conjugacy class. Then, the set of50primes not dividing ∆(K :Q) has density #C/#G.As a consequence of Chebotarev density theorem, for some polynomials, thereexist infinitely many primes p so that the polynomial does not factor over the finitefields Fp.Theorem 4.2.3. If n is an integer which is not a cube, then for f (x) = x3−n,1. f (x) is irreducible over Q.2. Gal( f )' S3Theorem 4.2.4. (Cauchy’s Theorem) Let p be a prime. Let G be a finite group andlet p divide |G|. Then G has an element of order p and, consequently, a subgroupof order p.Theorem 4.2.5. Let f (x) ∈ Z[x] be a monic polynomial, irreducible over Q, andhaving prime degree p. Then there exists a prime q such that f (x) is irreduciblemod qProof. Let deg( f (x)) = p, where p is a prime. Let L be a splitting field of f (x). Itis clear thatp | |Gal(L :Q)|By Theorem 4.2.4, Gal(L : Q) has an element (a1,a2, ...,ap) of order p and con-sequently, a subgroup of order p. This shows that there exists a prime q such thatf (x) is irreducible modulo q.We immediately obtain the following corollary.Corollary 4.2.6. Let n be an integer which is not a cube in Z. Then there existsinfinitely many primes q such thatf (x) = x3−nis irreducible mod q.51Chapter 5Minimal Indices in Pure CubicFieldsIn this Chapter, we will focus on analyzing pure cubic fields. In the first threesections, we define pure cubic fields, and calculate an integral basis for them withindex forms. Then using the index form we prove the unboundedness of its minimalindex via Hall[1]. Furthermore, we construct infinitely many families of pure cubicfields whose index equals a particular positive integer.5.1 Pure Cubic FieldsWe begin with the definition of pure cubic fields.Definition 5.1.1. (Pure cubic field) A field K is said to be pure cubic field if thereexists a rational integer d, which is not a perfect cube, such that K =Q( 3√d).Generally, we write pure cubic fields as K = Q( 3√ab2), where a,b are square-free, and gcd(a,b) = 1. It is easy to see ab2 is cubefree, and that the minimalpolynomial of 3√ab2 is m(x) = x3−ab2 ∈ Z[x] so that 3√ab2 is an algebraic integer.Let a= 2, and b= 1. Then, we have the pure cubic field K =Q( 3√2). Let α = 3√2.Then, its minimal polynomial m(x) ism(x) = x3−2 ∈ Z[x].52Alaca (pg. 175, [2]) shows that K = Q(α) = Q(θ), where θ ∈ R is a root of theirreducible polynomial given byf (x) = x3+6x+2.This shows the existence of pure cubic fields of the form K = Q(θ), where θ 3 +aθ + b = 0, a,b ∈ Z. We end this section by showing that the discriminant ofany polynomial of the form f (x) = x3 + px+ q is −4p3− 27q2, where p,q ∈ Z.Furthermore, we show that if K =Q(θ), where θ is a root of f (x), is a pure cubicfield, then the discriminant of f (x) equals −3c2 for some positive integer c.Example 5.1.1. We show that the discriminant of f (x) = x3+ px+q is−4p3−27q2.Let α,β ,γ be the three roots of f (x). Then,f (x) = (x−α)(x−β )(x− γ),where,α+β + γ = 0αβ +βγ+αγ = pαβγ = −q.This is equivalent tof (x) = x3− (α+β + γ)x2+(αβ +βγ+αγ)x−αβγ.By Definition 3.3.1, the discriminant of the polynomial is∆( f ) = (α−β )2(β − γ)2(α− γ)2√∆( f ) = (α−β )(β − γ)(α− γ).Note that√∆( f ) can be computed by using the determinant of the following Van-53dermonde matrix:M = 1 1 1α β γα2 β 2 γ2 .Thus, ∆( f ) is equal to the determinant of MMT , where MT is the transpose of M.That is,det(MMT ) =∣∣∣∣∣∣∣1 1 1α β γα2 β 2 γ2∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 α α21 β β 21 γ γ2∣∣∣∣∣∣∣ .In Section 4.1, we discussed symmetric polynomials of a cubic polynomial. Wedefines1 = α+β + γs2 = αβ +αγ+βγs3 = αβγ.We already mentioned that α + β + γ = 0 in the previous page. Thus, s1 = 0.Similarly, we have s2 = p and s3 = −q. Using Newton’s identities, MMT can beexpressed as followsMMT =3 a ba b cb c d ,wherea = s1 = 0b = s21−2s2 =−2pc = s1b− s2a+3s3 =−3qd = s1c− s2b+ s3a = 2p2.Therefore, the determinant of MMT isdet(MMT ) =∣∣∣∣∣∣∣3 0 −2p0 −2p −3q−2p −3q 2p2∣∣∣∣∣∣∣=−4p3−27q2.54Now, we prove that if θ is a root of f (t), and K = Q(θ) is a pure cubic field,then its discriminant is −3c2 for some positive integer c.Proof. Assume L = Q(θ) is a pure cubic field. Then there exist relatively primeintegers h,k such that K =Q( 3√hk2). By Definition 3.5.1, we get the following twoexpressions for d(K).d(K) =−4p3−27q2(ind(θ))2,d(L) =−27(hk2)2(ind( 3√hk2))2.Since the two fields are equal, we obtain the following equality.−4p3−27q2(ind(θ))2=−27(hk2)2(ind( 3√hk2))2,it follows that−4p3−27q2 = −27(hk2)2(ind(θ))2(ind( 3√hk2))2= −3(3ind(θ)(hk2)ind( 3√hk2))2.Using a result of Dedekind (see [10]), the index of 3√hk2 is either 1,k, or 3k. In anycase, we takec =3ind(θ)(hk2)ind( 3√hk2)∈ Z,which completes the proof.5.2 Computing Index FormsIn 1900, Dedekind [10] was the first one to generalize the integral basis for the purecubic field K =Q( 3√ab2).Theorem 5.2.1. Let d be a cubefree integer. Set d = ab2, where a,b are squarefreeintegers and gcd(a,b) = 1. Let θ = 3√d and K =Q(θ). Then an integral basis for55K is {1,θ ,θ 2b}, if d2 6≡ 1 (mod 9){1,θ ,b2±b2θ +θ 23b}, if d ≡±1 (mod 9)..The field discriminant d(K) of K is given byd(K) =−27a2b2, if d2 6≡ 1 (mod 9)−3a2b2, if d ≡±1 (mod 9).Using this result, we derive the index form of pure cubic fields due to Hall.Note that θ is a root of an irreducible monic polynomial x3− ab2 ∈ Z[x]. Also,by Theorem 2.1.6 (Euler’s Theorem), a2 6≡ b2 (mod 9) corresponds to d2 6≡ 1(mod 9). We obtain the following equations.θ +θ ′+θ ′′ = 0θθ ′+θθ ′′+θ ′θ ′′ = 0 (∗)θθ ′θ ′′ = ab2.Consider the case a2 6≡ b2 (mod 9). Let α ∈ OK be such thatα = x+ yθ + zθ 2bα ′ = x+ yθ ′+ zθ ′2bα ′′ = x+ yθ ′′+ zθ ′′2b,where x,y,z ∈ Z, and α,α ′,α ′′ are the conjugates of α with respect to K. Note thatα−α ′ = (θ −θ ′)(y− zθ′′b)α−α ′′ = (θ −θ ′)(y− zθ′b)α ′−α ′′ = (θ −θ ′)(y− zθb).56By Definition 3.3.3, the discriminant of α is given byD(α) = (α−α ′)2(α−α ′′)2(α ′−α ′′)2=[(θ −θ ′)(y− zθ′′b)(θ −θ ′)(y− zθ′b)(θ −θ ′)(y− zθb)]2= D(θ)[(y− zθ′′b)(y− zθ′b)(y− zθb)]2= D(θ)[y3− y2zb(θ +θ ′+θ ′′)+yz2b(θθ ′+θθ ′′+θ ′θ ′′)− z3θθ′θ ′′b]2= D(θ)[y3− z3 ab2b3]2=D(θ)b2(by3−az3)2.From Definition 3.5.1, we know thatindα =√D(α)d(K).The field discriminant d(K) is −27a2k2. The index of α isindα =√−27a2b4−27a2b4 (by3−az3)2=∣∣az3−by3∣∣ .We may denote the index form of the field K as I(x,y)I(x,y) =∣∣ax3−by3∣∣for x,y∈Z. Repeating the similar process for the case a2 ≡ b2 (mod 9), we obtainthe corresponding index form as follows.I(x,y) =∣∣∣∣ax3−by39∣∣∣∣ .575.3 Hall’s TheoremWe begin by considering a few examples.Example 5.3.1. Let K = Q( 3√ab2) be a pure cubic field. Let a = 5,b = 1. Since52 6≡ 12 (mod 9), the index form of this field isI(x,y) = |2x3− y3|.Choose x= 0,y=±1. The minimal index m(K)= 1. Then by Theorem 3.5.2, K pos-sesses a power basis, {1, 3√5,( 3√5)2}, which verifies the result in Example 3.4.1.The next example shows that the minimal index can be greater than 1.Example 5.3.2. Let K =Q( 3√3 ·112). Since 32 6≡ 112 (mod 9), the correspondingindex form of K isI(x,y) = |3x3−11y3|.Clearly, the minimal index m(K) = 3 6= 1.In fact, the minimal index of a pure cubic field can be arbitrarily large. Thefollowing is a theorem due to Hall (see [1]).Theorem 5.3.1. Given a large positive integer n, it is possible to find a cubic fieldK =Q( 3√ab2) in which every integer has an index greater than n.Proof. Let θ = 3√ab2, where a,b are relatively prime and squarefree. By Theorem5.2.1, an integral basis for K is{1,θ ,θ 2b}, if a2 6≡ b2 (mod 9){1,θ ,b2±b2θ +θ 23b}, if a2 ≡ b2 (mod 9).The corresponding index forms areI(x,y) =∣∣ax3−by3∣∣ , if a2 6≡ b2 (mod 9)|ax3−by3|9, if a2 ≡ b2 (mod 9).58It is sufficient to show that I(x,y)> 9n. Considera≡ 2 (mod 7) and b≡ 0 (mod 7).ThenI ≡ 2x3 ≡ 0,±2 (mod 7).This shows that I(x,y) 6=±1. We eliminate the possibility of I(x,y) =±2 by choos-inga≡ 1 (mod 13) and b≡ 0 (mod 13).ThenI(x,y) ≡ x3 ≡ 0,±1,±5 (mod 13).This shows that I(x,y) 6=±2,±3,±4. We eliminate the possibility of I(x,y) =±5by choosinga≡ 1 (mod 19) and b≡ 0 (mod 19).ThenI(x,y) ≡ x3 ≡ 0,±1,±7,±8 (mod 19).This shows that I(x,y) 6=±5,±6. Continuing this process, the index is unbounded.Take a sequence of primesp1, p2, ..., p9nof the form 3k+ 1, where k ∈ Z. Due to the choice of these primes, we chooseintegers a1, ...,a9n so that ai and n are not equivalent modulo pi as described inDefinition 2.1.7. That is,a1x3 6≡ n (mod p1)a2x3 6≡ n (mod p2).........a9nx3 6≡ n (mod p9n).Defineb = p1 p2 · · · p9n.59We may find a by solving the following system of modular equations using theChinese Remainder Theorem.a ≡ a1 (mod p1)a ≡ a2 (mod p2)...a ≡ a9n (mod p9N).Then, I(x,y) 6= ±N for all 1 ≤ n ≤ 9N. Thus, the minimal index of a pure cubicfield can be arbitrarily large.5.4 Main ResultMotivated by Hall, we evaluate the minimal index for infinitely many pure cubicfields. Before we state the main result, we may borrow a theorem from Erdo¨s[11].Let f (x) be a polynomial having integer coefficients with greatest common divisor1. We assume that the coefficient of the leading term in f (x) is positive.Theorem 5.4.1. If n≥ 3 and f (x) satisfies the conditions stated above, then thereare infinitely many positive integers x for which f (x) is (n−1)-th power free.Example 5.4.1. Let p be a prime of the form 3k+ 1, where k is an integer and nbe an integer not divisible by p. We show thatf (x) = 27px3+27px2+9px+(p+9n)is squarefree for infinitely many positive integers x.Since gcd(27p,27p,9p, p+ 9n) = 1, and 27p is positive, we conclude that thereare infinitely many positive integers x for which f (x) is squarefree by Theorem5.4.1.We use the result from Section 5.2 about integral bases and the index form ofpure cubic fields except we give a small restriction on a,b. When we consider thecase a2 ≡ b2 (mod 9), we choose the signs of a and b so that a≡ b≡ 1 (mod 3).60Then the corresponding integral basis is:{1,θ ,θ 2+ab2θ +b23b}.Consequently, the index form in this case is:I(x,y) =a(3x+ y)3−by39.We are now ready to state and prove the main result.Theorem 5.4.2. Let n be a cubefree positive integer. Then there exist infinitelymany pure cubic fields with minimal index equal to n.Proof. Let K = Q( 3√ab2) be a pure cubic field, where a, and b are squarefreeand such that gcd(a,b) = 1. Let θ = 3√ab2. Recall the index form I(x,y) for K. Ifa2 6≡ b2 (mod 9), thenI(x,y) = ax3−by3. (5.1)If a2 ≡ b2 (mod 9), thenI(x,y) =a(3x+ y)3−by39. (5.2)Suppose that n = 1. Choose a = 3p for any prime p > 3. Then the family of purecubic fieldsK =Q( 3√3p)has minimal index m(K) = 1 since the index form of K isI(x,y) = 3px3− y3,so thatI(0,−1) = 1.Now, suppose that n > 1 and cubefree. It is easy to see that the sequencen2k for k = 1,2, ...,n−1,61does not contain a perfect cube. Thus the cubic polynomialsfk(x) = x3−n2k, k = 1,2, ...,n−1,are irreducible over Q. The Galois groups of the cubic polynomials fk(x) containan element of order 3 so that by the Chebotarev Density Theorem [9] and Theorem4.2.5, we may select prime numbers pk, k = 1,2, ...,n− 1 such that fk(x) is irre-ducible modulo pk. Suppose pk = 3. Then, fk(x) is clearly reducible over F3. Nowsuppose, pk are of the form 3e+2, where e is an integer. ThenX3 ≡ n2k (mod pk)is solvable by Theorem 2.1.7, which contradicts our choice of pk. Thus, pk areof the form 3e+ 1. Define the positive integer b to be the product of the distinctprimes in the sequencep1, p2, ..., pn−1.By the choice of pk, gcd(n,b) = 1, and b is squarefree. Further, we haveb ≡ 1 (mod 3).Let z be an integer. We define the integer a = a(z) bya(z) = b(3z+1)3+9n. (5.3)We have shown that there are infinitely many integers z for which a is squarefreein Example 5.4.1. We now have a family of pure cubic fieldsK =Q( 3√ab2).We will show that these fields have the property that m(K) = n. By (5.3), it is easyto see thata ≡ b (mod 9),from which we deduce that we are using the index form given by equation (5.2).62As b ≡ 1 mod 3, we clearly have a ≡ 1 (mod 3), so that we use the index formI(x,y) =(b(3z+1)+9n)(3x+ y)3−by39.A calculation shows thatI(−z,3z+1) = n.If any of the equationsI(x,y) =±k, k = 1,2, ...,n−1are solvable for integers x,y then at least one of the congruencesI(x,y) ≡ ±k (mod pk), k = 1,2, ...,n−1is solvable. These congruences reduce ton(3x+ y)3 ≡ ±k (mod pk),implying that the congruenceX3 ≡ ±n2k (mod pk)is solvable for X modulo pk for some k, which contradicts the choice of the pk.Thus each of the infinitely many pure cubic fields K in this case satisfym(K) = n,completing the proof.We now give examples illustrating our main result.Example 5.4.2. We show there exists infinitely many pure cubic fields K withm(K) = 4.63Begin by choosing primes pk, k = 1,2,3 such that the cubic polynomialsfk(x) = x3−42k, k = 1,2,3are irreducible modulo pk. We find that we may choose p1 = p2 = 7 and p3 = 13.Thus b = 7 ·13. Next we choose a positive integer z so that7 ·13 · (3z+1)3+4 ·9is squarefree. We find the value z = 0 yields the squarefree integer 127. As in theproof of our theorem, we havea = 127 and b = 91.The pure cubic fieldQ(3√127 ·912)has index formI(x,y) =127(3x+ y)3−91y39.The cubic index form equationsI(x,y) =±k, k = 1,2,3are insolvable by construction, butI(0,1) = 4,so thatm(K) = 4.Next we consider an example which is not covered by our theorem.Example 5.4.3. We give a pure cubic field K with m(K) = 8. SetK =Q(3√23 ·152).64The index form for K isI(x,y) = 23x3−15y3.Using Magma, we find that the cubic index form equationsI(x,y) =±k, k = 1,2, ...,7are all insolvable. HoweverI(1,1) = 8,so thatm(K) = 8.65Chapter 6Conclusion and Future Work6.1 ConclusionThis thesis extends Hall’s idea on the unboundedness of minimal indices of purecubic fields. Our main goal was to construct a family of infinitely many pure cubicfields with the minimal index equal an arbitrary cubefree integer n. ChebotarevDensity Theorem held an important role in performing this construction. In fact,we needed to choose proper primes so that we find the appropriate expressions forthe index forms for these fields and show that the solvability of the index I(x,y) =1,2, ...,n−1 is impossible.In Chapter 4, we observed that there was a special relationship between theconjugacy class of a Galois group G and the factorization of polynomials over somefinite fields. By the Chebotarev Density Theorem, it turns out that the factorizationof polynomials of the form f (x) = x3−ab2 depends on the density |C|/|G| whereC is a conjugacy class of G. Further, the irreducibility of the polynomials f (x) overFp occurred for some primes of the form p = 3k+1.In Chapter 5, we first introduced Dedekind’s result on integral bases of pure cu-bic fields. Once the index forms were set up, we showed that the minimal indicesof pure cubic fields are unbounded by Hall [1]. In section 5.4, We have constructedinfinitely many pure cubic fields with minimal index equal to one. Then, we usedErdo¨s’ theorem on squarefree integers to set up appropriate expressions for eachparameters a,b of the standard form K = Q( 3√ab2). Finally, we used the Cheb-66otarev Density Theorem to show that the index form does not equal any positiveintegers less than the arbitrary cubefree integer n.6.2 Future WorkWe can apply our result to different algebraic number fields. Dummit and Kisilevsky[7] and Huard [8] showed that the minimal indices of cyclic cubic fields are un-bounded. Funakura[12] computed integral bases of pure quartic fields. Gaa´l, andPetra´nyl [13] calculated the elements of minimal index in an infinite parametricfamily of simplest quartic fields. Nakahara [14, 15] proved that the minimal indexis unbounded for bicyclic and cyclic quartic fields. The idea presented in this the-sis can still be applied to these fields mentioned above. Our next goal is to extendour result to some of these fields mentioned above. We finish our discussion withproviding some examples of the minimal indices of pure quartic and cyclic cubicfields.Example 6.2.1. Let K = Q( 4√2). Then θ = 4√2 is a root of the monic polynomialf (x) = x4−2. The discriminant of θ isD(θ) =−2048.By the result in Funakura [12], an integral basis for K is{1,θ ,θ 2,θ 3}.Thus, the most general element α in the ring of integers OK is expressed byα = a+bθ + cθ 2+dθ 4, where a,b,c,d ∈ Z.By using Maple, we find thatD(α) =−2048{(−2d2+b2)(4d4−16bc2d+8c4+b4+4b2d2)}2.The Diophantine equation(−2d2+b2)(4d4−16bc2d+8c4+b4+4b2d2) = 167is solvable if (b,c,d) = (1,0,0). Hence, the minimal index m(K) = 1. Therefore, Kis a monogenic field.Example 6.2.1 considers one particular pure quartic field, which possesses apower basis. We immediately recognize that the index form is more complicatedcompared to the case of pure cubic fields even though we are considering the sim-plest pure quartic field. Finding a family of infinitely many of these fields withminimal index equal to an arbitrary integer can be a challenging work. The firststep towards this problem would be to generalize the index forms of pure quarticfields by considering all the necessary cases for integral basis computed by Fu-nakura [12]. Lastly, we look at an example of cyclic cubic fields.Example 6.2.2. Let K =Q(c,d) be a cyclic cubic field. We need to choose appro-priate integers c and d so that every representation of4m = c2+27d2characterizes a unique cyclic cubic field, where m is a product of distinct primes ofthe form 3k+1. Let θ be a root of f (x) = x3−6717x−203749. An integral basisfor K is {1,2+θ3,37+7θ +θ 245}.By Example 4.1.3, we have shown thatGal(K/Q)' Z3as expected. The index form of K isI(x,y) = 5x3−371y3+7x2y−146xy2.Magma shows that there is no integers x,y such that I(x,y) = 1,2,3, or 4. Sincethe equation I(1,0) = 5,m(K) = 5.We made decent progress on constructing a family of infinitely many cycliccubic fields of index equal to n. Example 6.2.2 only shows one particular cyclic68cubic field with the case n = 5 to confirm the progress. A lot of work has beendone with algebraic number fields of degree 3. However, there is a lot of workwaiting for us for higher degree number fields as mentioned above. There is hugeresearch potential exploring the minimal indices of these number fields.69Bibliography[1] M. Hall, Indices in cubic fields, Bull. Amer. Math. Soc., 43, pp. 104-108,(1937) → pages 2, 5, 52, 58, 66[2] S. Alaca and K. S. Williams, Introductory Algebraic Number Theory, Cam-bridge University Press, New York (2004) → pages 19, 32, 53[3] J. B. Fraleigh, A First Course in Abstract Algebra, Pearson, New York, 7th Ed,(2002) → pages[4] K. H. Rosen, Elementary Number Theory and Its Application, Pearson,Boston, 6th Ed, (2010) → pages[5] H. T. 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