On Solvable Septics

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On Solvable Septics

Lau Jing Feng

A THESIS SUBMITTED FOR THE DEGREE OF MASTER OF SCIENCE.

Supervisor : A/P. Lang Mong Lung

Department of Mathematics National University of Singapore 2004 Acknowledgements

I am grateful to my supervisor A/P Lang Mong Lung for his patient guidance and for giving me this the opportunity to self propose my own thesis topic. He has never fail to inspire me and keep my interest in Mathematics ‘alive’ despite all the disappointment and unhappiness I face in 2002 and 2003. I also wish to thank Sze Ling, Chee Peng and my sincere friends in Gakkai for encouraging me to move on during my most diﬃcult time in 2003.

i Contents

Acknowledgements i

Summary iii

1 Introduction 1 1.1 Historical Development ...... 1 1.2 Recent Advancements ...... 4

2 Solving Solvable Polynomials of prime degree 6 2.1 Solvable Galois Groups of irreducible polynomials of prime degree p ...... 6 2.2 Fixed Fields of the Frobenius group Fp(p−1) ...... 8 2.3 Setting Up the Calculation ...... 14

3 Solvable Septics 20 3.1 Lagrange Resolvents for Septic Polynomials ...... 20 2π 3.2 Expressing cos 29 in radicals ...... 24 Bibliography 30

Appendix 32

A Solution of Polynomials by Real Radicals 33

B lagres.nb 37

ii Summary

This thesis seeks to examine the computational aspects in solving solvable septics. An account of the historical background is provided in Chapter 1. In Chapter 2, the approach in [D] is generalized to lay down the qualitative theory behind solving polynomials of an arbitrary prime degree p. The main problem is broken down into four points of consideration and the results and drawbacks of applying Dummit’s approach to each of the four points are outlined. Chapter 3 presents the approach of Lagrange resolvents to solving solvable septics. Technicalities faced in adopting this methodology are highlighted and explicit calculations are performed to solve for two roots of a particular 2π septic polynomial associated to cos 29 .

iii Chapter 1

Introduction

The purpose of this chapter is to expound on what has been done on the subject of solving polynomials in one variable by expressing the roots in radicals. Due to the technicalities involved, we shall omit the details and present the main ideas and results.

1.1 Historical Development

1. The Babylonians, Greeks and Arabs were known to be the ﬁrst to solve quadratic equations. Motivated by plane geometry, the method of “completing the square” yields the solution √ −a ± a2 − 4b x = 2 to the quadratic equation

x2 + ax + b = 0 (cf. [T]) .

2. The algebraic solution of x3+mx = n was first obtained around 1515 by Scipione del Ferro, Professor of Mathematics in Bologna. In 1535, An- tonio Mario Fior, a student of Scipione del Ferro challenged Tartaglia, who had previously attempted to solve certain types of cubic equations in a problem-solving contest to solve about thirty problems on equations of the form x3 + mx = n. Tartaglia succeeded in finding the solution to win the challenge and when news of this reached Jerôme Cardan, the latter asked Tartaglia to reveal his solution. Cardan pub- lished all these solutions in his book Ars Magna which resulted into a bitter quarrel between Tartaglia and Cardan, the former claiming that

1 Cardan had solemnly sworn never to publish Tartaglia’s solution, while the latter countered that there had never been any question of secrecy (cf. [E]). Using Cardan’s method, the general cubic equation

x3 + ax2 + bx + c = 0

a is reduced by the change of variable y = x + 3 to the form y3 + py + q = 0

where 3b − a2 9c − 3ab + 2a3 p = and q = . 3 27 Setting y := u + v, 3uv := −p and ω as a ﬁxed primitive cube root of unity, the roots of the last cubic polynomial are

u + v , ωy + ω2z , ω2y + ωz

where s s r 3 2 r 3 2 3 q p q 3 q p q u = − + + , v = − − + (cf. [DF]) . 2 27 4 2 27 4

Cardan’s cubic formula led to the discovery of “Casus Irreducibilis”, the impossibility of expressing the real roots of cubic equations in real radicals. Even though attempts to rewrite speciﬁc formulas to eliminate non-real complex numbers failed, they prompted greater understanding and usage of complex numbers (cf. [R2] and [I]).

3. The solution of quartic equations was found shortly after that of cubic equations. Cardan provided a method of solving such equations in the “Ars Magna” and he attributed it to his student Ludovico Ferrari (cf. [E] and [T]). For the general quartic equation

x4 + ax3 + bx2 + cx + d = 0 ,

a under the change of variable y = x + 4 transforms the equation to y4 + py2 + qy + r = 0

with 8b − 3a2 8c − 4ab + a3 256d − 64ac + 16a2b − 3a4 p = , q = , r = . 8 8 256

2 4 2 Let the roots of y + py + qy + r be y1, y2, y3 and y4. Deﬁne the resolvent cubic to be x3 − 2px2 + (p2 − 4r)x + q2 which has roots

θ1 = (y1 + y2)(y3 + y4) , θ2 = (y1 + y3)(y2 + y4) , θ3 = (y1 + y4)(y2 + y3) .

Then

1 p p p 1 p p p y1 = 2 ( −θ1 + −θ2 + −θ3) , y2 = 2 ( −θ1 − −θ2 − −θ3) ,

1 p p p 1 p p p y3 = 2 (− −θ1 + −θ2 − −θ3) , y4 = 2 (− −θ1 − −θ2 + −θ3) , p p p where −θ1 −θ2 −θ3 = −q (cf. [DF]). 4. By introducing the Lagrange resolvent in 1770, Lagrange proposed an- other alternative method of solving cubic equations. Moreover, he showed that polynomials of degree five or more cannot be solved by the methods used for cubics and quartics. Assuming without justifica- tion that the radicals can be rationally expressed in terms of the roots of the initial equation, Paolo Ruffini gave a proof on the impossibility to solve general equations of degree higher than 4 in his book General Theory of Equations (cf. [PS]). It was not till 1826 that Niels Henrik Abel provided the first complete proof on the unsolvability of the general equation of degree higher than 4. From the theoretical viewpoint, it was Evariste Galois who made the major breakthrough by drawing correspondence between equations and groups. With subsequent re- finements by Ludwig Sylow in 1871 and Emil Artin in 1938 on Galois’s results which culminated into what is known today as Galois Theory, these led to the development of the modern theory of algebraic equations (cf. [V]). Among these lies Galois epoch-making theorem which asserts that the solvability by radicals of a polynomial is equivalent to solvability of its Galois group.

5. Some progress were made in the direction of solving algebraic equations using elliptic functions. It ﬁrst started out in 1829 with Carl Gustav Jacobi studying modular equations for elliptic functions which are fundamental for Hermite’s solution of quintics in 1858. This is sub- sequently followed by the advancements of Camille Jordan in 1870 who showed that algebraic equations of any degree can be solved in terms of modular functions and Ferdinand von Lindemann expressing the roots of an arbitrary polynomial in terms of theta functions (cf. [V]).

3 1.2 Recent Advancements

1. Having showed that every quintic can be transformed to the form

x5 + ax + b = 0 (1.2.1)

by Erland Samuel Bring and George Birch Jerrad in 1786 and 1834 respectively (cf. [V]), Spearman and Williams [SW] showed that an irreducible quintic of the form (1.2.1) having rational coeﬃcients is solvable by radicals if and only if there exist rational numbers = ±1, c ≥ 0 and e 6= 0 such that 5e4(3 − 4c) −4e5(11 + 2c) a = , b = . c2 + 1 c2 + 1 When this is the case, the roots are given by

j 2j 3j 4j xj = e(ω u1 + ω u2 + ω u3 + ω u4)

where v2v 1/5 v2v 1/5 v2v 1/5 v1v 1/5 u = 1 3 , u = 3 4 , u = 2 1 , u = 4 2 , 1 D2 2 D2 3 D2 4 D2 √ q √ √ q √ v1 = D + D − D , v2 = − D − D + D, √ q √ √ q √ 2 v3 = − D + D + D , v4 = D − D − D,D = c + 1 , and ω is a ﬁxed primitive 5th root of unity.

2. David Dummit [D] and (independently) Sigeru Kobayashi and Hiroshi Nakagawa [KN] gave methods for finding the roots of a general solvable quintic in radicals. Dummit employed techniques from function field theory to give an explicit criterion for the solvability of a quintic in terms of the existence of a rational root θ of a certain sextic resolvent. Denoting the function field Q(x1, ··· , x5) as K and the Frobenius group of degree 5 by F20, the rational coefficients of certain invariants are viewed as elements of the fixed field KF20 which is of degree 6 over KS5 so that they can be written as a linear combination of 1, θ, ··· , θ5 and calculated by solving linear systems of 6 equations in 6 unknowns. Not only are these invariants used to compute other intermediate invariants, they are also used to determine uniqueness of some other invariants up to some permutation and the choice of the 5th root to take in order to obtain each of the Lagrange resolvents.

4 3. Even with the advent of modern computers, it took 9 years before the case for solvable degree 6 polynomials was settled by Thomas R. Hagedorn [H] in 2000. In [H], Hagedorn discusses the 16 transitive subgroups of S6 up to isomorphism and information about the Galois group of the original sextic f(x) is obtained by factoring resolvents of degree 2, 10 and 15 and computing the discriminant. Depending on whether the Galois group of f(x) is a subgroup of G48 or G72, the author designed algorithms which uses the rational roots of the resolvents of degree 10 and 15 to define new resolvent polynomials and other polynomials. The roots of these resolvents are used to define other polynomials and Galois resolvents and the previous step are repeated until enough polynomials are defined so that the Galois group and the roots of f(x) can be calculated.

5 Chapter 2

Solving Solvable Polynomials of prime degree

In this chapter, we shall identify the irreducible polynomials of prime degree p over Q which are solvable by radicals and outline both the qualitative theory and computational aspects in solving polynomials of prime degree via the method of Lagrange resolvents.

2.1 Solvable Galois Groups of irreducible polynomials of prime degree p

In this section, we shall present a classical result which was ﬁrst proved by Galois that provides a necessary and suﬃcient condition for an irreducible polynomial of prime degree p to be solvable by radicals.

We begin with some terminology and notations. Deﬁnition 2.1.1. Let G be a group. A G-set X is transitive if for every x, y ∈ X, there exists σ ∈ G with y = σx. In this case, the group G is said to act transitively on X. Deﬁnition 2.1.2. A group G is solvable if it has a normal series

{1G} = G0 ⊆ G1 ⊆ · · · ⊆ Gn = G whose factors Gi/Gi−1 are abelian for all 1 ≤ i ≤ n.

Deﬁnition 2.1.3. If X is a G-set, then the stabilizer of x, denoted by Gx, is the subgroup Gx := {g ∈ G : gx = x} ≤ G.

6 Next, we state the following result from [R1] without proof. Proposition 2.1.4. Let X be a transitive G-set, and let x, y ∈ X. If gx = y −1 for some g ∈ G, then Gy = g Gx g . Lemma 2.1.5. Let X be a transitive G-set and α ∈ X be arbitrary. If N is a normal subgroup of G contained in Gα, then N is a subgroup of \ Gβ . β∈X

Moreover, if G also acts faithfully on X, then Gα contains no non-trivial normal subgroups of G. Proof. If N is normal in G,

\ −1 \ −1 \ N = gNg ⊆ gNg ⊆ Gβ g∈G g∈G β∈X g(α)=β by Proposition 2.1.4. Hence the conclusion follows. Before we state and prove the main result of this section as promised, we will need to introduce some new notation. We shall denote the cyclic group of order n as Zn, the dihedral group of order 2n as D2n, the semidirect product of K by Q as K o Q and the Frobenius group of degree p as Fp(p−1) or Zp o Zp−1. Theorem 2.1.6. (Galois) An irreducible polynomial f(x) ∈ Q[x] of degree p is solvable by radicals if and only if its Galois group is isomorphic to a transitive subgroup of Zp o Zp−1, the Frobenius group of degree p. Proof. Let f(x) ∈ Q[x] be an irreducible polynomial of degree p with Galois group G. Suppose G is a subgroup of Zp o Zp−1. Since Zp o Zp−1 is solvable and every subgroup of a solvable group is itself solvable, G is solvable and so f(x) is solvable by radicals. Conversely, suppose f(x) is solvable by radicals. Thus G is solvable. Let G(n−1) be the last nontrivial subgroup in the normal series for G. Then G(n−1) as a solvable minimal normal subgroup of G have no non-trivial proper characteristic subgroups and so must be an elementary (n−1) abelian p-group. Since G is isomorphic to a subgroup of Sp, G is isomor- (n−1) (n−1) phic to Zp. Denoting the image of G in Sp as H, G being normal in

G implies that G is isomorphic to a transitive subgroup of NSp (H). From the above theorem, we deduce the following corollary. Corollary 2.1.7. Let f(x) ∈ Q[x] be an irreducible polynomial of degree 7. Then f(x) is solvable by radicals if and only if its Galois group is a transitive subgroup of F42. In particular, if G is the Galois group of f(x), then G is isomorphic to either Z7, D14, Z7 o Z3 or F42.

7 2.2 Fixed Fields of the Frobenius group Fp(p−1) In this section, we shall analyze closely the relationship between the intermediate fields of Q, the splitting field of a given solvable irreducible polynomial f(x) of prime degree p and the radical extension obtained by adjoining a primitive pth root of unity. Since all such polynomials have Galois groups which are isomorphic to transitive subgroups of Fp(p−1), we shall be working mainly with the function field Q(x1, ··· , xp) over the fixed field of Fp(p−1). For convenience, we shall denote the normal subgroup of Fp(p−1) which is isomorphic to Zp by N and its complement which is isomorphic to Zp−1 by C. We let σ and τ denote fixed generators of N and C respectively.

Deﬁnition 2.2.1. Let p be an odd prime, x1, ··· , xp be p indeterminates over Q, Y ∆ := (xi − xj) i

Fp(p−1) N Deﬁne K := Q(x1, ··· , xp), k := Q(s1, ··· , sp), F := K , E := K , F 0 := Ehτ 2i and L := K(ζ).

Proposition 2.2.2.

∼ ∗ Gal(L/F ) = Fp(p−1) × Zp .

∗ Proof. Since Q(ζ)/Q is Galois with Galois group isomorphic to Zp and Q(ζ)∩ F = Q, the same conclusion holds for the extension F (ζ)/F by natural irrationalities. This together with the fact that K/F is Galois with Fp(p−1) as Galois group implies our desired conclusion.

Remark 2.2.3. Keeping Proposition 2.2.2 and our description of σ as the generator of N in mind, σ is deﬁned to be the automorphism on L over k that permutes the set {x1, ··· , xp} cyclically. For our convenience, we shall denote σ by σ := (1, 2, ··· , p) .

8 (i). Recall that Fp(p−1) is isomorphic to a semidirect product of N by C. Consequently we have a homomorphism φ : C −→ Aut(N) deﬁned by φ(g):≡ φg where for all g ∈ C, a ∈ N,

−1 φg(a) := gag .

Note that kerφ is trivial because CSp (σ) = N. Hence φ as an injection from C to Aut(N) with |C| = |Aut(N)| = p − 1 must be an isomor- s phism. Therefore for any primitive root s of Zp , the map σ 7→ σ induces an automorphism of N of order p − 1 which corresponds to a generator of C. In view of this, the automorphism

τ := (2, s + 1, s2 + 1, ··· , sp−2 + 1)

acting trivially on constants is a generator of C.

2 2 Nohτ |K i (ii). We can deduce from σ, τ ∈ Ap that ∆ ∈ K .

In the following result, we shall state without proof that Fp(p−1) is maximal in Sp.

Proposition 2.2.4. Fp(p−1) is a maximal subgroup of Sp.

Next we will determine all subgroups of Fp(p−1) and among these subgroups identify those which are normal.

Proposition 2.2.5. Let G := hσi o hτi be a Frobenius group acting on X := {1, 2, ··· , p}. Denote by Gm the stabilizer of m ∈ X.

(p−1)/d (i). For each divisor d of p − 1, hσi o hτ i is the only subgroup of G of order pd.

(ii). Gm ∩ Gn = 1 for 1 ≤ m, n ≤ p, m 6= n.

∗ ∗ 0 (iii). G is the disjoint union of G1, ··· , Gp and G .

(p−1)/d (iv). The collection { hσi o hτ i | d|(p − 1) } are precisely all the non- trivial normal subgroups of G.

(v). Let H 6= 1 be a subgroup of Gm. Then NG(H) = Gm and for each i −i −1 p−1 0 ≤ i ≤ p − 1, σ Hσ ⊆ Gσim and H, σHσ , ··· , σ Hσ are precisely all the distinct conjugates of H.

(vi). Every non-trivial subgroup H of order d dividing p−1 must be contained in some unique Gn. In particular, the subgroups G1, G2,··· , Gp are precisely all the distinct copies of Zp−1 in G.

9 Proof. (i). Let S be a subgroup of order pd. By Sylow’s theorem, hσi ⊆ S. Hence S/hσi is the unique cyclic subgroup of G/hσi of order (p−1)/d. (p−1)/d As a consequence, S = hσi o hτ i by Correspondence theorem. (ii). (ii) follows from the fact that elements in G ﬁx at most 1 letter.

(iii). (ii) together with N ∩ Gn = 1 for all 1 ≤ n ≤ p implies (iii). (iv). Let A be a normal subgroup of G. Suppose that gcd(|A|, p) = 1. Then hσi × A admits an element of order pk where k > 1, a contradiction. Thus p||A| from which (iv) follows from (i).

(v). Suppose p||NG(H)|, then hσi ⊆ NG(H). It follows that hσi × H has an element of order pk, k > 1, a contradiction. This implies that NG(H) = Gm.

(vi). Since gcd(p, d) = 1 and hσi ¡ G, we have hσiH = hσi o H. Since ∼ H = (hσi o H) /hσi

where (hσi o H) /hσi is isomorphic to a subgroup of hτi, H is cyclic. This together with (iii) implies that H ⊆ Gm for some m.

By Galois correspondence, we have the following description for the subﬁelds Fp(p−1) of K := Q(x1, ··· , xp) containing F := K .

Fp(p−1) Theorem 2.2.6. Let K := Q(x1, ··· , xp), F := K and k := Q(s1, ··· , sp). (i). F is a minimal subﬁeld of K that contains k.

(ii). For each divisor d of p−1, there exists a unique subfield E0 := KZpo Z(p−1)/d of K containing F such that [E0 : F ] = d which is normal over F . Fur- thermore, every normal extension E0 of F contained in K is the fixed field of some normal subgroup containing G0 with [E0 : F ]|(p − 1).

(iii). For each d dividing p − 1, there exist p distinct subfields of K of degree pd over F and each of these subfields contains a unique fixed field KGm for some m. Moreover, for two subfields F1 and F2 which contains the Gm same unique fixed field K , F1 ⊆ F2 if and only if [F1 : F ]|[F2 : F ]. (iv). Any intermediate field of K and F arises from either (ii) or (iii). We shall devote the rest of this section to introduce the Lagrange resolvents and other associated invariants. Our final objective is to provide the global view on the subfields of L that contains F .

10 Deﬁnition 2.2.7. A Galois extension K/E is said to be abelian (respectively cyclic) if its Galois group G is abelian (respectively cyclic).

Definition 2.2.8. Let K/E be a cyclic extension over a field E of characteristic not dividing n which contains the nth roots of unity. Let σ be a generator for the cyclic group Gal(K/E). For α ∈ K and any nth root of unity ζ, define the Lagrange resolvent (α, ζ) ∈ K by

n−1 X (α, ζ) := ζjσj(α) . j=0

j p Deﬁnition 2.2.9. For each 0 ≤ j ≤ p−1, deﬁne rj := (x1, ζ ) and Rj := rj . p Expanding r1 yields

p−1 p p X j R1 := r1 = (x1, ζ) = ljζ j=0

p i where lj is the sum of the terms in (x1, z) involving powers z with i ≡ j (mod p).

Example 2.2.10. For the particular case when p = 7, we carry out an explicit calculation using Mathematica and saved this in a ﬁle named lagres.nb. 7 We expand r1 and collect terms with the same exponent for ζ. From this multinomial expansion, we express each lj as a polynomial in the variables x1, ··· , x7 and state this form for lj in the appendix. We quote the next result from [L] without proof.

Theorem 2.2.11. Let α1, ··· , αn be distinct non-zero elements of a ﬁeld K. If a1, ··· , an are elements of K such that for all integers v ≥ 0 we have

v v a1α1 + ··· + anαn = 0 then ai = 0 for all i. Corollary 2.2.12. Let ψ : ζ 7→ ζs be the automorphism of L that acts trivially on x1, ··· , xp. If a0, ··· , ap−1 are elements of L such that (i). a0 + ··· + ap−1 = 0 ,

(ii). ψ(ai) = ai for all 0 ≤ i ≤ p − 1 and

11 (iii). p−1 a0 + a1ζ + ··· + ap−1ζ = 0 , then ai = 0 for each 0 ≤ i ≤ p − 1. Proof. We argue that p−1 X jv ajζ = 0 j=0 for all integers v ≥ 0. When v is a multiple of p,

p−1 p−1 X jv X ajζ = aj = 0 . j=0 j=0

∗ n For v ∈ Zp, v = s for some 1 ≤ n ≤ p − 1. Therefore

p−1 p−1 X jv X n j ajζ = ajψ (ζ ) = 0 . j=0 j=0 Invoking Theorem 2.2.11 then yields our desired conclusion. The next remark will prove to be useful in the succeeding lemma.

Remark 2.2.13. Taking j = 0 in Deﬁnition 2.2.9, we have r0 = s1 and

p−1 X p lj = s1 . j=0 Consequently for any automorphism γ that acts trivially on

p−1 X lj , j=0

γ(lj) − lj (j = 0, 1, ··· , p − 1) satisﬁes hypothesis (i) of Corollary 2.2.12. Adopting our deﬁnition of σ, τ and ψ as before, by direct calculation, the action of σ, τ and ψ on rj, Rj and lj is as follows: Lemma 2.2.14. For 0 ≤ j ≤ p − 1,

−j −1 σ(rj) = ζ rj , τ(rj) = ψ (rj) = rs−1j

−1 where s is the multiplicative inverse of s in Zp. Hence Rj ∈ E(ζ), lj ∈ E and τ(lj) = lsj for each 0 ≤ j ≤ p − 1 by Corollary 2.2.12.

12 Applying Galois correspondence to Lemma 2.2.14 and Remark 2.2.3(ii) yields the following description of the ﬁelds K, k, F , E and L.

Theorem 2.2.15. The lattice diagram for the ﬁelds K, k, F , F (∆), E, F 0 and L is as follows: L

E(ζ)

@@ E

@@ 0 F

F (∆)

@@ F

Moreover, l0 ∈ F and l1, ··· , lp−1 are the roots of a polynomial g(x) of degree p − 1 over F and the field E := F (l1) is a cyclic extension of F of degree p − 1 with Gal(E/F ) = hτ|Ei. The unique quadratic subfield of E/F is the field F (∆).

Notice that E being generated by l1 over F shows that the other invari- 2 p−2 ants lj, 2 ≤ j ≤ p − 1 are F-linear combinations of 1, l1, l1, ··· , l1 which in principle permits us to write down each of the Rj explicitly in radicals provided we can express a primitive pth root of unity in radicals.

To motivate the use of Lagrange resolvents, we shall quote the next result from [T] without proof.

13 Theorem 2.2.16. Let x1, ··· , xn be roots of a irreducible polynomial of degree n over Q. Then for k = 1, ··· , n,

n−1 1 X x = ζj(1−k)r . k n j j=0

2.3 Setting Up the Calculation

In the previous section, we have provided the necessary qualitative data one needs to solve the roots of a irreducible solvable polynomial f(x) of degree p. Having done all these, we shall consider the main problem of doing the explicit calculation to express the roots in radicals and highlight the technicalities involved. Since the case for p = 7 still remains open and the case p = 3 is well known, we shall not be overly ambitious and just be concerned with the quintic case here and leave the septic case for performing explicit calculations to Section 3.1. Keeping these in mind, we shall break down the main problem into the following points of consideration:

(i). Finding the coeﬃcients of the degree p − 1 polynomial g(x) with lj, 1 ≤ j ≤ p − 1 as roots. This amounts to expressing the symmetric functions of the lj in terms of the symmetric functions of the roots for f(x).

(ii). Solving g(x) (not necessarily irreducible) for lj and labelling the roots 0 obtained lj correctly. (iii). Expressing a ﬁxed primitive pth root of unity in terms of radicals.

(iv). Taking the appropriate pth root of each Rj formed to obtain rj. We begin with the following observation.

Remark 2.3.1. Since Gal(E/F ) is cyclic of order p − 1, l1, ··· , lp−1 are the roots of a polynomial of degree p − 1 over F which factors over F (∆) into the product of two conjugate polynomials of degree (p − 1)/2:

 (p−3)/2   (p−3)/2  (p−1)/2 X m (p−1)/2 X m x + T2m+1 + T2m+2∆ x  x + T2m+1 − T2m+2∆ x  m=0 m=0 with Tj ∈ F for all 1 ≤ j ≤ p − 1. We see that under the action of the subgroup hτ 2i, the roots of one of these factors are the elements of the orbit containing l1: l1, ls2 , ··· , lsp−3 and the roots of the other factor are the elements of the orbit containing ls: ls, ls3 , ··· , lsp−2 .

14 For the case when p = 5, instead of finding the coefficients of g(x), Dummit (cf. [D]) factorized this quartic into the product of two conjugate quadratics over F (∆) (see Remark 2.3.1) and computed the coefficients of these two quadratics. This is done by first writing each invariant in F as a linear combination of 1, θ, ··· , θ5 over k where θ is a fixed invariant in F and applying the automorphisms generated by (1, 2, 3) and (1, 2) to generate complements of F20 in S5. The coefficients of the resulting 6 × 6 system of linear equations are then solved using Cramer’s rule.

Before we proceed to discuss step (ii), we need the following result which follows from Theorem 2.2.15. Theorem 2.3.2. Suppose f(x) is a irreducible solvable polynomial of degree p and g(x) is the degree p−1 polynomial with l1, ··· , lp−1 as roots where lj is deﬁned as in Deﬁnition 2.2.9. Then the Galois group of f(x) G is isomorphic to Zp o Zd if and only if g(x) factorizes into a product of (p−1)/d irreducibles each of degree d over Q.

Proof. Since [K : E] = p, G is isomorphic Zp o Zd if and only if [E : Q] = d. Because

(i). hτ|Ei being the Galois group of E over Q acts on the collection of generators {l1, ··· , lp−1} of E over F and (ii). elements in the same orbit are conjugates of the same minimal polynomial, [E : Q] = d is equivalent to the fact that g(x) factorizes into a product of (p − 1)/d irreducibles each of degree d over Q. In the case for quintics, Dummit’s approach to calculate the coeﬃcients of the two quadratic factors directly has several advantages as compared to computing the coeﬃcients of g(x). It facilitated greater ease in solving for the lj and obtaining the factorization of g(x) into irreducibles when ∼ Gal(f(x)) = D10. In addition, since the roots of the quadratics give {l1, l4} and {l2, l3} up to a permutation of the 2 pairs, more information is known to identify the lj correctly.

0 To label the lj correctly for p = 5, a square root of the discriminant ∆ is ﬁxed and assuming knowledge beforehand whether this corresponds to the 0 ∆ determined by the initial labelling of the roots xj, the lj obtained are arranged so that (l − l )(l − l ) (l0 − l0 )(l0 − l0 ) = Θ∆0 where Θ = 1 4 2 3 ∈ K. 1 4 2 3 ∆

15 0 0 0 0 This restriction narrows down to the 4 legitimate labelling of l1, l2, l3, l4 which are l1 , l2 , l3 , l4 ,

l4 , l3 , l2 , l1 ,

l2 , l4 , l1 , l3 ,

l3 , l1 , l4 , l2 .

We next characterize all the legitimate labelling of lj for a ﬁxed odd prime p.

Remark 2.3.3. Let X := {1 , ··· , p} and Y := {lj | 1 ≤ j ≤ p − 1}.

(i). Suppose X is permuted by some γ ∈ Sp. For u, v ∈ X such that γ(u) = v, the corresponding p-cycle that we have ﬁxed in Remark 2.2.3 is σ0 : v 7→ γ(u+1). Hence σ0 = (1, γ(γ−1(1)+1), ··· , γ(γ−1(1)+p−1)). 0 For each 1 ≤ j ≤ p−1, let rj denote the respective Lagrange resolvents, 0 0 p 0 0 p Rj := rj and lj be the sum of the terms in (x1, z) involving powers zi with i ≡ j (mod p). By deﬁnition,

0 j(1−γ−1(1)) rj = ζ rj .

0 0 Thus Rj = Rj and lj = lj for all 1 ≤ j ≤ p − 1.

(ii). Lemma 2.2.14 implies that for any γ ∈ Fp(p−1), γ induces an automorphism of Y which coincides with the automorphism on Y induced by τ n for some 1 ≤ n ≤ p − 1. In particular, this induced automorphism of Y is uniquely determined by its image on l1 and any permutation of Y that arises this way must be one of the following:

id(Y ): l1 , ··· , lp−1 ,

τ(Y ): ls , ··· , l−s , ··· , ··· , ··· , ··· , ··· , p−2 τ (Y ): lsp−2 , ··· , l−sp−2 .

0 Remark 2.3.3(ii) says that we may choose any of the lj to be l1 and the 0 rest of the lj is uniquely determined by this choice theoretically. From the computational point of view, this does not seem to be efficient as there will 0 be (p − 2)! ways to permute Y − {l1}. In practice, after obtaining the radical 0 expression of each lj by solving g(x), we approximate the roots of f(x) to suf- 0 ficiently high precision and fix a permutation of X. The lj are then computed

16 numerically using their formal deﬁnition and compared to the numerical val- 0 ues of each radical expression of lj to ensure that the radical expressions of 0 each lj are labelled correctly.

We begin our discussion of (iii) by stating the next result from [L] without proof. Theorem 2.3.4. Let ω be a primitive nth root of unity and σ denote the j automorphism of Q(ω) over Q such that σ(ω) = ω . The map σ 7→ j gives an isomorphism ∗ Gal(Q(ω)/Q) −→ Zn . In particular, [Q(ω): Q] = ϕ(n) where ϕ denotes the Euler-phi function. For our purpose, we restrict our attention to the case when n is an odd prime p and let ω := e2πi/p.

Remark 2.3.5. Let s be a primitive root of Zp. It follows from Theorem 2.3.4 s that the map Ψ : ω 7→ ω is a generator of Gal(Q(ω)/Q). Therefore for every divisor of p − 1, hΨ(p−1)/di is the unique subgroup of order d. Furthermore, given any two divisors d and d0 of p − 1, hΨ(p−1)/di ⊆ hΨ(p−1)/d0 i if and only 0 if d|d . Translating this information on the subgroups of Gal(Q(ω)/Q) to the corresponding fixed fields by Galois correspondence, for every divisor d of p − 1, there exists a unique subfield of Q(ω) of degree d over Q that corresponds to the subgroup hΨdi. In addition, for two such subfields E and F of degree d and d0 over respectively, E ⊇ F if and only if d0|d. When Q 0 this is the case, E/F is Galois with Gal(E/F ) ∼= hΨd i/hΨdi.

2π Proposition 2.3.6. (i). Q(cos p ) is the unique subﬁeld of Q(ω) of degree 2π (p − 1)/2 over Q. In addition, Q(cos p )/Q is Galois with 2π ∼ (p−1)/2 Gal(Q(cos p )/Q) = hΨi/hΨ i .

(ii). The elements of S := { ωj + ω−j | 1 ≤ j ≤ (p − 1)/2 } are precisely all −1 the conjugates of ω + ω over Q. Proof. (i). 2π 1 −1 cos p = 2 (ω + ω ) remains invariant under a non identity automorphism σ ∈ Gal(Q(ω)/Q) if and only if σ(ω) = ω−1, i.e. σ is an involution. (i) then follows by Remark 2.3.5.

17 (ii). By Artin’s theorem, it suﬃces to show that S := { Ψj(ω+ω−1) | 1 ≤ j ≤ (p − 1)/2 }. Since the automorphisms Ψj/hΨ(p−1)/2i are all distinct for 1 ≤ j ≤ (p − 1)/2, it follows that (Ψ(ω + ω−1), ··· , Ψ(p−1)/2(ω + ω−1)) are all distinct and thus diﬀers from (ω + ω−1, ··· , ω(p−1)/2 + ω(1−p)/2) by a permutation.

Since it is easy to solve the quadratic equation x2 − (ω + ω−1)x + 1 = 0 for ω, the problem of solving ω reduces to solving the minimal polynomial −1 p−1 of ω + ω which is of degree 2 over Q and can be computed from the pth p−1 1 cyclotomic polynomial x +···+1 by performing the substitution y = x+ x . Example 2.3.7. Applying the technique outlined above to calculate e2πi/5, we obtain √ √ 2π −1 + 5 4π −1 − 5 2 cos = , 2 cos = . 5 2 5 2

It remains to consider the choice of the pth roots of the Rj to obtain the Lagrange resolvents rj. Before we state and prove the next result which p asserts that, given R1 = r1, each of the p possible choices for r1 uniquely determines rj as a pth root of Rj (2 ≤ j ≤ p − 1), we shall need to make the following observations.

Remark 2.3.8. Viewing x1, ··· , xp as the roots of f(x), recall that K is hσi defined as the splitting field of f(x), k := Q, E := K , L := K(ζ) and s is ∗ a fixed primitive generator of Zp. (i). We deduce that the automorphism ψ : ζ 7→ ζs that acts trivially on x1, ··· , xp exists by natural irrationalities since the restriction of each automorphism in Gal(L/K) to k(ζ) gives an isomorphism of Gal(L/K) onto the Galois group of k(ζ) over k(ζ) ∩ K.

(ii). rj ∈/ Q for all 1 ≤ j ≤ p − 1. In particular, rj is non-vanishing for each 1 ≤ j ≤ p − 1. Suppose otherwise rj ∈ Q for some 1 ≤ j ≤ p − 1. Applying ψ iteratively p − 2 times, we see that rj = r1 for all 2 ≤ j ≤ p − 1. It follows from Theorem 2.2.16 and Remark 2.2.13 that

p−1 1 X 1 x = r = s + (p − 1)r ∈ . 1 p j p 1 1 Q j=0

This contradicts the fact that [Q(x1): Q] = p > 1. We now state and prove the result formally.

18 Theorem 2.3.9. Given r1, for each 2 ≤ j ≤ p − 1, there is a unique choice p of rj such that the equation Rj := rj is satisﬁed. Proof. Since

(i). the action of ψ on {r1, ··· , rp−1} is transitive and

(ii). rj 6= 0 for all 2 ≤ j ≤ 6 by Remark 2.3.8(ii), the theorem follows. For the sake of performing explicit calculations, we derive the following proposition which follows from Lemma 2.2.14. 0 p−1 Proposition 2.3.10. Let d > 1 be a divisor of p − 1 and d = d . For 1 ≤ j ≤ p − 1, d Y rjsnd0 n=1 is fixed by σ, τ d0 and τ d0−1ψ−1 and so is in the fixed field of hσ , τ d0 , τ d0−1ψ−1i in L which is of degree d0 over F . In particular, (i). 2 Y rjsn(p−1)/2 n=1 is fixed by σ, τ (p−1)/2/ and τ (p−3)/2ψ−1. (ii). (p−1)/2 Y rjs2n n=1 q is fixed by σ, τ 2 and τψ−1 and hence lies in F ∆ (−1)(p−1)/2p .

Using Proposition 2.3.10, Dummit created the following invariants √ 2 2 2 2 r1r4 , r2r3 , r1r2 + r4r3 , r3r1 + r2r4 ∈ F (∆ 5) and used them to determine uniqueness for r2, r3 and r4. For completeness, we shall state this result without proof here and refer the interested reader to [D] for the the proof.

Theorem 2.3.11. Given r1, there is a unique choice of r2, r3, r4 such that 2 2 2 2 r1r4 , r2r3 , r1r2 + r4r3 , r3r1 + r2r4 √ are ﬁxed invariants of F (∆ 5).

19 Chapter 3

Solvable Septics

After giving a brief survey for the general case when p is an odd prime, we shall now focus on the case for p = 7. We will also devote this chapter to giving an explicit example that exhibits calculations in solving solvable polynomials of degree 7 with Galois group isomorphic to Z7 by applying the core results in Chapter 2.

2πi/7 Throughout this chapter, we let ζ := e , x1, x2, x3, x4, x5, x6, x7 denote the roots of the general septic polynomial

6 5 4 3 2 f(x) := x7 − s1x + s2x − s3x + s4x − s5x + s6x − s7 where the s are the symmetric functions in the roots as deﬁned in Deﬁnition j √ 2.2.1. For simplicity, we let n denote the principal nth root, i.e. for z = reiθ 6= 0 where θ ∈ (−π, π] is the principal argument of z, √ n z = r1/neθi/n .

3.1 Lagrange Resolvents for Septic Polyno- mials

Throughout this section, we adopt Deﬁnition 2.2.9 to state the following Lagrange resolvents:

r0 := (x1, 1) := x1 + x2 + x3 + x4 + x5 + x6 + x7 = s1 ,

2 3 4 5 6 r1 := (x1, ζ) := x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ , 2 2 4 6 3 5 r2 := (x1, ζ ) := x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ ,

20 3 3 6 2 5 4 r3 := (x1, ζ ) := x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ , 4 4 5 2 6 3 r4 := (x1, ζ ) := x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ , 5 5 3 6 4 2 r5 := (x1, ζ ) := x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ , 6 6 5 4 3 2 r6 := (x1, ζ ) := x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ . It follows from Theorem 2.2.16 that

x1 = (r0 + r1 + r2 + r3 + r4 + r5 + r6)/7 ,

6 5 4 3 2 x2 = (r0 + ζ r1 + ζ r2 + ζ r3 + ζ r4 + ζ r5 + ζr6)/7 , 5 3 6 4 2 x3 = (r0 + ζ r1 + ζ r2 + ζr3 + ζ r4 + ζ r5 + ζ r6)/7 , 4 5 2 6 3 x4 = (r0 + ζ r1 + ζr2 + ζ r3 + ζ r4 + ζ r5 + ζ r6)/7 , 3 6 2 5 4 x5 = (r0 + ζ r1 + ζ r2 + ζ r3 + ζ r4 + ζr5 + ζ r6)/7 , 2 4 6 3 5 x6 = (r0 + ζ r1 + ζ r2 + ζ r3 + ζr4 + ζ r5 + ζ r6)/7 , 2 3 4 5 6 x7 = (r0 + ζr1 + ζ r2 + ζ r3 + ζ r4 + ζ r5 + ζ r6)/7 . Similarly, we have

7 2 3 4 5 6 R1 := r1 = l0 + l1ζ + l2ζ + l3ζ + l4ζ + l5ζ + l6ζ ,

7 2 3 4 5 6 R2 := r2 = l0 + l4ζ + l1ζ + l5ζ + l2ζ + l6ζ + l3ζ , 7 2 3 4 5 6 R3 := r3 = l0 + l5ζ + l3ζ + l1ζ + l6ζ + l4ζ + l2ζ , 7 2 3 4 5 6 R4 := r4 = l0 + l2ζ + l4ζ + l6ζ + l1ζ + l3ζ + l5ζ , 7 2 3 4 5 6 R5 := r5 = l0 + l3ζ + l6ζ + l2ζ + l5ζ + l1ζ + l4ζ , 7 2 3 4 5 6 R6 := r6 = l0 + l6ζ + l5ζ + l4ζ + l3ζ + l2ζ + l1ζ .

When f(x) is solvable, its Galois group is a transitive subgroup of F42 by Corollary 2.1.7. Adopting the same notations as in Remark 2.2.3, we ﬁx 3 as our choice for s, a ﬁxed primitive root of Z7 and set σ := (1, 2, 3, 4, 5, 6, 7), τ := (2, 4, 3, 7, 5, 6) acting trivially on ζ and ψ : ζ 7→ ζ3 acting trivially on x1, ··· , x7.

By Lemma 2.2.14, the action of σ, τ and ψ on rj and lj (1 ≤ j ≤ 6) is as follows: 6 5 σ(r1) = ζ r1, τ(r1) = ψ (r1) = r5, 5 5 σ(r2) = ζ r2, τ(r2) = ψ (r2) = r3, 4 5 σ(r3) = ζ r3, τ(r3) = ψ (r3) = r1,

21 3 5 σ(r4) = ζ r4, τ(r4) = ψ (r4) = r6, 2 5 σ(r5) = ζ r5, τ(r5) = ψ (r5) = r4, 5 σ(r6) = ζr6, τ(r6) = ψ (r6) = r2 . We shall now review the four problems that were proposed in the previous section for p = 7. In solving quintics, recall that Dummit first calculate those elements in the fixed field of the Frobenius group by generating complements of F20 in S5 and solving a 6×6 system of linear equations. The equivalent computation for the case p = 7 would not be computationally feasible as it requires us to solve a 120 × 120 system of linear equations. For the case of sextic polynomials, it was cited in [H] that the evaluation of a determinant for a 15 × 15 matrix symbolically is still not known. Theorem 3.1.1. Suppose f(x) is a irreducible solvable septic and g(x) is the degree 6 polynomial with l1, ··· , l6 as roots where lj is defined as in Definition 2.2.9. Then the Galois group of f(x) is Z7 o Zd if and only if g(x) factorizes into a product of 6/d irreducibles each of degree d over Q. Recall from Corollary 2.1.7 that a solvable septic f(x) can only have Ga- ∼ lois group isomorphic to either Z7, D14, Z7 o Z3 or F42. When Gal(f(x)) = Z7, we multiply g(x) by a suitable positive positive rational so that the resulting polynomial h(x) have coprime integral coefficients. Since all lj are rational with numerator and denominator dividing the constant term and leading term of h(x) respectively, an exhaustive search can be implemented ∼ on the computer to solve for the lj. When Gal(f(x)) = F42, g(x) is irreducible of degree 6 with Galois group isomorphic to Z6. Hence we may appeal to ∼ the results in [H] to obtain the lj. When Gal(f(x)) = D14, we know that g(x) factorizes into 3 irreducible quadratics over Q by Theorem 3.1.1 but to obtain this factorization explicitly is non-trivial in general. The case for ∼ solving lj from g(x) when Gal(f(x)) = Z7 o Z3 faces a similar technicality.

It is not clear how we can design an algorithm to label the lj correctly for p = 7. We now restate the analogous statements in Remark 2.3.3(ii).

Remark 3.1.2. For any γ ∈ F42, γ induces an automorphism of Y which coincides with the automorphism on Y induced by τ n for some 1 ≤ n ≤ 6. In particular, this induced automorphism of Y is uniquely determined by its image on l1 and any permutation of Y that arises this way must be one of the following: id(Y ): l1 , l2 , l3 , l4 , l5 , l6 ,

22 τ(Y ): l3, l6, l2, l5, l1, l4 , 2 τ (Y ): l2, l4, l6, l1, l3, l5 , 3 τ (Y ): l6, l5, l4, l3, l2, l1 , 4 τ (Y ): l4, l1, l5, l2, l6, l3 , 5 τ (Y ): l5, l3, l1, l6, l4, l2 . We adopt the method mentioned in Section 2.3 to perform the following calculation which expresses the non-trivial 7th roots of unity in radicals.

1 Example 3.1.3. By making the substitution y := x + x to x6 + x5 + x4 + x3 + x2 + x + 1 = 0 , we see that ζ + ζ−1 is a root of

y3 + y2 − 2y − 1 = 0 . (3.1.1)

Using Proposition 2.3.6, we deduce that y3 + y2 − 2y − 1 is the minimal −1 2π −1 4π polynomial of ζ +ζ and its roots are precisely 2 cos 7 = ζ +ζ , 2 cos 7 = 2 −2 6π 3 −3 6π 4π 2π ζ + ζ and 2 cos 7 = ζ + ζ . Since cos 7 < cos 7 < cos 7 , solving equation (3.1.1) by direct calculation using Cardan’s formula, we conclude that " s # 2π 1 98 r7 √ cos = −1 + 3 √ + 3 1 + 3 3i , 7 6 1 + 3 3i 2 " s # 4π 1 √ 98 √ r7 √ cos = − 2 + (1 − 3i) 3 √ + (1 + 3i) 3 1 + 3 3i , 7 12 1 + 3 3i 2 " s # 6π 1 √ 98 √ r7 √ cos = − 2 + (1 + 3i) 3 √ + (1 − 3i) 3 1 + 3 3i . 7 12 1 + 3 3i 2 √ Implementing the relation sin θ = 1 − cos2 θ for 0 ≤ θ ≤ π, we have s 2π 1 1r7 h √ √ √ i sin = 3 2v + (1 − 3i)w + 2 3 2(−2 + 3i) , 7 2u 3 4 s 4π 1 1 r7 h √ √ √ i sin = 3 2v + (1 + 3i)w + 3 2(5 + 3i) , 7 2u 3 4 s 6π 1 1 r7 h √ √ i sin = 3 2v − 2w − 3 2(1 + 3 3i) , 7 2u 3 4

23 where

q3 √ q √ q √ u := 1 + 3 3i , v := 3 49(−13 + 3 3i) , w := 3 7(1 + 3 3i) .

Since we are unable to express ﬁxed invariants in terms of the symmetric functions of the roots explicitly in general to show uniqueness of rj (2 ≤ j ≤ 6 given r1, we shall only restate Theorem 2.3.9 for the septic case here.

Theorem 3.1.4. Given r1, for each 2 ≤ j ≤ 6, there is a unique choice of p rj such that the equation Rj := rj is satisﬁed.

2π 3.2 Expressing cos 29 in radicals

2π It is well known that for each natural number n, cos n is algebraic. Moreover, in the quest of determining which n-sided regular polygon is constructible by 2π straightedge and compass, Gauss showed how to express cos 17 in radicals. The main purpose of this section is to perform an analogous calculation to 2π express cos 29 in radicals. Throughout this section, all technical calculations performed using Math- ematica are saved in a file named lagres.nb. Due to the overwhelming length of the output, we shall only present the commands of this program in the appendix. 2π Before we formally embark on the task of expressing cos 29 in radicals, 2π we investigate the possibility of expressing cos 29 completely in real radicals. Definition 3.2.1. Define α ∈ R to be a real radical element if it lies in some repeated radical extension of Q that is contained in R. The following result is taken from [I]. For the convenience of the readers, the proof is included in the appendix.

Theorem 3.2.2. Suppose f(x) ∈ Q[x] is an irreducible polynomial which splits over R. If f(x) has any root which is a real radical element, then the degree of f(x) is a power of 2 and the Galois group of f(x) over Q is a 2-group.

2π Proposition 3.2.3. cos p is a real radical element if and only if p is a Fermat prime.

2m 2π Proof. Necessity follows at once, for if p = 2 +1 for some m ≥ 0, Q(cos p ) ⊆ R can be obtained at the end of a ﬁnite tower where each step of the tower is a radical extension of degree 2. To show suﬃciency, since the minimal

24 2π polynomial of cos p is of degree (p − 1)/2 and splits over R by Proposition 2.3.6, we invoke Theorem 3.2.2 and see that (p − 1)/2 = 2n for some n ≥ 0. p being prime then forces n to be a power of 2.

2π It follows from Proposition 3.2.3 that cos 29 cannot be completely expressed in real radicals. Remark 3.2.4. We consider the case when p = 29. Since 2 is a primitive 2 root of Z29, we may fix the automorphism Ψ : ω 7→ ω . By Proposition 2π 2.3.6 and Remark 2.3.5, Q(cos 29 ) is of degree 14 over Q and contains a 2π unique subfield K of degree 7 over Q. In view of this, cos 29 satisfies a monic quadratic polynomial h(x) ∈ K[x] as a root. To determine h(x) explicitly, 7 14 2π we apply the non identity automorphism Ψ /hΨ i of Gal(Q(cos 29 )/K) and see that the non trivial conjugate of ω + ω28 over K is ω12 + ω17. Hence

h(x) = x2 − (ω + ω12 + ω28 + ω17)x + (ω16 + ω18 + ω13 + ω11) .

2π With this conclusion, we are able to express cos 29 in radicals if we can do the same to ω + ω12 + ω28 + ω17 and ω16 + ω18 + ω13 + ω11. Applying the 7 12 28 17 generator Ψ/hΨ i of Gal(K/Q) to ω + ω + ω + ω cyclically and using Maple, we see that

12 28 17 x1 := ω + ω + ω + ω , 2 24 27 5 x2 := ω + ω + ω + ω , 4 19 25 10 x3 := ω + ω + ω + ω , 8 9 21 20 x4 := ω + ω + ω + ω , 16 18 13 11 x5 := ω + ω + ω + ω , 3 7 26 22 x6 := ω + ω + ω + ω , 6 14 23 15 x7 := ω + ω + ω + ω , are the roots of the septic polynomial x7+x6−12x5−7x4+28x3+14x2−9x+1.

Recall our deﬁnition for rj and Rj for 0 ≤ j ≤ 6. Motivated by Theorem 2.2.16, we shall proceed as follows.

Example 3.2.5. We substitute the actual values of x1, ··· , x7 into the polynomial expression obtained for each lj in Example 2.2.10. The values for l3, l2, l6, l4, l5 are veriﬁed in this order by applying τ := (2, 4, 3, 7, 5, 6) repeatedly to x1, ··· , x7. Our computation yields l0 = −45410,

l1 = −37058 , l2 = −46396 , l3 = 63224 ,

l4 = 33383 , l5 = −26096 , l6 = 58352 .

25 Consequently,

2 3 4 5 6 R1 = 8352ζ − 986ζ + 108634ζ + 78793ζ + 19314ζ + 103762ζ ,

2 3 4 5 6 R2 = 78793ζ + 8352ζ + 19314ζ − 986ζ + 103762ζ + 108634ζ , 2 3 4 5 6 R3 = 19314ζ + 108634ζ + 8352ζ + 103762ζ + 78793ζ − 986ζ , 2 3 4 5 6 R4 = −986ζ + 78793ζ + 103762ζ + 8352ζ + 108634ζ + 19314ζ , 2 3 4 5 6 R5 = 108634ζ + 103762ζ − 986ζ + 19314ζ + 8352ζ + 78793ζ , 2 3 4 5 6 R6 = 103762ζ + 19314ζ + 78793ζ + 108634ζ − 986ζ + 8352ζ .

p7 For each 1 ≤ j ≤ 6, the ratio of rj to Rj is an integral power of ζ which we will denote by nj. Hence

nj p7 rj = ζ Rj (3.2.1) where nj is some element of Z7. Since the principal argument of a product diﬀers from the sum of the principal arguments of the factors by an integral multiple of 2π, we have 7Arg r − Arg R j j ≡ n (mod 7) . (3.2.2) 2π j Evaluating the left hand side of equation (3.2.2) numerically to 400 decimal places using a for loop with index 1 ≤ j ≤ 6 incrementing by one unit with each iteration yields an approximation for each nj having absolute error less than 5 × 10401. Therefore we can deduce

n1 = 1, n2 = 3, n3 = 1, n4 = 6, n5 = 4, n6 = 6 .

Thus, assigning the variables a, b, c, d, e and f with the expression of ζ, ζ2, 3 4 5 6 ζ , ζ , ζ and ζ in radicals respectively, for 1 ≤ j ≤ 6, we obtain each Rj and rj in radicals. Finally, appealing to the relations

6 6 1 X 1 X x = r , x = ζ3jr 1 7 j 5 7 j j=0 j=0 from Theorem 2.2.16 and applying the formula

2π 1 q cos = −x + x2 − 4x 29 4 1 1 5

26 yields what we sought out to compute initially. The interested reader is referred to the program lagres.nb in the appendix for the actual radical expression of Rj, rj (1 ≤ j ≤ 6), x1 and x5. Explicitly,

1 2 x1 = −72a1 + 2a2a3a4a31 − 2a24a3a1a31 + 2a29a3a15a a31 504a1 1 2 −6ia30a32a1a37a31 + 2a2a3a4a38 − 2a24a3a1a38 + 2a29a3a15a1a38

+6ia30a32a1a37a38 − 6ia34a32a4a43 − 2a34a3a4a43 − 2a41a3a1a43 2 2 +3ia42a44a15a1a43 − a42a3a15a1a43 + 6ia45a44a1a48a43 + 2a2a3a4a52 2 −2a24a3a1a52 + 2a29a3a15a1a52 + 6ia30a44a1a37a52 + 2a2a3a4a58 2 −2a24a3a1a58 + 2a29a3a15a1a58 − 6ia30a32a1a37 − 6ia34a32a4a59 2 2 −2a34a3a4a59 − 2a41a3a1a59 + 3ia42a32a15a1a59 − a42a3a15a1a59

−6ia45a32a1a48a59)

27 and

1 2 x5 = 2 −a57a1 − 18ia29a44a15a31 − 2a29a3a15a31 − 2a2a3a4a1a31 1008a1 2 −4a24a3a1a31 − ia60a32a15a48a31 + 3a60a3a15a48a31 2 −2ia61a32a4a1a48a31 + 2ia30a32a1a48a31 + 5ia60a44a15a37a31

−a60a3a15a37a31 + ia61a32a4a1a37a31 − a61a3a4a1a37a31 2 2 +2ia30a32a1a37a31 − a24a3a1a68a31 − 18ia29a32a15a38 − 2a29a3a15a38 2 −2a2a3a4a1a38 − 4a24a3a1a38 + ia60a44a15a48a38 − 3a60a3a15a48a38 2 +2ia61a32a4a1a48a38 − 2ia30a32a1a48a38 − 5ia60a32a15a37a38

+a60a3a15a37a38 − ia61a44a4a1a37a38 + a61a3a4a1a37a38 2 2 −2ia30a32a1a37a38 − a24a3a1a68a38 − 18ia42a44a15a43 − 2a42a3a15a43 2 −4a34a3a4a1a43 − 4a41a3a1a43 + ia67a32a15a48a43 − 3a67a3a15a48a43 2 +2ia70a44a4a1a48a43 − 2ia45a44a1a48a43 + 5ia67a44a15a37a43

−a67a3a15a37a43 + ia70a44a4a1a37a43 − a70a3a4a1a37a43 2 2 +2ia45a32a1a37a43 + a41a3a1a68a43 + 12ia29a44a15a52 − 8a29a3a15a52 2 −3ia2a32a4a1a52 + a2a3a4a1a52 − 4a24a3a1a52 + ia60a44a15a69a52 2 −3a60a3a15a69a52 + 2ia61a44a4a1a69a52 − 2ia30a32a1a69a52

+4ia60a44a15a37a52 + 2a60a3a15a37a52 − ia61a32a4a1a37a52 2 2 −a61a3a4a1a37a52 − 2ia30a32a1a37a52 − a24a3a1a73a52

+12ia29a32a15a74 − 8a29a3a15a74 − 3ia2a32a4a1a74 + a2a3a4a1a74 2 −4a24a3a1a74 − ia60a44a15a69a74 + 3a60a3a15a69a74 2 −2ia61a32a4a1a69a74 + 2ia30a32a1a69a74 − 4ia60a44a15a37a74

−2a60a3a15a37a74 + ia61a44a4a1a37a74 + a61a3a4a1a37a74 2 2 +2ia30a32a1a37a74 − a24a3a1a73a74 − 18ia42a44a15a59 − 2a42a3a15a59 2 −4a34a3a4a1a59 − 4a41a3a1a59 − ia67a44a15a48a59 + 3a67a3a15a48a59 2 −2ia70a32a4a1a48a59 + 2ia45a32a1a48a59 − 5ia67a44a15a37a59

+a67a3a15a37a59 − ia70a32a4a1a37a59 + a70a3a4a1a37a59 2 2 −2ia45a32a1a37a59 + a41a3a1a68a59 where √ 1/3 19/21 6/7 2/3 a1 := (1 + 3 3i) , a2 := 2 , a3 := 3 , a4 := 7 , a5 := 29 , √ √ 7/3 1/3 2/3 a6 := 7 , a7 := 2(14 ) , a8 := 13 − 119 3i , a9 := (2 + 6 3i) ,

a10 := a8 , a11 := 21922i , a12 := 1029 , a13 := 700 , a14 := 3290 , √ 1/3 2/3 3 1/3 a15 := 7 , a16 := 2 , a17 := −a1 , a18 := (14 + 42 3i) ,

28 s 2 3 6[14a1 + a15(−a16a1 − 2a18)] a19 := 2 , a1 √ √ √ 4/7 a20 := 5 + 3i , a21 := 1 + 3i , a22 := a21 , a23 := 3 + 2i , a24 := 2 ,

s 2 6[14a1 + a15(a16a20 + a21a18)] a25 := 2 , a1

s 2 6[14a1 + a15(a22a18 + 2ia16a23)] a26 := 2 , a1 5/21 1/14 a27 := a12a19 − a13a25 − a14a26 , a28 := −a11 + a27 , a29 := 2 , a30 := 2 , r 7 1 a31 := [a5 (a6(a7a8 + a9a10) − 2ia1a28)] , a1 √ 5/14 1/21 a32 := 3 , a33 := a11 + a27 , a34 := 2 , a35 := 86 − 33 3i , a36 := a35 ,

s 2 14a1 + a15(a22a18 + 2ia16a23) a37 := 2 , a1 r 7 1 a38 := [a5 (a6(a7a8 + a9a10) + 2ia1a33)] , a1 5/7 8/21 a39 := a13a19 − a14a25 − a12a26 , a40 := a11 + a39 , a41 := 2 , a42 := 2 , r 7 1 a43 := [a5 (a6(a9a35 + a7a36) + ia1a40)] , a1 √ 5/14 3/14 a44 := 3 , a45 := 2 , a46 := 29i , a47 := −53 3 + 185i ,

s 2 3 14a1 − a15(a16a1 + 2a18) a48 := 2 , a1 √ a49 := 53 3 + 185i , a50 := a14a19 − a12a25 − a13a26 , a51 := −a11 + a50 , r 7 1 a52 := − [a46 (−a6(a47a9 + a7a49) + 2a1a51)] , a1 a53 := a11 + a50 , a54 := 21922 , a55 := −ia39 , a56 := −a54 + a55 , a57 := 144 , r 7 1 a58 := [a46 (−a6(a47a9 + a7a49) + 2a1a53)] , a1 r 7 1 a59 := [a5 (a6(a9a35 + a7a36) + a1a56)] , a1 31/42 17/42 2 3 a60 := 2 , a61 := 2 , a62 := 14a1 , a63 := a15(a16a1 + 2a18) ,

29 37/42 a64 := a62 −a63 , a65 := a15(a22a18 +2ia16a23) , a66 := a62 +a65 , a67 := 2 , s r 2 a64a66 14a1 + a15(a16a20 + a21a18) a68 := 4 , a69 := 2 , a1 a1

23/42 a70 := 2 , a71 := a15(a16a20 + a21a18) , a72 := a62 + a71 , r r a72a66 7 1 a73 := 4 , a74 := [a46 (a6(a47a9 + a7a49) + 2a1a53)] . a1 a1

30 Bibliography

[D] D. S. Dummit, Solving Solvable Quintics, Mathematics of Computation, Volume 57, Issue 195, 387-401, 1991.

[DF] D. S. Dummit & Richard M. Foote, Abstract Algebra, Prentice Hall, 1999.

[E] J.-P. Escoﬁer, Galois Theory, Springer-Verlag, 2002.

[H] T. R. Hagedorn, General Formulas for Solving Solvable Sextic Equations, Journal of Algebra, Volume 233, Issue 2, 704-757, 2000.

[I] I. M. Isaacs, Solution of Polynomials by Real Radicals, The American Mathematical Monthly, Volume 92, Issue 8, 571-575, 1985.

[K] R. B. King, Beyond the Quartic Equation, Birkh¨auser, 1996.

[KN] S. Kobayashi & H. Nakagawa, Resolution of solvable quintic equation, Mathematica Japonica, Volume 37, Issue 5, 1992.

[L] S. Lang, Algebra, Springer-Verlag, 2002.

[PS] V. Prasolov and Y. Solovgev, Elliptic Functions and Elliptic Integrals, American Mathematical Theory, 1977.

[R1] J. J. Rotman, An Introduction to the Theory of Groups, Springer-Verlag, 1994.

[R2] J. J. Rotman, Galois Theory, Springer-Verlag, 1990.

[SW] B. Spearman and K. S. Williams On solvable Quintics x5 + ax + b and x5 + ax2 + b, Rocky Mountain Journal of Mathematics, Volume 26, Number 2, 753-772, 1996.

[T] J.-P. Tignol, Galois’ Theory of Algebraic Equations, World Scientiﬁc, 2001.

31 [V] T. M. Bˇosel, http://www.vimagic.de/hope/index.html?/hope/inhalt.html, University of Adelaide, 2002.

[W] E. Weisstein, Eric Weisstein’s World of Mathematics, http://mathworld.wolfram.com/QuinticEquation.html

32 Appendix A

Solution of Polynomials by Real Radicals

In section 1.3, we have considered the following problem: Given a solvable polynomial with all its roots real, what are the necessary and suﬃcient con- ditions for one of the roots to be completely expressible in real radicals? Necessity is resolved aﬃrmatively by the key result in [I] and we shall give an alternative proof of it here.

We shall ﬁrst make the notion of real radical element precise.

Deﬁnition A.1. Deﬁne α ∈ R to be a real radical element if it lies in some repeated radical extension of Q that is contained in R. We quote the following result from [L] without proof.

Theorem A.2. (natural irrationalities) Let K be a Galois extension of k, let F be an arbitrary extension and assume that K, F are subﬁelds of some other ﬁeld. Then the compositum KF is Galois over F , and K is Galois over K ∩ F . Let H be the Galois group of KF over F , and G the Galois group of K over k. If σ ∈ H then the restriction of σ to K is in G, and the map σ 7→ σ|K gives an isomorphism of H on the Galois group of K over K ∩ F . In partic-

33 ular, [KF : F ] = [K : K ∩ F ].

@@ K ∩ F

k We will need the following lemma to prove the main theorem of this section.

n Lemma A.3. Let k and k[γ] be subﬁelds of R such that γ ∈ k for some n ≥ 0. Denote the normal closure of k[γ] over k by K. If K ⊆ R, then [k[γ]: k] ≤ [K : k] ≤ 2. Proof. Let m(x) be the minimal polynomial of γ over k. For σ ∈ Gal(K/k),

(γσ)n = (γn)σ = γn ,

σ and so γ = γζ where ζ ∈ K is some nth root of unity. K ⊆ R forces ζ = ±1 and thus γσ = ±γ for every element σ ∈ Gal(K/k). Since the action of Gal(K/k) on the roots of m(x) is transitive, it follows that ±γ are the only possible conjugates of γ over k. Hence m(x) being separable over k implies [K : k] ≤ 2! as desired.

Theorem A.4. Suppose f(x) ∈ Q[x] is an irreducible polynomial which splits over R. If f(x) has any root which is a real radical element, then the degree of f(x) is a power of 2 and the Galois group of f(x) over Q is a 2-group.

34 Proof.

SEm = Fm @@

S ∩ Em

Q(α)

Q Let α be the real radical element with f(α) = 0 and S ⊆ R be the splitting ﬁeld for f(x). By hypothesis, there exists a tower of ﬁelds

Q = E0 ⊆ E1 ⊆ · · · ⊆ Em

nj with Ej = Ej−1[γj] such that γj ∈ Ej−1 for some nj ≥ 0 for all 1 ≤ j ≤ m. Without loss of generality, we may assume α ∈ Em − Em−1. For 1 ≤ j ≤ m, let Fj denote the normal closure of Ej over Ej−1. Since deg[f(x)] divides [S : Q] = |Gal(S/Q)| and [S : Q] divides [SEm : Q], it suﬃces to show that [SEm : Q] is a power of 2. We ﬁrst claim that SEm = Fm. Because Fm is normal over Em−1, Fm is also normal over Q(α). Thus S ⊆ Fm and SEm ⊆ Fm. On the other hand, by natural irrationalities applied to S and Em, SEm is Galois over Em. Therefore Fm ⊆ SEm and so SEm = Fm as claimed. Note that S being normal over Q is also normal over Ej for 1 ≤ j ≤ m − 1. Consequently, Fj ⊆ S ⊆ R for all 1 ≤ jm − 1. Also S, Em ⊆ R implies Fm = SEm ⊆ R. Hence applying Lemma A.3 by taking k = Ej−1, γ = γj, n = nj for each 1 ≤ j ≤ m yields [Ej : Ej−1] ≤ 2 for all 1 ≤ j ≤ m − 1 and [Fm : Em−1] ≤ 2. Therefore

m−1 Y u [Fm : Q] = [Fm : Em−1] [Ej : Ej−1] = 2 j=1

35 for some u ≤ m as required.

36 Appendix B lagres.nb

2 3 4 5 6 7 Collect[Expand[(x1 + x2ζ + x3ζ + x4ζ + x5ζ + x6ζ + x7ζ ) ] /. ζ → Exp[2 ∗ Pi ∗ I / 7], Exp[2 ∗ Pi ∗ I / 7]]

12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ 1 Simplify[l0[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l0 ] 12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l1[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l1]

1In practice, we shall use the cut and paste commands from the output of the ﬁrst ˆ command to generate the symbolic expressions for lj which we denote by lj, 0 ≤ j ≤ 6.

37 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l3[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l3] 12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l1[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l1]

12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l2[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l2] 12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l1[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l1]

12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l6[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l6] 12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29]

38 8 9 21 20 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l1[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l1]

12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l4[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l4] 12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l1[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l1]

12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l5[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l5] 12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 4 19 25 10 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ˆ Simplify[l1[x1 , x2 , x3 , x4 , x5 , x6 , x7 ] → l1]

12 28 17 x1 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 2 24 27 5 x2 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29]

39 4 19 25 10 x3 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 8 9 21 20 x4 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 16 18 13 11 x5 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 3 7 26 22 x6 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] 6 14 23 15 x7 = ω + ω + ω + ω /. ω → Exp[2 ∗ Pi ∗ I / 29] ζ = Exp[2 ∗ Pi ∗ I / 7] 2 3 4 5 6 R1 = 8352ζ − 986ζ + 108634ζ + 78793ζ + 19314ζ + 103762ζ 2 3 4 5 6 R2 = 78793ζ + 8352ζ + 19314ζ − 986ζ + 103762ζ + 108634ζ 2 3 4 5 6 R3 = 19314ζ + 108634ζ + 8352ζ + 103762ζ + 78793ζ − 986ζ 2 3 4 5 6 R4 = −986ζ + 78793ζ + 103762ζ + 8352ζ + 108634ζ + 19314ζ 2 3 4 5 6 R5 = 108634ζ + 103762ζ − 986ζ + 19314ζ + 8352ζ + 78793ζ 2 3 4 5 6 R6 = 103762ζ + 19314ζ + 78793ζ + 108634ζ − 986ζ + 8352ζ 2 3 4 5 6 Lr[ z ] := x1 + x2z + x3z + x4z + x5z + x6z + x7z j For[j = 1, j < 7, j + +, nj = N[(7 ∗ Arg [Lr[ζ ]]-Arg [Rj])/(2 ∗ Pi), 400]; Print[nj]] ! 1 1 72/3 1 7 √ 1/3 a = − + √ + (1 + 3i 3) + 2 3 1 1/3 3 2 3( 2 (1 + 3i 3)) s √ √ √ √ ! 1 14(1 + 3i 3)2/3 + 71/3((1 − i 3)(14 + 42i 3)1/3 + 2i22/3(2i + 3)) i √ 2 6(1 + 3i 3)2/3 √ ! 1 1 ( 7 )2/3(1 − i 3) 1 √ 7 √ 1/3 b = − − 2 √ − (1 + i 3) (1 + 3i 3) + 2 3 3(1 + 3i 3)1/3 6 2

s √ √ √ √ ! 1 14(1 + 3i 3)2/3 + 71/3(22/3(5 + i 3)1/3 + (1 + i 3)(14 + 42i 3)1/3) i √ 2 6(1 + 3i 3)2/3 √ ! 1 1 ( 7 )2/3(1 + i 3) 1 √ 7 √ 1/3 c = − − 2 √ − (1 − i 3) (1 + 3i 3) + 2 3 3(1 + 3i 3)1/3 6 2 s √ √ √ ! 1 14(1 + 3i 3)2/3 − 71/3(22/3(1 + 3i 3) + 2(14 + 42i 3)1/3) i √ 2 6(1 + 3i 3)2/3 √ ! 1 1 ( 7 )2/3(1 + i 3) 1 √ 7 √ 1/3 d = − − 2 √ − (1 − i 3) (1 + 3i 3) − 2 3 3(1 + 3i 3)1/3 6 2 s √ √ √ ! 1 14(1 + 3i 3)2/3 − 71/3(22/3(1 + 3i 3) + 2(14 + 42i 3)1/3) i √ 2 6(1 + 3i 3)2/3

40 √ ! 1 1 ( 7 )2/3(1 − i 3) 1 √ 7 √ 1/3 e = − − 2 √ − (1 + i 3) (1 + 3i 3) − 2 3 3(1 + 3i 3)1/3 6 2

s √ √ √ √ ! 1 14(1 + 3i 3)2/3 + 71/3(22/3(5 + i 3)1/3 + (1 + i 3)(14 + 42i 3)1/3) i √ 2 6(1 + 3i 3)2/3 ! 1 1 72/3 1 7 √ 1/3 f = − + √ + (1 + 3i 3) − 2 3 1 1/3 3 2 3( 2 (1 + 3i 3)) s √ √ √ √ ! 1 14(1 + 3i 3)2/3 + 71/3((1 − i 3)(14 + 42i 3)1/3 + 2i22/3(2i + 3)) i √ 2 6(1 + 3i 3)2/3

Simplify[Expand[R1 = 8352a−986b+108634c+78793d+19314e+103762f]] Simplify[Expand[R2 = 78793a+8352b+19314c−986d+103762e+108634f]] Simplify[Expand[R3 = 19314a+108634b+8352c+103762d+78793e−986f]] Simplify[Expand[R4 = −986a+78793b+103762c+8352d+108634e+19314f]] Simplify[Expand[R5 = 108634a+103762b−986c+19314d+8352e+78793f]] Simplify[Expand[R6 = 103762a+19314b+78793c+108634d−986e+8352f]]

p7 Simplify[Expand[r1 = a (R1)]] p7 Simplify[Expand[r2 = c (R2)]] p7 Simplify[Expand[r3 = a (R3)]] p7 Simplify[Expand[r4 = f (R4)]] p7 Simplify[Expand[r5 = d (R5)]] p7 Simplify[Expand[r6 = f (R6)]] x1 = Simplify[Expand[(1/7) ∗ (−1 + r1 + r2 + r3 + r4 + r5 + r6)]] x5 = Simplify[Expand[(1/7)∗(−1+c∗r1 +f ∗r2 +b∗r3 +e∗r4 +a∗r5 +d∗r6)]]

p 2 c = Simplify[Expand[(1/4) ∗ (x1 + (x1) − 4x5)]]

41 Explicitly, 6 6 4 2 2 2 3 3 4 5 l1 = 7x1x2 + 7x2x3 + 105x1x2x3 + 210x1x2x3 + 35x1x3 + 42x1x2x4 2 3 3 2 6 3 2 2 4 2 +420x1x2x3x4 + 420x1x2x3x4 + 7x3x4 + 210x1x2x4 + 105x1x3x4 4 2 2 2 3 3 3 3 4 4 +105x2x3x4 + 210x2x3x4 + 140x1x3x4 + 35x2x4 + 210x1x2x3x4 2 5 2 4 3 2 4 2 5 +21x1x4 + 105x1x2x5 + 420x1x2x3x5 + 105x1x3x5 + 42x2x3x5 4 2 3 4 3 2 +210x1x2x4x5 + 420x2x3x4x5 + 210x1x3x4x5 + 420x2x3x4x5 2 2 2 3 2 3 6 5 2 +1260x1x2x3x4x5 + 420x1x2x4x5 + 420x1x3x4x5 + 7x4x5 + 21x1x5 3 2 2 3 2 4 2 2 2 +210x2x3x5 + 420x1x2x3x5 + 105x2x4x5 + 1260x1x2x3x4x5 2 2 2 2 2 2 4 2 3 3 +630x1x3x4x5 + 630x1x2x4x5 + 105x3x4x5 + 140x1x2x5 2 3 3 3 2 2 3 3 3 3 4 +420x1x2x3x5 + 140x1x4x5 + 210x3x4x5 + 140x2x4x5 + 35x3x5 4 2 4 2 5 5 3 3 +210x2x3x4x5 + 105x1x4x5 + 21x2x5 + 42x1x3x5 + 140x1x2x6 4 2 4 5 5 3 2 +210x1x2x3x6 + 105x2x3x6 + 42x1x3x6 + 42x1x4x6 + 420x2x3x4x6 3 4 2 2 2 2 2 2 +840x1x2x3x4x6 + 105x2x4x6 + 1260x1x2x3x4x6 + 630x1x3x4x6 2 3 5 4 2 2 +420x1x2x4x6 + 42x3x4x6 + 210x2x3x5x6 + 1260x1x2x3x5x6 2 3 3 2 3 2 +420x1x3x5x6 + 840x1x2x4x5x6 + 2520x1x2x3x4x5x6 + 420x1x4x5x6 2 3 4 2 2 2 3 2 +420x3x4x5x6 + 210x2x4x5x6 + 630x1x2x5x6 + 420x1x3x5x6 3 2 2 2 3 2 2 3 +420x3x4x5x6 + 1260x2x3x4x5x6 + 420x1x4x5x6 + 420x2x3x5x6 2 3 3 4 6 5 2 +420x2x4x5x6 + 840x1x3x4x5x6 + 210x1x2x5x6 + 7x5x6 + 21x2x6 3 2 2 2 2 2 2 2 3 2 +420x1x2x3x6 + 630x1x2x3x6 + 630x1x2x4x6 + 420x1x3x4x6 3 2 2 3 2 4 2 3 2 4 2 +210x3x4x6 + 420x2x3x4x6 + 105x1x4x6 + 420x1x2x5x6 + 105x3x5x6 2 2 2 2 2 2 2 2 2 2 +1260x2x3x4x5x6 + 630x2x4x5x6 + 1260x1x3x4x5x6 + 630x2x3x5x6 2 2 2 2 2 2 3 2 4 2 +630x1x3x5x6 + 1260x1x2x4x5x6 + 210x1x5x6 + 105x4x5x6 4 3 3 3 2 3 2 3 2 3 +35x1x6 + 140x2x3x6 + 420x2x3x4x6 + 420x1x3x4x6 + 420x1x2x4x6 3 3 3 2 3 2 2 3 +140x2x5x6 + 840x1x2x3x5x6 + 420x1x4x5x6 + 210x4x5x6 3 3 2 4 2 4 3 4 4 +140x3x5x6 + 105x1x2x6 + 105x1x3x6 + 35x4x6 + 210x3x4x5x6 2 4 2 5 5 5 4 2 +105x2x5x6 + 21x3x6 + 42x2x4x6 + 42x1x5x6 + 105x1x2x7 5 3 3 4 4 +42x1x3x7 + 140x2x3x7 + 210x1x2x3x7 + 210x2x3x4x7 2 2 2 3 3 2 2 2 +1260x1x2x3x4x7 + 420x1x3x4x7 + 420x1x2x4x7 + 1260x1x2x3x4x7 3 3 2 4 5 5 3 +140x1x4x7 + 105x3x4x7 + 42x2x4x7 + 42x2x5x7 + 840x1x2x3x5x7 2 2 2 2 3 3 2 +1260x1x2x3x5x7 + 1260x1x2x4x5x7 + 840x1x3x4x5x7 + 420x3x4x5x7 3 4 3 2 4 2 +840x2x3x4x5x7 + 210x1x4x5x7 + 420x1x2x5x7 + 105x3x5x7 2 2 2 2 2 2 2 2 3 +1260x2x3x4x5x7 + 630x2x4x5x7 + 1260x1x3x4x5x7 + 420x2x3x5x7 2 3 3 2 4 5 +420x1x3x5x7 + 840x1x2x4x5x7 + 105x1x5x7 + 42x4x5x7 4 2 2 3 2 3 +210x1x2x6x7 + 1260x1x2x3x6x7 + 420x1x3x6x7 + 840x1x2x4x6x7 4 2 2 2 3 3 +210x3x4x6x7 + 1260x2x3x4x6x7 + 420x2x4x6x7 + 840x1x3x4x6x7 42 4 3 2 +210x1x5x6x7 + 840x2x3x5x6x7 + 2520x2x3x4x5x6x7 2 2 3 2 +2520x1x3x4x5x6x7 + 2520x1x2x4x5x6x7 + 420x2x5x6x7 2 2 2 2 3 4 +2520x1x2x3x5x6x7 + 1260x1x4x5x6x7 + 420x4x5x6x7 + 210x3x5x6x7 2 2 2 3 2 3 2 2 +630x2x3x6x7 + 420x1x3x6x7 + 420x2x4x6x7 + 2520x1x2x3x4x6x7 2 2 2 2 2 2 2 3 2 +630x1x4x6x7 + 1260x1x2x5x6x7 + 1260x1x3x5x6x7 + 420x4x5x6x7 2 2 3 2 2 3 2 3 +1260x3x4x5x6x7 + 420x2x5x6x7 + 420x1x2x6x7 + 420x3x4x6x7 2 3 3 2 3 4 +420x3x5x6x7 + 840x2x4x5x6x7 + 420x1x5x6x7 + 210x2x3x6x7 4 6 2 3 2 3 2 5 2 +210x1x4x6x7 + 7x6x7 + 210x1x2x7 + 420x1x2x3x7 + 21x3x7 4 2 3 2 2 2 2 2 2 2 +105x1x4x7 + 420x2x3x4x7 + 630x2x3x4x7 + 630x1x3x4x7 3 2 2 2 2 3 2 3 2 +420x1x2x4x7 + 630x2x3x5x7 + 420x1x3x5x7 + 420x2x4x5x7 2 2 2 2 2 2 2 2 2 2 +2520x1x2x3x4x5x7 + 630x1x4x5x7 + 630x1x2x5x7 + 630x1x3x5x7 3 2 2 3 2 4 2 3 2 +210x4x5x7 + 420x3x4x5x7 + 105x2x5x7 + 420x2x3x6x7 2 2 2 2 2 2 4 2 +1260x1x2x3x6x7 + 1260x1x2x4x6x7 + 1260x1x3x4x6x7 + 105x4x6x7 2 2 2 2 2 2 2 2 2 +1260x1x2x5x6x7 + 1260x3x4x5x6x7 + 630x3x5x6x7 + 1260x2x4x5x6x7 3 2 3 2 2 2 2 2 2 2 2 +420x1x5x6x7 + 210x1x6x7 + 630x3x4x6x7 + 630x2x4x6x7 2 2 2 2 2 3 2 3 2 +1260x2x3x5x6x7 + 1260x1x4x5x6x7 + 210x2x6x7 + 420x1x3x6x7 4 2 4 3 2 3 2 2 3 2 3 +105x5x6x7 + 35x2x7 + 420x1x2x3x7 + 210x1x3x7 + 420x1x2x4x7 3 3 3 3 2 3 2 3 2 3 +140x3x4x7 + 140x1x5x7 + 420x3x4x5x7 + 420x2x4x5x7 + 420x2x3x5x7 2 3 3 3 3 2 3 +420x1x4x5x7 + 140x3x6x7 + 840x2x3x4x6x7 + 420x1x4x6x7 2 3 3 2 3 2 2 3 3 3 +420x2x5x6x7 + 840x1x3x5x6x7 + 420x1x2x6x7 + 210x5x6x7 + 140x4x6x7 2 4 2 4 4 4 3 4 +105x2x3x7 + 105x2x4x7 + 210x1x3x4x7 + 210x1x2x5x7 + 35x5x7 2 4 4 2 4 2 5 5 +105x1x6x7 + 210x4x5x6x7 + 105x3x6x7 + 21x4x7 + 42x3x5x7 5 6 +42x2x6x7 + 7x1x7 , l2 = (2, 3, 5)(4, 7, 6)l1 , l3 = (2, 4, 3, 7, 5, 6)l1 , l4 = (2, 5, 3)(4, 6, 7)l1 , 6 X l5 = (2, 6, 5, 7, 3, 4)l1 , l6 = (2, 7)(4, 5)(3, 6)l1 , l0 = s1 − lj . j=1