On the Galois Groups of Sextic Trinomials by Stephen Christopher Brown B.Sc., The University of British Columbia, 2008 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in The College of Graduate Studies (Mathematics) THE UNIVERSITY OF BRITISH COLUMBIA (Okanagan) August 2011 c© Stephen Christopher Brown 2011 Abstract It is well known that the general polynomial anx n + an−1xn−1 + · · ·+ a1x+ a0 cannot be solved algebraically for n ≥ 5; that is, it cannot be solved in terms of a finite number of arithmetic operations and radicals. We can, however, associate every irreducible sextic polynomial with a Galois group. The Ga- lois group of a given polynomial can give us a great deal of information about the nature of the roots of a polynomial and it can also tell us if the polynomial itself is algebraically solvable. This leads to the typical problem in Galois theory: finding the Galois group of a given polynomial. In this thesis, we investigate the inverse problem: for a specific Galois group, what irreducible polynomials occur. More specifically, we look at monic trinomials – polynomials with only three terms, having 1 as the lead- ing coefficient. The first unresolved case of trinomials are of degree six and we will look specifically at trinomials of the form x6 + ax+ b. We begin by investigating families of these trinomials that will result in Galois groups having a particular structure. From these families of trino- mials, we can then make a final determination of individual Galois groups after eliminating any reducible possibilities. In the main calculations of this thesis, we investigate two parametric families of trinomials, one of which is given in [1]. From these families we completely characterize five of the possible sixteen Galois groups that ii Abstract can occur for sextic polynomials. In the notation of Butler and Mckay [2], these groups are 6T1, 6T2, 6T4, 6T5, and 6T6. In the final determination of these polynomials, rational points are found on genus 2 curves using a method known as elliptic Chabauty. We give an introduction to Galois theory followed by a brief explanation of the methods used to attain our results. We then discuss our results and proceed to prove them through the use of powerful software such as MAPLETMand the Magma algebra system [3]. iii Preface The main results discussed in Chapter 5 and Chapter 6 are from col- laborative research done with Dr. Blair Spearman and Dr. Qiduan Yang. The results discussed in Chapter 5 are from the paper [4] and all authors listed contributed to this paper as well as the results discussed in Chapter 6, which are intended to be published in a subsequent paper. In both of these chapters I have been responsible for finding rational solutions to genus 2 curves using software such as MAPLETMand the Magma algebra system [3]. I have also contributed to the editing of these papers. iv Table of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . v List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . ix 1 Algebraic Preliminaries . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Homomorphisms, Quotient Rings, and Ideals . . . . . . . . . 1 1.2 The Ring of Polynomials . . . . . . . . . . . . . . . . . . . . 4 1.3 Factorization of Polynomials . . . . . . . . . . . . . . . . . . 5 1.4 Algebraic Numbers . . . . . . . . . . . . . . . . . . . . . . . 11 2 Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.1 Introduction to Galois Theory . . . . . . . . . . . . . . . . . 16 2.2 The Idea Behind Galois Theory . . . . . . . . . . . . . . . . 17 2.3 The Galois Group of a Polynomial . . . . . . . . . . . . . . . 18 2.4 Determining the Galois Group of a Given Polynomial . . . . 20 3 Polynomials and their Known Galois Groups . . . . . . . . 24 3.1 Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . 24 3.2 Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . 25 3.2.1 The Klein-4 Group V4 . . . . . . . . . . . . . . . . . . 26 v Table of Contents 3.2.2 The Dihedral Group D4 . . . . . . . . . . . . . . . . . 27 3.2.3 The Cyclic Group C4 . . . . . . . . . . . . . . . . . . 27 3.2.4 The Alternating Group A4 and Symmetric Group S4 28 3.3 Quintic Trinomials . . . . . . . . . . . . . . . . . . . . . . . . 28 3.4 Sextic Trinomials . . . . . . . . . . . . . . . . . . . . . . . . 31 4 Elliptic Curves, Genus 2 Curves, and Elliptic Chabauty . 33 4.1 Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.2 Genus 2 Curves and Elliptic Chabauty . . . . . . . . . . . . 37 5 Sextic Trinomials x6 + Ax+ B Defining Sextic Fields with a Cyclic Cubic Subfield . . . . . . . . . . . . . . . . . . . . . . . 43 5.1 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 43 5.2 A Parametric Family . . . . . . . . . . . . . . . . . . . . . . 44 5.3 Proof of Theorem . . . . . . . . . . . . . . . . . . . . . . . . 54 6 Sextic Trinomials x6 + Ax + B Defining Normal Sextic Ex- tensions of Number Fields . . . . . . . . . . . . . . . . . . . . 56 6.1 Main Theorem and Corollaries . . . . . . . . . . . . . . . . . 56 6.2 Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . 58 6.3 Proof of Theorem and Corollaries . . . . . . . . . . . . . . . 67 7 Results and Future Work . . . . . . . . . . . . . . . . . . . . . 69 7.1 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 7.2 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 vi Table of Contents Appendices A Common Group Names and Structures . . . . . . . . . . . . 75 A.1 The Symmetry Group Sn . . . . . . . . . . . . . . . . . . . . 75 A.2 The Alternating Group An . . . . . . . . . . . . . . . . . . . 76 A.3 The Dihedral Group Dn . . . . . . . . . . . . . . . . . . . . . 76 A.4 The Cyclic Group Cn . . . . . . . . . . . . . . . . . . . . . . 77 A.4.1 The Group 〈Zn,+〉 . . . . . . . . . . . . . . . . . . . 78 A.5 Other Groups Gn, Fn and Direct Products . . . . . . . . . . 78 B Magma Code Related to Chapter 5 . . . . . . . . . . . . . . 80 C Magma Code Related to Chapter 6 . . . . . . . . . . . . . . 88 vii List of Tables 5.1 Local Solvability of the Quadratic . . . . . . . . . . . . . . . 50 6.1 Local Solvability of the Quadratic . . . . . . . . . . . . . . . 66 7.1 List of Findings . . . . . . . . . . . . . . . . . . . . . . . . . . 70 viii Acknowledgements I am very grateful for all the encouragement I have received from my family and from a network of friends. I thank these amazing people for all the support they have given me. I would also like to thank my wonderful committee of professors: Dr. Blair Spearman, Dr. Qiduan Yang, Dr. Shawn Wang, and Dr. Rebecca Tyson. They have contributed to the completion of this thesis through their guidance and exceptional teaching, which have helped me get through my education this far. I would especially like to recognize my supervisor, Dr. Blair Spearman, who has always given me encouragement and confidence in myself. You have played an integral role in my decision to continue my education and you inspire me for what I could achieve in the future. ix Chapter 1 Algebraic Preliminaries 1.1 Homomorphisms, Quotient Rings, and Ideals We begin with some basic definitions regarding rings and maps. Definition 1.1. A ring is a triple 〈R,+, ·〉, where R is a set with two binary operations + and · defined on R such that i. 〈R,+〉 is an abelian group; ii. · is associative; and iii. for all a, b, c,∈ R, the left distributive law a · (b+ c) = (a · b) + (a · c) and the right distributive law (a+ b) · c = (a · c) + (b · c) hold. We call + addition and · multiplication. We shall write ab instead of a · b. Definition 1.2 (Homomorphism). For rings R and S, a map φ : R→ S is a homomorphism if the following two conditions are satisfied for all a, b ∈ R: i. φ(a+ b) = φ(a) + φ(b) ii. φ(ab) = φ(a)φ(b). Definition 1.3 (Isomorphism). An isomorphism φ : R→ S from a ring R to a ring S is a homomorphism that is one to one and onto S. 1 1.1. Homomorphisms, Quotient Rings, and Ideals If an isomorphism exists from a ring R to a ring S, then R is said to be isomorphic to S and we write R ∼= S. Definition 1.4 (Subring). A subring of a ring R is a subset S of R which is itself a ring under the operations it inherits from R. Definition 1.5 (Ideal). A subset I of a ring R is an ideal of R if i. 〈I,+〉 is a group under the addition operation defined in R and ii. for all x ∈ I and for all r ∈ R, xr ∈ R. With the idea of ideals, we can now define a quotient ring. First we must understand the concept of cosets. Definition 1.6 (Cosets). Let H be a subgroup of G. The subset aH = {ah | h ∈ H} is called the left coset of H containing a, whereas the subset Ha = {ha | h ∈ H} is called the right coset of H containing a. In the case where addition is the group operation, we write a + H = {a+ h | h ∈ H} and H + a = {h+ a | h ∈ H} as the left and right cosets of H containing a, respectively. Remark 1.1. In the case where G is an abelian group, the left and right cosets of H containing a are equal. Definition 1.7 (Quotient ring). If I is an ideal of a ring R, we can form the quotient ring R/I, consisting of the cosets of I in R considered as a group under addition, having the properties i. (I + r) + (I + s) = I + (r + s) ii. (I + r)(I + s) = I + (rs). 2 1.1. Homomorphisms, Quotient Rings, and Ideals Definition 1.8 (Kernel of a homomorphism). Let a map φ : R → S be a ring homomorphism. The subring φ−1(0s) = {r ∈ R | φ(r) = 0s} is the kernal of φ where 0s is the zero element in S. We denote the kernel of φ by Ker(φ). It should also be noted that the kernel Ker(φ) of a ring homomorphism φ : R → S is an ideal of R. The concept of an integral domain and a field will also be useful for later definitions. An integral domain is a ring D with an additional three properties. Definition 1.9 (Integral Domain). An integral domain is a ring 〈D,+, ·〉 such that i. · is commutative; ii. there exists an element 1 ∈ D such that a1 = 1a = a for all a ∈ D; and iii. if ab = 0 for a, b,∈ D then either a = 0 or b = 0. Finally, we can build onto this concept one step further to define a field. Definition 1.10 (Field). A field is a ring 〈F,+, ·〉 such that F\{0} is an abelian group under multiplcation. In this definition, f\{0} means “all non-zero elements of F”. Since 〈f\{0}, ·〉 is an abelian group, for every a ∈ F , we use a−1 ∈ F to denote the multiplicative inverse of a. 3 1.2. The Ring of Polynomials 1.2 The Ring of Polynomials We can express a polynomial in x (called an indeterminate) with coeffi- cients in a ring R as a finite sum n∑ i=0 aix i = a0 + a1x+ · · ·+ anxn, where ai ∈ R and n is the degree of the polynomial. We say that two polyno- mials are equal if and only if the corresponding coefficients are equal (where any omitted powers of x can be taken to have a coefficient of zero). We define the addition and multiplication operations on polynomials as follows: If f = n∑ i=0 aix i and g = n∑ i=0 bix i, then we define f + g = n∑ i=0 (ai + bi)x i fg = n∑ i=0 cix i where ci = n∑ j+k=i ajbk. Under these operations, the set of all polynomials with coefficients in R and an indeterminate x forms a ring, which we denote by R[x]. Lemma 1.1. If D is an integral domain and x is an indeterminate, then D[x] is an integral domain. 4 1.3. Factorization of Polynomials Proof. Let f = a0 + a1x+ · · ·+ anxn and g = b0 + b1x+ · · ·+ bmxm where an 6= 0 and bm 6= 0 and all the coefficients are in D. The coefficient of tm+n in fg is anbm, which is non-zero since D is an integral domain. Therefore if f, g are non-zero, then fg is also non zero. Thus, D[x] is an integral domain. In particular, if F is a field, then F [x] is an integral domain. Remark 1.2. F [x] is not a field, as x is does not have a multiplicative inverse. That is, there is no polynomial f(x) ∈ F [x] such that xf(x) = 1. 1.3 Factorization of Polynomials Not all polynomials can be factored over a given integral domain or field, so let us first define what what it means for a polynomial to be reducible. Definition 1.11 (Reducible Polynomial). A non-constant polynomial f(x) is said to be reducible over a ring R if it can be expressed as the product of two or more non-constant polynomials of lesser degree in R[x]. That is, f(x) = g(x)h(x) f(x), g(x), h(x) ∈ R[x]. If no such factorization is possible, then f(x) is said to be irreducible. All polynomials of degree 0 or 1 are irreducible, because they cannot be expressed as a product of polynomials of lesser degree. 5 1.3. Factorization of Polynomials Definition 1.12 (Roots of a function). Given a function f(x), a root of f(x) is any value r that gives f(x) = 0 when x = r. Corollary 1.1 (Factor theorem). Let f(x) be a polynomial in Q[x]. f(x) has a root a ∈ Q if and only if x− a is a factor of f(x) in Q[x]. Proof. Suppose that f(a) = 0 for some a ∈ Q. By the division algorithm, f(x) = (x− a)q(x) + r(x) for some q(x), r(x) ∈ Q[x] with either r(x) = 0 or r(x) having degree less than 1. In either case we have r(x) = c for some c ∈ F . Thus, f(x) = (x− a)q(x) + c. Evaluating at x = a, f(x) = (a− a)q(a) + c 0 = 0q(a) + c 0 = c Then f(x) = (x− a)q(x) and x− a is a factor of f(x). Conversely, if x−a is a factor of f(x) ∈ F [x] where a ∈ Q, it is obvious that evaluating at x = a will give f(a) = 0 and therefore a is a root of f(x). We can now establish some criterion that will help determine whether a specific polynomial is reducible over Q. The Rational Root Test or Rational Root Theorem gives candidates for the roots of a polynomial (pg 214–215, [5]). Theorem 1.1 (Rational Root Test). Let f(x) be a polynomial f(x) = anx n + an−1xn−1 + · · ·+ a1x+ a0 6 1.3. Factorization of Polynomials with integer coefficients. Let an and a0 be nonzero. Then any root of f(x) in the rational numbers Q can be expressed as x = pq for p, q ∈ Z, q 6= 0, where p and q satisfy the following two properties: i. p divides the constant term a0 ii. q divides the leading coefficient an. We now state Gauss’ Lemma (Proposition 2.4, pg. 19, [6]). Theorem 1.2. Let f(x) be a polynomial in Z[x] be irreducible over Z. Then f(x) is also irreducible over Q. Proof. By way of contradiction, we assume that f(x) ∈ Z[x] is irreducible over Z, but reducible over Q. That is, f(x) = g(x)h(x) where g(x), h(x) are polynomials in Q[x] of smaller degree. By multiplying by the product of denominators of the coefficients of g and h, we can rewrite this equation as nf = g′h′ where n ∈ Z and g′, h′ are polynomials in Z[x]. We will now cancel out the prime factors of n one by one, while staying within Z[x]. Suppose that p is a prime factor of n. If we define g′ = g0 + g1x+ · · ·+ grxr h′ = h0 + h1x+ · · ·+ hsxs then we claim that either p divides all of the coefficients gi or p divides all the coefficients hj . Otherwise, there must exist smallest values i and j such that p - gi and p - hj . However, p does divide the coefficient of xi+j in the polynomial g′h′, which is h0gi+j + h1gi+j−1 + · · ·+ hjgi + · · ·+ hi+jg0. 7 1.3. Factorization of Polynomials By the choice of i and j, the prime p divides every term of this expression, perhaps with the exception of hjgi. However, we know that p divides the whole expression, so p|hjgi. But p - hj and p - gi, which gives a contradiction. This establishes our claim. Without loss of generality, we may now assume that p divides every coefficient gi. Then g ′ = pg′′ where g′′ is a polynomial in Z[x] of the same degree as g′ or g. Let n = pn1. Then pn1f = pg ′′h′ so that n1f = g ′′h′. Repeating this process, we can remove all of the prime factors of n and we arrive at an equation f = ḡh̄ where ḡ and h̄ are polynomials in Z[x] that are rational multiples of the original g and h. But, this contradicts the irreducibility of f(x) over Z. Hence our assumption that f(x) is reducible over Q is false, and f(x) must be irreducible over Q. Another important test for irreducibility is Eisenstein’s Criterion (The- orem 2.5, pg. 20, [6]) Theorem 1.3 (Eisenstein’s Criterion). Let f(x) be a polynomial f(x) = anx n + an−1xn−1 + · · ·+ a1x+ a0 with integer coefficients. Suppose that there exists a prime p ∈ Z such that i. p divides each ai for i 6= n, 8 1.3. Factorization of Polynomials ii. p does not divide an, and iii. p2 does not divide a0. If such a p exists, then f(x) is irreducible over Q. Proof. Suppose that such a polynomial f(x) exists, which satisfies the above conditions. By way of contradiction, assume that f is reducible overQ. Then f(x) = g(x)h(x) for some non-constant polynomials g(x), h(x) ∈ Q[x]. The reduction map mod p given by φp : Z[x]→ Zp[x] is a homomorphism and thus, φp (f(x)) = φp (g(x))φp (h(x)) . But φp (f(x)) = φp ( anx n + an−1xn−1 + · · ·+ a1x+ a0 ) = αxn where an ≡ α mod p because p | ai for i 6= n. So φp (g(x))φp (h(x)) = αx n and φp (g(x)) = βx k, φp (h(x)) = γx n−k for some 0 < k < n and β, γ such that βγ ≡ α mod p. Since Zp[x] is a unique factorization domain, then p must divide each of the non-leading 9 1.3. Factorization of Polynomials coefficients of g(x) and h(x). Looking at the constant terms b0 and c0 of g(x) and h(x) respectively, b0 = pr c0 = ps for some r, x ∈ Z. Then the constant term of f(x) must be a0 = b0c0 = (pr)(ps) = p2(rs), which is a contradiction because p2 - a0. A brief definition and Theorem are also provided in this section that will serve as tools when looking at polynomials later in this thesis. Definition 1.13 (Discriminant of a polynomial). Let f(x) be a polynomial with roots r1, . . . , rn ∈ C (not necessarily distinct). The discriminant of f(x) is defined by ∏ 1≤i 1, we have to use alternate methods. This is why we use elliptic Chabauty. Elliptic Chabauty methods work even if the rank is greater than 1 and theoretically applies to any rank of the Jacobian. Another major advantage of elliptic Chabauty methods is that we do not need to use the Jacobian in calculations. Before using the method of elliptic Chabauty in the following two chap- ters, we will give a brief description of the method. We start with a hyper- elliptic curve y2 = f(x) of genus 2 where f(x) ∈ Q(x) is a monic polynomial of degree six with rational coefficients. We then factor f(x) over a number field K to obtain y2 = F1(x)F2(x), where F1(x) is a quadratic in K[x] and F2(x) is a quartic in K[x]. Looking at these two polynomial factors, we find the greatest common divisor of F1(x) and F2(x) in K[x] modulo squares. If this gcd is equal to 1 modulo squares, then both F1(x) = gU 2 and F2(x) = gV 2 for some polynomials U and V , since each of the factors must then be equal to a square. In these equations, g = (−1)i0i11 · · · inn , i0, . . . , in = 0, 1, where each  is a fundamental unit of K. This gives a list of paired equations where possible rational solutions can be found. Each of these equations is a 41 4.2. Genus 2 Curves and Elliptic Chabauty simpler curve than the original problem and many of the possible values of g can be ruled out through local solvability of the associated two curves. This test for local solvability can be done through the Magma algebra system [3] for each of the curve’s “bad primes”. The IsLocallySolvable routine in Magma then looks for a possible solution through p-adic analysis. After local solvability, there will likely be only very few remaining pos- sible values for g that result in equations which are locally solvable. Of these resulting equations we then pick one for each value of g that gives the equation of an elliptic curve. If we are fortunate enough at this point to have the rank of the Mordell-Weil group of this elliptic curve less than the degree [K : Q], we can then apply elliptic Chabauty to the curve. Once again, Magma provides routines PseudoMordellWeilGroup and Chabauty for the calculations. If Magma is successful, we find only finitely many rational points on these elliptic curves. These rational points from Magma give us finitely many candidates for x at which rational points on the original hyperelliptic curve y2 = f(x) may occur. We must then see if any of these candidates actually translate to rational points on the original curve. This completes our method. 42 Chapter 5 Sextic Trinomials x6 + Ax +B Defining Sextic Fields with a Cyclic Cubic Subfield 5.1 Main Theorem Let f(x) be a polynomial with rational coefficients which is irreducible over the rational numbers Q. Let Gal(f) denote the Galois group of f(x). In this chapter, we characterize irreducible trinomials of the form f(x) = x6 +Ax+B, having a Galois group isomorphic to either A4 the alternating group on four letters, or A4 × C2 where C2 is the cyclic group of order 2. These polynomials define sextic fields (as discussed in chapter 1) having a cyclic cubic subfield and they occur in a parametric family, which enables an analysis of their Galois groups. While there are 16 possible Galois groups for irreducible sextic polynomials in Q[x], [9, pp. 323-325], basic group theory shows that only these three can occur if f(x) defines a sextic field with a cyclic cubic subfield. These groups are C6 the cyclic group of order 6, A4 or A4 ×C2. In the notation of Butler and Mckay [2], these three groups are 6T1, 6T4, and 6T6 respectively. It has already been shown in [20] that up to scaling, there exists a single, unique sextic trinomial with Galois group isomorphic to C6, which was given as x6 + 133x+ 209. 43 5.2. A Parametric Family For the trinomials discussed in this chapter, we show that only A4 × C2 occurs. Our main theorem is the following, which appears in [4]. Theorem 5.1. Let A and B denote nonzero rational numbers and set f(x) = x6 +Ax+B i. f(x) is irreducible over Q and Gal(f) ' A4 × C2 ⇔ A = 4u(u2 + 3)(3u2 + 1)(3u2 + 25)2v5, B = (u2 − 5)(3u2 + 1)(u4 + 10u2 + 5)(3u2 + 25)2v6, for rational numbers u and v with u 6= 0,±5 and v 6= 0. ii. If f(x) is irreducible over Q then Gal(f) ' A4 does not occur. In the following section, we describe the parametric family of sextic tri- nomials x6 +Ax+B that define sextic fields with cyclic cubic subfields and assess the irreducibility of these trinomials. In section 3, we will prove our theorem characterizing the Galois groups of these polynomials. 5.2 A Parametric Family We begin with the family of sextic trinomials which appears in [1]. Proposition 5.1. Let A and B denote nonzero rational numbers such that f(x) = x6 + Ax + B is irreducible over Q. Then f(x) defines a sextic field containing a cyclic cubic subfield if and only if there exist rational numbers u and v such that A = 4u(u2 + 3)(3u2 + 1)(3u2 + 25)2v5, B = (u2 − 5)(3u2 + 1)(u4 + 10u2 + 5)(3u2 + 25)2v6. 44 5.2. A Parametric Family These restrictions on A and B have been carefully expressed in this way so as to parametrize A and B in terms of a single variable u with scaling factor. In order to completely determine the Galois groups of the polynomials in this family, we will first need to investigate which of these polynomials, if any, are irreducible. Such a study frequently requires an analysis of an algebraic curve as for example in [21]. For our determination, we will require the study of a genus 2 curve and an application of elliptic Chabauty as discussed in Chapter 4. We begin with a Lemma establishing a condition for reducibility. Lemma 5.1. Let A and B denote rational numbers such that A = 4u(u2 + 3)(3u2 + 1)(3u2 + 25)2v5, B = (u2 − 5)(3u2 + 1)(u4 + 10u2 + 5)(3u2 + 25)2v6. for rational numbers u and v. Let f(x) = x6 + Ax+ B. If f(x) is reducible over Q, then the cubic polynomial g(x) = x3 − (3u2 + 25)(3u2 + 1)x− 4u(3u2 + 25)(3u2 + 1) has a rational root. Proof. Let θ denote a root of f(x) and set K = Q(θ). It was shown in the proof of the main theorem in [1] that K contains a subfield E defined by g(x) = x3 − (3u2 + 25)(3u2 + 1)x− 4u(3u2 + 25)(3u2 + 1). Suppose that f(x) is reducible over Q, yet g(x) is irreducible over Q. Since the discriminant of g(x) is equal to 4(3u2 + 1)2(3u2 − 5)2(3u2 + 25)2, 45 5.2. A Parametric Family a perfect square in Q, we see that E is a cyclic cubic field and hence E ⊆ R. The degree of K over Q is divisible by 3 by Theorem 1.7 and less than 6. It follows that [K : Q] = 3 and so K = E, which implies θ ∈ R for any root θ of f(x). However, the trinomial f(x) clearly has complex roots, which can be deduced from Rolle’s Theorem (Theorem 1.4). This contradiction shows that g(x) must be reducible over Q. Lemma 5.2. The projective curve y2 = x6 − 3x4 + 51x2 + 15 has the six points ∞+,∞−, (1, 8), (−1, 8), (1,−8), (−1,−8). Proof. The method of elliptic Chabauty as discussed in chapter 5 is used for this determination. We work in the number field defined by a root of x3 − 3x2 + 51x + 15. This field is K = Q(t), where t3 + 3t + 1 = 0. The maximal order OK = Z[t], and there is one fundamental unit,  = t, which has norm −1. The class number of OK is 1. We have the following prime ideal factorizations in OK , which are confirmed by MAPLE TM: 〈2〉 = ℘2, 〈3〉 = ℘33, 〈5〉 = ℘251℘52 where ℘2 = 〈2〉, ℘3 = 〈t+ 1〉, ℘51 = 〈1− t〉, and ℘52 = 〈 t2 − t+ 3〉. A scripted ℘ notation is used to distinguish prime ideals from a prime integer p. In the equation specified in this Lemma, we make the substitution (x, y) = (X/Z, Y/Z3), where X,Y, Z ∈ Z and gcd(X,Z) = 1, to obtain Y 2 = X6 − 3X4Z2 + 51X2Z4 + 15Z6. (5.1) This substitution now allows us to use many useful properties of integers. Next we show that X must be odd, for if X were even, then Z would be odd and equation (5.1) would reduce to Y 2 ≡ 3(mod4), 46 5.2. A Parametric Family which is impossible. Furthermore, we can see that X is not divisible by 3, for otherwise Z would not be divisible by 3 and equation (5.1) would reduce to Y 2 ≡ 6(mod9), which is impossible. Finally we can see that X 6≡ ±2Z(mod5) or else (5.1) reduces to Y 2 ≡ ±10(mod25), which is impossible. We now factor the right hand side of (5.1) over K to obtain Y 2 = ( X2 − (4t+ 1)Z2) (X4 + 2(2t− 1)X2Z2 + (16t2 − 4t+ 49)Z4). (5.2) We will now calculate the ideal gcd of the two factors on the right of (5.2) modulo squares. For notation, we use F2 for the quadratic factor and F4 for the quartic factor. We begin with 2 identities F4 − (X2 + (8t− 1)Z2)F2 = 48Z4(1 + t2) and (1 + 4t)2F4 − ((−32t2 + 8t− 47)X2 + 15Z2)F2 = 48X4(1 + t2), which can be obtained through the division algorithm for polynomials. Since gcd(X,Z) = 1 and as ideals we have ℘51 = 〈 1 + t2 〉 , we see that the ideal gcd of F2 and F4 divides 〈48〉℘51 47 5.2. A Parametric Family and therefore consists of some powers of (2), ℘3, and ℘51. We will consider these cases of ideals to determine the power of each ideal in the gcd modulo squares. Case i. The power of 〈2〉 in the gcd. We expand F2 to get F2 = X 2 − Z2 − 4tZ2. We already know that X is odd. Therefore 〈2〉 divides F2 if and only if Z is odd, in which case we have 8 | (X2 − Z2) so that 〈2〉2 ‖ F2. Turning to F4, we note that since F2F4 = Y 2, then 〈2〉0 or 2 ‖ F2 and we conclude that 〈2〉even ‖ F4. This eliminates the presence of 〈2〉 in the gcd modulo squares. Case ii. The power of ℘3 in the gcd. Since ℘3 = 〈t+ 1〉 , we have t ≡ −1(mod℘3). Hence F2 = X 2 − (4t+ 1)Z2 ≡ X2 + 3Z2(mod℘3) and so ℘3 | F2 if and only if 3 | X which was ruled out earlier. Therefore ℘3 - gcd(F2, F4) and is also eliminated from the gcd modulo squares. Case iii. The power of ℘51 in the gcd. Since we know that 1 + t2 ≡ 0(mod℘51), 48 5.2. A Parametric Family we must have either t ≡ 2(mod℘51) or t ≡ 3(mod℘51). From t3 + 3t+ 1 ≡ 0(mod℘51), we are left only with the possibility that t ≡ 2(mod℘51). Using this congruence and recalling our expressions for F2 and F4, we deduce that F2 ≡ X2 + Z2(mod℘51) ≡ (X + 3Z)(X + 2Z)(mod℘51) and F4 ≡ X4 +X2Z2(mod℘51) ≡ X2(X + 3Z)(X + 2Z)(mod℘51). These two identities show that ℘51 | gcd(F2, F4)⇔ X ≡ ±2Z(mod5), but these possibilities were already ruled out earlier. Therefore, ℘51 - gcd(F2, F4) and ℘51 is not a factor in the gcd modulo squares. Through these three cases, we can see that the ideal gcd of the factors F2 and F4 is equal to 1 modulo squares. Therefore we can deduce that F2 and F4 must each be equal to a square multiplied by some product of fundamental units. This gives rise to four pairs of element equations F2 = gU 2, F4 = gV 2 49 5.2. A Parametric Family Table 5.1: Local Solvability of the Quadratic (i0, i1) locally insolvable at (0, 1) ℘3 (1, 0) ℘3 with Y = gUV and g = (−1)i0(t)i1 , i0, i1 = 0, 1. Next we turn to local solvability. We can rule out two of the above pairs by testing the quadratic. The results are given in a table. This leaves only the possibilities (i0, i1) = (0, 0), (1, 1). Case 1. (i0, i1) = (0, 0) In this case, the quartic Y 2 = X4 + 2(2t− 1)X2 + (16t2 − 4t+ 49) defines an elliptic curve of rank 1 over K, confirmed by the PseudoMordell- WeilGroup command in Magma [3]. Using the elliptic Chabauty routines in Magma with p = 17, we find two points at infinity as well as X = 0. This value X = 0 does not yield a point on the original sextic y2 = x6 − 3x4 + 51x2 + 15. Case 2. (i0, i1) = (1, 1). In this case, the quartic Y 2 = −t(X4 + 2(2t− 1)X2 + (16t2 − 4t+ 49)) also defines an elliptic curve of rank 1 over K, once again using Pseu- doMordellWeilGroup in Magma. Using the elliptic Chabauty routines in 50 5.2. A Parametric Family Magma again with p = 17, we find two choices for X, namely X = ±1. These lead to the points (x, y) = (±1,±8) Together Case 1 and Case 2 yield all 6 points in the statement of the Lemma. We can now use this lemma to assess the reducibility of our family of trinomials. Lemma 5.3. Let f(x) = x6 +Ax+B where the nonzero rational numbers A and B are given by A = 4u(u2 + 3)(3u2 + 1)(3u2 + 25)2v5 B = (u2 − 5)(3u2 + 1)(u4 + 10u2 + 5)(3u2 + 25)2v6 for nonzero rational numbers u and v. Then f(x) is irreducible over Q. Proof. Suppose that f(x) is reducible over Q for some nonzero value of u. Since v is a nonzero scaling factor, we may assume that v = 1. By Lemma 5.1, g(x) must also be reducible over Q where g(x) is given by g(x) = x3 − (3u2 + 25)(3u2 + 1)x− 4u(3u2 + 25)(3u2 + 1). Reducibility of g(x) implies the existence of linear factor in Q[x] and there- fore a rational root. Thus, there exists a rational number r such that r3 − (3u2 + 25)(3u2 + 1)r − 4u(3u2 + 25)(3u2 + 1) = 0. (5.3) Viewed as an algebraic curve, (5.3) has genus 2 and is birationally equivalent to y2 = x6 − 3x4 + 51x2 + 15 51 5.2. A Parametric Family via the transformations: x = 3u2 − 5 r + 6u , y = −−75r 2 + 3r4 − 2000− 12ur3 − 234r2u2 − 6240u2 − 27u4r2 − 720u4 2r3 , r = 8x5 − 48x3 − 216x− (24 + 8x2)y 2(−3x4 − 3 + 6x2) , u = −x3 + 9x+ y −3x2 + 3 . These transformations can easily be found using MAPLETMand confirmed by substitution. Since u 6= 0, a simple calculation using (5.3) confirms that we must have r 6= 0, r 6= −6u so that any point on the first curve contributes a point on the second curve. By Lemma 5.2, the only finite rational points on the second curve are (±1,±8). None of these points transfers back to the first curve as the denominators in the transformations for r and u are equal to 0 for x = ±1. Hence g(x) has no rational root and therefore f(x) is irreducible over Q. Having eliminated the possibility of reducible trinomials in our paramet- ric family, we can introduce a Lemma which will assist in our determination of specific Galois groups. Lemma 5.4. Let u denote a nonzero rational number. Then the quantity −(3u2 + 1)(u6 − 15u4 − 225u2 − 225) is not equal to a square in Q. Proof. Suppose by way of contradiction that for some nonzero rational num- ber u, the given quantity is equal to a square in Q. We may assume that u is positive. In this expression we make a substitution u = a/b where a, b 52 5.2. A Parametric Family are nonzero integers with gcd(a, b) = 1. By clearing the denominator b6, we have −G1G2 = a square in Q where G1 = (3a 2 + b2), G2 = (a 6 − 15a4b2 − 225a2b4 − 225b6). Using division of polynomials we obtain the identities −27G2 + (9a4 − 138a2b2 − 1979b4)G1 = 212b6 and G2 − (−1365a4 + 450a2b2 − 225b4)G1 = 212a6. Using these identities and recalling that gcd(a, b) = 1, we see that if we define d = gcd(G1, G2), then d = 2 k for some nonnegative integer k. If k > 0, then 2 | G1 so we must have a and b odd and a ≡ b ≡ 1(mod2), in which case a calculation shows that 22 ‖ G1 and 24 ‖ G2. Therefore, gcd(G1, G2) = 1 or 2 2 and in either case is equal to a square. It follows that since G1 > 0, both of G1 and −G2 are squares in Q. Continuing, since a 6= 0, we have 1 a6 G2 = a square in Q. 53 5.3. Proof of Theorem In this last equation we set x = b2 a2 . We can now conclude that y2 = 225x3 + 225x2 + 15x− 1 with y ∈ Q. This last equation can be viewed as an elliptic curve over Q and using MAPLETM, we find that it is birationally equivalent to Y 2 = X3 − 13500X + 540000 via the transformations: X = 225x+ 75, Y = 225y, x = X 225 − 1 3 , y = Y 225 . Using Magma we find that the rank of this curve is equal to zero and it has no finite rational points. Therefore we conclude that −G1G2 cannot equal a square in Q, completing the proof of the Lemma. 5.3 Proof of Theorem We give the proof of Theorem 5.1. Proof. As noted at the beginning of the chapter, irreducible trinomials x6 + Ax + B ∈ Q[x] with Galois group isomorphic to either A4 or A4 × C2 over Q define sextic fields with a cyclic cubic subfield. By Lemma 5.1, A and B are given by A = 4u(u2 + 3)(3u2 + 1)(3u2 + 25)2v5, B = (u2 − 5)(3u2 + 1)(u4 + 10u2 + 5)(3u2 + 25)2v6, 54 5.3. Proof of Theorem for rational numbers u and v with u 6= 0 and v 6= 0. It was shown in [20] that the Galois group of f(x) is C6 if and only if u = ±5. These values of u yield the unique cyclic sextic trinomial up to scaling which was given in the introduction of this chapter as x6 + 133x+ 209. The remaining choices of u and v, namely u 6= 0, u 6= ±5, v 6= 0, will produce a trinomial with Galois group isomorphic to either A4 or A4 × C2. Fortunately, we can distinguish between these possibilities with a useful fact regarding the the polynomial discriminant of f(x), as given by [9, p. 327]. Cohen states that the Galois group of f(x) is isomorphic to A4 if and only if its discriminant is equal to a square in Q. The discriminant of f(x) given by MAPLETM, is equal to − 26(3u2 + 1)5(u6 − 15u4 − 225u2 − 225)(3u2 + 25)10× (27u12 + 540u10 + 6075u8 + 29600u6 + 20625u4 + 22500u2 + 5625)2. The squarefree part of this discriminant is equal to −(3u2 + 1)(u6 − 15u4 − 225u2 − 225). It was shown in Lemma 5.4 that for u 6= 0, this quantity is not equal to a square in Q. Hence the Galois group of x6 +Ax+B cannot be isomorphic to A4. The only remaining possibility for the Galois group is A4×C2. This completes the proof of our theorem. 55 Chapter 6 Sextic Trinomials x6 + Ax +B Defining Normal Sextic Extensions of Number Fields 6.1 Main Theorem and Corollaries Let f(x) be a polynomial with coefficients in an algebraic number fieldK. Assume that f(x) is irreducible over K. Let Gal(f) denote the Galois group of f(x). We are interested in determining those irreducible trinomials f(x) = x6 +Ax+B, with A,B nonzero elements of K, which define normal sextic extensions of K. A discussion of the Galois group of a sextic polynomial and the theory of resolvents used to calculate them is available in Cohen, [9, p. 323], or Jensen, Ledet and Yui [22]. Irreducible polynomials, f(x) ∈ K[x] defining normal extensions of K can only have two possible Galois groups, namely C6 the cyclic group of order 6 or S3 the symmetric group on three letters. These Galois groups are denoted by 6T1 or 6T2, respectively in the notation of Butler and Mckay [2]. Our determination will establish a correspondence between these trinomials and the K− rational points on a genus two curve. As a consequence of Faltings’ theorem [16], there are finitely many K− rational points on this curve hence finitely many normal extensions of the type we are interested in. As a bonus, our method will determine trinomial sextic extensions with Galois group C3 × D3, or 6T5 again referring to [2]. We apply our method to find all trinomials x6+Ax+B 56 6.1. Main Theorem and Corollaries with one of these three Galois groups over Q. Using the method of elliptic Chabauty, we determine the rational points on our genus 2 curve in the case K = Q and find that there is a unique normal trinomial sextic extension and no occurrence of the Galois groups S3 or C3×D3. The details of elliptic Chabauty are given by Bruin [23]. Our main results are the following. Theorem 6.1. Let K be an algebraic number field. If f(x) = x6+Ax+B ∈ K[x] is irreducible over Q and has Galois group C6, S3, or C3 × D3, then there exist elements u, v and w ∈ K with u 6= 0 and v 6= 0, such that A = 4u(3u+ 1)v5 B = −u(1− 18u+ u2)v6 (6.1) and (3w2 + 144)u2 + (−672 + 4w4 + 210w2)u+ 784 + 27w2 = 0. (6.2) Corollary 6.1. There is a single normal sextic extension of Q defined by an irreducible trinomial with rational coefficients. It is Q(θ) where θ is a root of x6 + 133x+ 209. Corollary 6.2. There do not exist trinomials x6 + Ax + B ∈ Q[x] with Galois group S3 or G18. Corollary 6.3. Let K be an algebraic number field and suppose that A and B are nonzero elements of K such that x6 + Ax+B is irreducible over K. The set of trinomials x6 +Ax+B with Galois group C6, S3, or C3 ×D3 is finite. In section 2 of this chapter, some preliminary Lemmas are given to assist in our determination. In section 3 we study the rational points on a genus 2 curve. In section 4 we prove our theorem and corollaries. 57 6.2. Preliminary Results 6.2 Preliminary Results We begin with a family of trinomials defining sextic fields which contain a quadratic subfield. This family would seem to be well known, but unable to find a reference for it we give a short proof. Lemma 6.1. Let A and B denote nonzero rational numbers such that f(x) = x6 + Ax+ B is irreducible over K. Then f(x) defines a sextic field containing a quadratic subfield if and only if there exist rational numbers u and v with u not equal to a square in Q such that A = 4u(3u+ 1)v5, B = −u(1− 18u+ u2)v6. (6.3) The quadratic subfield is Q ( √ u) . Proof. Let θ be a root of f(x) and set L = K(θ) and suppose that L has a quadratic subfield E containing K. Since the degree [L : E] = 3, the minimal polynomial g(x) of θ in E[x] has degree 3 and hence has the form g(x) = x3 + (m+ n √ t)x2 + (p+ q √ t) + (r + s √ t) for elements m,n, p, q, r, s, t ∈ K with t not equal to a square in K. Clearly t 6= 0. If we define g(x) by g(x) = x3 + (m− n√t)x2 + (p− q√t) + (r − s√t) we see that θ is a root of h(x) = g(x)g(x) ∈ K[x]. The polynomial h(x) is given explicitly by h(x) = x6 + 2mx5 + (m2 − n2t+ 2p)x4 + (2mp− 2nqt+ 2r)x3 + (2mr − 2nst+ p2 − q2t)x2 + (2pr − 2qst)x+ r2 − s2t. 58 6.2. Preliminary Results By uniqueness of the minimal polynomial of θ over K we deduce that f(x) = h(x). Thus we can equate coefficients of f(x) − h(x) to zero to solve for A and B. Begin with the coefficient of x5, which yields the equation 2m = 0, so that m = 0. Substituting m = 0 into f(x) − h(x) and examining the coefficient of x4, we obtain the equation n2t 2 = p. Substituting both m = 0 and p = n2t 2 into f(x)− h(x) = 0 and examining the coefficient of x3 gives the equation −2nqt+ 2r = 0, so that r = nqt. Substituting all of the previous equations into f(x)−h(x) = 0, we note that the coefficient of x2 yields the equation t(n4t− 8ns− 4q2) 4 = 0. Since t 6= 0 we deduce that (n4t− 8ns− 4q2) 4 = 0. If n = 0, then it would follow from the previous equation that q = 0. This would imply from equating the coefficient of x to zero that A = 0, 59 6.2. Preliminary Results which contradicts the assumption that A 6= 0. Therefore n 6= 0. Hence we may solve the above equation for s giving s = n4t− 4q2 8n . Substitution of this value of s into the coefficients of x and the constant term yields the equations −3n4t2q − 4q3t+ 4An 4n = 0. and −72n4q2t2 + n8t3 + 16n4t+ 64Bn2 64n2 = 0. Since n was shown to be nonzero, and A 6= 0 by assumption, the first equation shows that q 6= 0. Applying a scaling on A and B and a change of variable on q and t, specifically q → n2q, t→ 4q2u,A→ n5q5A,B → n6q6B yields the equations −n5q5(−12u2 − 4u+A) = 0 and −n6q6(−18u2 + u3 + u+B) = 0. Cancelling the nonzero factors in each equation and solving for A and B gives the values of A and B stated in the theorem. Finally we insert a scaling factor v to get the general polynomial. Conversely, a simple calculation shows that f(x) factors into a pair of conjugate cubics over K ( √ u), showing that L contains a quadratic subfield. Next we give a condition which enables us to determine when x6+Ax+B defines a normal extension of K, which is given in Dummit and Foote [11, p.525]. 60 6.2. Preliminary Results Proposition 6.1 (Dummit and Foote p. 525). Let K be a field of charac- teristic zero and f(x) ∈ K[x] with degree n. Then the Galois group of f(x) is a subgroup of An if and only if the discriminant of f(x) is a square in K. The previous proposition is now specialized to trinomials x6 +Ax+B. Lemma 6.2. Let A and B denote nonzero rational numbers such that f(x) = x6 +Ax+B is irreducible over K. If f(x) defines a relative normal cubic extension field over a quadratic extension field of K then there exist nonzero elements u, v in K with u not equal to a square in K such that A = 4u(3u+ 1)v5 B = −u(1− 18u+ u2)v6 and −(−112√u+ 210u+ 48u√u+ 3u2 + 27) = (a+ b√u)2 for some elements a and b, belonging to K. Proof. If f(x) = x6 +Ax+B defines a relative cubic extension field L over a quadratic extension field of K then by Lemma 6.1, there exist nonzero elements u, v in K such that A = 4u(3u+ 1)v5 B = −u(1− 18u+ u2)v6. We may assume that v = 1, replacing f(x) as necessary by 1 v6 f(vx). One of the factors of f(x) over K ( √ u) is x3 + 2 √ ux2 + (2 √ u+ 2u)x+ 4u+ u √ u−√u. If in addition L/K( √ u) is normal and of degree 3, we know that the Galois group of L/K( √ u) must be isomorphic to A3. Proposition 6.1 shows that 61 6.2. Preliminary Results the discriminant of this cubic must be a square in K ( √ u). This discriminant is −u(−112√u+ 210u+ 48u√u+ 3u2 + 27). Since u is obviously a square in K( √ u), we deduce that there must exist elements a, b ∈ K such that −(−112√u+ 210u+ 48u√u+ 3u2 + 27) = (a+ b√u)2 thus establishing the Lemma. The condition given in the previous Lemma can be converted to a genus 2 curve as is shown in the following Lemma. Lemma 6.3. If f(x) = x6+Ax+B defines a relative normal cubic extension field over a quadratic extension field of K so that A and B are given by 6.1 with u not equal to a square in K and a, b are elements of K such that −(112√u+ 210u− 48u√u+ 3u2 + 27) = (a+ b√u)2, then (3w2 + 144)u2 + (−672 + 4w4 + 210w2)u+ 784 + 27w2 = 0, for some element w ∈ K. Proof. Under the assumptions stated in the lemma, we have −(−112√u+ 210u+ 48u√u+ 3u2 + 27) = (a+ b√u)2. Equating the coefficients of 1 and √ u we deduce the pair of equations 48u+ 2ab− 112 = 0 3u2 + (b2 + 210)u+ 27 + a2 = 0. 62 6.2. Preliminary Results If b = 0 then the first equation yields u = 7/3, which corresponds to w = 0 in the equation stated in this lemma. If b 6= 0, we use a resultant to eliminate the variable a to obtain an equation in b and u, then set b = 2w and divide by 16, obtaining the result stated in this Lemma. The algebraic curve (3w2 + 144)u2 + (−672 + 4w4 + 210w2)u+ 784 + 27w2 = 0 from Lemma 6.3 is birationally equivalent to y2 = x6 + 105x4 + 2400x2 − 19200 via the transformations: x = w, y = (3w2 + 144)u− 336 + 2w4 + 105w2 2w , w = x, u = 336− 2x4 − 105x2 + 2xy 3x2 + 144 . These transformations can easily be found through MAPLETMand can be confirmed by substitution. Lemma 6.4. The projective curve y2 = x6 + 105x4 + 2400x2 − 19200 has the six points ∞+, ∞−, (4, 224), (−4, 224), (4,−224), (−4,−224). Proof. This proof uses the method known as elliptic Chabauty. Computa- tions involved are supported by Magma [3]. We work in the number field defined by a root of x3 + 105x2 + 2400x − 19200. This field is K = Q(t), 63 6.2. Preliminary Results where t3 − 15t+ 20 = 0. The maximal order OK = Z[t], and there are two fundamental units which we can take to be 1 = 8t 2 + 22t− 59, 2 = −13t2 − 21t+ 161, both of norm 1.The class number of OK is 1. We have the following prime ideal factorizations in OK : 〈2〉 = ℘21℘222 = 〈t− 2〉 〈t− 3〉2 , (6.4) 〈3〉 = ℘33 = 〈 t2 + 3t− 7〉3 , (6.5) 〈5〉 = ℘35 = 〈 t2 + t− 15〉3 . (6.6) Once again, a scripted notation is used to differentiate prime ideals from prime integers. Assuming a solution (x, y) to the equation specified in this Lemma, put (x, y) = ( X/Z, Y/Z3 ) , where X,Y, Z ∈ Z, gcd(X,Z) = 1, giving Y 2 = X6 + 105X4Z2 + 2400X2Z4 − 19200Z6. (6.7) We note that X is not divisible by 3, for otherwise Z would not be divisible by 3 and equation (6.7) would reduce to Y 2 ≡ 6(mod9), which is impossible. Factoring the right hand side of (6.7) over K gives Y 2 = F2F4 (6.8) where F2 = X 2 − (7t2 + 20t− 105)Z2 64 6.2. Preliminary Results and F4 = X 4 + (7t2 + 20t)X2Z2 + (1120t+ 400t2 − 3200)Z4 We will calculate the ideal gcd of the two factors F2 and F4. Begin with the following two identities. F4 = (X 2 + (14t2 + 40t− 105)Z2)F2 + 15(31t2 + 84t− 225)Z4, and 4F4 = (−46t2 − 129t+ 332)F2 + (46t2 + 129t− 328)X4. As ideals, the factorization of the remainders is ℘1222℘ 3 3℘ 4 5 〈 Z4 〉 and℘221℘3 〈 X4 〉 . Since the integers X and Z are relatively prime, the only possible prime ideal divisor of both F2 and F4 is ℘3. From the remark at the start of this proof, we know that 3 - X so that ℘3 - F2. Hence the ideal gcd of F2 and F4 modulo squares is equal to 1. Thus we can deduce a pair of element equations X2 − (7t2 + 20t− 105)Z2 = gU2, X4 + (7t2 + 20t)X2Z2 + (1120t+ 400t2 − 3200)Z4 = gV 2, with Y = gUV and g = (−1)i0i11 i22 , i0, i1, i2 = 0, 1. Using local solvability in Magma, we can rule out 6 of the above pairs of equations by testing the quadratic Y = g(x2−(7t2+20t−105)). We present the results in a table. 65 6.2. Preliminary Results Table 6.1: Local Solvability of the Quadratic (i0, i1, i2) locally insolvable at (0, 0, 1) ℘22 (0, 1, 1) ℘22 (1, 0, 0) ℘21 (1, 0, 1) ℘21 (1, 1, 0) ℘21 (1, 1, 1) ℘21 For the remaining cases (i0, i1, i2) = (0, 0, 0) and (0, 1, 0) working with the quartics V 2 = X4 + (7t2 + 20t)X2Z2 + (1120t+ 400t2 − 3200)Z4 and (1V ) 2 = 1(X 4 + (7t2 + 20t)X2Z2 + (1120t+ 400t2 − 3200)Z4), A rational point X/Z satisfying either of these quartics contributes a point (x, y) with x ∈ Q on one of the elliptic curves y2 = x(x2 + (7t2 + 20t)x+ (1120t+ 400t2 − 3200)) or y2 = 1x(x 2 + (7t2 + 20t)x+ (1120t+ 400t2 − 3200)) We find using the PseudoMordellWeilGroup routine in Magma that the ranks of these elliptic curves are both equal to 1. Using the elliptic Chabauty procedure in Magma, with prime p = 17 in both cases, successfully deter- mines the points (x, y) on these curves with x ∈ Q, which give all of the points on the genus 2 curve stated in this lemma. 66 6.3. Proof of Theorem and Corollaries 6.3 Proof of Theorem and Corollaries We give the proof of Theorem 6.1, then the proofs of our corollaries. Proof. If f(x) = x6 + Ax + B ∈ K[x] is irreducible over K and has Galois group C6, S3 or C3×D3 then f(x) defines a normal relative cubic extension of a quadratic extension of K so that A and B are given by 6.1 by Lemma 6.1 and the quadratic extension field of K is K ( √ u). This statement is obvious for the Galois groups C6 and S3 while for C3×D3 the splitting field of f(x) over K ( √ u) has degree 9, so is an abelian extension of K ( √ u). We may assume that the nonzero scaling factor v = 1. From Lemma 6.2 we see that −(112√u+ 210u− 48u√u+ 3u2 + 27) = (a+ b√u)2 for some rational numbers a and b. Lemma 6.3 shows that (3w2 + 144)u2 + (−672 + 4w4 + 210w2)u+ 784 + 27w2 = 0. for some element w ∈ K, which establishes (6.2). Next we prove Corollary 6.1. Proof. Suppose that L/Q is a normal extension defined by an irreducible trinomial x6 + Ax + B ∈ Q[x]. By Theorem 6.1 equations (6.1) and (6.2), we have A = 4u(3u+ 1)v5 B = −u(1− 18u+ u2)v6 and (3w2 + 144)u2 + (−672 + 4w4 + 210w2)u+ 784 + 27w2 = 0 67 6.3. Proof of Theorem and Corollaries for rational numbers u, v and w. From (6.2) we calculate the discriminant with respect to u and deduce that 16w2(w6 + 105w4 + 2400w2 − 19200) is equal to a square in Q. One possibility is w = 0, in which case (6.3) reduces to 16(3u− 7)2 = 0 so that u = 7/3. Using (6.2) and setting the scaling factor v = 1, we obtain the trinomial x6 + 224/3x+ 2240/27, which has Galois group D3 × D3 (or 6T9 in the notation of Butler and McKay [2]) and does not define a normal extension of Q. If w 6= 0 then we must have w6 + 105w4 + 2400w2− 19200 equal to a square in Q and we can invoke Lemma 6.4, which shows that w = ±4. Appealing to (6.3), we obtain u = −1/3 or u = −19. The first value of u yields a reducible polynomial, while the second value of u gives x6 + 133x+ 209 after scaling, which defines a normal extension with Galois group C6 as seen in Chapter 5, completing the proof. Since all trinomials with Galois group C6, S3, or C3×D3 were determined in the proof of Corollary 6.1, the proof of Corollary 6.2 is immediate. For Corollary 6.3, the finitude of sextic trinomials under study is guaranteed by their correspondence with the K− rational points on the genus 2 curve and the theorem of Faltings. 68 Chapter 7 Results and Future Work 7.1 Results We give our findings from the theorems proved in Chapter 5 and Chapter 6 in the form of a table. Here “T-notation” refers to the notation of Butler and Mckay [2]. A total of five solvable Galois groups have been completely categorized for trinomials x6 +Ax+B with A,B ∈ Q. In the case of trinomials x6 +ax+ b with a, b ∈ Q defining normal sextic extensions of Q, the trinomial x6 + 224 3 x+ 2240 27 was discovered to have Galois group G36 ∼= D3×D3 in Chapter 6. This did not define a normal sextic extension of Q and we do not claim it to be the only trinomial of this form to have a Galois group of G36. Also recall that in the case of our trinomials f(x) = x6 +Ax+B, that f(x) has Galois group A4 × C2 if and only if A = 4u(u2 + 3)(3u2 + 1)(3u2 + 25)2v5, B = (u2 − 5)(3u2 + 1)(u4 + 10u2 + 5)(3u2 + 25)2v6. for some u, v ∈ Q with u 6= 0,±5 and v 6= 0. This generates infinitely many trinomials with Galois group A4×C2 as given in the above table. However, it must be noted that the occurrence of such a Galois group amongst all 69 7.1. Results Table 7.1: List of Findings T-notation Group notation For trinomials x6 +Ax+B, A,B ∈ Q 6T1 C6 x 6 + 133x+ 209 (up to scaling) 6T2 S3 Does not occur 6T4 A4 Does not occur 6T5 G18 ∼= C3 ×D3 Does not occur 6T6 A4 × C2 Infinitely many 6T10 G36 ∼= D3 ×D3 At least x6 + 224/3x+ 2240/27 (up to scaling) trinomials is very rare. In the case of A and B both integers we find x6 + 49x− 49 for (u, v) = (1, 1/4), x6 + 10647x− 13013 for (u, v) = (3, 1/4), and x6 + 996632x− 1085617 for (u, v) = (2, 1) all have Galois group A4 × C2, showing that an extensive search is necessary for even a few example polynomials. As mentioned in Chapter 3, a trinomial x6 + ax+ b with a, b ∈ Q chosen at random will almost certainly have a Galois group of S6, which can be calculated in MAPLETM. Any case that yields a result other than S6 (even if there are infinitely many) is an exceptional result, particularly when it implies that the polynomial itself is solvable. 70 7.2. Future Work 7.2 Future Work The methods and procedures used in obtaining the main results in this thesis lend themselves to many future extensions of this work. Firstly and most obviously, a complete determination of all sixteen possible Ga- lois groups for trinomials x6 + Ax + B with A,B ∈ Q would provide a great deal of future work. In order to accomplish this, further criteria would be needed to establish parametric families of polynomials that would give a desired selection of Galois groups. From this, further investigation into these parametrizations through the use of elliptic Chabauty could yield a complete determination of each individual Galois group. Secondly, an investigation intro trinomials of other forms, namely x6 + Ax2 + B, would require similar calculations as those used in this thesis. These results would translate directly to trinomials of the forms x6+Ax4+B and x6 +Ax5 +B through a simple transformation. Trinomials of the form x6+Ax3+B could also be looked at, though they could easily be considered as quadratics in x3. It would also be theoretically possible to look at sextic polynomials with an additional number of terms. Finally, another possible extension of this work would be to analyze sextic trinomials over fields other than Q. Much of theory used to create the restrictions on the coefficients of these trinomials can be applied to a field extension K of Q. 71 Bibliography [1] B. K. Spearman, Trinomials x6 + Ax + B defining sextic fields with a cyclic cubic subfield, J. Appl. Algebra Discrete Struct. 4 (2006), 149–155. [2] G. Butler and J. McKay, The transitive groups of degree up to eleven, Comm. Algebra 11 (1983), 863–911. [3] W. Bosma, J. Cannon, and C. Playoust, The Magma algebra system. I. The user language, J. Symbolic Comput. 24 (1997), 235–265 (Com- putational algebra and number theory (London, 1993)). [4] S. C. Brown, B. K. Spearman, and Q. Yang, On the Galois groups of sextic trinomials, JP J. Algebra Number Theory Appl. 18 (2010), 67–77. [5] A. Marshall, Precalculus: functions and graphs, Addison-Wesley Pub., 1990. [6] I. Stewart, Galois theory, Second edition, Chapman and Hall Ltd., London, 1989. [7] J. Stewart, Calculus: Early Transcendentals Single Variable, Fifth edition, Brooks/Cole, 2003. [8] J. B. Fraleigh, A first course in abstract algebra, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1967. [9] H. Cohen, A course in computational algebraic number theory, Vol. 138 of Graduate Texts in Mathematics, Springer-Verlag, Berlin, 1993. 72 Bibliography [10] F. Seidelmann, Die gesamtheit der kubischen und biquadratischen gle- ichungen mit affekt bei beliebigem rationalittsbereich, Math. Annalen 78 (1918), 230–233. [11] D. S. Dummit, Solving solvable quintics, Math. Comp. 57 (1991), 387–401. [12] B. K. Spearman and K. S. Williams, On solvable quintics X5 + aX + b and X5 + aX2 + b, Rocky Mountain J. Math. 26 (1996), 753–772. [13] ———, Characterization of solvable quintics x5 + ax+ b, Amer. Math. Monthly 101 (1994), 986–992. [14] J. Silverman, The arithmetic of elliptic curves, Graduate texts in mathematics, Springer, 2009. [15] C. Chabauty, Sur les points rationnels des courbes algébriques de genre supérieur à l’unité, C. R. Acad. Sci. Paris 212 (1941), 882–885. [16] G. Faltings, Endlichkeitssätze für abelsche Varietäten über Zahlkörpern, Invent. Math. 73 (1983), 349–366. [17] R. F. Coleman, Effective Chabauty, Duke Math. J. 52 (1985), 765–770. [18] J. W. S. Cassels and E. V. Flynn, Prolegomena to a middlebrow arith- metic of curves of genus 2, Vol. 230 of London Mathematical Society Lecture Note Series, Cambridge University Press, Cambridge, 1996. [19] L. C. Washington, Elliptic curves, Discrete Mathematics and its Ap- plications (Boca Raton), second edition, Chapman & Hall/CRC, Boca Raton, FL, 2008 (Number theory and cryptography). 73 Bibliography [20] A. Bremner and B. K. Spearman, Cyclic sextic trinomials x6+Ax+B, Int. J. Number Theory 6 (2010), 161–167. [21] M. J. Lavallee, B. K. Spearman, and K. S. Williams, Reducibility and the galois group of a parametric family of quintic polynomials, Missouri Journal of Mathematical Sciences 19 (2007), 2–10. [22] C. U. Jensen, A. Ledet, and N. Yui, Generic polynomials, Vol. 45 of Mathematical Sciences Research Institute Publications, Cambridge University Press, Cambridge, 2002 (Constructive aspects of the inverse Galois problem). [23] N. Bruin, Chabauty methods using elliptic curves, J. Reine Angew. Math. 562 (2003), 27–49. 74 Appendix A Common Group Names and Structures A.1 The Symmetry Group Sn The symmetry group Sn is also known as the symmetric group on n letters. This group consists of all the n! permutations of n elements. Any two of these permutations can then be combined together in order to form a single composite permutation. For instance, the group S5 consists of the 5! = 120 permutations of the set {1, 2, 3, 4, 5}. The identity permutation would be I : (1, 2, 3, 4, 5)→ (1, 2, 3, 4, 5). Other permutations would include σ : (1, 2, 3, 4, 5)→ (2, 3, 4, 5, 1), τ : (1, 2, 3, 4, 5)→ (2, 1, 3, 4, 5). These permutations could then be combined: στ : (1, 2, 3, 4, 5)→ (1, 3, 4, 5, 2), σ3 : (1, 2, 3, 4, 5)→ (4, 5, 1, 2, 3). It should be noted that the order in which the permutations should be applied is important. In the case of στ , τ should be applied before σ. 75 A.2. The Alternating Group An A.2 The Alternating Group An To understand the concept of the alternating group on n letters, consider that any permutation as described in the previous section can be expressed as a product of transpositions. A transposition is a permutation that leaves all but two elements fixed; that is, a transposition only interchanges two elements and leaves all other elements unchanged. If a permutation can be expressed as a composition of an even number of transpositions, then it is called an even permutation. A permutation that is not even is called odd; a permutation must either be called even or odd. With this concept, we can construct a subgroup of Sn of only the even permutations. Like in the case of integers, an even permutation combined with another even permutation gives an even permutation. This subgroup of Sn is exactly the alternating group An. Another interesting property of even and odd permutations is that for a given set of n unique elements, there is an equal number of even and odd permutations. Thus, the number of permutations contained in An is exactly n! 2 . A.3 The Dihedral Group Dn The dihedral group Dn is also frequently known as the symmetry group of a regular n−sided polygon. Using this idea, we can understand the group structure of the dihedral group. We begin by visualizing two identical copies of a regular n−sided polygon with each vertex uniquely labeled 1, 2, etc. The second copy can then be manipulated through rotations about its centre and through mirror images over a bisecting line so that when the two copies 76 A.4. The Cyclic Group Cn are placed on top of each other, every vertex is covered by another vertex. This process creates a mapping all of the labeled vertices on the first copy of the polygon to the vertices on the second copy. In fact, these mappings are a subset of the total n! permutations included in Sn and once again form a group under composition. In the case of an equilateral triangle, a rotation could be viewed as ρ : (1, 2, 3)→ (2, 3, 1) and a mirror image over the line bisecting the angle at vertex 3 could be viewed as µ3 : (1, 2, 3)→ (2, 1, 3). Both of these mappings belong to the group D3. A.4 The Cyclic Group Cn A cyclic group has the unique property of being generated by a single element. Following the same idea of permutations, a cyclic group can be generated by taking any permutation α on m elements other than the iden- tity mapping and applying it until the composition results in the identity mapping. If this α requires n applications to produce the identity map- ping, then there are a total of n unique permutations generated by α. The notation for a group generated by alpha is typically 〈α〉. In the case of the dihedral groupD3, we can visualize the group generated 77 A.5. Other Groups Gn, Fn and Direct Products by a rotation of an equilateral triangle ρ : (1, 2, 3)→ (2, 3, 1) ρ2 : (1, 2, 3)→ (3, 1, 2) 1 = ρ3 : (1, 2, 3)→ (1, 2, 3). This produces a cyclic group C3 consisting of {1, ρ, ρ2} under composition of maps. A.4.1 The Group 〈Zn,+〉 For positive n, the quotient ring Z/nZ ∼= Zn consisting of the n cosets of Z under addition modulo n is in fact a cyclic group. For all n we can consider the group 〈1̄〉 where 1̄ denotes the coset containing all integers with a remainder of 1 when divided by n. It is obvious to see that adding 1̄ to itself n times results in the coset 0̄, the additive identity. There is then a one–to–one correspondence between the cyclic group Cn and Zn and we say that Cn ∼= Zn. A.5 Other Groups Gn, Fn and Direct Products Other groups in this thesis will typically be given the notation Gn or Fn. These groups typically have a more complex structure than the specific examples given above. In these cases the n denotes the number of distinct elements in the group. For example, G72 has 72 elements and we write |G72| = 72. Some of these groups can also be written using direct product notation G ×H. We can define elements in this direct of groups as an ordered pair 78 A.5. Other Groups Gn, Fn and Direct Products consisting of one element from the group G and one element from the group H. The operation between two of these pairs must then be defined as (g, h)(g′, h′) = (g · g′, h ∗ h′) where g, g′ ∈ G, h, h′ ∈ H and · and ∗ are the binary operations defined in G and H, respectively. It is not necessary for the operations in G and H to be distinct. 79 Appendix B Magma Code Related to Chapter 5 print "Initialization"; print "***********************************"; _:=PolynomialRing(Rationals()); PS:=ProjectiveSpace(Rationals(),1); print ""; print "Set up the number field"; print "***********************************"; K:=NumberField(x^3+3*x+1); OK:=MaximalOrder(K); Basis(OK, NumberField(OK)); // [ 1, t, t^2 ] ClassNumber(OK); // 1 U,mU:=UnitGroup(OK); U; // Abelian Group isomorphic to Z/2 + Z // Defined on 2 generators // Relations: // 2*U.1 = 0 [mU(U.2)]; // [ [0, 1, 0] ] 80 Appendix B. Magma Code Related to Chapter 5 [mU(U.1), mU(U.2)]; // [ [-1, 0, 0], [0, 1, 0] ] e1:=t; [Norm(e1)]; // [ -1 ] p2:=Factorization(2*OK); p2; // [ ] p21:=p2[1][1]; p3:=Factorization(3*OK); p3; // [ ] p31:=p3[1][1]; p5:=Factorization(5*OK); p5; // [ , // ] p51:=p5[1][1]; p52:=p5[2][1]; 81 Appendix B. Magma Code Related to Chapter 5 print ""; print "Set up the sextic"; print "***********************************"; _:=PolynomialRing(K); Factorization(X^6-3*X^4+51*X^2+15); // [ , // , // ] print ""; print "Local Solvability"; print "***********************************"; print ""; print "Testing the Quadratic"; print "*********************"; for i0 in [0..1] do for i1 in [0..1] do bool:=true; C:=HyperellipticCurve(g*(X^4+2*(2*t-1)*X^2+(16*t^2-4*t+49))) where g is (-1)^i0*t^i1; bp:=BadPrimes(C); print [i0,i1]; for i in [1..4] do bool:=bool and IsLocallySolvable(C,bp[i]); if not IsLocallySolvable(C,bp[i]) then print "Fails at bad prime:"; 82 Appendix B. Magma Code Related to Chapter 5 print bp[i]; end if; end for; print bool; print ""; end for; end for; // [ 0, 0 ] // true // // [ 0, 1 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [3, 0, 0] // [1, 1, 0] // false // // [ 1, 0 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [3, 0, 0] // [1, 1, 0] // false // // [ 1, 1 ] // true 83 Appendix B. Magma Code Related to Chapter 5 print ""; print "Testing the Quartic"; print "*********************"; for i0 in [0..1] do for i1 in [0..1] do bool:=true; C:=HyperellipticCurve(g*(X^4+2*(2*t-1)*X^2+(16*t^2-4*t+49))) where g is (-1)^i0*t^i1; bp:=BadPrimes(C); print [i0,i1]; for i in [1..4] do bool:=bool and IsLocallySolvable(C,bp[i]); if not IsLocallySolvable(C,bp[i]) then print "Fails at bad prime:"; print bp[i]; end if; end for; print bool; print ""; end for; end for; // [ 0, 0 ] // true // // [ 0, 1 ] // Fails at bad prime: // Prime Ideal of OK 84 Appendix B. Magma Code Related to Chapter 5 // Two element generators: // [3, 0, 0] // [1, 1, 0] // false // // [ 1, 0 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [3, 0, 0] // [1, 1, 0] // false // // [ 1, 1 ] // true print ""; print "Case g=1"; print "*********************"; C1:=HyperellipticCurve(X^4+2*(2*t-1)*X^2+(16*t^2-4*t+49)); E1, C1toE1 := EllipticCurve(C1); boo,G1,m1:=PseudoMordellWeilGroup(E1); boo; // true G1; // Abelian Group isomorphic to Z/2 + Z/2 + Z // Defined on 3 generators 85 Appendix B. Magma Code Related to Chapter 5 // Relations: // 2*G1.1 = 0 // 2*G1.2 = 0 print ""; print "Chabauty"; print "*************"; C1toPS:=mapPS|[C1.1,C1.3]>; E1toPS:=Expand(Inverse(C1toE1)*C1toPS); N1,V1,R1,W1:=Chabauty(m1,E1toPS,17); N1;V1; // 4 // { 0, G1.1 + G1.2, G1.2, G1.1 } R1; // 4 [EvaluateByPowerSeries(E1toPS,m1(v)): v in V1]; // [ (1 : 0), (0 : 1), (1 : 0), (0 : 1) ] print ""; print "Case g=-t"; print "*********************"; C2:=HyperellipticCurve(-t*(X^4+2*(2*t-1)*X^2+(16*t^2-4*t+49))); print "Find rational points:"; RationalPoints(C2: Bound := 10); // {@ (-1 : -4 : 1), (1 : -4 : 1), (2*t^2 + 7 : 8*t^2 + 28 : 1), // (-2*t^2 - 7 : 8*t^2 + 28 : 1) @} pt:=C2![2*t^2+7,8*t^2+28]; E2, C2toE2 := EllipticCurve(C2,pt); 86 Appendix B. Magma Code Related to Chapter 5 boo,G2,m2:=PseudoMordellWeilGroup(E2); boo; // true G2; // Abelian Group isomorphic to Z/2 + Z/2 + Z // Defined on 3 generators // Relations: // 2*G2.1 = 0 // 2*G2.2 = 0 print ""; print "Chabauty"; print "*************"; C2toPS:=mapPS|[C2.1,C2.3]>; E2toPS:=Expand(Inverse(C2toE2)*C2toPS); N2,V2,R2,W2:=Chabauty(m2,E2toPS,17); N2;V2; // 4 // { G2.2 - G2.3, G2.1 + G2.2 - G2.3, G2.1 + G2.2, G2.2 } R2; // 32 [EvaluateByPowerSeries(E2toPS,m2(v)): v in V2]; // [ (1 : 1), (-1 : 1), (-1 : 1), (1 : 1) ] 87 Appendix C Magma Code Related to Chapter 6 print "Initialization"; print "***********************************"; clear; _:=PolynomialRing(Rationals()); PS:=ProjectiveSpace(Rationals(),1); print ""; print "Set up the number field"; print "***********************************"; print "** Number field defined by K=Q(t) where t^3-15*t+20=0"; K:=NumberField(x^3 - 15*x + 20); OK:=MaximalOrder(K); print "** Has basis:"; Basis(OK, NumberField(OK)); // [ 1, t, t^2 ] print "** Class number is:"; ClassNumber(OK); // 1 print "** Find unit group:"; U,mU:=UnitGroup(OK); U; // Abelian Group isomorphic to Z/2 + Z + Z 88 Appendix C. Magma Code Related to Chapter 6 // Defined on 3 generators // Relations: // 2*U.1 = 0 [-mU(U.-3), mU(U.-2)]; // [ [-59, 22, 8], [161, -21, -13] ] e1:=-59 + 22*t + 8*t^2; e2:=161 - 21*t - 13*t^2; print "** Two fundamental units are"; print [e1,e2]; // [ 8*t^2 + 22*t - 59, // -13*t^2 - 21*t + 161 ] print "** and have norms:"; [Norm(e1),Norm(e2)]; // [ 1, 1 ] print "** Ideal factorization of <2> in OK:"; p2:=Factorization(2*OK); p2; // [ , // ] p21:=p2[1][1]; p22:=p2[2][1]; print "** Ideal factorization of <3> in OK:"; p3:=Factorization(3*OK); p3; 89 Appendix C. Magma Code Related to Chapter 6 // [ ] print "** Ideal factorization of <5> in OK:"; p5:=Factorization(5*OK); p5; // [ ] print ""; print "Set up the sextic"; print "***********************************"; _:=PolynomialRing(K); print "** The sextic X^6 + 105*X^4 + 2400*X^2 - 19200 factors over K as:"; Factorization(X^6 + 105*X^4 + 2400*X^2 - 19200); // [ , // ] print ""; print "Local Solvability"; print "***********************************"; print ""; print "Testing the Quadratic"; print "*********************"; print "** Test hyperelliptic curve g*(X^2 - 7*t^2 - 20*t + 105)"; 90 Appendix C. Magma Code Related to Chapter 6 print "** where g is (-1)^i0*e1^i1*e2^i2"; print "** [i0,i1,i2]"; i0:=0; i1:=0; i2:=0; for i0 in [0..1] do for i1 in [0..1] do for i2 in [0..1] do bool:=true; C:=HyperellipticCurve(g*(X^2 - 7*t^2 - 20*t + 105)) where g is (-1)^i0*e1^i1*e2^i2; bp:=BadPrimes(C); print [i0,i1,i2]; for i in [1..2] do bool:=bool and IsLocallySolvable(C,bp[i]); if not IsLocallySolvable(C,bp[i]) then print "Fails at bad prime:"; print bp[i]; end if; end for; print bool; print ""; end for; end for; end for; // [ 0, 0, 0 ] // true // // [ 0, 0, 1 ] // Fails at bad prime: // Prime Ideal of OK 91 Appendix C. Magma Code Related to Chapter 6 // Two element generators: // [2, 0, 0] // [1, 1, 0] // false // // [ 0, 1, 0 ] // true // // [ 0, 1, 1 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [2, 0, 0] // [1, 1, 0] // false // // [ 1, 0, 0 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [2, 0, 0] // [2, 1, 0] // false // // [ 1, 0, 1 ] // Fails at bad prime: // Prime Ideal of OK 92 Appendix C. Magma Code Related to Chapter 6 // Two element generators: // [2, 0, 0] // [2, 1, 0] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [2, 0, 0] // [1, 1, 0] // false // // [ 1, 1, 0 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [2, 0, 0] // [2, 1, 0] // false // // [ 1, 1, 1 ] // Fails at bad prime: // Prime Ideal of OK // Two element generators: // [2, 0, 0] // [2, 1, 0] // Fails at bad prime: // Prime Ideal of OK // Two element generators: 93 Appendix C. Magma Code Related to Chapter 6 // [2, 0, 0] // [1, 1, 0] // false print "** This eliminates any cases where i0=1"; print ""; print "Testing the Quartic"; print "*********************"; print "** Test hyperelliptic curve g*(X^4 + (7*t^2 + 20*t)*X^2 + 400*t^2 + 1120*t - 3200)"; print "** where g is (-1)^i0*e1^i1*e2^i2"; print "** [i0,i1,i2]"; i0:=0;i1:=0;i2:=0; for i0 in [0..0] do for i1 in [0..1] do for i2 in [0..1] do bool:=true; C:=HyperellipticCurve(g*(X^4 + (7*t^2 + 20*t)*X^2 + 400*t^2 + 1120*t - 3200)) where g is (-1)^i0*e1^i1*e2^i2; bp:=BadPrimes(C); print [i0,i1,i2]; for i in [1..#bp] do bool:=bool and IsLocallySolvable(C,bp[i]); if not IsLocallySolvable(C,bp[i]) then print "Fails at bad prime:"; print bp[i]; 94 Appendix C. Magma Code Related to Chapter 6 end if; end for; print bool; print ""; end for; end for; end for; // [ 0, 0, 0 ] // true // // [ 0, 0, 1 ] // true // // [ 0, 1, 0 ] // true // // [ 0, 1, 1 ] // true print "** Therefore g=1 or g=e1=8*t^2 + 22*t - 59"; print ""; print "Case g=1"; print "*********************"; C1:=HyperellipticCurve(X^4 + (7*t^2 + 20*t)*X^2 + 400*t^2 + 1120*t - 3200); print "** Find rational points:"; RationalPoints(C1: Bound := 10); // {@ (4 : 4*t^2 + 20*t - 16 : 1), (-4 : 4*t^2 + 20*t - 16 : 1), (1 : -1 : 0) @} E1, C1toE1 := EllipticCurve(C1); 95 Appendix C. Magma Code Related to Chapter 6 boo,G1,m1:=PseudoMordellWeilGroup(E1); boo; // false print "** Try multiplying quartic by X^2 and substituting X=sqrt(X) to get a cubic"; C1:=HyperellipticCurve(X^3+(7*t^2+20*t)*X^2+(400*t^2+1120*t-3200)*X); print "** Find rational points:"; RationalPoints(C1: Bound := 10); // {@ (0 : 0 : 1), (1 : 0 : 0) @} E1, C1toE1 := EllipticCurve(C1); boo,G1,m1:=PseudoMordellWeilGroup(E1); boo; // true G1; // Abelian Group isomorphic to Z/2 + Z // Defined on 2 generators // Relations: // 2*G1.1 = 0 C1toPS:=mapPS|[C1.1,C1.3]>; E1toPS:=Expand(Inverse(C1toE1)*C1toPS); N1,V1,R1,W1:=Chabauty(m1,E1toPS,17); N1;V1; // 4 // { 0, 2*G1.2, G1.1, -2*G1.2 } R1; // 1 [EvaluateByPowerSeries(E1toPS,m1(v)): v in V1]; 96 Appendix C. Magma Code Related to Chapter 6 // [ (1 : 0), (16 : 1), (0 : 1), (16 : 1) ] print ""; print "Case g=8*t^2 + 22*t - 59"; print "*********************"; C2:=HyperellipticCurve(e1*(X^4 + (7*t^2 + 20*t)*X^2 + 400*t^2 + 1120*t - 3200)); print "** Find rational points:"; RationalPoints(C2: Bound := 10); // {@ (0 : -48*t^2 - 140*t + 320 : 1) @} pt:=C2![0,-48*t^2-140*t+320]; E2, C2toE2 := EllipticCurve(C2,pt); boo,G2,m2:=PseudoMordellWeilGroup(E2); boo; // true G2; // Abelian Group isomorphic to Z/2 + Z // Defined on 2 generators // Relations: // 2*G2.1 = 0 C2toPS:=mapPS|[C2.1,C2.3]>; E2toPS:=Expand(Inverse(C2toE2)*C2toPS); N2,V2,R2,W2:=Chabauty(m2,E2toPS,17); N2;V2; // 2 // { 0, G2.1 } R2; // 6 97 Appendix C. Magma Code Related to Chapter 6 [EvaluateByPowerSeries(E2toPS,m2(v)): v in V2]; // [ (0 : 1), (0 : 1) ] 98