Intersective Polynomials and TheirConstructionbyPaul David LeeB.Sc., The University of British Columbia, 2009M.Sc., The University of British Columbia, 2011A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE COLLEGE OF GRADUATE STUDIES(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Okanagan)September 2016c© Paul David Lee, 2016The undersigned certify that they have read, and recommend to theCollege of Graduate Studies for acceptance, a thesis entitled: IntersectivePolynomials and Their Construction submitted by Paul David Leein partial fulfilment of the requirements of the degree of Doctor of PhilosophyDr. Blair Spearman - Science/MathematicsSupervisor, Professor (please print name and faculty/school above the line)Dr. Qiduan Yang - Science/MathematicsSupervisory Committee Member, Professor (please print name and faculty/schoolabove the line)Dr. Javad Tavakoli - Science/MathematicsSupervisory Committee Member, Professor (please print name and faculty/schoolabove the line)Dr. Paul Shipley - Science/ChemistryUniversity Examiner, Professor (please print name and faculty/school above theline)Dr. Yang Zhang - Science/Mathematics - University of ManitobaExternal Examiner, Professor (please print name and faculty/school above the line)September 20, 2016(Date Submitted to Grad Studies)Additional Committee Members include:(please print name and faculty/school above the line)(please print name and faculty/school above the line)iiAbstractA monic polynomial with integer coefficients is called intersective if ithas no root in the rational numbers, but has a root modulo m for all positiveintegers m > 1. Equivalently, the polynomial has a root in each p-adic fieldQp. Using three different methods for forming these intersective polynomi-als, we produce an infinite family with Galois group A4, an infinite familywith Galois group D5, and classify intersective polynomials with holomorphGalois group Z2e o Z∗2e .iiPrefaceThis thesis is primarily based on the following papers:Published:1. P. D. Lee, B. K. Spearman, and Q. Yang. A parametric family of inter-sective polynomials with galois group A4. Communications in Algebra,43(5), 2015, 1784-1790.Accepted:2. P. D. Lee, B. K. Spearman, and Q. Yang. Covering a semi-direct prod-uct and intersective polynomials. Manuscripta Mathematica, 2015,10.1007/s00229-015-0811-1.Submitted:3. P. D. Lee, B. K. Spearman, and Q. Yang. Intersective polynomialsand dihedral quintic trinomials.For each of these papers with multiple authors, each author contributedequally in terms of acquisition and analysis of data and preparation of papersfor publishing purposes.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiList of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . ixChapter 1: Preliminaries . . . . . . . . . . . . . . . . . . . . . . 11.1 Basic Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Dedekind Domains . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Field and Galois Theory . . . . . . . . . . . . . . . . . . . . . 61.4.1 Field Theory . . . . . . . . . . . . . . . . . . . . . . . 61.4.2 Field Extensions as Vector Spaces . . . . . . . . . . . 71.4.3 The Group of Permutations of a Polynomial . . . . . . 81.4.4 Normality, Separability, and the Galois Correspondence 91.5 Algebraic Number Theory . . . . . . . . . . . . . . . . . . . . 121.5.1 Field Extensions and Algebraic Number Fields . . . . 131.5.2 Conjugates and Conjugate Fields of an Algebraic Num-ber Field . . . . . . . . . . . . . . . . . . . . . . . . . 151.5.3 Discriminants . . . . . . . . . . . . . . . . . . . . . . . 191.5.4 Ideals in Algebraic Number Theory . . . . . . . . . . . 211.5.5 Ideals in a Dedekind Domain . . . . . . . . . . . . . . 231.5.6 Norm of a Prime Ideal . . . . . . . . . . . . . . . . . . 241.5.7 Factoring Primes in a Quadratic Field . . . . . . . . . 26ivTABLE OF CONTENTS1.5.8 Factoring Primes in a Monogenic Number Field . . . . 28Chapter 2: Group Coverings and Intersective Polynomials . . 302.1 Group Coverings with Examples . . . . . . . . . . . . . . . . 302.1.1 2-Coverable Groups . . . . . . . . . . . . . . . . . . . 322.1.2 The Holomorph Z32 o Z∗32 . . . . . . . . . . . . . . . . 372.2 Intersective Polynomials . . . . . . . . . . . . . . . . . . . . . 392.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 392.2.2 The Frobenius Group . . . . . . . . . . . . . . . . . . 402.2.3 Hensel’s Lemma . . . . . . . . . . . . . . . . . . . . . 412.2.4 Decomposition Groups . . . . . . . . . . . . . . . . . . 42Chapter 3: Intersective Polynomials with Specified Galois Group 443.1 Intersective Polynomials with Galois Group A4 . . . . . . . . 443.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 443.1.2 Proof of Theorem . . . . . . . . . . . . . . . . . . . . . 463.2 Intersective Polynomials with Galois Group D5 . . . . . . . . 523.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 523.2.2 The Decomposition Groups . . . . . . . . . . . . . . . 563.2.3 Proof of Theorem . . . . . . . . . . . . . . . . . . . . . 593.2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 61Chapter 4: The Holomorph Z2e o Z∗2e . . . . . . . . . . . . . . . 644.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.2 Binomials x2e − a with Galois group G . . . . . . . . . . . . . 664.3 n-cover of G ' Z2e o Z∗2e . . . . . . . . . . . . . . . . . . . . . 684.4 Proof of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . 704.5 Proof of Theorem 4.2 . . . . . . . . . . . . . . . . . . . . . . . 73Chapter 5: Future Work and Conclusion . . . . . . . . . . . . . 785.1 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.1.1 Covering Dihedral Groups Dn . . . . . . . . . . . . . . 785.1.2 Covering Semi-Direct Products ZmoZ∗m . . . . . . . . 785.1.3 Product of Two Simplest Cubics . . . . . . . . . . . . 79vTABLE OF CONTENTS5.1.4 Hilbert Class Polynomials . . . . . . . . . . . . . . . . 795.2 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81viList of TablesTable 1.1 Possible Galois groups for polynomials from degree 2to 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Table 3.1 Decomposition groups of the ramified primes in L . . . 51Table 3.2 Examples of intersective polynomials with Galois groupA4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Table 3.3 Parameter Values and Their Associated Polynomials . 61Table 3.4 Intersective Polynomials Produced Using Corollary . . 63Table 4.1 Intersective polynomials with Galois group G ' Z2eoZ∗2e 77viiList of SymbolsZ Set of integersQ Set of rationalsR Set of real numbersC Set of complex numbersQp p-adic field for prime integer p > 1R∗ Set of non-zero real numbersQ∗ Set of non-zero rational numbersQ∗2 Set of square-free rational numbersZn Set of integers modulo nQ(α) Field extension of α 6∈ Q over QZ[x] Set of all polynomials in x with integer coefficientsQ[x] Set of all polynomials in x with rational coefficientsK[x] Set of all polynomials in x with coefficients in any number field KD(α) Discriminant of the algebraic number αD(f) Discriminant of a polynomial f(x)deg(f), ∂f Degree of a polynomial f(x)OK Ring of integers of a number field, KU(OK) Group of units of the ring of integers of a number field, KviiiAcknowledgementsI would like to thank my family and friends for their support and love,especially my parents. Without them, none of this would be possible.I would also like to express my deepest gratitude to my supervisor, Dr.Blair Spearman, for his constant support and advice. Your patience, under-standing, and ability to teach is one of my core inspirations for becominga mathematician and teacher. I would also like to thank my co-supervisorDr. Qiduan Yang for all his support and for instilling a broad knowledge ofmathematics into me throughout my undergraduate and graduate work.I would also like to thank all of my colleagues and professors from UBCOkanagan that have been such a great source of support and encouragement.ixChapter 1PreliminariesWhile the content in this preliminary chapter is well known in algebraand number theory, it should be noted that we have followed the contentand structure contained within Alaca and Williams [AW04] and Fraleigh[Fra02] quite closely with minor notation and wording changes.1.1 Basic AlgebraIn this section, we present some fundamental theory in algebra that willserve as a precursor in knowledge for the rest of the thesis.Definition 1.1. A group (G, ∗) is a set G, together with a binary operation∗, such that G satisfies the following:− Closure: G is closed under the operation ∗, that is for all a, b ∈ G,a ∗ b ∈ G.− Associativity: For all a, b, and c in G, (a ∗ b) ∗ c = a ∗ (b ∗ c).− Identity element: There exists an element e ∈ G such that for everyelement a ∈ G, the equation e ∗ a = a ∗ e = a holds. Such an elementis unique.− Inverse element: For each a ∈ G, there exists an element b ∈ G suchthat a ∗ b = b ∗ a = e where e is the identity element.When a ∗ b = b ∗ a for all a, b ∈ G, then we call G an abelian group.Definition 1.2. A subset H of a set G is called a subgroup of G if H formsa group under the operation ∗ of G.11.1. Basic AlgebraDefinition 1.3. A subgroup H of a group G is a normal subgroup if andonly if gH = Hg for all g in G.Definition 1.4. Let (G, ∗) and (H, ?) be groups with binary operations ∗and ?. A map φ : G→ H such thatϕ(x ∗ y) = ϕ(x) ? ϕ(y), for all x, y ∈ Gis called a homomorphism. If the group operations of G and H are notwritten explicitly, we just write ϕ(xy) = ϕ(x)ϕ(y).Definition 1.5. The map ϕ : G→ H is called an isomorphism and G andH are said to be isomorphic if the two following hold:1. ϕ is a homomorphism and2. ϕ is a bijection (both onto and one-to-one).Definition 1.6. Let G be a group. An isomorphism from G onto itself iscalled an automorphism of G. The set of all automorphisms of G is denotedby Aut(G).Definition 1.7. Let H and K be groups and let ϕ be a homomorphismfrom K into Aut(H). The semidirect product of H by K via ϕ is the setof ordered pairs {(h, k) | h ∈ H, k ∈ K}, together with the binary operationdefined by(h1, k1)(h2, k2) = (h1ϕk1(h2), k1k2).We write H oϕ K for the semidirect product of H and K.Definition 1.8. A ring is a set R equipped with binary operations + and ·(addition and multiplication) that satisfies the following three sets of axioms:− R is an abelian group under addition.− Multiplication in R is associative.− Multiplication in R is distributive with respect to addition, i.e. a · (b+c) = a · b+ a · c and (b+ c) · a = b · a+ c · a.21.1. Basic AlgebraIf multiplication is also commutative, then R is a “commutative ring”. If Rhas a multiplicative identity, 1, then we say that R is a “ring with 1”.If all of the above holds true, then we call R a “commutative ring with 1”.Definition 1.9. An element a of a ring R is called a zero divisor if thereexists a nonzero element x such that ax = xa = 0.Example 1.10. Consider the ring of elements modulo 6. The elements 2and 3 are zero divisors since 2 · 3 = 3 · 2 = 0.Definition 1.11. An integral domain D is a commutative ring that has amultiplicative identity but no zero divisors.Example 1.12. The set of all integers Z is an integral domain.Definition 1.13. A field is an integral domain where for each a ∈ D, a 6= 0,there exists a b ∈ D such that ab = 1. That is, all elements in D have amultiplicative inverse.Definition 1.14. Let D be an integral domain. Then there exists a field F ,called the field of quotients of D (or the quotient field of D) that containsan isomorphic copy D′ of D. That is, the quotient field of D is the smallestfield containing D as a subring.Definition 1.15. An integral domain D is said to be integrally closed if theonly elements of its quotient field that are integral over D are those of Ditself.Definition 1.16. An element a of an integral domain D is called a unit ifa | 1, that is 1 = ad for some d ∈ D. The set of units of D is denoted byU(D).Theorem 1.17. The set of units U(D) of an integral domain D forms anAbelian group with respect to multiplication.Definition 1.18. A nonzero, nonunit element a of an integral domain D iscalled an irreducible if a = bc, where b, c ∈ D implies that either b or c is aunit.31.2. IdealsDefinition 1.19. A nonzero, nonunit element p of an integral domain D iscalled a prime if p | ab, where a, b ∈ D, implies that either p | a or p | b.Theorem 1.20. In any integral domain D a prime is irreducible.1.2 IdealsDefinition 1.21. An ideal I of an integral domain D is a nonempty subsetD having the following two properties:− a ∈ I, b ∈ I ⇒ a+ b ∈ I− a ∈ I, r ∈ D ⇒ ra ∈ IThe following are some properties of ideals:− If a1, . . . , an ∈ I, then so are all r-linear combinations of these ai.That is, for all r1, . . . , rn ∈ D, r1a1 + · · ·+ rnan ∈ I.− If a ∈ I and b ∈ I, then −a ∈ I and a− b ∈ I.− 0 ∈ I, and if 1 ∈ I then I = D for an integral domain D.Example 1.22. If {a1, . . . , an} is a set of elements of an integral domainD, then the set of all finite linear combinations of a1, . . . , an{n∑i=1riai | r1, . . . , rn ∈ D}is an ideal of D, which we denote by 〈a1, . . . , an〉.Definition 1.23. An ideal I of an integral domain D is called a principalideal if there exists an element a ∈ I such that I = 〈a〉 = {ra | r ∈ D}. Theelement a is called a generator of the ideal I. The princpal ideal 〈0〉 is justthe set {0} and the ideal 〈1〉 is all of D.Definition 1.24. An ideal I of an integral domain D is called a properideal of D if I 6= 〈0〉 , 〈1〉.41.3. Dedekind DomainsDefinition 1.25. A proper ideal P of an integral domain D is called a primeideal ifa, b ∈ D and ab ∈ P implies a ∈ P or b ∈ P.Theorem 1.26. Let D be an integral domain. Let a ∈ D be such that a 6= 0and a 6∈ U(D). Then〈a〉 is a prime ideal of D ⇐⇒ a is prime in D.Definition 1.27. A proper ideal M of an integral domain D is called amaximal idea if whenever I is an ideal of D such that M ⊆ I ⊆ D impliesthat either I = M or I = D.1.3 Dedekind DomainsStill building structure on top of integral domains, we will define aDedekind domain, which has important structure for our algebraic numbertheory, namely with respect to the ring of integers of an algebraic numberfield.Definition 1.28. An infinite sequence of ideals {In : n = 1, 2, . . .} in anintegral domain D is said to be an ascending chain ifI1 ⊆ I2 ⊆ . . . ⊆ In ⊆ . . .The chain is said to be a strictly ascending chain ifI1 ⊂ I2 ⊂ . . . ⊂ In ⊂ . . .Definition 1.29. We say that an integral domain satisfies the ascendingchain condition (ACC) if there exists a positive integer r such that Ir = Isfor all s ≥ r.Definition 1.30. An integral domain that satisfies the ascending chaincondition above is called a Noetherian domain. More generally, a Noetherian51.4. Field and Galois Theoryring is a ring R in which every ascending chain of (two-sided) ideals in Rterminates.Definition 1.31. An integral domain D that satisifies the following threeproperties:− D is a Noetherian domain,− D is integrally closed, and− each prime ideal of D is a maximal ideal,is called a Dedekind domain.1.4 Field and Galois Theory1.4.1 Field TheoryWe introduce some basic field theory that will allow us to define fieldextensions and introduce some Galois theory. We have defined a field in theprevious section, but for clarity, we will present the definition of a field withall of its core axioms:Definition 1.32. F is a field if all of the following hold:1. F is closed under addition.2. Addition and multiplication are commutative and associative in F .3. F contains an additive and multiplicative identity (usually denoted as0 and 1) that are unique.4. Every element a in F has an additive inverse (denoted by −a) andevery nonzero element a in F has a multiplicative inverse (denoted bya−1).5. For every a ∈ F , a0 = 0a = 0.6. Addition and multiplication are distributive, that is for every a, b, c ∈F , (a+ b)c = ab+ bc and c(a+ b) = ca+ ba.61.4. Field and Galois TheoryDefinition 1.33. The characteristic char(F ) of the field F is the smallestpositive integer n such that n · 1 = 0 ∈ F , or 0 if no such n exists.Lemma 1.34. Let F be a field. Then char(F ) is either 0 or a prime.Definition 1.35. If a subset E of the elements of a field F satisfies the fieldaxioms with the same operations of F , then E is called a subfield of F . Ina finite field of order pn, where p is a prime, there exists a subfield of orderpm for every m dividing n.Definition 1.36. A field E is an extension of the field F if F is a subfieldof E. We denote this as E/F .Definition 1.37. Let E be a field extension of F . Then α ∈ E is algebraicover F if α is a root of some polynomial f(x) that has coefficients in F .Definition 1.38. We say that E is an algebraic extension of F (or E/F isalgebraic) if every α ∈ E is algebraic over F .Lemma 1.39. The following are equivalent:1. α is algebraic over F .2. F (α)/F is finite.3. F (α) = {polynomials in α with coefficients in F}.4. α is in B for some finite extension B of F .1.4.2 Field Extensions as Vector SpacesDefinition 1.40. Let L/K be a field extension, and suppose that α ∈ L isalgebraic over K. Then the minimial polynomial of α over K is the uniquemonic polynomial f over K of smallest degree such that f(α) = 0.A very useful consequence of forming a field extension over a field Kis that with certain operations, the extension forms a vector space over K.This allows us to associate the dimension of the vector space over the fieldK with the minimial polynomial of the α ∈ L of the extension.71.4. Field and Galois TheoryTheorem 1.41. If L/K is a field extension, then the operations(λ, u) 7→ λu (λ ∈ K,u ∈ L)(u, v) 7→ u+ v (u, v ∈ L)define on L the structure of a vector space over K.Definition 1.42. The degree [L : K] of a field extension L/K is the dimen-sion of L considered as a vector space over K.Theorem 1.43. If K0 ⊂ K1 ⊂ · · · ⊂ Kn are subfields of C with [Kn : K0] <∞, then[Kn : K0] = [Kn : Kn−1][Kn−1 : Kn−2] · · · [K1 : K0]Proposition 1.44. Let K(α)/K be a simple extension and let f be theminimal polynomial of α over K. If the extension is algebraic, then [K(α) :K] = δf where δf is the polynomial degree of f .1.4.3 The Group of Permutations of a PolynomialNow that we have talked about field extensions, we introduce the conceptof the group of permutations of the roots of a given polynomial. We use anexample to illustrate this idea.Consider the polynomial f(t) = t4 − 4t2 − 5 which factorizes asf(t) = (t2 + 1)(t2 − 5).From this, we can see that f(t) has four roots: t = ±i,±√5, where i and−i are conjugates and √5 and −√5 are conjugates. We letα = i, β = −i, γ =√5, δ = −√5.If we now consider a set of polynomial equations that satisfy the abovefour roots, we can find valid algebraic equations that may either stay valid,or become invalid, depending on how we interchange the roots. A set of81.4. Field and Galois Theoryequations for which the above are satisfied are stated next. Note that thereare infinitely many such equations.α2 + 1 = 0 α+ β = 0 δ2− 5 = 0 γ + δ = 0 αγ − βδ = 0The next step is to decide which roots we can interchange. For example,if we interchange α and γ, we obtain the equation γ2 + 1 = 0, which is false.We can interchange α to β only and keep γ, δ fixed, interchange γ to δ onlyand keep α, β fixed, interchange both at once, or don’t interchange them atall. This gives us four possible permutations of the roots of f(t):I = (1)R = (αβ)S = (γδ)T = (αβ)(γδ)These four permutations form a subgroup of S4, namely the Klein-4group C2 × C2 as all non-identity elements have order 2.1.4.4 Normality, Separability, and the GaloisCorrespondenceDefinition 1.45. If K is a subfield of C and f is a polynomial over K, thenf splits over K if it can be expressed as a product of linear factorsf(t) = k(t− α1) · · · (t− αn)where k, α1, . . . , αn ∈ K.Definition 1.46. Let K ⊆ C be a field. A subfield Σ of C is a splittingfield for the polynomial f over the subfield K of C if K ⊂ Σ and1. f splits over Σ.2. If K ⊂ Σ′ ⊂ Σ and f splits over Σ′, then Σ′ = Σ.91.4. Field and Galois Theory3. Σ = K(σ1, . . . , σn) where σ1, . . . σn are the zeros of f in Σ.Definition 1.47. A field extension L/K is normal if every irreducible poly-nomial f over K that has at least one zero in L splits in L.Definition 1.48. Let L be a finite extension of K. A normal closure ofL/K is an extension N of L such that1. N/K is normal.2. If L ⊆M ⊆ N and M/K is normal, then M = N .Thus N is the smallest extension of L that is normal over K.Normality is important as non-normal extensions have Galois groups thatdon’t behave in a straight-forward way. Normal extensions have a Galoiscorrespondence that is a bijection.Theorem 1.49. A field extension L/K is normal and finite if and only ifL is a splitting field for some polynomial over K.Definition 1.50. An algebraic field extension L/K that is normal and sep-arable is called a Galois extension of L over K; or equivalently, L/K isalgebraic and the field fixed by the automorphism group Aut(L/K) is pre-cisely the base field K.Definition 1.51. An irreducible polynomial f over a subfield K of C isseparable over K if it has simple zeros in C, or equivalently, simple zeros inits splitting field.Proposition 1.52. If K is a subfield of C, then every irreducible polynomialover K is separable.The connection between field theory and group theory made by Galoisallows us to describe how the roots of a polynomial equation are related toeach other. For a given polynomial with roots over Q, we want to find theGalois group that contains the permutations of that root within that field.The Fundamental Theorem of Galois Theory provides the link between thesetwo ideas.101.4. Field and Galois TheoryTheorem 1.53. Let E be a finite extension of a field F , G = Gal(E/F ),and GB = {g ∈ G | g(b) = b ∀b ∈ B}.1. There is a one-to-one correspondence between intermediate fields E ⊇B ⊇ F and subgroups {1} ⊆ GB ⊆ G given byB = Fix(GB)where Fix(GB) is the fixed field of GB, that is the set of all elementsx ∈ E such that x is fixed under the permutations α ∈ B.2. B is a normal extension of F if and only if GB is a normal subgroupof G. This is the case if and only if B is a Galois extension of F . Inthis caseGal(B/F ) ' G/GB3. For each E ⊇ B ⊇ F , [B : F ] = [G : GB] and [E : B] = |GB|The inverse Galois problem poses another interesting question: given afinite group G, is it possible to find a polynomial of specified degree over Qwhose Galois group is G? Moreover, can we find an infinite family of thesepolynomials?We are interested in finite groups and want to find parametric families ofpolynomials that have a prescribed Galois group. A partial existence resultis the following theorem.Theorem 1.54. Every finite abelian group is the Galois group of a Galoisextension E of Q.The possible Galois groups of irreducible polynomials of degrees from 2to 5 are listed in a table below:We can approach this table two ways:1. Given a polynomial of say degree 4, there are only five possible Galoisgroups that this polynomial can belong to, or2. Given a particular Galois group, say A4, we know that an irreduciblepolynomial with this Galois group must be a quartic.111.5. Algebraic Number TheoryTable 1.1: Possible Galois groups for polynomials from degree 2 to 5Quadratic Cubic Quartic QuinticC2 C3 C4 C5S3 S4 S5A4 A5D4 D5V = C2 × C2 F201.5 Algebraic Number TheoryIn this section, we provide a brief overview of the necessary algebraicnumber theory used in the rest of the thesis.Definition 1.55. An algebraic number is an element α ∈ C that is a rootof a monic polynomial with coefficients in Q. That is, α is a root of apolynomialxn + an−1xn−1 + · · ·+ a1x+ a0,where ai ∈ Q, i = 1, 2, . . . , n− 1.Definition 1.56. An algebraic integer is an element β ∈ C that is a rootof a monic polynomial with coefficients in Z. That is, β is a root of thepolynomialxn + an−1xn−1 + · · ·+ a1x+ a0,where ai ∈ Z, i = 1, 2, . . . , n− 1.Example 1.57. The element√22 is an algebraic number because it is a rootof the monic polynomial x2 − 12 ∈ Q[x].Example 1.58. The element√2 is an algebraic integer because it is a rootof the monic polynomial x2 − 2 ∈ Z[x].Example 1.59. The element√2 +√3 is an algebraic integer because it isa root of the monic polynomial x4 − 10x2 + 1 ∈ Z[x].In general, any algebraic number is a root of infinitely many polynomialsin Q[x]. For example,√2 is also a root of the monic polynomials x3−2x, x3−121.5. Algebraic Number Theoryx2 − 2x + 2, x5 − 2x4 − 7x3 + 10x2 + 10x − 12, etc. We want to work withthe polynomial of least degree with the algebraic integer/number as a root.This leads us into the following definition:Definition 1.60. Let K be a subfield of C. Let α ∈ C be algebraic overK. Then the monic polynomial f(x) ∈ K[x] such that f(α) = 0 with leastdegree is the minimal polynomial of α over K.The minimal polynomial defined above is unique and also irreducible,which we will restate from Definition 1.18 in the specific case of polynomialirreducibility.Definition 1.61. A polynomial f(x) ∈ K[x] is called irreducible over Kif deg(f) > 0 and it cannot be factored into the product of two non-constant polynomials over K[x]. That is, if f(x) = g(x)h(x) for polynomialsg(x), h(x) ∈ K[x] then either g(x) or h(x) is a unit in K[x]. Otherwise, f(x)is reducible over K[x].Definition 1.62. Let K be a subfield of C. Let α ∈ C be algebraic over K.Then the degree of α over K, written degK(α), is defined bydegK(α) = deg(irrK(α)),where irrK(α) is the irreducible, minimal polynomial of α over K. If K = Q,we drop the subscript K and write degQ(α) = deg(α).1.5.1 Field Extensions and Algebraic Number FieldsDefinition 1.63. Let α ∈ C be algebraic over K. We define K(α) to be theintersection of the subfields of C containing K and α. We say that K(α) isformed from K by adjoining α.Definition 1.64. A subfield L of C for which there exists α ∈ C such thatL = K(α) is called a simple extension of K.131.5. Algebraic Number TheoryTheorem 1.65. Let K be a subfield of C. Let α ∈ C be algebraic over K.Let n = deg(irrK(α)). ThenK(α) = {a0 + a1α+ · · ·+ an−1αn−1 | a0, . . . , an−1 ∈ K}.This allows us to view K(α) as an n-dimensional vector space over Kwith basis {1, α, . . . , αn−1}.Definition 1.66. Let K be a subfield of C. Let α ∈ C be algebraic overK of degree n (so that n = degK(α) = deg(irrK(α))). The degree of theextension K(α) over K, written [K(α) : K], is defined by[K(α) : K] = n.In the same fashion, let α1, . . . , αk ∈ C be algebraic over K and defineK(α1, . . . , αk) to be the smallest field containing K and the α1, . . . , αk.Remarkably, fields constructed in this fashion are always simple extensionsas the next theorem illustrates:Theorem 1.67 (Primitive Element Theorem). Let K be a subfield of C. Letα1, α2, . . . , αn be algebraic over K. Then there exists α ∈ C that is algebraicover K such thatK(α1, α2, . . . , αn) = K(α).We now look at algebraic number fields, which are formed by adjoiningfield elements onto a base field, usually Q.Definition 1.68. An algebraic number field is a subfield of C of the formQ(α1, . . . , αn) where α1, . . . , αn are algebraic numbers.By Theorem 1.67, an algebraic number field can always be obtained byadjoining a single algebraic number θ to Q. The following theorem gives therepresentation of the elements of an algebraic number field Q(θ).Theorem 1.69. Let K = Q(θ) be an algebraic number field, where θ is analgebraic number. Let the degree of the polynomial irrQ(θ) be n. Then every141.5. Algebraic Number Theoryelement of K is expressible uniquely in the formc0 + c1θ + · · ·+ cn−1θn−1,where c0, . . . , cn−1 ∈ Q, and every such quantity c0 + c1θ + · · · + cn−1θn−1(c0, . . . , cn−1 ∈ Q) belongs to K.We now formally introduce a concept touched upon in the last section,the integers that lie inside a number field K. This set will be very importantin the theory of algebraic numbers.Definition 1.70. Let Ω be the set of all algebraic integers. The set of allalgebraic integers that lie in the algebraic number field K is denoted by OK ;that is,OK = Ω ∩K.We call OK the ring of integers of the algebraic number field K.Theorem 1.71. Let K be an algebraic number field. Then OK is an integraldomain.1.5.2 Conjugates and Conjugate Fields of an AlgebraicNumber FieldIn this section, we will discuss conjugates of a number field K that willbe necessary for our later chapters.Consider the element α ∈ K and let f(x) ∈ Q[x] be the minimal poly-nomial of α of degree n where α1, α2, . . . αn are the complex roots of f . Foreach k = 1, . . . , n, the map σk : α 7→ αk induces a field homomorphismσk : Q(α) −→ Q(αk) ⊂ C.This map above is well-defined and we call the maps σ1, . . . σn the distinctfield embeddings K → C.Definition 1.72. Let α ∈ C be algebraic over a subfield K of C. Theconjugates of α over K are the roots in C of irrK(α).151.5. Algebraic Number TheoryExample 1.73. Consider the element α =1 + i√2. The minimal polynomialof α is f(x) = x4 + 1.Sincex4+1 =(x−(1 + i√2))(x−(1− i√2))(x+(1 + i√2))(x+(1− i√2)),the conjugates of (1 + i)/√2 over Q are1 + i√2,1− i√2,−1− i√2,−1 + i√2.Theorem 1.74. If α is an algebraic integer then its conjugates over Q arealso algebraic integers.Theorem 1.75. If α is an algebraic integer then irrQ(α) ∈ Z[x].Theorem 1.76. Let K be an algebraic number field of degree n over Q.Then there are exactly n distinct field monomorphisms σk : K → C (k =1, . . . , n).Definition 1.77 (Conjugate fields of an algebraic number field). LetK be an algebraic number field. Let θ be an algebraic number such thatK = Q(θ). Letθ1 = θ, θ2, . . . , θnbe the conjugates of θ over Q. Then the fieldsQ(θ1) = Q(θ) = K,Q(θ2), . . . ,Q(θn)are called the conjugate fields of K.Let K be an algebraic number field of degree n over Q. Let θ ∈ K besuch that K = Q(θ) and let θ1 = θ, θ2, . . . , θn be the conjugates of θ overQ. Recall Theorem 1.69 that states for α in K there exist unique rationalnumbers c0, c1, . . . , cn−1 such thatα = c0 + c1θ + · · ·+ cn−1θn−1.161.5. Algebraic Number TheoryFor k = 1, 2, . . . , n setαk = c0 + c1θk + · · ·+ cn−1θn−1k ∈ Q(θk).Definition 1.78. The K-conjugates of α, or the complete set of conjugatesof α relative to K, are the set of algebraic numbers {α1 = α, α2, . . . , αn}.The K-conjugates of the algebraic number α are actually the roots ofirr(α), the minimal polynomial of α over K.Definition 1.79. Let K be an algebraic number field of degree n. Letα ∈ K. Let α = α1, α2, . . . , αn be the K-conjugates of α. Then the fieldpolynomial of α over K is the polynomialfldK(α) =n∏k=1(x− αk).Theorem 1.80. Let K be an algebraic number field of degree n. Let α ∈ K.ThenfldK(α) ∈ Q[x].Definition 1.81. Let K = Q(θ) be an algebraic number field of degree n.Let θ1 = θ, θ2, . . . , θn be the conjugates of θ over Q. If Q(θ1) = Q(θ2) =· · · = Q(θn) = K, the field K is said to be a normal or Galois extension ofQ.If all the roots of the minimal polynomial of θ are in K, then K is anormal extension over Q.Example 1.82. Let K = Q(√2,√3) = Q(√2 +√3). The conjugates of√2 +√3 are ±√2±√3 and the conjugate fields of K all coincide with K asQ(±√2±√3) = Q(√2 +√3) = K.Thus K is a normal extension.Example 1.83. Let K = Q( 3√2) so that K ⊆ R. The conjugates of 3√2 are3√2, ω3√2, ω23√2,171.5. Algebraic Number Theorywhere ω = exp(2pii/3) and ω2 = exp(4pii/3) are the two complex cube rootsof unity, sinceirrQ(3√2) = x3 − 2 = (x− 3√2)(x− ω 3√2)(x− ω2 3√2).The conjugate fields of K areK1 = Q(3√2) = K, K2 = Q(ω3√2), K3 = Q(ω23√2).We want to show that the conjugate fields are distinct, showing that Q( 3√2)is not a normal extension. Clearly as K1 is a real field and K2 and K3 arenot, we have K1 6= K2 and K1 6= K3. Therefore it remains to show thatK2 6= K3.By way of contradiction, assume K2 = K3. Then ω2 3√2 ∈ K2. Thisimplies that there exist a, b, c ∈ Q such thatω23√2 = a+ bω3√2 + c(ω3√2)2.Taking complex conjugates we obtainω3√2 = a+ bω23√2 + cω(3√2)2since ω¯ = ω2. Subtracting the second equation from the first equation above,(ω2 − ω) 3√2 = −b(ω2 − ω) 3√2 + c(ω2 − ω)( 3√2)2,so that3√2 = −b 3√2 + c(3√2)2.Cancelling 3√2 from this equation, we get1 + b = c3√2.181.5. Algebraic Number TheoryBut since 3√2 6∈ Q, we must have 1 + b = c = 0 soω23√2 = a− ω 3√2.Thus(ω2 + ω)3√2 = aand since ω2 + ω = −1,3√2 = −a ∈ Q,which is a contradiction. Therefore all the conjugate fields of Q( 3√2) aredistinct, and Q( 3√2) is not a normal field.1.5.3 DiscriminantsDefinition 1.84. Let K be an algebraic number field of degree n. Letω1, . . . , ωn be n elements of the field K. Let σk (k = 1, 2, . . . , n) denote then distinct monomorphisms: K → C. For i = 1, 2, . . . , n letω(1)i = σ1(ωi) = ωi, ω(2)i = σ2(ωi), . . . , ω(n)i = σn(ωi)denote the conjugates of ωi relative toK. 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.In a similar fashion, we define the discriminant of a single element α ofK as follows:Definition 1.85. Let K be an algebraic number field of degree n. Letα ∈ K. Then we define the discriminant of α byD(α) = D(1, α, α2, . . . , αn−1).Theorem 1.86. Let K be an algebraic number field of degree n. Let α ∈ K.191.5. Algebraic Number TheoryThenD(α) =∏1≤i<j≤n(α(i) − α(j))2,where α(1) = α, α(2), . . . , α(n) are the conjugates of α with respect to K. Thediscriminant above is actually the determinant of a Vandermonde matrix.We now give some theory that links the discriminant of a polynomialwith the discriminant of an element α of an algebraic number field K.Definition 1.87 (Discriminant of a polynomial). Letf(x) = anxn + an−1xn−1 + · · ·+ a1x+ a0 ∈ C[x],where n ∈ N and an 6= 0. Let α1, . . . , αn ∈ C be the roots of f(x). Thediscriminant of f(x) is the quantitydisc(f(x)) = a2n−2n∏1≤i<j≤n(αi − αj)2 ∈ C.Theorem 1.88. Let K be an algebraic number field of degree n. Let α ∈ K.ThenD(α) = disc(fldK(α)).Theorem 1.89. Let K be an algebraic number field of degree n. Let α ∈ K.ThenK = Q(α) if and only if D(α) 6= 0.Theorem 1.90. Let K be an algebraic number field of degree n.1. If ω1, . . . , ωn ∈ K thenD(ω1, . . . , ωn) ∈ Q.2. If ω1, . . . , ωn ∈ OK thenD(ω1, . . . , ωn) ∈ Z.201.5. Algebraic Number Theory3. If ω1, . . . , ωn ∈ K thenD(ω1, . . . , ωn) 6= 0 if and only if ω1, . . . , ωn are linearly independent over Q.1.5.4 Ideals in Algebraic Number TheoryTheorem 1.91. Let K be an algebraic number field of degree n. Let I be anonzero ideal in OK . Then there exist η1, . . . , ηn ∈ I such thatD(η1, . . . , ηn) 6= 0.Theorem 1.92. Let K be an algebraic number field of degree n. Let I bea nonzero ideal of OK . There there exist elements η1, . . . , ηn of I such thatevery element α of I can be expressed uniquely in the formα = x1η1 + · · ·+ xnηn,where x1, . . . , xn ∈ Z.Theorem 1.93. Let K be an algebraic number field. Then OK is a Noethe-rian domain.Definition 1.94 (Basis of an ideal). Let K be an algebraic number field ofdegree n. Let I be a nonzero ideal of OK . If {η1, . . . , ηn} is a set of elementsof I such that every element α ∈ I can be expressed uniquely in the formα = x1η1 + · · ·+ xnηn (x1, . . . , xn ∈ Z)then {η1, . . . , ηn} is called a basis for the ideal I.Definition 1.95 (Discriminant of an ideal). Let K be an algebraic numberfield of degree n. Let I be a nonzero ideal of OK . Let {η1, . . . , ηn} be a basisof I. Then the discriminant D(I) of the ideal I is the nonzero integer givenbyD(I) = D(η1, . . . , ηn).Definition 1.96 (Integral basis of an algebraic number field). Let K be analgebraic number field. A Z-basis for OK is called an integral basis for K.211.5. Algebraic Number TheoryTheorem 1.97. Let K be a quadratic field. Let m be the unique squarefreeinteger such that K = Q(√m). Then {1,√m} is an integral basis for K ifm 6≡ 1 (mod 4) and{1, 1+√m2}is an integral basis for K if m ≡ 1 (mod 4).Definition 1.98 (Discriminant of an algebraic number field). Let K be analgebraic number field of degree n. Let {η1, . . . , ηn} be an integral basis forK. Then D(η1, . . . , ηn) is called the discriminant of K and is denoted byd(K).Theorem 1.99. Let K be a quadratic field. Let m be the unique squarefreeinteger such that K = Q(√m). Then the discriminant d(K) of K is givenbyd(K) =4m, if m 6≡ 1 (mod 4),m, if m ≡ 1 (mod 4).Definition 1.100 (Norm of an ideal). Let K be an algebraic number fieldof degree n. Let I be a nonzero ideal of OK . Then the norm of the ideal I,written N(I), is the positive integer defined byN(I) =√D(I)d(K).Definition 1.101 (Index of θ). Let K be an algebraic number field. Letθ ∈ OK be such that K = Q(θ). Then the index of θ, written indθ, is thepositive integer given byD(θ) = (indθ)2d(K).Theorem 1.102. Let K be an algebraic number field of degree n. Let θ ∈OK be such that K = Q(θ). Then {1, θ, θ2, . . . , θn−1} is an integral basis forK if and only if indθ = 1.Theorem 1.103. Let K be an algebraic number field of degree n. Let θ ∈OK be such that K = Q(θ). If D(θ) is squarefree then {1, θ, . . . , θn−1} is anintegral basis for K.221.5. Algebraic Number TheoryTheorem 1.104 (Stickelberger’s Theorem). Let K be an algebraic numberfield. Thend(K) ≡ 0 or 1 (mod 4).1.5.5 Ideals in a Dedekind DomainIt’s known that the ring of algebraic integers OK of an algebraic numberfield K is a Dedekind domain. Therefore we are interested in how ideals actin this sort of structure. The motivation for the study of ideal theory is torestore unique factorization when working in the ring of integers of K.Definition 1.105 (Fractional Ideal). Let D be an integral domain. Let Kbe the quotient field of D. A nonempty subset A of K with the followingthree properties:1. α ∈ A, β ∈ A =⇒ α+ β ∈ A,2. α ∈ A, r ∈ D =⇒ rα ∈ A, and3. there exists γ ∈ D with γ 6= 0 such that γA ⊆ Dis called a fractional ideal of D. An ideal in the ordinary sense (let γ = 1)is a fractional ideal, and is often referred to as an integral ideal.Theorem 1.106. If D is a Dedekind domain, every proper integral ideal isa product of prime ideals and this factorization is unique in the sense thatifP1P2 · · ·Pk = Q1Q2 · · ·Ql,where the Pi and Qj are prime ideals, then k = l, and after relabeling (ifnecessary)Pi = Qi, i = 1, 2, . . . , k.Recall that the fundamental theorem of arithmetic for integers statesthat every integer can be written uniquely as a product of primes. Thisnext theorem for ideals is analogous.231.5. Algebraic Number TheoryCorollary 1.107. Let K be an algebraic number field. Then every properintegral ideal of OK can be expressed uniquely up to order as a product ofprime ideals.Definition 1.108. Let D be a Dedekind domain. Let A and B be nonzerointegral ideals of D. We say that A divides B, written A | B, if there existsan integral ideal C of D such that B = AC.1.5.6 Norm of a Prime IdealTheorem 1.109. Let K be an algebraic number field. Let P be a primeideal of OK . Then there exists a unique rational (integer) prime p such thatP | 〈p〉.The rational prime p in the above theorem is called the prime lying belowP since P ⊇ 〈p〉. In this case, we would likewise say that the prime ideal Pis lying above the rational prime p.Definition 1.110. Let K be an algebraic number field of degree n. Let Pbe a prime ideal of OK . Let p be a rational prime lying below P . Then theunique positive integer e such thatP e | 〈p〉, P e+1 - 〈p〉is called the ramification index of P in K and is written eK(P ).Theorem 1.111. Let K be an algebraic number field with [K : Q] = n. LetP be a prime ideal of OK . Let p be the rational prime lying below P . ThenN(P ) = pffor some integer f ∈ {1, 2, . . . , n}.Definition 1.112. Let K be an algebraic number field with [K : Q] = n.Let p be the rational prime lying below P . Then the positive integer f such241.5. Algebraic Number TheorythatN(P ) = pfis called the inertial degree of P in OK and is denoted by fK(P ).Theorem 1.113. Let K be an algebraic number field with [K : Q] = n. Letp be a rational prime. Suppose that the principal ideal 〈p〉 factors in OK inthe form〈p〉 = P e11 · · ·P egg ,where P1, . . . , Pg are distinct prime ideals of OK and e1, . . . , eg are positiveintegers. Suppose that fi is the inertial degree of Pi (i = 1, 2, . . . , g) in K,that is, fi = fK(Pi). Thene1f1 + · · ·+ egfg = n.Definition 1.114. The positive integer g in the above theorem is called thedecomposition number of p in K and is written gK(p) with gK(p) ≤ n..Definition 1.115. Let K be an algebraic number field of degree n and pbe a rational prime. Let〈p〉 = P e11 · · ·P errbe the prime ideal factorization of 〈p〉 in K. If ei > 1 for some i = 1, 2, . . . , rthen p is said to ramify in K. If ei = 1 for i = 1, 2, . . . , r then p is said tobe unramified in K.Proposition 1.116. [Nar90, p. 159] If p is a prime ideal of a Dedekinddomain R unramified in both K1/K and K2/K then it is also unramified inthe composite extension K1K2/K.Theorem 1.117. Let K be an algebraic number field. Then the rationalprime p ramifies in K if and only if p | d(K).Theorem 1.118. Let K be an algebraic number field. Let I be an nonzeroideal of OK .(a) If N(I) = p, where p is a prime, then I is a prime ideal.(b) N(I) ∈ I.251.5. Algebraic Number Theory1.5.7 Factoring Primes in a Quadratic FieldLet p be a rational prime and let K be a quadratic field. Since the degreeof K over Q is 2, we know that the decomposition number g = gK(p) ≤ 2so that g = 1 or 2.If g = 2, then by Theorem 1.113e1f1 + e2f2 = 2so thate1 = f1 = e2 = f2 = 1.If g = 1, thene1f1 = 2so that(e1, f1) = (2, 1) or (1, 2).Therefore, we have three different cases:1. g = 2, e1 = f1 = e2 = f2 = 1,2. g = 1, e1 = 2, f1 = 1,3. g = 1, e1 = 1, f1 = 2.In other words,1. 〈p〉 = P1P2, N(P1) = N(P2) = p, P1 6= P2,2. 〈p〉 = P 2, N(P ) = p,3. 〈p〉 = P, N(P ) = p2.It is important to note that the ideal norm is multiplicative. This nexttheorem gives necessary and sufficient conditions for the above cases to oc-cur. It is the essential theorem to use when deciding how ideals factor intoprimes in quadratic fields.261.5. Algebraic Number TheoryDefinition 1.119. If m is a positive integer, we say that the integer a is aquadratic residue of m if gcd(a,m) = 1 and the congruence x2 ≡ a (mod m)has a solution. If the congruence has no solution, then a is a quadraticnonresidue of m.Definition 1.120. Let p be an odd prime and a be an integer not divisibleby p. The Legendre symbol(ap)is defined by(ap)= 1 if a is a quadratic residue of p−1 if a is a quadratic nonresidue of pTheorem 1.121. Let K be a quadratic field so that there exists a squarefreeinteger m such that K = Q(√m). Let p be a rational prime.1. If p > 2,(mp)= 1 or p = 2, m ≡ 1 (mod 8) then〈p〉 = P1P2,where P1 and P2 are distinct prime ideals with N(P1) = N(P2) = p.2. If p > 2, p | m or p = 2,m ≡ 2 or 3 (mod 4) then〈p〉 = P 2,where P is a prime ideal with N(P ) = p.3. If p > 2,(mp)= −1 or p = 2, m ≡ 5 (mod 8) then〈p〉 is a prime ideal of OK .Using the previous theorem, we can now express the factorizations ofthe ideals 〈p〉 into prime ideals of OK in the quadratic field K = Q(√m) asfollows:271.5. Algebraic Number Theory〈2〉 =〈2〉, if m ≡ 5 (mod 8),〈2, 12(1 +√m)〉〈2, 12(1−√m)〉, if m ≡ 1 (mod 8),〈2, 1 +√m〉2, if m ≡ 3 (mod 4),〈2,√m〉2, if m ≡ 2 (mod 4),and for p > 2,〈p〉 =〈p〉, if p - m and x2 ≡ m (mod p) is insolvable,〈p, x+√m〉〈p, x−√m〉 if p - m and x2 ≡ m (mod p) is solvable,〈p,√m〉2, if p | m.1.5.8 Factoring Primes in a Monogenic Number FieldDefinition 1.122. Let K be an algebraic number field. K is monogenic ifthere exists θ ∈ OK such that {1, θ, θ2, . . . , θn−1} is an integral basis for K.The following theorem shows how to factor the ideal 〈p〉 (p a rationalprime) into prime ideals in a monogenic number field.Theorem 1.123 (Dedekind’s Theorem). Let K = Q(θ) be an algebraicnumber field of degree n with θ ∈ OK . Let p be a rational prime. Letf(x) = irrQ(θ) ∈ Z[x].If p is not a divisor of the index of θ, andf(x) ≡ g1(x)e1 · · · gr(x)er (mod p),is the decomposition of f(x) (mod p) where g1(x), . . . , gr(x) are distinct,monic, irreducible polynomials in Zp[x] and e1, . . . , er are positive integers,then there exists r distinct prime ideals of OK , P1, . . . , Pr with〈p〉 = P e11 · · ·P err281.5. Algebraic Number TheoryandN(Pi) = pdeg gi , i = 1, 2, . . . , r.Furthermore,Pi = 〈p, gi(θ)〉, i = 1, 2, . . . , r.Example 1.124. Let K = Q( 3√2). We want to factor the ideal 〈11〉 as aproduct of prime ideals in OK . Let θ =3√2 so that the minimal polynomialof θ is x3 − 2. An integral basis for K = Q(θ) is {1, θ, θ2} so that K ismonogenic. Factoring x3 − 2 modulo 11, we getx3 − 2 = (x+ 4)(x2 + 7x+ 5) (mod 11),where x+ 4 and x2 + 7x+ 5 are both irreducible modulo 11. Therefore, byTheorem 1.123, we have〈11〉 = PQwhereP = 〈11, θ + 4〉, Q = 〈11, θ2 + 7θ + 5〉are distinct prime ideals withN(P ) = 11, N(Q) = 112 = 121.29Chapter 2Group Coverings andIntersective Polynomials2.1 Group Coverings with ExamplesIn this section, we discuss group covers, which will be integral in findingour intersective polynomials.Definition 2.1. Let H be a subgroup of a group G. Let g be a fixedelement of G that is not a member of H. Then the elements ghig−1 for allhi in H, i = 1, 2, . . . generates the conjugate subgroup gHg−1. If for all g,gHg−1 = H, then H is a normal (or self-conjugate or invariant) subgroup.Definition 2.2. Let G be a group. Two subgroups H1 and H2 of G arecalled conjugate subgroups if there is an element g in G such that gH1g−1 =H2.Proposition 2.3. Let G be a group and R be a relation on G defined bya ∼ b if a is conjugate to b. Then a ∼ b if there is a g in G such thata = gbg−1. Then R is an equivalence relation, that is, conjugation adheresto the three following properties:1. a ∼ a (Reflexive)2. If a ∼ b then b ∼ a (Symmetric)3. If a ∼ b and b ∼ c then a ∼ c (Transitive)Definition 2.4. A group is n-coverable if it is a union of conjugates of nproper subgroups, whose total intersection is trivial. More explicitly, let302.1. Group Coverings with ExamplesG be a finite group. Let H1, . . . ,Hn be the proper subgroups of G, eachhaving k1, . . . kn distinct conjugates (including the subgroup itself), notatedas H(j)i = {gHig−1 | g ∈ G} for 1 ≤ j ≤ ki.Now let Hi =ki⋃j=1H(j)i . The set-theoretical union of the Hi is called acovering for G if G =n⋃i=1Hi, where all of the conjugates of all the subgroupsin the union have trivial intersection.When we talk about n-coverable groups we will always assume that n > 1as the next lemma establishes that no group is 1-coverable.Lemma 2.5. Let G be a finite group. Then G is not 1-coverable.Proof. Let G be a finite group of order |G| and let H be a proper subgroupof G. The conjugates of H we denote as Hi, for i = 1, 2, · · · k.Now let G =k⋃i=1Hi and let NG(H) = {g ∈ G | gHg−1 = H} be thenormalizer of H in G. Note that H ( NG(H) since H is not a normalsubgroup of G. The number of conjugates of H is the index of NG(H) in G,which is less than or equal to |G|/|H|, i.e.k ≤ |G|/|H|.However, if you only count the identity once, then the number of elementsin the union of the conjugates of H is at most|H| · |G||H| − (k − 1) = |G| − k + 1.Since H is a proper subgroup of G, k is at least 2. Thus the numberof elements in the union of the conjugates of H is less than |G|, and thus a1-cover is impossibleLemma 2.6. A cyclic group G cannot be n-coverable for n > 1.312.1. Group Coverings with ExamplesProof. Suppose that G is cyclic with G = 〈g〉 for some g ∈ G and supposethat H is a proper subgroup of G.Now suppose that G is n-coverable for n > 1. Then the element g mustbe contained in one of the conjugates of H since G is composed of the unionof the conjugates. Then this subgroup contains 〈g〉 = G. However theconjugate of a proper subgroup is proper, contradicting that it contains G.Thus G is not n-coverable.2.1.1 2-Coverable GroupsGroups that are 2-coverable (n = 2) are of particular interest. It has beenproven in Bubboloni [Bub98] that the symmetric group Sm is 2-coverable ifand only if 3 ≤ m ≤ 6 and the alternating group Am is 2-coverable if andonly if 4 ≤ m ≤ 8.The Frobenius GroupDefinition 2.7. A finite group G is said to be a Frobenius group if there isa non-trivial subgroup H of G such that H ∩gHg−1 = {1} whenever g 6∈ H.This gives a decompositionG =⋃gH∈G/H(gHg−1\{1}) ∪Kwhere K is the subset defined as the identity element 1 together with allthe non-identity elements that are not conjugate to any element of H. Thesubset K is called the Frobenius kernel of G and H is called the Frobeniuscomplement.Interestingly enough, there is a great amount of structure on K and thuson G, shown in a theorem of Frobenius himself.Theorem 2.8. Let G be a Frobenius group with Frobenius complement Hand Frobenius kernel K. Then K is a normal subgroup of G and thereforeG is the semidirect product K oH of H and K.322.1. Group Coverings with ExamplesThe Frobenius group is of interest because it has been proven in Sonn[Son08] that all Frobenius groups are 2-coverable. An example of an inter-sective polynomial with a Frobenius Galois group is given in Section 2.2.2.A4, The Alternating Group of 4 LettersConsider A4, the alternating group on 4 letters. We will show that A4is 2-coverable with subgroups isomorphic to V4 and Z/3Z respectively. NowA4 = {1, (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)}.ConsiderH1 = {1, (12)(34), (13)(24), (14)(23)}.Then H1 is isomorphic to V4, the Klein-4 group, and further [A4 : H1] = 3.H1 is normal in A4 and so the only conjugate of H1 in A4 is itself. NowconsiderH2 = {1, (123), (132)}, H3 = {1, (124), (142)}, H4 = {1, (134), (143)}, H5 = {1, (234), (243)}.Each of these subgroups are isomorphic to Z/3Z. Furthermore, they are allconjugate to one another.We show that H2, H3, H4, H5 are conjugate subgroups by finding ele-ments in A4 such that all of the Hi transitively conjugate with each other.This will finalize the result. Note that the conjugation of the identity ele-ment 1 is always the identity element. The necessary conjugations are givenbelow:(124)H2(142) = H5(124)H5(142) = H4(123)H4(132) = H3Therefore, H2 ∼ H5, H5 ∼ H4, and H4 ∼ H3 and thus Hi ∼ Hj for2 ≤ i, j ≤ 5.Taking the union of all 5 groups above, with H1 = H1 and H2 = H2 ∪332.1. Group Coverings with ExamplesH3 ∪H4 ∪H5, we conclude thatA4 = H1 ∪H2with trivial intersection. Thus {H1,H2} is a 2-cover of A4.S3, The Symmetric Group on 3 LettersWe now consider S3, the symmetric group on 3 letters. The two sub-groups of S3, A3 and C2, the cyclic group of order 2 (and its conjugates)form a cover. RecallS3 = {1, (12), (13), (23), (123), (132)}.Now consider the subgroupsH1 = {1, (123), (132)}andH2 = {1, (12)}, H3 = {1, (13)}, H4 = {1, (23)}.The subgroup H1 has index 2 and is therefore normal. This is the sub-group A3. We claim that the three subgroups H2, H3, and H4 are conjugatesubgroups. Indeed, letting g1 = (23) and g2 = (12), it is easy to see thatg1(12)g1−1 = (13)g2(13)g2−1 = (23)which verifies the claim so that H2 ∼ H3 and H3 ∼ H4. Then by transitivity,H2 ∼ H4 and they are all conjugate subgroups. Taking the union of all fourgroups above, we can show that for H1 = H1 and H2 = H2 ∪H3 ∪H4,S3 = H1 ∪H2with trivial intersection. Thus {H1,H2} is a 2-cover of S3.342.1. Group Coverings with ExamplesD5, The Dihedral Group of Order 10Recall that D5 is the group of symmetries of a regular pentagon withorder 10. Interestingly enough, the dihedral group can be viewed as a semi-direct product with D5 = C5 o C2. The subgroups of D5 either have order5 or order 2, thus D5 has only cyclic subgroups.We will work with the following group presentation of D5:D5 = {1, r, r2, r3, r4, f, fr, fr2, fr3, fr4} =〈f, r | r5 = f2 = 1, frf = r−1〉where r represents the rotation of the pentagon and f represents the flipof the pentagon over its vertical axis. The subgroups of D5 are H1 = 〈r〉,which is normal because it has index 2 in D5 and isomorphic to C5, and thesubgroups H2 = 〈f〉 , H3 = 〈fr〉 , H4 =〈fr2〉, H5 =〈fr3〉, and H6 =〈fr4〉,all isomorphic to C2. Notice that H1, . . . ,H6 are conjugate subgroups viathe relationsrH2r−1 = H5r2H5r−2 = H6fH6f−1 = H3rH3r−1 = H4Therefore, for H1 = H1 and H2 = H2 ∪H3 ∪H4 ∪H5 ∪H6, a 2-cover for D5is {H1,H2}.Dp, The Dihedral Group of Order 2pWe now generalize a cover for the dihedral group of order 2p. We will useSylow theory to show that all dihedral groups of order 2p are 2-coverable.Theorem 2.9 (Lagrange’s Theorem). If G is a finite group and H is asubgroup of G, then the order of H divides the order of G (that is |H| | |G|)and the number of left cosets of H in G equals |G|/|H|.352.1. Group Coverings with ExamplesWhile the consequence of this theorem is prevalent in group theory, theconverse isn’t always true: for any given divisor m of |G|, there doesn’tnecessarily exist a subgroup H of G such that |H| = m. For example, thealternating group on 4 letters, A4, has order 12 but has no subgroup oforder 6. We can, however, use a partial converse to Lagrange’s theorem wellknown as Sylow’s Theorem.Definition 2.10. Let G be a group and let p be a prime.1. A group of order pn for some n ≥ 1 is called a p-group. Subgroups ofG which are p-groups are called p-subgroups.2. If G is a group of order pnm, where p - m, then a subgroup of orderpn is called a Sylow p-subgroup of G.Theorem 2.11. For every prime factor p with multiplicity n of the orderof a finite group G, there exists a Sylow p-subgroup of G, of order pn.Theorem 2.12. Given a finite group G and a prime number p, all Sylowp-subgroups of G are conjugate to each other, i.e. if H and K are Sylowp-subgroups of G, then there exists an element g in G with g−1Hg = K.Theorem 2.13. Let p be a prime factor with multiplicity n of the order ofa finite group G, so that the order of G can be written as pnm, where n > 0and p does not divide m. Let np be the number of Sylow p-subgroups of G.Then the following hold:1. np divides m, which is the index of the Sylow p-subgroup in G.2. np ≡ 1 (mod p).3. np = |G : NG(P )|, where P is any Sylow p-subgroup of G and NGdenotes the normalizer of P in G.Lemma 2.14 (Cavior’s Theorem). If n ≥ 3, the number of subgroups of Dnis τ(n)+σ(n), where τ(n) denotes the number of divisors of n and representsthe number of cyclic subgroups of Dn, and σ(n) denotes the sum of divisorsof n and represents the number of noncyclic subgroups of Dn.362.1. Group Coverings with ExamplesTheorem 2.15. There are two kinds of subgroups for a dihedral group Dngiven byDn =〈r, f : rn = f2 = e, frf = r−1〉(see [Con]):1. Subgroups of the form〈rd〉, where d | n. There is only one suchsubgroup for each d. The total number of such subgroups is τ(n),where τ(n) denotes the number of positive divisors of n.2. Subgroups of the form〈rd, rkf〉where d | n and 0 ≤ k < d. There ared such subgroups for each such divisor d. The total number of suchsubgroups is σ(n), the sum of positive divisors of n.Because the order of Dp is 2p, by Theorem 2.11 there exists a Sylow 2-subgroup and a Sylow p-subgroup of Dp. The index of the Sylow p-subgroupis 2, and is therefore normal. Since np divides 2 and is also congruent to 1mod p, it must be true that np = 1, implying that there exists only 1 Sylowp-subgroup. By the first condition above, there are subgroups 〈r〉 and 〈rp〉of Dp. Because rp = e, we have 〈rp〉 = {e}. As | 〈r〉 | = p, we see that 〈r〉 isa Sylow p-subgroup of Dp, hence is unique by the above argument.Condition 2 also implies the existence of subgroups of the form〈rd, frk〉where d | p and 0 ≤ k < d. The case where d = 1 yields the entiregroup. For d = p there are p such subgroups, all of order 2. These p groupsare all conjugate groups by Theorem 2.12. Therefore for H1 = 〈r〉 andH2 =p−1⋃k=0〈frk〉, {H1,H2} forms a cover for Dp.2.1.2 The Holomorph Z32 o Z∗32We will introduce some theorems that are proven in general in Chapter4 for the holomorph Z2e o Z∗2e . This will allow us to state a 3-cover for thesingle case where e = 5. Let G be the groupG ' Z32 o Z∗32.372.1. Group Coverings with ExamplesConsider the subgroups {H1, H2, H3} of G defined byH1 = {(b, 5c) : b = 0, 1, . . . , 31, c = 0, 1, . . . , 7} ,H2 = {(0, d) : d = 1, 3, . . . , 31} ,H3 = 〈(1,−1), (3,−5)〉 .We will show that these three subgroups form a 3-cover for G.Lemma 2.16. The conjugates of the subgroupH2 = {(0, d) : d = 1, 3, . . . , 31}of the groupG ' Z32 o Z∗32.have the form{(n(d− 1), d) : d = 1, 3, . . . , 31} ,where n is a fixed integer modulo 32.Lemma 2.17. The conjugates of the subgroupH3 = 〈(1,−1), (3,−5)〉of the groupG ' Z32 o Z∗32.have the form〈(m,−1), (3m,−5)〉 ,for a fixed odd integer m modulo 32.We begin by showing G is equal to the union of the conjugates of thesubgroups Hi, i = 1, 2, 3. For notation, Hci denotes a conjugate of Hi. Weconsider a typical element(k, j) ∈ G382.2. Intersective Polynomialswhere k is any integer modulo 32 and j is any odd integer modulo 32. Wemay write j = ±52a or ±52a+1 for a nonnegative integer a. Some of the casesrequire that we set k = 2w or k = 2w + 1. The following table summarizesall of the containments.Type 1 2 3 4Element (k, 5a) (2w,−(5)a) (2w + 1,−(5)2a+1) (2w + 1,−(5)2a)Belongs to H1 Hc2 Hc3 Hc3It can easily be shown that the four above inclusions are satisfied, that is,all elements of G are contained in some conjugate of the Hi.All of these subgroups/containments have trivial intersection, and thus{H1, H2, H3} forms a cover for G.2.2 Intersective Polynomials2.2.1 IntroductionIn this section, we introduce intersective polynomials and provide anoverview of the theory used to show that a family of polynomials is inter-sective.Definition 2.18 (Intersective Polynomial). A monic polynomial f(x) withinteger coefficients is called intersective if it has no root in the rationalnumbers Q but has a root modulo m for all positive integers m > 1.The reason why intersective polynomials are interesting is that they pro-vide counterexamples to the local global principle. This principle states thatthe existence or non-existence of solutions in Q (global) of a diophantineequation can be detected by studying, for each p ≤ ∞, the solutions of theequation in the p-adic field Qp (local) [Gou93].Intersective polynomials have applications in combinatorial number the-ory, intersective sets, multiple recurrence in ergodic theory and Diophantineapproximation. The reader should consult Leˆ [L1ˆ4], Bergelson, Leibman392.2. Intersective Polynomialsand Lesigne [BLL08] and Leˆ and Spencer [LS14]. Intersective polynomialsalso arise in the context of number fields and function fields. Informationon these topics can be found Bergelson and Robertson [BR] and Yamagishi[Yam].Sonn [Son08, Thm 2.2] proved that every finite, noncyclic, solvable groupG can be realized as the Galois group over Q of an intersective polynomial,with the noncyclic condition being necessary. The same statement was alsoestablished by Sonn [Son09] for realizable nonsolvable Galois groups. Thesepapers also explicitly give a method of constructing intersective polynomials.Intersective polynomials are challenging to construct. Single examplesof intersective polynomials with Galois group isomorphic to the alternatinggroup An, 4 ≤ n ≤ 8 or to the symmetric group Sn, 3 ≤ n ≤ 6 are givenby Rabayev and Sonn [RS13]. These groups are 2-coverable so that theintersective polynomial will have two irreducible factors overQ. For example,in the case where the Galois group is A4, they prove that the polynomial(x4 − 10x3 − 7x2 + 3x+ 2)(x3 + 89x2 + 2586x+ 24649)is intersective, and in the case where the Galois group is S4, they prove thatthe polynomial(x4 − 5x2 + x+ 4)(x3 + 10x2 + 9x+ 1)is intersective.Candidates for intersective polynomials are constructed by forming theproduct of the defining polynomials of the fixed fields corresponding to thesubgroups in the n-cover via Galois theory. The only infinite families ofintersective polynomials that we are aware of appear in [LSY14] and in thepapers published that appear in this thesis.2.2.2 The Frobenius GroupThe Frobenius group is of interest because it has been proven in Sonn[Son08] that all Frobenius groups are 2-coverable and that there exists a402.2. Intersective Polynomialspolynomial f(x) which is the product of two irreducible polynomials in Q[x]with Frobenius Galois group G and having a root modulo m for all m > 1.A polynomial of particular interest with Frobenius Galois group waspresented in [Bra01] that has the formf(x) = (xp − 2)Φp(x)where Φp(x) denotes the pth cyclotomic polynomial with p an odd prime.The proof that f(x) is intersective is as follows:Let q be a prime. If p does not divide q − 1, then 2 is a pth power inthe p-adic integers Zq, and so xp − 2 has a root mod q. If p | q − 1, thenZq contains primitive pth roots of unity, and thus Φp(x) has a root mod q.Therefore f(x) has a root mod q for all primes q.Of course, this is a special case that is easy to show because of the con-struction of the Frobenius group. For other groups, more advanced methodsneed to be employed.2.2.3 Hensel’s LemmaShowing that a polynomial f(x) has a solution mod m for every integerm > 1 is equivalent to showing that it has a solution mod pj for each primep and positive integer j. The equivalence is a consequence of the ChineseRemainder Theorem. An infinite family of intersective polynomials has beenfound in Hyde, Lee, and Spearman [HLS14] where we use Hensel’s Lemmaand a refined version of Hensel’s Lemma to lift certain polynomial solutionsmodulo p to arbitrarily high powers of p. We state the two lemmas belowand then give the family of polynomials in question.Theorem 2.19 (Hensel’s Lemma). (see [INM95, p. 87]) Suppose that f(x)is a polynomial with integral coefficients. If f(a) ≡ 0(mod pj) and f ′(a) 6≡0(mod p), then there is a unique t(mod p) such that f(a + tpj) ≡ 0(modpj+1).If the condition f ′(a) 6≡ 0(mod p) holds, then the root a is called nonsingular.By repeated application of Hensel’s Lemma, a nonsingular root a of f(x) ≡412.2. Intersective Polynomials0(mod p) may be lifted to a root modulo pj , for j = 2, 3, . . . . The refinedversion of Hensel’s Lemma which, in the case of a singular root, enables usto lift their solutions modulo arbitrarily high prime powers. This version isas follows:Theorem 2.20 (Hensel-Rychlik Lemma). (see [INM95, p. 89]) Let f(x) bea polynomial with integral coefficients. Suppose that f(a) ≡ 0(mod pj), thatpτ ‖ f ′(a) and that j ≥ 2τ+1. If b ≡ a(mod pj−τ ) then f(b) ≡ f(a)(mod pj)and pτ ‖ f ′(b). Moreover there is a unique t(mod p) such that f(a+tpj−τ ) ≡0(mod pj+1).Using these lemmas, we were able to show that for n, a cubefree integernot equal to 1, the polynomialf(x) = (x3 − n)(x2 + 3)is intersective if and only if the prime factors of n are of the form 3k+1 andn ≡ 1(mod 9).2.2.4 Decomposition GroupsWhen elementary methods cannot be applied to intersective polynomialproblems, more advanced theory must be introduced to establish the inter-sective property. One method is to use decomposition groups, which will beused extensively later in this thesis. We give their definition now.Definition 2.21. The decomposition group G(p) is the set of elements σ ∈Gal(L/Q) such that σ(p) = p, where p is a prime ideal in L and L/Q is theextension of L over Q.To form intersective polynomials, we first need to discover whether ornot intersective polynomials can be formed with the group we are workingwith. Once we know that the conditions are satisfied, we can use Galoisand field theory to form the factors that our intersective polynomial will becomposed of. The method is outlined in detail next.422.2. Intersective PolynomialsLet G be a finite, noncyclic group. We first have to show that G is n-coverable for some n > 1, that is to show G is the union of conjugates of nproper subgroups, the intersection of whose conjugates is trivial. Supposethat the Galois group of f(x), denoted Gal(f), is isomorphic to G andlet L denote the splitting field of f(x) so that Gal(f) = Gal(L/Q). Ifevery decomposition group G(p) for p a prime ideal in L is contained in aconjugate of one of the proper subgroups in the n-cover then we know thatan intersective polynomial can be constructed.Showing that every decomposition group for the prime ideals in the split-ting field are contained in the proper subgroups is where the bulk of the worklies. The methodology changes depending on the Galois group you are work-ing within but for A4 and D5, we took advantage of prime ideal factorizationtheory to deduce which subgroups the decomposition group lay within. ForA4, we use Dedekind’s theorem for prime ideal factorization in a monogenicfield. For D5, we use known theory on dihedral trinomials to deduce primeideal factorization.Using the subgroup-subfield correspondence in Galois theory, form theproduct of the set of nmonic polynomials with integer coefficients that definethe subfields of L corresponding to the n subgroups forming the cover. Theresulting polynomial is intersective.In practice, when constructing intersective polynomials, it can happenthat a decomposition group is cyclic so that containment in the n-cover isautomatic.43Chapter 3Intersective Polynomialswith Specified Galois Group3.1 Intersective Polynomials with Galois GroupA43.1.1 IntroductionRecall that a monic polynomial f(x) with integer coefficients is calledintersective if it has no root in the rational numbers Q but has a root modulom for all positive integers m > 1. Sonn [Son08, Thm 2.2] proved that everyfinite, noncyclic, solvable group G can be realized as the Galois group over Qof an intersective polynomial, with the noncyclic condition being necessary.The same statement was also established by Sonn [Son09] for realizablenonsolvable Galois groups. We will use the method described in Section2.2.4 to find an infinite family of intersective polynomials with Galois groupA4.We first begin by utilizing the knowledge that G ' A4 is 2-coverable.Using Sonn’s theory, we will show that every decomposition group G(p) forp a prime ideal in L, the splitting field, is contained in a conjugate of one ofthe proper subgroups in the 2-cover.Using the subgroup-subfield correspondence in Galois theory, we thenform the product of the set of two monic polynomials with integer coefficientsthat define the subfields of L corresponding to the two subgroups formingthe cover. The resulting polynomial is intersective.In this chapter, we give infinitely many intersective polynomials withGalois group A4 and nonisomorphic splitting fields. Next we will give some443.1. Intersective Polynomials with Galois Group A4preliminaries for this family and in Section 3.1.2, we prove our theorem andgive some examples. We now state our main theorem.Theorem 3.1. There are infinitely many positive integers t such that t(t2+108) is squarefree. Let t be such an integer and define ft(x) and gt(x) byft(x) = x4 + 18x2 − 4tx+ t2 + 81 (3.1)andgt(x) = x3 − (t2 + 108)x+ 4t2 + 432.Then the polynomialft(x)gt(x)is intersective and has Galois group A4. Furthermore for these values of tthe splitting fields of ft(x)gt(x) are nonisomorphic.We begin by recalling from Rabayev and Sonn [RS13] that a 2-cover ofA4 consists of the Sylow 2-subgroup and a Sylow 3-subgroup of A4. TheSylow 2-subgroup is normal in A4 and isomorphic to the Klein 4-group. Wewill construct the family of intersective polynomials from the polynomialsgiven by Spearman [Spe06]. The following proposition summarizes theirrelevant properties.Proposition 3.2. Suppose that t is a positive integer and that t(t2 + 108)is squarefree. Let θt denote a root of ft(x) given by (3.1). Then Kt = Q (θt)is a quartic field with field discriminant d(Kt) = 28t2(t2 + 108)2, whose ringof integers has a power integral basis, namely {1, θt, θ2t , θ3t }. Furthermore,the Galois group of ft(x) is isomorphic to A4 and the fields Kt are distinct.The resolvent cubic is a cubic polynomial defined from a monic quarticpolynomial q(x), where the coefficients of the resolvent cubic can be obtainedfrom the coefficients of q(x) using only basic arithmetic operations. Knowingthe roots of the resolvent cubic is useful for finding the roots of q(x). We re-mark that the resolvent cubic of ft(x) is x3−18x2−4(t2+81)x+56t2+5832,which after translating to eliminate the x2 term and scaling, simplifies to453.1. Intersective Polynomials with Galois Group A4gt(x). The polynomials ft(x) and gt(x) correspond to the subgroups giv-ing the 2-cover of A4 using the subfield-subgroup correspondence of Galoistheory.3.1.2 Proof of TheoremProof. The fact that there are infinitely many integers t such thatt(t2 + 108)is squarefree follows from a theorem of Erdo¨s [Erd53]. We note that t mustbe squarefree and must not be divisible by 2 or 3. The fact that the Galoisgroup of ft(x)gt(x) is isomorphic to A4 follows from Proposition 3.2 and theremark after it. The confirmation that ft(x)gt(x) is intersective requires usto show that the decomposition group G(p) is contained in either the Sylow2-subgroup or a Sylow 3-subgroup of A4 for any prime ideal p in the splittingfield L of ft(x). The groups in the cover are the Klein 4 group, Z/3Z, andthe Sylow subgroups.If p is unramified in L then G(p) is cyclic as noted by Sonn [Son09]so the containment is automatic. Now let p be a prime ideal in L lyingabove the ramified rational prime p. It suffices to show that G(p) is a propersubgroup of A4. As Gal(ft) acts transitively on the set of prime ideals inL lying above p (see Rosen [Ros02], Proposition 9.2) we deduce that G(p)is a proper subgroup of A4 if there are at least two prime ideals in L lyingabove p. If αt denotes a root of gt(x), then L is the compositum of Q (θt) andQ (αt) (because the degrees must multiply to 12) so that p must ramify inat least one of these fields by Proposition 1.116. We recall from Proposition3.2 that the field discriminant of Q (θt) is equal to 28t2(t2 + 108)2 whilethe polynomial discriminant of gt(x) is equal to 22t2(t2 + 108)2. Since thesediscriminants contain the same prime factors we deduce that p must ramifyin Q (θt).Now consider all primes that divide the discriminant. We note that theconditions of Theorem 1.123 are satisfied for ft(x) and its root θt as ft(x)is the miniminal polynomial for θt and ind(θt) 6= 0 (mod p). We will also463.1. Intersective Polynomials with Galois Group A4make use of the following theorem.Theorem 3.3. [Nar90, p. 262] If L/Q is a normal extension of an algebraicnumber field, G is its Galois group, and p is a prime ideal of OK , then theindex of the decomposition group in Gal(L/Q) equals the number of primeideals lying above p in L.Before we begin, we will also state a theorem by Llorente and Nart[LN83] that gives the decomposition of primes in cubic fields K defined byf(x) = x3 − ax+ b, where a, b ∈ Z.Theorem 3.4. Let K be a cubic field and let K = Q(θ), where θ is a rootof an irreducible polynomial of the formf(X) = X3 − aX + b, a, b ∈ ZThe discriminant of f(X) is ∆ = 4a3 − 27b2 and if D(K) is the field dis-criminant of K, then ∆ = ind(θ)2 ·D(K). Let sp = vp(∆) and ∆p = ∆/pspfor every prime p. The primes of Q decompose in K as follows:Decomposition of 2:a, b even:1 ≤ v2(b) ≤ v2(a)1 = v2(a) < v2(b)2 = P 32 = P ·Q2a even, b odd: .............................................. 2 = P ·Qa odd, b even:s2 odd:s2 even:∆2 ≡ 3 (mod 4)∆2 ≡ 5 (mod 8)∆2 ≡ 1 (mod 8)2 = P ·Q22 = P ·Q22 = P ·Q2 = P ·Q ·Ra, b odd: .............................................. 2 = PDecomposition of 3:473.1. Intersective Polynomials with Galois Group A43 | a, 3 | b:1 ≤ v3(b) ≤ v3(a)1 = v3(a) < v3(b) 3 = P33 = P ·Q23 - a:a ≡ −1 (mod 3)a ≡ 1 (mod 3)3 - b3 | b3 = P ·Q3 = P3 = P ·Q ·R3 | a, 3 - b:a 6≡ 3 (mod 9)b2 ≡ a + 1 (mod 9)b2 6≡ a + 1 (mod 9)a ≡ 3 (mod 9)b2 ≡ a + 1 (mod 27)s3 odd:s3 even:∆ ≡ −1 (mod 3)∆ ≡ 1 (mod 3)s3 = 6s3 > 6b2 6≡ a + 1 (mod 27)3 = P ·Q23 = P 33 = P ·Q23 = P ·Q3 = P3 = P ·Q ·R3 = P 3Decomposition of p > 3:p | a, p | b:1 ≤ vp(b) ≤ vp(a)1 = vp(a) < vp(b)p = P 3p = P ·Q2p | a, p - b:p ≡ −1 (mod 3)p ≡ 1 (mod 3)(b/p)3 = 1(b/p)3 6= 1p = P ·Qp = P ·Q ·Rp = Pp - a, p | b:(a/p) = 1(a/p) = −1 p = P ·Q ·Rp = P ·Qp - ab:sp odd :sp even :(∆p/p) = 1f(X) has some root (mod p)f(X) has no roots (mod p)(∆p/p) = −1p = P ·Q2p = P ·Q ·Rp = Pp = P ·QCase 1: First we consider p = 2, and its factorization in Q (θt). We haveft(x) ≡ x4(mod 2),483.1. Intersective Polynomials with Galois Group A4so that 〈2〉 = ℘4 for some prime ideal ℘ in Q (θt) by Theorem 1.123. Thefactorization of 〈2〉 in the cubic field defined by gt(x) can be deduced fromTheorem 3.4. We find that〈2〉 = P1P2P3for prime ideals P1, P2, and P3 in Q(αt). Combining these two factorizationsshows that in L we have〈2〉 = (p1p2p3)4for prime ideals p1, p2, and p3 in L. Recall by Theorem 3.3 that the index ofthe decomposition group is equal to the number of prime ideals lying above〈2〉 in L so that the order of the decomposition group is 4. This impliesthat the decomposition group of a prime ideal in L lying above 〈2〉 is iso-morphic to the Sylow 2-subgroup which is of course contained in the 2-cover.Case 2: Next we treat primes p dividing t. If only one prime ideal p inL lies above p it follows that only one prime ideal of Q (θt) lies above p.Factoring ft modulo p yieldsft(x) ≡ (x2 + 9)2(mod p).The factor x2+9 is reducible mod p if and only if p ≡ 1(mod 4) so that −1 isa square modulo p. If p ≡ 3(mod 4), then x2 + 9 is irreducible mod p. SinceQ (θt) is monogenic with ring of integers Z [θt] we have ind(θt) 6≡ 0(mod p).Applying Theorem 1.123 we find that 〈p〉 = ℘2 if p ≡ 3(mod 4) while〈p〉 = ℘21℘22 if p ≡ 1(mod 4), where ℘, ℘1, and ℘2 are prime ideals.Turning to the cubic field defined by gt(x) and again using Theorem 3.4,we see that the prime ideal factorization of 〈p〉 has the form〈p〉 = P1P2P3.493.1. Intersective Polynomials with Galois Group A4Thus in L we have the possible factorizations〈p〉 = (p1p2p3)2or〈p〉 = (p1p2p3p4p5p6)2 ,according to whether p ≡ 3(mod 4) or p ≡ 1(mod 4). Therefore, dependingon whether p ≡ 3(mod 4) or p ≡ 1(mod 4), the decomposition group of aprime ideal in L lying above 〈p〉 is isomorphic to either the Sylow 2-subgroupor the cyclic group of order 2, both of which are contained in the 2-cover.Case 3: Finally we treat the prime divisors of t2 + 108 which we recall arenot equal to 2 or 3. The identity432ft(x)− (2x+ t)(6x− t)3 = (t2 + 108)(72x2 − 16xt+ t2 + 324)shows that modulo p, ft(x) has two roots of multiplicities 1 and 3. UsingTheorem 1.123 we see that the prime ideal factorization of 〈p〉 in Q (θt) hasthe form〈p〉 = ℘1℘32. (2)Theorem 3.4 implies that the prime ideal factorization of 〈p〉 in the cubicfield defined by gt(x) has the form〈p〉 = P 3. (3)The prime ideal factors of 〈p〉 in L each have the same ramification indexe and inertial degree (see Narkiewicz [Nar90, Theorem 4.6]). We see that emust be divisible by 3 from equation (3) forcing ℘1 in equation (2) to havethe factorization℘1 = p31in the normal relative cubic extension L/Q (θt) . Equality of inertial de-grees ensures that ℘2 does not remain prime in L/Q (θt) , while equality of503.1. Intersective Polynomials with Galois Group A4ramification indices implies that ℘2 does not ramify in L/Q (θt) . The onlyremaining possibility for the prime ideal factorization of ℘2 in L/Q (θt) is℘2 = p2p3p4so that the final factorization of 〈p〉 is〈p〉 = (p1p2p3p4)3Thus in this case the decomposition group of a prime ideal lying above 〈p〉is isomorphic to the Sylow 3-subgroup which is contained in the 2-cover.Since the decomposition groups of the prime ideals in L lying above 〈p〉 arecontained in the 2-cover for each of the three possible cases, we concludethat f(x) is intersective.We finish by summarizing the decomposition groups in the followingtable, using the notation Cn to denote the cyclic group of order n.Table 3.1: Decomposition groups of the ramified primes in Lp (prime) in N Factorization of 〈p〉 in L Decomposition Groupp = 2 〈2〉 = (p1p2p3)4 C2 × C2p | tp ≡ 1(mod 4) 〈p〉 = (p1p2p3p4p5p6)2 C2p | tp ≡ 3(mod 4) 〈p〉 = (p1p2p3)2 C2 × C2p | t2 + 108 〈p〉 = (p1p2p3p4)3 C3Below are some examples of intersective polynomials obtained from ourtheorem.513.2. Intersective Polynomials with Galois Group D5Table 3.2: Examples of intersective polynomials with Galois group A4t t(t2 + 108) Intersective Polynomial1 109 (x4 + 18x2 − 4x+ 82)(x3 − 109x+ 436)5 5 · 7 · 19 (x4 + 18x2 − 20x+ 106)(x3 − 133x+ 532)7 7 · 157 (x4 + 18x2 − 28x+ 130)(x3 − 157x+ 628)11 11 · 229 (x4 + 18x2 − 44x+ 202)(x3 − 229x+ 916)3.2 Intersective Polynomials with Galois GroupD53.2.1 IntroductionIn this section, we study dihedral quintic trinomials and give necessaryand sufficient conditions under which they give rise to intersective polyno-mials. Infinite families of intersective polynomials with Galois group D5,the dihedral group of order 10 are given in [LSY14]. In the previous chap-ter, the number field was monogenic and we were able to take advantage ofa theorem of Dedekind (Theorem 1.123) for ideal factorization. However,the number fields defined by dihedral quintic trinomials we consider in thischapter are not monogenic and hence we will need a different approach.For the family of polynomials in this section, the intersective propertyis investigated by appealing to the formula for the field discriminant ofthe quintic number fields defined by these trinomials, as given in [SW02].We will also employ the theory of ideal factorization in dihedral fields ofprime degree as described in [Coh99]. In this section, we briefly summarizethe information we need on field discriminants, while in Section 3.2.2, westudy the decomposition groups associated with our family of trinomials.In Section 3.2.3, we prove our theorem, and in Section 3.2.4, we provide amethod of construction for intersective polynomials and show that we canobtain infinitely many such polynomials.523.2. Intersective Polynomials with Galois Group D5Algebraic PreliminariesIn this section, we give the algebraic preliminaries and results on fielddiscriminants required to study our intersective polynomials. We recall theparametrization of trinomials x5 +ax+ b with a solvable Galois group givenin [RYZ82] or [Web79, p. 376]. Explicitly, if a and b are rational numbersand the Galois group of x5 + ax+ b is solvable, then a and b are given bya =55λµ4(λ− 1)4 (λ2 − 6λ+ 25)b =55λµ5(λ− 1)4 (λ2 − 6λ+ 25)with λ, µ ∈ Q, λ 6= 1, µ 6= 0. Parametrizing the values of λ for which thediscriminant of x5 + ax + b is equal to a square restricts us to trinomialswith Galois group D5. Making the change of variablesλ = 5u+ 1u− 1 ,5µλ− 1 = v,now makes the discriminant of x5 + ax+ b equal toD =56(u+ 1)4(2u3 + 4u2 + 11u+ 8)224(u2 + 4)5v20.This discriminant is a perfect square if and only if u2 +4 is a perfect square.Therefore, setting u = β − 1/β for some β ∈ Q and setting α = v/(β2 + 1),then letting β = m/n for m,n ∈ Z and d = 2n2/α, we obtain the followingpresentation up to scaling.For the irreducible dihedral quintic x5+ax+b ∈ Z [x], there exist coprimeintegers m and n, and integers i, j = 0 or 1, such thata = 22−4i51−4jd2(m2 −mn− n2)E2F, (2)b = 24−5i5−5jd1(2m− n)(m+ 2n)E3Fwhere d21 is the largest square dividing m2 +n2, d52 is the largest fifth power533.2. Intersective Polynomials with Galois Group D5dividing m2 +mn− n2, andE = (m2 + n2)/d21, F = (m2 +mn− n2)/d52.The choice of i and j, as well as the values of E and F , ensures that theresulting polynomial x5 + ax+ b satisfies the condition that there does notexist a prime number p such thatp4 | a and p5 | b.This condition is required, for example, when calculating field discriminantsof trinomials which can be seen in more detail in Llorente, Nart, and Vila[LNV84]. If it were the case that p4 | a and p5 | b, then we could rewrite ourquintic in the formx5 + p4cx+ p5dfor integers c and d and substitute x = px to obtainp5x5 + p5cx+ p5dwhich, after dividing by p5, becomesx5 + cx+ d.For a dihedral quintic polynomial f(x) of prime degree with root θ andnumber field K = Q(θ), the field discriminant has the form(d(k)f2)(p−1)/2(3)where k is the quadratic subfield of the splitting field L of f(x) while theinteger f is called the conductor of L/k [Coh99, p. 491]. We require somefacts about the discriminants of the quintic fields defined by the trinomialsdefined earlier in this chapter. We recall from [SW02] the following twoPropositions giving the discriminant of K.Proposition 3.5. If θ is a root of a dihedral quintic trinomial as parametrized543.2. Intersective Polynomials with Galois Group D5by (2) and K = Q (θ), then the discriminant of the field K is given byd(K) = 2α5β∏p 6=2,5p|Ep2∏p 6=2,5p|Fp4 (4)whereα =4 if m ≡ n+ 1 (mod 2)6 if m ≡ n ≡ 1 (mod 2)andβ =0 m ≡ 3n (mod 5), E ≡ 0 (mod 5)orm ≡ 2n (mod 5),m ≡ 57n (mod 125), E ≡ 0 (mod 5).2 m ≡ 3n (mod 5), E 6≡ 0 (mod 5)orm ≡ 2n (mod 5),m ≡ 57n (mod 125), E 6≡ 0 (mod 5)6 m 6≡ 2n, 3n (mod 5)orm ≡ 2n (mod 5),m 6≡ 57n (mod 125), E 6≡ 0 (mod 5)8 m ≡ 2n (mod 5),m 6≡ 57n (mod 125), E ≡ 0 (mod 5)Proposition 3.6. With the same notation as the previous proposition, wehaved(K) = d(k)2f4 (5)wheref = 5θ∏1≤vp(b)≤vp(a)p,553.2. Intersective Polynomials with Galois Group D5andθ =0 if 5 - a or 52 ‖ a, 53 | b.1 if 5 ‖ a, 5 - b or 52 ‖ a, 52 ‖ b.2 if 54 ‖ a, 54 ‖ b.3.2.2 The Decomposition GroupsIn this section, we treat four different cases (in order) of the decompo-sition groups of the prime ideals lying above the rational primes: p = 2,p = 5, p | E and p | F. Proposition 3.5 shows that these cover all cases oframified primes. If we intend to show that a particular decomposition groupis cyclic, then it suffices to show that it is a proper subgroup of D5 since thetwo proper cyclic subgroups of D5 form our 2-cover. Equivalently, we showthat there is more than one prime ideal lying above p. This is deduced by atheorem of Narkiewicz [Nar90, Theorem 6.5]. By knowing the factorizationof the prime ideals in the splitting field, we can use the theorem below tofind the index of the decomposition group and thus know whether or notthe decomposition group lies within our cover.Theorem 3.7. If L/K is a normal extension of an algebraic number field,G its Galois group, p a prime ideal of OK , and P a prime ideal of lyingabove p, then the index of the decomposition group of the ideal P is equal tothe number of prime ideals lying above p in L.We now state some theory from Cohen [Coh99], where we have alteredthe notation to match our own. Recall that L is the normal closure of thequintic, dihedral extension K and the quadratic extension k = Q(√t), andthe integer f(L/k) is the conductor of L/k, which provides a quantitativemeasure of the ramification within the extension. Also, the odd prime num-ber l is the degree of the number field extension K/Q (in our case, l = 5).Proposition 3.8. [Coh99, Prop. 10.1.26(9)] If p is totally ramified in L/Q,in other words if 〈p〉 = P 2l for some prime ideal P , then p is totally ramifiedin K/Q, and in addition, p | l.563.2. Intersective Polynomials with Galois Group D5Proposition 3.9. [Coh99, Prop. 10.1.28(2)] The ideal p is totally ramifiedin K/Q if and only if p | f .Lemma 3.10. Let p be a prime ideal lying above 2 in the splitting field ofa dihedral quintic trinomial. Then the decomposition group G(p) is cyclic.Proof. By way of contradiction, assume that there is only one prime ideallying above 2. It was shown in [SWY07] that the prime 2 is a commonindex divisor of K and that 2 ramifies in the unique quadratic subfield of L.This implies that 2 ramifies in K as well since by Proposition 3.6, 2 | d(K).Because there is only one prime ideal lying above 2, we deduce that 2 istotally ramified in L. This contradicts Proposition 3.8 as 2 - 5. Hence G(p)is cyclic.Lemma 3.11. Let K be the number field defined by a dihedral quintic tri-nomial. Let p be a prime ideal lying above 5 in the splitting field of thetrinomial. Set β = v5(d(K)) and recall that k = Q(√t)is the quadraticsubfield of the splitting field L of f(x). Then the decomposition group G(p)is cyclic if and only ifβ = 2, 4orβ = 8 and(t5)= +1.Proof. We treat cases according to the power β of 5 in the discriminant ofK as given in Proposition 3.5. According to Proposition 3.5, β ∈ {0, 2, 6, 8}.Case 1: If β = 0 then 5 is unramified in L as it is unramified in bothK and the quadratic field k. Thus for a prime ideal p lying above 5, thedecomposition group G(p) is cyclic and contained in the 2-cover.Case 2: Next suppose β = 2. It was shown in the proof of Proposition 4.2of [SW02] that the prime ideal decomposition of 5 in K is given by〈5〉 = P1P 22P 23 .573.2. Intersective Polynomials with Galois Group D5Thus more than one prime ideal lies above 5 in L so that for a prime idealp lying above 5, the decomposition group G(p) is cyclic.Case 3: Next we treat the case β = 6. Referring to (5) with p = 5, since56 || d(K) and d(k) is free of odd squares, we deduce that 5 | f . UsingProposition 3.9, we see that 5 is totally ramified in K. Further, from Propo-sition 3.6, we deduce that 5 | d(k) so that 5 ramifies in k. Hence 5 is totallyramified in L so that there is only one prime ideal lying above 5 in L. Thedecomposition group for this prime ideal is all of D5, which is not cyclic andnot contained in the 2-cover.Case 4: The remaining case is β = 8. The discriminant formula in Propo-sition 3.6 implies that5 - d(k), 5 | f,so that 5 is unramified in k and by Proposition 3.9, 5 is totally ramified inK. Thus one prime ideal in L lies over 5 if and only if the ideal 〈5〉 remainsprime in k. By Theorem 1.121, this occurs precisely when(t5)= −1,showing that the associated decomposition group is all of D5, so is not cyclic.In the situation where 〈5〉 splits in k, showing that two prime ideals lie above5 in L, the decomposition groups are cyclic. This occurs precisely when(t5)= +1,proving the lemma.For the next two lemmas, we note the easily proved statement that ifp 6= 2, 5 is a prime, then p cannot divide both E and F.Lemma 3.12. Suppose that p 6= 2, 5 is a prime such that p | E. Then thedecomposition group of a prime p lying above p is cyclic.583.2. Intersective Polynomials with Galois Group D5Proof. The hypothesis in this lemma implies that p ramifies in k so that〈p〉 = ℘2,and it follows from Proposition 3.5 that p | d(K) so p ramifies in K. If onlyone prime ideal lies above p in L then this is also true in K so that p is totallyramified in K. In conclusion, p is totally ramified in L which is impossibleby Proposition 3.8 as p 6= 5. Hence the factorization of ℘ in L contains morethan one prime ideal, implying that the decomposition group of each suchprime is a proper subgroup of D5 and is hence cyclic.Lemma 3.13. Suppose that p 6= 2, 5 is a prime such that p | F. Then thedecomposition group of a prime p lying above p is cyclic if and only if(tp)= +1.Proof. Certainly p is not ramified in k so it either splits or remains prime.However we see from Proposition 3.5 and 3.6 that p | f , so that by Propo-sition 3.9, p is totally ramified in K. Thus more than one prime ideal in Llies over p if and only if p splits in k, or equivalently(tp)= +1.In this case, the decomposition group is contained in a conjugate of the2-cover of D5 so it is cyclic.3.2.3 Proof of TheoremFor a rational prime p and a nonzero integer n, the notation vp(n) = ameans that we have pa ‖ n, that is, pa | n but pa+1 - n. Utilizing thisnotation, we now state our main theorem.Theorem 3.14. Let f(x) = x5 + ax + b ∈ Z[x] have Galois group D5. Letθ be a root of f(x). Set K = Q(θ). Let d(K) denote the field discriminantof K. Let t be the unique, squarefree integer such that k = Q(√t)is the593.2. Intersective Polynomials with Galois Group D5quadratic subfield of the splitting field L of f(x). Then the polynomial(x5 + ax+ b)(x2 − t) (1)is intersective with Galois group D5 if and only ifv5(d(K)) 6= 6andif v5(d(K)) = 8 then(t5)= +1,and for all primes p > 5if p4 | d(K) then(tp)= +1.Proof. Suppose that (x5 + ax + b)(x2 − t) is intersective. Then for anyprime ideal in L, the decomposition group is cyclic. Lemma 3.11 shows thatv5(d(K)) 6= 6 and if v5(d(K)) = 8 then(t5)= +1.For any prime with p4 | d(K), we have p | F and Lemma 3.13 shows that ifa prime ideal lying above p in L has a cyclic decomposition group, then(tp)= +1.Conversely, we show that all decomposition groups are cyclic. This iscertainly the case for prime ideals in L lying above an unramified rationalprime p. If p = 2, then Lemma 3.10 shows that any associated decompositiongroups are cyclic. If p = 5, then Lemma 3.11, combined with the conditionson 5 stated in the theorem, show that any associated decomposition groupsare cyclic. Lemma 3.12 shows that for any prime p > 5 with p | E, allassociated decomposition groups are cyclic. Finally, Lemma 3.13, combinedwith the stated conditions on primes dividing F , equivalently p > 5 and603.2. Intersective Polynomials with Galois Group D5p4 | d(K), ensures that all decomposition groups are cyclic. Thus (x5+ax+b)(x2 − t) is intersective and clearly has Galois group D5.3.2.4 ExamplesWe first begin by giving some examples illustrating our main theorem inthe following table. Note that the chosen values of a and b satisfy (2).Table 3.3: Parameter Values and Their Associated Polynomialsm n d(K)(x5 + ax+ b)(x2 − t) IntersectiveReasona b t Yes/No?1 1 26 · 56 -5 12 -10 No v5(d(K)) = 6Insolvable in Q53 1 26 · 114 11 44 -2 Yes v5(d(K)) = 0,(−211)= +1Has a root in Qp ∀ prime p1 0 24 · 56 20 32 -5 No v5(d(K)) = 6Insolvable in Q52 11 24 · 194 10564 51072 -1 No v5(d(K)) = 0, but(−119)= −1Insolvable in Q191 7 26 · 52 · 414 11275 61500 -10 Yes v5(d(K)) = 2,(−1041)= +1Has a root in Qp ∀ prime p11 3 26 · 58 · 132 · 294 241986875 51448247500 -26 No v5(d(K)) = 8,(−265)= +1, but(−2629)= −1Insolvable in Q296 13 24 · 58 · 114 · 412 9754002500 242601920000 -41 Yes v5(d(K)) = 8,(−415)= +1,(−4111)= +1Has a root in Qp ∀ prime pWe now show that infinitely many intersective polynomials can be con-structed using our theorem. In [RYZ82, Example 4], Roland, Yui and Zagierderived an infinite set of dihedral quintic trinomials with F = f = 1. Usingformula (5) in Proposition 3.6 with f = 1 and the fact that vp(d(k)) ≤ 1 foran odd prime, we can easily deduce that the conditions of our main theoremare satisfied. This gives the following corollary to our theorem.Corollary 3.15. Suppose that m,n are given bym = f(r, s), n = f(s,−r),orm = −f(r, s) + 2f(s,−r), n = 2f(r, s) + f(s,−r),613.2. Intersective Polynomials with Galois Group D5for integers r, s withf(r, s) = 2r5 − 5r4s+ 10r3s2 + 5rs4 + s5and∆(r, s) = (f(r, s)2 + f(s,−r)2)/5.Then with a and b given by (2) we have(x5 + ax+ b)(x2 + ∆(r, s))is intersective ifr 6≡ 2s(mod 5),while(x5 + ax+ b)(x2 + 5∆(r, s))is intersective ifr ≡ 3s(mod 5).Using the examples following Theorem 2 in [RYZ82], in the following ta-ble we give the values of r, s,m, and n, the corresponding quadratic subfieldk, the field discriminant d(K), and the intersective polynomial produced byour Corollary. The values in this table were produced with the help of thecomputer algebra software MapleTM.623.2. Intersective Polynomials with Galois Group D5Table 3.4: Intersective Polynomials Produced Using Corollaryr s m n k d(K)1 1 13 21 Q(√−2 · 61) (23 · 61)21 2 144 233 Q(√−5 · 3001) (22 · 5 · 3001)21 2 322 521 Q(√−3001) (22 · 3001)21 4 2446 3987 Q(√−17 · 257401) (22 · 17 · 257401)21 5 6477 10649 Q(√−13 · 1195021) (23 · 13 · 1195021)21 7 29269 49083 Q(√−2 · 13063261) (23 · 13063261)21 7 68897 107621 Q(√−2 · 13063261) (23 · 5 · 13063261)22 3 1597 2584 Q(√−13 · 141961) (22 · 13 · 141961)22 5 11039 17868 Q(√−29 · 41 · 74201) (22 · 29 · 41 · 74201)23 1 367 269 Q(√−2 · 5 · 41 · 101) (23 · 5 · 41 · 101)23 1 171 1003 Q(√−2 · 41 · 101) (23 · 41 · 101)23 2 1028 1591 Q(√−13 · 55201) (22 · 13 · 55201)24 3 5831 9242 Q(√−61 · 15661) (22 · 61 · 15661)24 3 12653 20904 Q(√−61 · 15661) (22 · 5 · 61 · 15661)25 1 4401 277 Q(√−2 · 13 · 101 · 1481) (23 · 13 · 101 · 1481)25 4 21174 33823 Q(√−41 · 821 · 9461) (22 · 41 · 821 · 9461)28 1 50217 -11606 Q(√−5 · 13 · 8173681) (22 · 5 · 13 · 8173681)28 1 -73429 88828 Q(√−13 · 8173681) (22 · 13 · 8173681)2(x5 + ax + b)(x2 − t)a b t-405589 9987164 -12212086953942100 -72540492158684000 -15005-12086737798076 72550005665852352 -750253313637326184074053956 -505677611146965620655322432 -4375817128783546396420228711099 -38406227348562407748428726140 -310705464182675207741220240349531 -2148795349854978602098040875540 -26126522-2492582075130905668889382725 9555956338886635211233133444653500 -130632610-22487550931996197164 16600257189596220550773792 -184549327032040902373203052323924 -86546738189288328981286881324640 -88224989-34325342094725 597650406547546500 -41410216596445368171 408729111659573348 -207050-14093859584995428524 463005250073880374039712 -717613-1460789163238843866236 164167653953588413305312960 -955321938754748284345136870100 -418035937984409159057821684000 -4776605396367183999732006139 49695536440652420291999740 -3889106-3321980232966402393993671356 15652268869254582426329901425453120 -31846672147615387577854206517299376900 290395831685848919478209698254032000 -531289265-12903086258609681658270011564 471538454128083451015339001654246304 -265644632563Chapter 4The Holomorph Z2e o Z∗2e4.1 IntroductionJacobson and Ve´lez [JV90] determined the Galois group of an irreduciblebinomial x2e−a over Q for integers e and a with e ≥ 3. This Galois group isa full subgroup of the holomorph of the cyclic group of order 2e. Explicitly,this holomorph is the set Z2e×Z∗2e with binary operation given by(α, u) (β, v) = (α+ uβ, uv) ,which we recognize as Z2e oZ∗2e , the semi-direct product of the cyclic groupZ2e with Z∗2e . A subgroup of Z2e o Z∗2e is full if the projections onto Z2eand Z∗2e are surjective. If θ is a root of x2e − a, ζ = exp(2pii/2e), and theGalois group of x2e − a is Z2e o Z∗2e , then the automorphism (α, u) of thesplitting field Q (θ, ζ) of x2e − a is defined by(α, u)(θ) = ζαθ and (α, u)(ζ) = ζu.The group Z∗2e is not cyclic but is generated by the residue classes {−1, 5}modulo 2e. We shall use G to denote this holomorph. There are two pur-poses to this paper. First we give an n-cover of G, that is a collection ofproper subgroups of G, the union of whose conjugates equals G, and whoseintersection is trivial. We show that G has a 3-cover. Then we use our 3-cover to produce a new family of intersective polynomials with Galois groupG. Such polynomials are monic, have integer coefficients, and have no ratio-nal root but have roots with respect to every modulus. Equivalently thesepolynomials have a root in every p-adic field Qp.644.1. IntroductionWithout loss of generality and for convenience, we shall impose a scalingcondition on our binomials; namely, we shall assume that our binomial x2e−ahas the property that for any prime number p we havep2e- a. (1)Scaling polynomials will not alter the property of their being intersectivenor of course their Galois group. We state our main theorems now.Theorem 4.1. Let e be an integer with e ≥ 3. Let G be the groupG ' Z2e o Z∗2e .The subgroups {H1, H2, H3} of G defined byH1 ={(b, 5c) : b = 0, 1, . . . , 2e − 1, c = 0, 1, . . . , 2e−2 − 1} ,H2 = {(0, d) : d = 1, 3, . . . , 2e − 1} ,H3 = 〈(1,−1), (3,−5)〉 ,form a 3-cover of G.We note that the elements (1,−1), (3,−5) commute in G and a simplecalculation shows that H3 has order 2e.Theorem 4.2. Let a be an integer satisfying the scaling assumption (1).The polynomial (x2e − a)(x2 + 1) (x2e−1 + a)is intersective and has Galois group G ' Z2e o Z∗2e if and only ifa 6= ±d2,±2d2for all integers d, andif p | a, p ≡ 3(mod 4) then p2e−1 ‖ a and(−a/p2e−1p)= 1, (a)654.2. Binomials x2e − a with Galois group Gand one of the following conditions holds.a ≡ 1(mod 2e+2), (b)a ≡ −1(mod 2e+1),or22e−1 ‖ a and a/22e−1 ≡ −1(mod 2e+1).For examples of polynomials that are intersective or non-intersective(with reason), refer to the table at the end of this chapter.In Section 4.2, we use a theorem of Jacobson and Ve´lez [JV90] to deter-mine binomials x2e − a with Galois group G. In Section 4.3, we derive theproperties of the subgroups for our 3-cover of G. Finally, in Sections 4.4 and4.5 we prove Theorems 4.1 and 4.2.4.2 Binomials x2e − a with Galois group GFor two algebraic number fields K and L we denote their compositumby KL.Proposition 4.3 ([DF04], Corollary 20, pg. 592). If K and L are numberfields and K/Q is Galois then[KL : Q] =[K : Q] · [L : Q][K ∩ L : Q] (2)Lemma 4.4. Let a, e be integers with e ≥ 3. The binomialx2e − ais irreducible over Q and has Galois groupG ' Z2e o Z∗2eif and only ifa 6= ±d2, ± 2d2664.2. Binomials x2e − a with Galois group Gfor all integers d.Proof. Let ζ = exp(2pii/2e) and θ be a root of x2e − a. Set K = Q (ζ) andL = Q (θ) . We note that K/Q is Galois. Suppose that x2e − a is irreducibleover Q and has Galois group G. Irreducibility implies that a 6= d2 for allintegers d. The splitting field of x2e −a is the compositum of K and L, thuswe have|G| = 22e−1 = [KL : Q] = 2e · 2e−1 = [K : Q] · [L : Q]. (3)If a = −d2, (respectively,−2d2, 2d2) then(θ2e−1d)2= −1, (respectively(θ2e−1d)2= −2,(θ2e−1d)2= 2),so thatK ∩ L ⊇ Q(√−1), (respectively Q(√−2), Q(√2)).Equation (2) now shows that (3) is impossible. Thus a 6= ±d2, ±2d2 for allintegers d.Now suppose that a 6= ±d2, ±2d2 for all integers d. Irreducibility isdeduced from the Vahlen-Capelli Theorem [Cap01], [Vah95], which statesthat xn−a is reducible over Q if and only if for some prime p > 1, p | n anda = bp or 4 | n and a = −4b4, for some integer b. Both possibilities are ruledout by assumption. We determine the Galois group of x2e − a by a degreeargument. Since √a = ±θ2e−1we deduce that √a ∈ L.Using [Wei09, Corollary 4.5.4], we see that the quadratic subfields of Q (ζ)are preciselyQ(√−1), Q(√−2), and Q(√2).Since a 6= ±d2, ±2d2 for all integers d, √a does not belong to any of these674.3. n-cover of G ' Z2e o Z∗2equadratic subfields, hence√a does not belong to Q (ζ). It follows from[JV90, Theorem A(b)] that the Galois group of x2e − a is G, completing theproof.4.3 n-cover of G ' Z2e o Z∗2eLemma 4.5. The groupG ' Z2e o Z∗2eis not 2-coverable.Proof. Suppose that for two proper subgroups H1 and H2 of G, {H1, H2} isa 2-cover for G. We write|H1| = 2a and |H2| = 2bfor positive integers a and b. Since these subgroups are proper subgroups ofG and |G| = 22e−1, we have1 ≤ a, b < 2e− 1. (4)The normalizer of H1 (respectively H2) strictly contains H1 (respectivelyH2) (see [Fra02, Cor. 36.7]). Consequently the number of conjugates of H1and H2 satisfy|G : N(H1)| ≤ 22e−12a+1= 22e−a−2 and |G : N(H2)| ≤ 22e−12b+1= 22e−b−2so that the number of elements in the union of the conjugates of H1 and H2,after removing one for the identity element which is counted at least twice,is at most2a22e−a−2 + 2b22e−b−2 − 1 ≤ 22e−1 − 1 < 22e−1using (4). Hence G is not 2-coverable.In order to give the proof of Theorem 4.1, we need to determine the684.3. n-cover of G ' Z2e o Z∗2econjugates of the subgroups specified in that theorem. The subgroup H1is normal, but H2 and H3 are not. The next two lemmas determine theconjugates of H2 and H3.Lemma 4.6. The conjugates of the subgroupH2 = {(0, d) : d = 1, 3, . . . , 2e − 1}of the groupG ' Z2e o Z∗2e .have the form{(n(d− 1), d) : d = 1, 3, . . . , 2e − 1} ,where n is a fixed integer modulo 2e.Proof. We begin by conjugating a typical element of H2 with an arbitraryelement (k, j) with inverse (−kj−1, j−1) in G which leads us to(−kj−1, j−1)(0, d)(k, j)= (kj−1(d− 1), d),so assuming thatkj−1 ≡ n(mod 2e)we have produced a conjugate of H2 with elements of the form(n(d− 1), d).Reversing this calculation proves the Lemma.Lemma 4.7. The conjugates of the subgroupH3 = 〈(1,−1), (3,−5)〉of the groupG ' Z2e o Z∗2e .694.4. Proof of Theorem 4.1have the form〈(m,−1), (3m,−5)〉 ,for a fixed odd integer m modulo 2e.Proof. Conjugating the generators of H3 with an arbitrary element (k, j)with inverse (−kj−1, j−1) in G gives(−kj−1, j−1)(3,−5)(k, j)= (3m,−5),and(−kj−1, j−1)(1,−1)(k, j)= (m,−1),where the odd integer m satisfiesm ≡ j−1(1− 2k) (mod 2e).Furthermore, as j and k can be freely chosen modulo 2e we conclude that ifm is an odd integer modulo 2e then〈(m,−1), (3m,−5)〉 ,is a conjugate of H3.Now we give the proof of Theorem 4.1.4.4 Proof of Theorem 4.1Proof. We begin by showing G is equal to the union of the conjugates of thesubgroups Hi, i = 1, 2, 3. For notation, Hci denotes a conjugate of Hi. Weconsider a typical element(k, j) ∈ G704.4. Proof of Theorem 4.1where k is any integer modulo 2e and j is any odd integer modulo 2e. Wemay write j = ±52a or ±52a+1 for a nonnegative integer a. Some of the casesrequire that we set k = 2w or k = 2w + 1. The following table summarizesall of the containments.Type 1 2 3 4Element (k, 5a) (2w,−(5)a) (2w + 1,−(5)2a+1) (2w + 1,−(5)2a)Belongs to H1 Hc2 Hc3 Hc3Type 1 inclusion is obvious from the definition of H1. To demonstrate Type2 inclusion we use Lemma 4.6 and determine n and d from(2w,−(5)a) = (n(d− 1), d).We set d = −(5)a and determine n from the linear congruencen(−(5)a − 1) ≡ 2w(mod 2e),noting thatgcd((5)a + 1, 2e) = 2.For the third type, we use Lemma 4.7 and determine nonnegative integersr and s together with a positive integer m such that(2w + 1,−(5)2a+1) = (m,−1)r(3m,−5)s.An easy calculation shows that(3m,−5)2a+1 =(−m((−5)2a+1 − 12), (−5)2a+1).We choose r = 0, s = 2a+ 1 and m satisfying the linear congruence−m((−5)2a+1 − 12)≡ 2w + 1 (mod 2e),714.4. Proof of Theorem 4.1noting that gcd(((−5)2a+1 − 12), 2e)= 1, to complete type 3 contain-ment. For the last case, using Lemma 4.7 and setting r = 1, s = 2a weget(m,−1)(3m,−5)2a =(m((−5)2a + 12),−(5)2a)and determining m from the solvable congruencem((−5)2a + 12)≡ 2w + 1(mod 2e)establishes the final inclusion.To establish the result that the intersection of the conjugates of the Hiis trivial, we prove the stronger statement that there are two conjugates ofH2 with trivial intersection. Recalling Lemma 4.6, we choose n = 1 andn = 2 and compare the conjugate subgroupsH2 = {((d− 1), d) : d = 1, 3, . . . , 2e − 1}andHc2 ={(2(d′ − 1), d′) : d′ = 1, 3, . . . , 2e − 1} .Since each second component of these ordered pairs occurs exactly once, anelement belonging to the intersection of these two conjugate subgroups mustsatisfy d′ = d. Further we deduce from the first component that2(d− 1) ≡ d− 1(mod 2e),leading to the conclusion thatd ≡ 1(mod 2e).Thus the only element in the intersection of these conjugates is the identity(0, 1), so that{H1, H2, H3}yields a 3-cover of G.724.5. Proof of Theorem 4.24.5 Proof of Theorem 4.2We begin by giving defining polynomials for the subfields of the splittingfield of x2e − a corresponding to the subgroups in the 3-cover via Galoistheory.Lemma 4.8. Suppose that the Galois group of x2e − a is isomorphic toZ2e o Z∗2e . Let θ be a root of x2e − a and suppose that ζ = exp(2pii/2e).Let H1, H2, and H3 be the subgroups specified in Theorem 4.1. Let K1,K2,and K3 be the fixed fields corresponding to these subgroups by Galois theory.ThenK1 = Q (i) with defining polynomial f1(x) = x2 + 1,K2 = Q (θ) with defining polynomial f2(x) = x2e − a,K3 = Q(ζθ2)with defining polynomial f3(x) = x2e−1 + a.Proof. We recall that for a typical element of (k, j) of the Galois group Gthat(k, j)(θ) = ζkθ and (k, j)(ζ) = ζj .An easy exercise now shows that the elements i (respectively, θ, ζθ2) arefixed by H1 (respectively H2, H3). Furthermore their degrees over Q areequal to, in each case, the index of the subgroup under which they are fixed.The result now follows.The following result characterizes k-th powers in the p-adic integers, Zp(see for example Castillo [Cas11, Lemma 2.7]). We let Skp denote the set ofnonzero k-th powers in Zp.Proposition 4.9. Let g be a primitive root modulo p. We haveSkp ={pkmgk`(1 + pε+1+vp(k)c) : m ≥ 0, ` ≥ 0, c ∈ Zp},whereε = 1 if p = 2,0 if p > 2.734.5. Proof of Theorem 4.2The following lemma is concerned with power congruences.Lemma 4.10. Let p be a prime satisfying p ≡ 3(mod 4) and suppose thatb is an integer not divisible by p. Then the congruencex2k ≡ b(mod p`)is solvable for all positive integers k and ` if and only if b is a quadraticresidue modulo p. Equivalently, for all fixed positive integers k, x2k − b hasa root in Qp if and only if b is a quadratic residue mod p.Proof. Setting k = ` = 1 in the assumed solvable congruence shows that bis a quadratic residue modulo p. Conversely, since p ≡ 3(mod 4) then thequadratic residues modulo p are quartic residues, octic residues and so on.Thusx2k ≡ b(mod p)is solvable for all positive integers k. Furthermore as p - b a standard appli-cation of Hensel’s Lemma for any fixed integer k now shows that.x2k ≡ b(mod p`)is solvable for all positive integers `. The equivalent statement that x2k − bhas a root in Qp is immediate.Now we give the proof of Theorem 4.2.Proof. Suppose that a is an integer satisfying the scaling assumption (1),and setF (x) = (x2e − a)(x2 + 1)(x2e−1 + a).It follows from Lemma 4.8 that the roots of the second and third factorsof F (x) can be expressed in terms of the roots of x2e − a so that F (x) andx2e − a have the same Galois group. This Galois group is G if and only ifa 6= ±d2, ±2d2 for all integers d by Lemma 4.4. To finish we must establishthe conditions for F (x) to be intersective. We study solvability of F (x) for744.5. Proof of Theorem 4.2various classes of prime powers by working in Qp.Case 1: p = 2. The first factor of F (x) is solvable in Q2 if and only if a isa 2e-th power in Q2. Using Proposition 4.9, with g = 1, we deduce thata = 22em(1 + 2e+2c)for c ∈ Z. By the scaling assumption (1), m = 0 so that we can write theequivalent statementa ≡ 1(mod 2e+2)The second factor x2 + 1 clearly has no root in Q2. We consider the thirdfactor which has a root in Q2 if and only if −a is a 2e−1-st power in Q2.Using Proposition 4.9, with m = 0 and g = 1, we deduce that−a = 22e−1m(1 + 2e+1c)where c ∈ Z. From the scaling assumption (1), we see that either m = 0 orm = 1 so that−a = (1 + 2e+1c) or − a = 22e−1(1 + 2e+1c)for c ∈ Z. Thus we have established all the conditions for p = 2.Case 2: p ≡ 1(mod 4). Since(−1p)= +1, a straightforward application ofHensel’s Lemma [INM95, p. 87] shows that x2 + 1 is solvable in Qp.Case 3: p ≡ 3(mod 4). Suppose that p - a. We begin by showing that thefirst or third factor of F (x) always has a root in Qp. By Lemma 4.10 withk = e, the first factor of F (x) is solvable in Qp if and only if(ap)= 1. (5)Again by Lemma 4.10, with k = e − 1, the third factor of F (x) has a root754.5. Proof of Theorem 4.2in Qp if and only if(−ap)= 1 so that(ap)= −1,establishing solvability in Qp in this case. Now suppose that p | a. Writinga = pta1where t is a positive integer and recalling the scaling assumption (1) wededuce that 0 ≤ t < 2e. The inequality on t combined with Proposition 4.9shows that a is not a 2e-th power in Qp so that the first factor of F (x) cannothave a root in Qp. The second factor of F (x) cannot have a root in Qp since−1 is not a square modulo p. Thus solvability of F (x) in Qp depends on thethird factor of F (x). This factorx2e−1+ ahas a root in Qp if and only if −a is a 2e−1-st power in Qp which impliesthatp2e−1 ‖ a.Further by Lemma 4.10 with k = e− 1 we must have(−a/p2e−1p)= 1establishing the conditions in the theorem.Below is a table of polynomials for different values of a that uses theconditions stated in Theorem 4.2. Some are intersective and some are not.For those that are not, the reason is provided to the reader.764.5. Proof of Theorem 4.2Table 4.1: Intersective polynomials with Galois group G ' Z2e o Z∗2ee a F (x) = (x2e − a)(x2 + 1)(x2e−1 + a)3 65 Intersective3 33 Not intersective, violates (a), no root in Q3 or Q11.3 −17 Intersective4 −38 · 97 Intersective4 41 Not intersective, violates (b), no root in Q2.4 −97 Intersective5 641 Intersective5 316 Not intersective, violates (a) and (b), no root in Q2 or Q3.5 −316 Intersective77Chapter 5Future Work and Conclusion5.1 Future Work5.1.1 Covering Dihedral Groups DnGiven a dihedral group Dp of order 2p, find intersective polynomials thatrepresent any dihedral group of order 2p. The coverings for these groups areeasy to find but the theory required to form intersective polynomials mayneed to be constructed or researched fully.A possible extension of the problem above could be given a dihedralgroup Dpq of order 2pq, find a cover for this dihedral group: first for smallordered groups, then perhaps a generalization. Once a cover is constructed,form families of intersective polynomials with this dihedral Galois group.Then, find out of if it’s possible to find a cover for small dihedral groupsof any order 2n. It may be helpful working with the case when n is odd orwhen n is even.5.1.2 Covering Semi-Direct Products ZmoZ∗mIt is well known that prime dihedral groups are semi-direct products of Zpwith Z2. This may lead to interesting conclusions about semi-direct productsin general. Or conversely, it may be possible to use theory about semi-directproducts to form intersective polynomials represented by dihedral Galoisgroups.Instead of the specific holomorph used in this thesis, find a cover formore generalized semi-direct products ZmoZ∗m, then construct intersectivepolynomials that represent these products.785.1. Future Work5.1.3 Product of Two Simplest CubicsFor a ∈ N, let f = x3 − ax2 − (a+ 3)x− 1 and let K = Ka be the cycliccubic number field generated by a root α of f . The splitting field for theproduct of two of these polynomials would be isomorphic to C3 × C3. Findintersective polynomials constructed as the product of these types of cubics.5.1.4 Hilbert Class PolynomialsGiven a number field K, there is a finite Galois extension L of K suchthat1. L is an unramified Abelian extension of K.2. Any unramified Abelian extension of K lies in L.The field L above is called the Hilbert class field of K. Now let K =Q(√D), where D is a negative fundamental discriminant (not equal to 1,not divisible by any square of any odd prime, and satisfies d ≡ 1 (mod 4) ord ≡ 8, 12 (mod 16)).Let C(D) be the set of all reduced quadratic forms [a, b, c] of the dis-criminant D. Then the Hilbert class polynomial is defined to beHD(X) =∏[a,b,c]∈C(D)(X − j(−b+ i√D2a))where j(α) is the j-invariant or j-value of α. This polynomial has degreeequal to the class number of the imaginary quadratic field defined by D.It is of interest that this polynomial is in fact the minimal polynomialfor the Hilbert class field defined above. Due to the nature of this field,all primes are unramified and thus the decomposition group for each primeideal is cyclic. Therefore, the decomposition group for every prime ideal inthe Hilbert class field L of K is contained in a proper subgroup of L, andintersective polynomials can automatically be constructed.795.2. Conclusion5.2 ConclusionThe purpose of this thesis was to provide methods to construct infinitefamilies of intersective polynomials with various Galois groups. The meth-ods used in Chapter 3 required relatively recent theory on decompositiongroups. In the case of polynomials with Galois group A4, we were able totake advantage of the monogeneity of the number field of the roots of thepolynomials we were studying. 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Intersective polynomials and their construction Lee, Paul David 2016
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Title | Intersective polynomials and their construction |
Creator |
Lee, Paul David |
Publisher | University of British Columbia |
Date Issued | 2016 |
Description | A monic polynomial with integer coefficients is called intersective if it has no root in the rational numbers, but has a root modulo m for all positive integers m > 1. Equivalently, the polynomial has a root in each p-adic field ℚp. Using three different methods for forming these intersective polynomials, we produce an infinite family with Galois group A₄, an infinite family with Galois group D₅ and classify intersective polynomials with holomorph Galois group ℤ₂e × ℤ*₂e |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2016-09-22 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivatives 4.0 International |
DOI | 10.14288/1.0314575 |
URI | http://hdl.handle.net/2429/59252 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Irving K. Barber School of Arts and Sciences (Okanagan) Mathematics, Department of (Okanagan) |
Degree Grantor | University of British Columbia |
GraduationDate | 2016-11 |
Campus |
UBCO |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/4.0/ |
AggregatedSourceRepository | DSpace |
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