School of Mathematics and Systems Engineering

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Solubility of Polynomials of the Form x6 ax + b

Jenny Löwerot

Oct MSI Report 06150 2006 Växjö University ISSN 1650-2647 SE-351 95 VÄXJÖ ISRN VXU/MSI/MA/E/--06150/--SE

Jenny Löwerot

Solubility of Polynomials of the Form x6 ax + b −

Master´s Thesis

Matematik

2006

Växjö University

Abstract

This thesis is an attempt to find a criterion of when a sixtic equation of the form x6 ax + b = 0, with a, b Z, is soluble. To do this we use Galois theory and a theorem− by Frobenius. We ∈state some different criteria of the coefficients, a and b, which make the sixtic soluble.

Sammanfattning

Det här arbetet är ett försök att finna ett kriterium för när en sjättegradsekvation på formen x6 ax + b = 0, med a, b Z är lösbar. För att göra detta använder − ∈ vi oss av Galoisteori och en sats av Frobenius. Vi visar på några olika kriterier på koefficienterna, a och b, vilka ger lösbarhet.

Acknowledgments

First of all, I would like to thank my supervisor Per-Anders Svensson for introducing me to the subject and for guiding me through the working process. I also want to thank Per-Anders for his help understanding the necessary commands in the programs Mathematica and GP/PARI. Talking about programs, I would like to thank Robert Nyqvist for his help with Mathematica and LATEX. I also want to thank him for his introduction of Chebotarëv’s Density Theorem and the theorem by Frobenius. I would like to thank associate professor Alexander Hulpke at Colorado State University for answering my e-mail quickly and for sorting out the which articles I might need. It helped me understand how humble my own knowledge of Galois groups is. I thank my fiancé Tomas Nilsson for helping me with the installation of the above mentioned programs. I also thank him for reading and commenting the thesis during the writing process. I thank my mother Margareta Löwerot for reading the thesis and helping me find and correct all the faults in the language. She did this without quite liking the subject, or even understanding the purpose of it. That is a fine proof of a mother’s love and support.

iii Contents

1 Introduction 1

2 Basic Group Theory 2

3 Basic Galois Theory 4

4 The Theorem of Frobenius 5

5 Procedure 6

6 Analysis 8 6.1 Types One, Two, and Three ...... 8 6.2 Types Four and Eight ...... 9 6.3 Types Five and Seven ...... 9 6.4 Type Six ...... 10 6.5 Type Nine ...... 11

7 Conclusions 11

A Symbol Index 14

B Program functions 15

C Polynomials and their Galois groups 17

D Output from Mathematica 19

E Graphs 23 E.1 Type One ...... 23 E.2 Type Two ...... 27 E.3 Type Three ...... 28

References 30

iv 1 Introduction The problem of solving polynomial equations has interested mathematicians for ages. According to Stewart [12], the Babylonians had methods for solving some quadratic equations in 1600 BC. The ancient Greeks had other methods for solving quadric equations and their geometric approach also gave them a tool for solving some cubic equations. In AD 1500 a formula for solving cubic equations was found, although it is uncertain who was first. The result was published by Cardano 1545 in Ars Magna, which also contained a method for solving the quartic equation. Algebraic notations of equations were introduced by Descartes in the 17th century ([13], page 96). With modern notation the equations of degree 4 have the solutions shown in table 1.1, where a, b, c, d C. ≤ ∈ Degree Equation Solution 1 ax + b = 0, a = 0 x = b 6 − a 2 a a2 2 x + ax + b = 0 x = 2 4 b −  q − 3 2 3 3 2 3 3 x3 + ax2 + bx + c = 0 y = q + q + p + q q + p r 2 4 27 r 2 4 27 − q 2 3 − − q3 a a −2a +3b 2a −9ab+27c x = y 3 , p = 3 , q = 27 4 3 2 − 4 x + ax + bx + cx + d = 0 too complicated for this presentation Table 1.1: Solutions for polynomial equations

Since it was now possible to solve all polynomial equations of degree 4 by radicals, the next problem was how to solve the quintic equation. In 1770 Lagrange≤ proved that the tricks used to solve equations of a lower degree do not work for the quintic. This arose the suspicion that the quintic equation may not be soluble by radicals. The first person to publish a proof for this was Ruffini. He made a first attempt in 1799 in his book Teoria Generale delle Equazioni and then tried again with a better, but still not accurate, proof in a journal in 1813. In 1824 Abel filled the gap in Ruffini´s work and came up with a less faulty but long proof. Kronecker published in 1879 a simpler proof that there is no formula for solving all quintic equations by radicals. But he did not prove that all quintics cannot be solved by a special formula for each and every one. This led to a new question: How can we see if a special equation can be solved by radicals? In 1843 Liouville wrote to the Academy of Science in Paris that, among the papers of the late Galois, he had found a proof that the quintic is insoluble by radicals. This was the origin of the Galois theory, which will be further explained in the next chapter. After this, mathematicians have tried to figure out which equations are soluble and which are not. This has resulted in a sufficient, but not necessary, condition for when quintics are insoluble. Theorem 1.1. Assume that f = x5 + ax4 + bx3 + cx2 + dx + e Q[x]. Then f is insoluble if f is irreducible over Q and has exactly three real zeros.∈ It has also resulted in theorems like the one on page 376 in Cox [2]. Theorem 1.2. Assume that f = x5 + ax + b F [x], where a = 0, is irreducible and that F has characteristic 0. Then f is soluble∈by radicals over6 F if and only if there are λ, µ F such that ∈ 3125λµ4 3125λµ5 a = , b = . (λ 1)4(λ2 6λ + 25) (λ 1)4(λ2 6λ + 25) − − − − 1 This thesis is an attempt to find something similar for the sixtic equation.

2 Basic Group Theory To fully understand the beauty of the Galois theory we need some basic knowledge of groups and fields and their different operations. This chapter will therefore be somewhat of an enumeration of definitions. Definition 2.1 (Group). A group (G, ) is a set G together with an operation ∗ ∗ such that (i) (G, ) is closed under the operation , meaning that for each pair g, h of elemen∗ ts of G, g h is an element of G∗ ∗ (ii) is associative, meaning that for all elements g, h, j of G, ∗ (g h) j = g (h j) ∗ ∗ ∗ ∗ (iii) there is an element e in G such that for all g in G

e g = g = g e, ∗ ∗ this e is called the identity element (iv) given an element g in G, there is an element g−1 in G such that

g g−1 = e = g−1 g, ∗ ∗ g−1 is called the inverse of g. We call a group abelian if the group operation is commutative, meaning that for ∗ all g and h in G, g h = h g. ∗ ∗ Example 2.1. (Z, +) is an abelian group because for all a Z there is an a−1 = a, ∈ − such that a + ( a) = 0, the identity element. But (N, +) is not a group, since 5 N but 5 / N. − ∈ − ∈ From now on we will use the notation G instead of (G, ) for a group. In abstract algebra there is also another interesting object that we need∗ to define. Definition 2.2 (Field). A field (F, +, ) is a set F with two operations + and , · · called addition and multiplication respectively, such that

(i) (F, +) is an abelian group with identity element 0F , called zero (ii) is associative · (iii) is commutative · (iv) the distributive laws are valid for all a, b, c F , meaning that ∈ a (b + c) = (a b) + (a c) · · · and (a + b) c = (a c) + (b c) · · · 2 (v) there is an identity element 1 = 0 , such that F 6 F 1 f = f 1 = f F · · F for all f F ∈

(vi) all elements except 0F have a multiplicative inverse, meaning that

if f = 0 then there exists f −1 F such that f f −1 = 1 . 6 F ∈ · F From now on we will use the notation F instead of (F, +, ) for a field. · Example 2.2. Q and R are fields. Z is not a field, since 2 Z, but 2−1 = 1 / Z. ∈ 2 ∈ Now we need something to make it easier to work with these new objects.

Definition 2.3 (Isomorphism). An isomorphism between two groups K and L is a map σ : K L such that → σ(a b) = σ(a) σ(b) ∗ ∗ for all a, b K, where σ is one-to-one and onto. If K and L are instead two fields, ∈ the map needs to fullfil σ(a + b) = σ(a) + σ(b) and σ(a b) = σ(a) σ(b) · · for all a, b K. ∈ An isomorphism from a group to itself is called an automorphism.

Definition 2.4 (Extension). Suppose that K and L are fields, such that K L. Then we say that L is a field extension of K and we write L : K. ≤

The next definition will show what this has got do to with polynomials.

Definition 2.5 (Splitting field). Let f F [x] have degree n > 0. Then an extension L : F is a splitting field of f over F if ∈

(i) f = c(x α1) . . . (x α ), where c F and α L, i = 1 . . . n, and − · · − n ∈ i ∈

(ii) L = F (α1, . . . , αn)

This means that all zeros of f lie in L and that L is the smallest extension of F , containing all the zeros of f.

Now we are getting close to the core of the Galois theory, the definition of the Galois group. But first we need another definition and a theorem.

Definition 2.6 (K-automorphism). Let L : K be a field extension, so that K is a subfield of the subfield L of C. A K-automorphism of L is an automorphism σ of L such that σ(k) = k for all k K. (2.1) ∈ We say that σ fixes k K if (2.1) holds. ∈ 3 Theorem 2.1. If L : K is a field extension, then the set of all K-automorphisms of L forms a group under composition of maps.

Proof of Theorem 2.1. Suppose that σ and ϕ are K-automorphisms of L. Then σ ϕ must be an automorphism, and if k K we have σ ϕ(k) = σ(k) = k, so σ ϕ◦is ∈ ◦ ◦ a K-automorphism. Composition of maps is associative. The identity map on L is clearly a K-automorphism. Obviously σ−1 is an automorphism of L, and for any k K we have ∈ −1 −1 k = σ σ(k) = σ (k), ◦ meaning that σ−1 is a K-automorphism. Thus, the set of all K-automorphisms of L is a group.

3 Basic Galois Theory Definition 3.1 (Galois group of extension). The Galois group Γ(L : K) of a field extension L : K is the group of all K-automorphisms of L under the operation of composition of maps.

Definition 3.2 (Galois group of polynomial). Let f F [x]. The Galois group of f over F is Γ(L : F ), where L is a splitting field of f ov∈er F .

Example 3.1. Let f Q[x] and f(x) = x2 3x + 2. The splitting field of f over Q is Q because f(x) = (x∈ 1)(x 2). The only−Q-automorphism that fixes all elements in Q is the trivial automorphism.− − This means that the Galois group Γ(Q : Q) of f over Q is the trivial group with only one element, namely the identity element.

Example 3.2. Let g Q[x] and g(x) = x2 2x + 3. The polynomial g does ∈ − not split in Q. Since g(x) = x (1 + i√2) x (1 i√2) , the splitting field − − − of g over Q is Q(i√2). There are two Q-automorphisms
of
Q(i√2). These are σ(a + i√2b) = a + i√2b and ϕ(a + i√2b) = a i√2b, where a, b Q. This means that the Galois group Γ(Q(i√2) : Q) of g ov−er Q has two elemen∈ts, actually Γ(Q(i√2) : Q) = S2.

According to Theorem 2.1 the Galois group is well defined. But what has this got do with polynomial equations? You have to wait just a little bit more to figure out that. Now we will look at an interesting concept.

Definition 3.3 (Solubility). A group G is soluble if it has a finite series of subgroups

1 = G0 G1 G = G { G} ⊆ ⊆ · · · ⊆ n such that

(i) G C G +1 for i = 0, , n 1 i i · · · −

(ii) G +1/G is abelian for i = 0, , n 1. i i · · · − Theorem 3.1. Let G be a soluble group. Then all subgroups of G are soluble.

Theorem 3.2. Let f be a polynomial over a subfield K of C. If and only if f is soluble by radicals, then the Galois group of f over K is soluble.

4 For proofs of Theorem 3.1 and Theorem 3.2 look in [12], pages 145 and 158. Now we can see the beauty of the Galois theory. Instead of looking at polynomials we can look at the polynomials’ Galois groups. If we can find one polynomial of degree 5 that is not soluble we know that the quintic equation is insoluble. The same is valid for the sixtic. Now it should be fairly easy to calculate the Galois groups of the sixtic equation and see when it is soluble? No, all known methods for calculating Galois groups depend on knowing the roots of the equation, se [12], and Hulpke [3] and [4]. But all we know are the coefficients. The sixtic equation with roots α1, α2, α3, α4, α5 and 6 5 4 3 2 α6 can be written as x s1x + s2x s3x + s4x s5x + s6 where − − −

s1 = α1 + α2 + α3 + α4 + α5 + α6

s2 = α1α2 + α1α3 + α1α4 + α1α5 + α1α6 + α2α3 + α2α4 + α2α5 +

+ α2α6 + α3α4 + α3α5 + α3α6 + α4α5 + α4α6 + α5α6

s3 = α1α2α3 + α1α2α4 + α1α2α5 + α1α2α6 + α1α3α4 + α1α3α5 + α1α3α6 +

+ α1α4α5 + α1α4α6 + α1α5α6 + α2α3α4 + α2α3α5 + α2α3α6 + α2α4α5 +

+ α2α4α6 + α2α5α6 + α3α4α5 + α3α4α6 + α3α5α6 + α4α5α6

s4 = α1α2α3α4 + α1α2α3α5 + α1α2α3α6 + α1α2α4α5 + α1α2α4α6 +

+ α1α2α5α6 + α1α3α4α5 + α1α3α4α6 + α1α3α5α6 + α1α4α5α6 +

+ α2α3α4α5 + α2α3α4α6 + α2α3α5α6 + α2α4α5α6 + α3α4α5α6

s5 = α1α2α3α4α5 + α1α2α3α4α6 + α1α2α3α5α6 +

+ α1α2α4α5α6 + α1α3α4α5α6 + α2α3α4α5α6

s6 = α1α2α3α4α5α6

If it had been easy to find the roots just by looking at the coefficients we would not need any Galois theory. To make it easier we instead look at equations of the form 6 x ax + b = 0, thus s1 = s2 = s3 = s4 = 0. Solving this system gives us −

5 4 3 2 2 3 4 5 a = α5 + α5α6 + α5α6 + α5α6 + α5α6 + α6 4 3 2 2 3 4 b = α5α6(α5 + α5α6 + α5α6 + α5α6 + α6).

Knowing one root will still mean solving a quintic equation, which may not be soluble, and it is not even sure that we can find that first root. This means we have to choose another way to solve the problem.

4 The Theorem of Frobenius We stated in the previous chapter that we need another way to see which sixtic equations are soluble. In this chapter we will explain this other way. First we need to define something called density.

Definition 4.1 (Density). Let S be a set of prime numbers. We say that S has the natural density δ if p x : p S | ≤ ∈ | δ for x . p x : p prime → → ∞ | ≤ | 5 If S has a natural density it also has an analytic density equal to δ if

−1 1 1 δ for s 1 + .  ps  ps  → → Xp∈S p Xprime

The opposite is not true.

We can now state the theorem by Frobenius.

Theorem 4.1 (Theorem of Frobenius). Let f(x) Z[x] and assume that f(x) is monic and that ∆(f) = 0. Let G be the Galois∈group of f over Q. Then the 6 density of the set of primes for which f has a given decomposition type n1, n2, . . . , nt exists, and is equal to 1/ G times the number of elements in G with cycle pattern | | (n1, n2, . . . , nt).

Because it is not obvious what the theorem means we will explain it. Suppose that f is a monic polynomial with integer coefficients and degree n. Suppose further that ∆(f) = 0, meaning that f has n distinct zeros α1, α2, . . . , α in an extension 6 n field K = Q(α1, α2, . . . , α ) of Q. This means that each σ Γ(K : Q) permutes the n ∈ zeros α1, α2, . . . , αn and is uniquely determined by the way it permutes these. Hence, Γ(K : Q) is a subgroup of Sn. If we write an element σ Γ(K : Q) as a product of disjoint cycles (including cycles of length 1) and look at∈the length of these cycles, we obtain the cycle pattern of σ, which is a partition n1, n2, . . . , nt of n. If p is a prime number not dividing ∆(f) we can write f modulo p as a product of irreducible factors over Zp. The degrees of these irreducible factors form the decomposition type of f modulo p and this is also a partition of n. What Theorem 4.1 says is that the number of primes with a given decomposition type is proportional to the number of elements in the Galois group with the same cycle pattern. All according to [6]. There is a somewhat similar theorem about integers that might be easier to comprehend.

Theorem 4.2 (Theorem of Dirichlet). Let n be a positive integer. Then for each integer a with gcd(a, m) = 1 the set of prime numbers p with p a mod m has density 1/φ(n). ≡

Theorem 4.2 of Dirichlet and Theorem 4.1 of Frobenius will combined lead to Chebotarëv’s Density Theorem and are therefore described in literature about this.

5 Procedure Because of Theorem 4.1 we can instead of studying just the polynomial take the polynomial modulo a large number of primes and by looking at the irreducible factors figure out which group is the polynomial’s Galois group. To do this we use a computer and an algorithm written in Mathematica. We first need an algorithm to calculate the discriminant of a polynomial. The function in Figure B.1 is taken from Weisstein [15]. We can now take a look at the function ”Frobenius” in Figure B.3. Under Module we have all the variables in the function. FactorInteger gives all the divisors of the discriminant, including 1. Therefore we need to substract the length with 1 to get the number of true divisors. We then need to know the number of primes among the n that do not divide the discriminant of the polynomial, this

6 is m. We then look at the n first primes p, and if p does not divide the discriminant we take the polynomials coefficients modulo p. We then look at the factors of this new polynomial and add 1 to the factor counting variables that they match. When we have gone through all of the n first primes we get a table as result. Now we can use this algoritm on different polynomials to get an idea of how their Galois groups look. But first we need some interesting candidates. We find these with PARI/GP and the function in Figure B.2. With the first if-statement we asort polynomials of the types x6 ax and x6 b because we know that these are soluble. − − The next if-statement is necessary because the command ”polgalois” only works on polynomials that are irreducible over Q. This is also why we need the lines 9 through 13. The only possible group with 720 elements is S6 ( Sn = n!, and 6! = 720), therefore we do not need to look at polynomials with a |Galois| group of order 720. We neither need to look at polynomials with an irreducible factor (over Q) with a group of order 120, because the only subgroup of S6 with 120 elements is S5 and this is not soluble. Notice that if S5 and S6 had been soluble all quintic and sixtic equations would be soluble by Theorem 3.1. As output we get ten different types of polynomials, they are listed in Table C.1 and Table C.2. GP/PARI gives us the names of the Galois groups but we can only be sure that they are correct for the polynomials that are irreducible over Q. When a polynomial splits in Q, GP/PARI gives us several Galois groups but these are the Galois groups of the factors of the polynomial. To get the Galois group of the original polynomial we need to use our Mathematica program. Because we want to know the properties of the different Galois groups we run our program on one polynomial from each group. The polynomials that are marked with bold letters are the ones that we try the Mathematica program on. Tables D.1 to D.10 show the results from the function ”Frobenius” with n = 10,000,000, meaning that the function iterates over the 10 million first prime num- bers. We need this many prime numbers to get an accurate result because the convergence of the function is very slow. Remember that we use the frequency ( R) of primes for the different factor patterns to get the number ( Z) of elements ∈ ∈ in the Galois group with a certain cycle pattern. Going from reals to integers we need great precision. We can see that the function is not giving us any Galois group to type six. This is because both x6 + 6x + 5 and x6 6x + 5 have double zeros. The polynomials x6 6x + 5 split into x4 2x3 + 3x−2 4x + 5 and (x 1)2 in Q. This however means∓ that they are soluble because quartics are soluble.∓To decide whether or not the other types have soluble groups we use a theorem from [5].

Theorem 5.1. Every group with less than 60 elements is soluble.

Theorem 5.1 tells us that all polynomials from types two, tree, four, five, seven, eight, and nine are soluble. The Galois group of type ten is A6 and this is not soluble, according to Theorem 5.3, which can be found in [12].

Theorem 5.2. A soluble group is simple if and only if it is cyclic of prime order.

Theorem 5.3. If n 5, then the alternating group A is simple. ≥ n The polynomials of type one have Galois groups of order 72. Theorem 5.1 gives us no help here. We need another theorem and the wellknown Theorem 5.4 is just what we are looking for.

7 Theorem 5.4 (Burnside’s pαqβ-theorem). Let G be a finite group with order pαqβ, where p and q are primes and α, β 0. Then G is soluble. ≥ Because 72 = 23 32 we have that any group of order 72 must be soluble according to · Theorem 5.4. Therefore the polynomials of type one are soluble. This means that all the polynomials of types one, two, tree, four, five, six, seven, eight, and nine are soluble. We have here 20, 2, 4, 22, 2, 2, 2, 2, and 2 polynomials respectively from each of these types. Meaning that out of 200 200 = 40,000 polynomials of the form x6 ax + b, a, b Z, not counting those with· a = 0 or b = 0, we have only 56 − ∈ polynomials that are soluble. This would mean that approximately 1.4 percent of all polynomials of the form x6 ax + b, a, b Z are soluble. Notice however that we here assume that neither a −nor b vanish. If∈ we let a = 0 or b = 0 we will get a lot more soluble polynomials. Notice also that the study is not statistically correct, since the samples are not random.

6 Analysis This chapter will show an attempt to analyse the different types of soluble sixtics of the form x6 ax + b and hopefully give some explanations of why they are soluble. − 6.1 Types One, Two, and Three We have listed the graphs of the polynomials of types one, two, and three in Ap- pendix E. As we can see, the polynomials of type one have either two real zeros or none, the polynomials of type two have two real zeros, and the polynomials of type three have no real zeros. Although, this is too little material to say anything definite. We can however do a general analysis. Since

f(x) = x6 ax + b , when x − → ∞ → ∞ and f 0(x) = 6x5 a, with one real zero, − f(x) has two real zeros if and only if f 0(β) = 0 and f(β) < 0 for some β R. If 0 ∈ f (β) = 0 and f(β) = 0 for some β R, then β is a multiple root of f. We have ∈ a a f 0(β) = 0 β5 = β = 5 , ⇔ 6 ⇔ r6 which leads to 6 5 a / f(β) = 5 + b − · 6 and 6 5 5 a / 6 √6 b f(β) < 0 b < 5 · · < (√5 a)6. ⇔ · 6 ⇔ 5 We also have b 0 f(β) 0. This gives us the following theorem. ≤ ⇒ ≤ Theorem 6.1. Let f(x) = x6 ax + b, where a, b Q and a = 0, b = 0. Then f(x) has exactly two real zeros if and− only if b < 0 or ∈6 √5 6 b <6 5 (√56 a)6. Otherwise f(x) has only complex zeros. · · ·

We give one example from each type.

8 Example 6.1. Let f(x) = x6 10x + 5. We have b = 5 and a = 10 > 6. This means that 6 √5 6 b < 5 a √5 −a and f(x) has two real zeros. · · · · Example 6.2. Let f(x) = x6 + 49x 49. We have b < 0 and f(x) has two real − zeros. Example 6.3. Let f(x) = x6 3x + 5. We have b = 5 and a = 3 < 6. This means 5 5 − that 6 √6 b > 5 a √a and f(x) has no real zeros. · · · · 6.2 Types Four and Eight The polynomials of type four split into a quadratic and a quartic. The polynomials of type eight split into a linear, a quadratic, and a cubic. If we disregard that the polynomials of type eight have an integer zero we can conclude that a polynomial p(x) belongs to either type four or type eight if it can be written in the form p(x) = x6 ax + b, where a = m5 + 4m3n 3mn2 and b = m4n 3m2n2 + n3, and − − 2 − 4 3 −2 m, n Z. We then have p(x) = (x + mx + n) (x + cx + dx + ex + f), where m/2∈ m2/4 n are the zeros of the quadratic,· and −  − p c = m −2  d = m n −3  e = m + 2mn  −4 2 2 f = m 3m n + n . −  m m2  m2 Notice that 2 4 n belong to C, but that they are real if n < 4 . On the −  q − m m2 other hand, 2 4 n / Z because then p(x) would belong to another type. −  q − ∈ 6.3 Types Five and Seven The polynomials of types five and seven have two integer zeros, neither of them equal to zero. This occurs when a = m5 + m4n + m3n2 + m2n3 + mn4 + n5 and b = m5n + m4n2 + m3n3 + m2n4 + mn5, where m, n Z. Then x6 ax + b = (x m)(x n)(x4 + cx3 + dx2 + ex + f), where ∈ − − − c = m + n  d = m2 + mn + n2  3 2 2 3  e = m + m n + mn + n f = m4 + m3n + m2n2 + mn3 + n4.  However, since m >2 or n > 2 would make a and b too high for our study, types five and sev| en| actually| only| includes two sp|ecial| cases.| | Namely what happens when n m = 1. || | − | || 6.3.1 Type Five Belonging to type five are the polynomials where the two integer zeros have different signs. Meaning that we can assume that n = (m + 1), where either m > 0 or m < 1. We then have x6 ax + b = (x + m)(x −(m + 1))(x4 + cx3 + dx2 + ex + f), where− − − a = 3m4 + 6m3 + 7m2 + 4m + 1  b = m6 3m5 6m4 7m3 4m2 m  − − − − − −  c = 1  2  d = m + m + 1 e = 2m2 + 2m + 1  4 3 2  f = m + 2m + 4m + 3m + 1.   9 6.3.2 Type Seven Belonging to type seven are the polynomials where the two integer zeros have the same sign. Meaning that if we have n = m + 1, then either m > 0 and n > 1, or m < 1 and n < 0. We then have − x6 ax + b = (x m)(x (m + 1))(x4 + cx3 + dx2 + ex + f), − − − where a = 6m5 + 15m4 + 20m3 + 15m2 + 6m + 1  b = 5m6 + 15m5 + 20m4 + 15m3 + 6m2 + m   c = 2m + 1  2  d = 3m + 3m + 1 e = 4m3 + 6m2 + 4m + 1  4 3 2  f = 5m + 10m + 10m + 5m + 1.   6.4 Type Six We have already mentioned that the polynomials of this type have double zeros in Q. If a polynomial has a double zero at some point, then its derivative has a zero at the same point. This means that for polynomials with double zeros we get the following system of equations α6 aα + b = 0 − 5  6α a = 0 − where α is the double zero. The solution to this system is a = 6α5  b = 5α6 and we know that α Q. ∈ Lemma 6.2. The α mentioned above is an integer. Proof of Lemma 6.2. We know that 6α5, 5α6 Z. Assume that α = m/n, where m, n Z, n = 1 and gcd(m, n) = 1. This means∈ that α5 = m5/n5 and α6 = m6/n6. Since∈gcd(m,6 n) = 1, n5 and n6 must be the lowest integers that multiplied with α5 and α6 respectively give an integer. This combined with what we know means that n5 6 and n6 5. This goes against the fact that n Z and n = 1. Our assumption is |false and |α must be an integer. ∈ 6  Now we know that sixtics of the form x6 6α5x + 5α6 equal p = (x α)2 times q = x4 + 2αx3 + 3α2x2 + 4α3x + 5α4, where− α Z. According to Mathematica,− q has only complex zeros. This is in accordance ∈with what we have seen earlier in Theorem 6.1. Thus polynomials of type six are reducible in Q and have two real zeros. Notice that there cannot be any sixtics of the form x6 ax + b, a = 0, b = 0 with triple zeros. A triple zero means − 6 6 α6 aα + b = 0 − 5  6α a = 0  30−α4 = 0 with the solution a = b = α = 0. It is however possible for the general sixtic to have multiple zeros of higher multiplicity than two.

10 6.5 Type Nine The polynomials of type nine split into two cubics. A polynomial p(x) belonging to 6 1 6 4 2 2 3 type nine can be written in the form x ax+b, where a = 2r (r 3r s+3r s 2s ) 1 8 6 4 2 2 3 − 4 − − and b = 4r2 (r 4r s + 3r s + 2r s s ), where r and s are even integers. We then have p(x) =− (x3 + rx2 + sx + 1 ( −r4 + 3r2s s2)) (x3 + cx2 + dx + e), where 2r − − · c = r −2  d = r s  −r4−+r2s+s2 e = 2r .  Actually, for the two polynomials that we know of type nine, s = 0. This means 3 that p(x) = x6 ax + b = (x3 + rx2 r ) (x3 + cx2 + dx + e), where − − 2 · r5 a = 2 r6  b = 4   c = r  −2  d = r r3  e = 2 .  −  7 Conclusions Now we can state a theorem that gives sufficient, but not necessary, conditions for solubility of sextics of the form x6 ax + b. − Theorem 7.1. A polynomial p(x) = x6 ax + b is soluble if it is reducible in Q and − (i)

a = m5 + m4n + m3n2 + m2n3 + mn4 + n5 and b = m5n + m4n2 + m3n3 + m2n4 + mn5,

where m, n Z, or ∈ (ii)

a = m5 + 4m3n 3mn2 − − and b = m4n 3m2n2 + n3, − where m, n Z, or ∈ (iii) 1 a = (r6 3r4s + 3r2s2 2s3) 2r − − and 1 b = (r8 4r6s + 3r4s2 + 2r2s3 s4), 4r2 − − where r and s are even integers. Proof of Theorem 7.1. We look at the different conditions one by one.

11 (i) We have two cases; if

m = n, we have 6 p(m) = m6 (m5 + m4n + m3n2 + m2n3 + mn4 + n5) m − · + m5n + m4n2 + m3n3 + m2n4 + mn5 = 0 and p(n) = n6 (m5 + m4n + m3n2 + m2n3 + mn4 + n5) n − · + m5n + m4n2 + m3n3 + m2n4 + mn5 = 0,

meaning that we have got two zeros, m and n, and can reduce p(x) to a quartic which we know is soluble m = n, we have p(x) = x6 6n5x + 5n6 with p(n) = n6 6n6 + 5n6 = 0, actually 2 −4 3 2 2 3 − 4 p(x) = (x n) (x + 2nx + 3n x + 4n x + 5n ) and we know that quartics are−soluble.·

(ii) We have again two cases; if

m m2 m = n, assume that A = 2 4 n, we then have 6 −  q − p(A) = A6 ( m5 + 4m3n 3mn2) A − − − · + m4n 3m2n2 + n3 − = 0,

2 2 meaning that we have got two zeros, m + m n and m m n, − 2 4 − − 2 − 4 − and can reduce p(x) to a quartic which weqknow is soluble q m = n, we have p(x) = x6 (n5 4n4 + 3n3) x + n5 3n4 + n3 and − − · − 6 n √n √n 4 n √n √n 4 p −  · − = −  · −  2   2  n √n √n 4 (n5 4n4 + 3n3) −  · − − − ·  2  + n5 3n4 + n3 = 0, −

meaning that we have got two zeros, namely 1 ( n + √n √n 4) and 2 − · − 1 ( n √n √n 4), and can reduce p(x) to a quartic which we know 2 − − · − is soluble.

(iii) Assume that

1 q(x) = x3 + rx2 + sx + ( r4 + 3r2s s2)  2r − −  and r4 + r2s + s2 t(x) = x3 rx2 + (r2 s)x + − .  − − 2r 

12 Then ( 2r7 + 6r5s 6r3s2 + 4rs3) x q(x) t(x) = x6 + − − · · 4r2 r8 4r6s + 3r4s2 s4 + 2r2s3 + − − 4r2 (r6 3r4s + 3r2s2 2s3) x = x6 − − · − 2r r8 4r6s + 3r4s2 s4 + 2r2s3 + − − 4r2 = p(x).

This means that p(x) splits into two cubics and we know that cubics are soluble. Therefore p(x) must be soluble.

We have now proved all conditions. We have not been able to find something similiar to Theorem 1.1 for the sixtic equation. A sixtic equation cannot have an odd number of real zeros because true complex zeros always come in pairs, one being the conjugate of the other. Actually we have seen that a sixtic of the form x6 ax+b, with a, b Z, might be irreducible over Q and have two or no real zeros, but−still be soluble, see∈ section 6.1. The reason why Theorem 1.1 can not so easily be changed to fit the sixtic is that in the proof of Theorem 1.1 the fact that 5 is a prime number is being used. Theorem 1.1 can be modified to work for all prime numbers (with another number of real zeros, of course) but it does not work with the sixtic.

13 A Symbol Index Symbol Meaning N the set of all nonnegative integers, N = 0, 1, 2, . . . Z the set of all integers, is also a group {under addition,} Z = . . . , 1, 2, 0, 1, 2, . . . { − − } Q the set/field of all rational numbers, Q = a/b : a, b Z, b = 0 R the set/field{ of ∈all real6 num}bers C the set/field of all complex numbers, C = a + bi : a, b R, i2 = 1 { ∈ − } (G, ) G is a group (F, +∗, ) F is a field σ, ϕ· isomorphisms K L K is a subfield of L; K L and K is a field under the ≤ operations of L ⊆ L : K L is a field extension of K F [x] the set of all polynomials in x with coefficients in the field F F (α1, . . . , αn) the field F with the elements α1, . . . , αn added Γ(L : K) Galois group of L : K N C G N is a normal subgroup of G, meaning that gN = Ng for all g G ∈ G/N quotient group of G modulo N S the number of elements in a set, group, or field S, also called | | the order of that set, group, or field ∆(f) the discriminant of the polynomial f S the permutation group of n symbols, S = n! n | n| Zp the set of all residues modulo p, meaning Z = 0, 1, 2, . . . , p 1 p { − } gcd(a, b) greatest common divisor of a and b, the largest integer that divides both a and b a b a divides b, there is an integer n such that a n = b | · a b mod m a is congruent to b modulo m, meaning m (a b) ≡ φ(n) the Euler φ-function of n, the number of p|ositiv−e integers not exceeding n that are relatively prime to n A the alternating group of degree n, A = 1 n! n | n| 2

14 B Program functions The functions used to calculate with polynomials in this thesis.

Discriminant [p_?PolynomialQ , x_] := With[ { n = Exponent [ p , x ] , k = Exponent [D[ p , x ] , x ] } , Cancel [(( 1)^( n (n 1)/2) Resultant−[ p , D[ p , −x ] , x ] ) Coefficient [ p , x , n ] ^ ( n k 2 ) ] − − ] Figure B.1: Function to calculate the discriminant of a polynomial

{ for ( a= 100,100, for (b=− 100,100, − i f (b!=0&a !=0 , i f ( p o l i s i r r e d u c i b l e ( x^6 a x+b ) , − ∗ i f ( p o l g a l o i s ( x^6 a x+b )[1] <720 , w r i t e ( i n t e r e s t i n−g ∗, x^6 a x+b , p o l g a l o i s ( x^6 a x+b ) ) ) , − ∗ − ∗ p=x^6 a x+b ; v=f a c−t o ∗r (p ) ; nxk=matsize ( v ) ; n=nxk [ 1 ] ; f a k t=v e c t o r (n , i , v [ i , 1 ] ) ; for ( i =1,n , i f (n>=2&p o l g a l o i s ( f a k t [ 2 ] ) [ 1 ] ! = 1 2 0 , w r i t e ( i n t e r e s t i n g , x^6 a x+b , p o l g a l o i s ( f a k t [ i ] ) ) − ∗ ) ) ) ) ) ) } Figure B.2: Function to calculate the Galois group of a polynomial

15 Frobenius [ a_, b_, n_] := Module [ { polynom , fact6 , fact51 , fact42 , fact411 , fact33 , fact321 , fact3111 , fact222 , fact2211 , fact21111 , fact111111 , p , table , m, d e v i s o r s , disc , exponents } , polynom = x^6 a x + b ; d i s c = Discrim−ina∗nt [ polynom , x ] ; d e v i s o r s = Length [ FactorInteger [ d i s c ] ] 1 ; m = n d e v i s o r s ; − f a c t 6 −= f a c t 5 1 = f a c t 4 2 = f a c t 4 1 1 = f a c t 3 3 = f a c t 3 2 1 = = f a c t 3 1 1 1 = f a c t 2 2 2 = f a c t 2 2 1 1 = = fact21111 = fact111111 = 0 ; Do[ p = Prime [ i ] ; I f [Mod[ disc , p ] != 0 , exponents = Exponent [ Drop[ FactorList [ polynom , Modulus > p ] − [ [ All , 1 ] ] , 1 ] , x ] ; Which[ exponents == {1 , 1 , 1 , 1 , 1 , 1} , fact111111++, exponents == {1 , 1 , 1 , 1 , 2} , fact21111 ++, exponents == {1 , 1 , 2 , 2} , f a c t 2 2 1 1++, exponents == {2 , 2 , 2} , f a c t 2 2 2++, exponents == {1 , 1 , 1 , 3} , f a c t 3 1 1 1++, exponents == {1 , 2 , 3} , f a c t 3 2 1++, exponents == {3 , 3} , f a c t 3 3 ++, exponents == {1 , 1 , 4} , f a c t 4 1 1++, exponents == {2 , 4} , f a c t 4 2 ++, exponents == {1 , 5} , f a c t 5 1 ++, exponents == {6} , f a c t 6 ++] ] , { i , n} ] ; t a b l e = {{6 , fact6 , f a c t 6 /m // N} , {51 , fact51 , f a c t 5 1 /m // N} , {42 , fact42 , f a c t 4 2 /m // N} , {411 , fact411 , f a c t 4 1 1 /m // N} , {33 , fact33 , f a c t 3 3 /m // N} , {321 , fact321 , f a c t 3 2 1 /m // N} , {3111 , fact3111 , f a c t 3 1 1 1 /m // N} , {222 , fact222 , f a c t 2 2 2 /m // N} , {2211 , fact2211 , f a c t 2 2 1 1 /m // N} , {21111 , fact21111 , fact21111 /m // N} , {111111 , fact111111 , fact111111 /m // N} } ; TableForm [ table , TableHeadings > {Automatic , − {" degree ␣ o f ␣ f a c t o r s " , "number" , " frequency " } } ] ] Figure B.3: ”Frobenius” function

16 C Polynomials and their Galois groups The result from running the function in Figure B.2.

Polynomial Galois group Type one 6 2 x + 65x + 13 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 56x + 62 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 45x + 55 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 44x + 55 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 40x + 82 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 30x + 93 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 10x + 5 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 8x + 20 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 8x + 89 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x + 3x + 3 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − 2 x − 3x + 3 [72, −1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 2 x 8x + 20 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 8x + 89 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 10x + 5 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 30x + 93 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 40x + 82 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 44x + 55 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 45x + 55 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 56x + 62 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] 6 − − 2 x 65x + 13 [72, 1, 1, ”F36(6) : 2 = [S(3) ]2 = S(3)wr2”] − − Type two 6 3 x + 49x 49 [24, 1, 2, ”2A4(6) = [2 ]3 = 2wr3”] 6 − − 3 x − 49x − 49 [24, −1, 2, ”2A4(6) = [2 ]3 = 2wr3”] Type tree 6 3 x + 14x + 35 [48, 1, 1, ”2S4(6) = [2 ]S(3) = 2wrS(3)”] 6 − 3 x + 3x + 5 [48, 1, 1, ”2S4(6) = [2 ]S(3) = 2wrS(3)”] 6 − 3 x − 3x + 5 [48, −1, 1, ”2S4(6) = [2 ]S(3) = 2wrS(3)”] 6 3 x 14x + 35 [48, 1, 1, ”2S4(6) = [2 ]S(3) = 2wrS(3)”] − − Table C.1: Output from GP/PARI

C.0.1 How to read the output GP/PARI gives us as output a vector with four elements. The first three elements are numbers while the fourth is a line of symbols. The first number is the order of the Galois group. The second number is 1 if the Galois group is a subgroup of An, otherwise it is 1. The third number is an internal code in GP/PARI. The fourth element is the name− of the Galois group.

17 Polynomial Galois groups Polynomial Galois group(s) Type four Type five x6 + 80x + 24 [2, 1, 1, ”S2”] x6 + 21x 22 [1, 1, 1, ”S1”] x6 + 80x + 24 [24−, 1, 1, ”S4”] x6 + 21x − 22 [1, 1, 1, ”S1”] 6 − 6 − x + 70x 29 [2, 1, 1, ”S2”] x + 21x 22 [24, 1, 1, ”S4”] x6 + 70x − 29 [24−, 1, 1, ”S4”] x6 − 21x−− 22 [1, 1−, 1, ”S1”] x6 + 45x − 44 [2, −1, 1, ”S2”] x6 − 21x − 22 [1, 1, 1, ”S1”] 6 − − 6 x + 45x 44 [24, 1, 1, ”S4”] x − 21x − 22 [24, −1, 1, ”S4”] x6 + 40x − 57 [2, −1, 1, ”S2”] Type six x6 + 40x − 57 [24−, 1, 1, ”S4”] x6 + 6x + 5 [1, 1, 1, ”S1”] 6 − − 6 x + 33x + 20 [2, 1, 1, ”S2”] x + 6x + 5 [24, 1, 1, ”S4”] x6 + 33x + 20 [24−, 1, 1, ”S4”] x6 − 6x + 5 [1, 1−, 1, ”S1”] 6 − 6 x + 22x 95 [2, 1, 1, ”S2”] x − 6x + 5 [24, −1, 1, ”S4”] x6 + 22x − 95 [24−, 1, 1, ”S4”] Type seven x6 + 16x +− 3 [2, −1, 1, ”S2”] x6 + 63x + 62 [1, 1, 1, ”S1”] 6 − 6 x + 16x + 3 [24, 1, 1, ”S4”] x + 63x + 62 [1, 1, 1, ”S1”] x6 + 10x 33 [2, −1, 1, ”S2”] x6 + 63x + 62 [8, 1, 1, ”D(4)”] x6 + 10x − 33 [24−, 1, 1, ”S4”] x6 − 63x + 62 [1,−1, 1, ”S1”] 6 − − 6 x + 8x 8 [2, 1, 1, ”S2”] x − 63x + 62 [1, 1, 1, ”S1”] x6 + 8x − 8 [24−, 1, 1, ”S4”] x6 − 63x + 62 [8, −1, 1, ”D(4)”] 6 − − x + 8x 5 [2, 1, 1, ”S2”] Type eight x6 + 8x − 5 [24−, 1, 1, ”S4”] x6 + 56x + 55 [1, 1, 1, ”S1”] x6 + 5x − 2 [2, −1, 1, ”S2”] x6 + 56x + 55 [2, 1, 1, ”S2”] 6 − − 6 − x + 5x 2 [24, 1, 1, ”S4”] x + 56x + 55 [6, 1, 1, ”A3”] x6 5x − 2 [2, −1, 1, ”S2”] x6 − 56x + 55 [1,−1, 1, ”S1”] 6 − − − 6 x 5x 2 [24, 1, 1, ”S4”] x − 56x + 55 [2, −1, 1, ”S2”] x6 − 8x − 8 [2, −1, 1, ”S2”] x6 − 56x + 55 [6, −1, 1, ”A3”] x6 − 8x − 8 [24−, 1, 1, ”S4”] Type nine 6 − − − 6 x 8x 5 [2, 1, 1, ”S2”] x + 16x + 16 [6, 1, 1, ”A3”] x6 − 8x − 5 [24−, 1, 1, ”S4”] x6 + 16x + 16 [6, −1, 1, ”A3”] x6 − 10x− 33 [2, −1, 1, ”S2”] x6 − 16x + 16 [6,−−1, 1, ”A3”] 6 − − − 6 x 10x 33 [24, 1, 1, ”S4”] x − 16x + 16 [6, −1, 1, ”A3”] x6 −− 16x−+ 3 [2, −−1, 1, ”S2”] Type ten x6 − 16x + 3 [24, −1, 1, ”S4”] x6 + 48x 80 [360, 1, 1, ”A6”] x6 22x 95 [2, 1, 1, ”S2”] x6 + 30x +− 25 [360, 1, 1, ”A6”] x6 − 22x − 95 [24−, 1, 1, ”S4”] x6 + 24x 20 [360, 1, 1, ”A6”] 6 − − − 6 − x 33x + 20 [2, 1, 1, ”S2”] x 24x 20 [360, 1, 1, ”A6”] x6 − 33x + 20 [24−, 1, 1, ”S4”] x6 −− 30x−+ 25 [360, 1, 1, ”A6”] x6 − 40x 57 [2, −1, 1, ”S2”] x6 48x 80 [360, 1, 1, ”A6”] 6 − − − − − x 40x 57 [24, 1, 1, ”S4”] x6 − 45x − 44 [2, −1, 1, ”S2”] 6 − − − x 45x 44 [24, 1, 1, ”S4”] x6 − 70x − 29 [2, −1, 1, ”S2”] x6 − 70x − 29 [24−, 1, 1, ”S4”] 6 − − − x 80x + 24 [2, 1, 1, ”S2”] x6 − 80x + 24 [24−, 1, 1, ”S4”] − −

Table C.2: Output from GP/PARI

18 D Output from Mathematica The output when using the function in Figure B.3 with n = 10,000,000.

Polynomial: x6 3x + 3 degree of factors frequency− number of elements 6 0.166654 12 5, 1 0.0 – 4, 2 0.250068 18 4, 1, 1 0.0 – 3, 3 0.0555712 4 3, 2, 1 0.166652 12 3, 1, 1, 1 0.0555191 4 2, 2, 2 0.0833274 6 2, 2, 1, 1 0.124987 9 2, 1, 1, 1, 1 0.0833658 6 1, 1, 1, 1, 1, 1 0.0138555 1 Order of Galois group: 72 Table D.1: Type one

Polynomial: x6 49x 49 − − degree of factors frequency number of elements 6 0.33336 8 5, 1 0.0 – 4, 2 0.0 – 4, 1, 1 0.0 – 3, 3 0.333315 8 3, 2, 1 0.0 – 3, 1, 1, 1 0.0 – 2, 2, 2 0.0416643 1 2, 2, 1, 1 0.125036 3 2, 1, 1, 1, 1 0.125031 3 1, 1, 1, 1, 1, 1 0.0415928 1 Order of Galois group: 24 Table D.2: Type two

D.0.2 Why are there no numbers in the right column of Table D.6? The program Frobenius does not work when the discriminant of the polynomial is zero. The discriminant of a polynomial is zero if and only if the polynomial has at least one multiple zero.

19 Polynomial: x6 3x + 5 degree of factors frequency− number of elements 6 0.166647 8 5, 1 0.0 – 4, 2 0.125014 6 4, 1, 1 0.125022 6 3, 3 0.166706 8 3, 2, 1 0.0 – 3, 1, 1, 1 0.0 – 2, 2, 2 0.145806 7 2, 2, 1, 1 0.187513 9 2, 1, 1, 1, 1 0.0625254 3 1, 1, 1, 1, 1, 1 0.020766 1 Order of Galois group: 48 Table D.3: Type three

Polynomial: x6 16x + 3 − degree of factors frequency number of elements 6 0.0 – 5, 1 0.0 – 4, 2 0.124972 6 4, 1, 1 0.125183 6 3, 3 0.0 – 3, 2, 1 0.166644 8 3, 1, 1, 1 0.166595 8 2, 2, 2 0.0624935 3 2, 2, 1, 1 0.187509 9 2, 1, 1, 1, 1 0.145815 7 1, 1, 1, 1, 1, 1 0.0207898 1 Order of Galois group: 48 Table D.4: Type four

Polynomial: x6 21x 22 − − degree of factors frequency number of elements 6 0.0 – 5, 1 0.0 – 4, 2 0.0 – 4, 1, 1 0.25002 6 3, 3 0.0 – 3, 2, 1 0.0 – 3, 1, 1, 1 0.333301 8 2, 2, 2 0.0 – 2, 2, 1, 1 0.125028 3 2, 1, 1, 1, 1 0.25004 6 1, 1, 1, 1, 1, 1 0.041611 1 Order of Galois group: 24 Table D.5: Type five

20 Polynomial: x6 6x + 5 degree of factors frequency− number of elements 6 0.0 – 5, 1 0.0 – 4, 2 0.0 – 4, 1, 1 0.0 – 3, 3 0.0 – 3, 2, 1 0.0 – 3, 1, 1, 1 0.0 – 2, 2, 2 0.0 – 2, 2, 1, 1 0.0 – 2, 1, 1, 1, 1 0.0 – 1, 1, 1, 1, 1, 1 0.0 – Order of Galois group: – Table D.6: Type six

Polynomial: x6 63x + 62 − degree of factors frequency number of elements 6 0.0 – 5, 1 0.0 – 4, 2 0.0 – 4, 1, 1 0.250044 2 3, 3 0.0 – 3, 2, 1 0.0 – 3, 1, 1, 1 0.0 – 2, 2, 2 0.0 – 2, 2, 1, 1 0.374991 3 2, 1, 1, 1, 1 0.250021 2 1, 1, 1, 1, 1, 1 0.124944 1 Order of Galois group: 8 Table D.7: Type seven

Polynomial: x6 56x + 55 − degree of factors frequency number of elements 6 0.0 – 5, 1 0.0 – 4, 2 0.0 – 4, 1, 1 0.0 – 3, 3 0.0 – 3, 2, 1 0.166737 2 3, 1, 1, 1 0.166619 2 2, 2, 2 0.0 – 2, 2, 1, 1 0.249961 3 2, 1, 1, 1, 1 0.333323 4 1, 1, 1, 1, 1, 1 0.0833598 1 Order of Galois group: 12 Table D.8: Type eight

21 Polynomial: x6 16x + 16 − degree of factors frequency number of elements 6 0.0 – 5, 1 0.0 – 4, 2 0.0 – 4, 1, 1 0.0 – 3, 3 0.11112 4 3, 2, 1 0.333382 12 3, 1, 1, 1 0.111113 4 2, 2, 2 0.0 – 2, 2, 1, 1 0.249969 9 2, 1, 1, 1, 1 0.166687 6 1, 1, 1, 1, 1, 1 0.0277288 1 Order of Galois group: 36 Table D.9: Type nine

Polynomial: x6 30x + 25 − degree of factors frequency number of elements 6 0.0 – 5, 1 0.400138 144 4, 2 0.249933 90 4, 1, 1 0.0 – 3, 3 0.111181 40 3, 2, 1 0.0 – 3, 1, 1, 1 0.111081 40 2, 2, 2 0.0 – 2, 2, 1, 1 0.124888 45 2, 1, 1, 1, 1 0.0 – 1, 1, 1, 1, 1, 1 0.0027803 1 Order of Galois group: 360 Table D.10: Type ten

22 E Graphs Some graphs of the polynomials belonging to types one, two and three.

E.1 Type One Graphs for half of the polynomials of type one.

200

150

100

50

-3 -2 -1 1 2 3

-50

-100

Figure E.1: x6 + 65x + 13

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50

-3 -2 -1 1 2 3

-50

-100

Figure E.2: x6 + 56x + 62

23 50

40

30

20

10

-3 -2 -1 1 2 3

-10

Figure E.3: x6 + 45x + 55

10

7.5

5

2.5

-3 -2 -1 1 2 3 -2.5

-5

-7.5

-10

Figure E.4: x6 + 44x + 55

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100

50

-3 -2 -1 1 2 3

Figure E.5: x6 + 40x + 82

24 200

150

100

50

-3 -2 -1 1 2 3

Figure E.6: x6 + 30x + 93

100

80

60

40

20

-3 -2 -1 1 2 3

Figure E.7: x6 + 10x + 5

200

150

100

50

-3 -2 -1 1 2 3

Figure E.8: x6 + 8x + 20

25 200

150

100

50

-3 -2 -1 1 2 3

Figure E.9: x6 + 8x + 89

10

7.5

5

2.5

-3 -2 -1 1 2 3 -2.5

-5

-7.5

-10

Figure E.10: x6 + 3x + 3

26 E.2 Type Two Graphs for the polynomials of type two.

200

150

100

50

-3 -2 -1 1 2 3 -50

-100

-150

-200

Figure E.11: x6 + 49x 49 −

200

150

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50

-3 -2 -1 1 2 3 -50

-100

-150

-200

Figure E.12: x6 49x 49 − −

27 E.3 Type Three Graphs for the polynomials of type three.

100

80

60

40

20

-3 -2 -1 1 2 3

Figure E.13: x6 + 14x + 35

100

80

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40

20

-3 -2 -1 1 2 3

Figure E.14: x6 + 3x + 5

28 100

80

60

40

20

-3 -2 -1 1 2 3

Figure E.15: x6 3x + 5 −

100

80

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40

20

-3 -2 -1 1 2 3

Figure E.16: x6 14x + 35 −

29 References [1] John H. Conway, Alexander Hulpke, and John McKay: On Transitive Permuta- tion Groups, LMS J. Comput. Math. 1, 1-8 (electronic), 1998, ISSN 1461-1570

[2] David A. Cox: Galois Theory, Wiley, 2004, ISBN 0-471-43419-1

[3] Alexander Hulpke: Determining the Galois group of a rational polynomial, http://www.math.colostate.edu/ hulpke/talks/galoistalk.pdf ∼ was valid on October 20, 2006

[4] Alexander Hulpke: Techniques for the Computation of Galois Groups, http://www.math.colostate.edu/ hulpke/paper/gov.pdf was valid on October 20, 2006 ∼

[5] John F. Humphreys: A Course in Group Theory, Oxford University Press, 1996, ISBN 0-19-853453-1

[6] P. Stevenhagen and H. W. Lenstra, Jr.: Chebotarëv and his Density Theorem, The Mathematical Intellegencer, Volume 18, Number 2, Springer-Verlag New York, 1996

[7] Hendrik Lenstra: The Chebotarev Density Theorem, http://websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf was valid on October 20, 2006

[8] Robert Nyqvist, Chebotarevs densitetssats, First Regional Conference of Math- ematics in South-East Sweden, Kalmar, June 3, 2004, this is a poster and has not been published, Language: Swedish

[9] Robert Nyqvist, Polynomial p-adic Dynamics, Licentiate Thesis, School of Mathematics and Systems Engineering, Växjö University, Sweden, 2004, ISSN 1650-2647

[10] Kenneth H. Rosen: Elementary Number Theory, Pearson International Edition, Fifth Edition, 2005, ISBN 0-321-26314-6

[11] Ian Stewart: Galois Theory, Chapman and Hall Mathematics, Second Edition, 1989, ISBN 0-412-34550-1

[12] Ian Stewart: Galois Theory, Chapman and Hall/CRC Mathematics, Third Edi- tion, 2004, ISBN 1-58488-393-6

[13] Dirk J. Struik: A concise History of Mathematics, Dover, Fourth Revised Edi- tion, 1987, ISBN 0-486-60255-9

[14] Per-Anders Svensson: Abstrakt algebra, Studentlitteratur, 2001, ISBN 91-44- 01262-4, Language: Swedish

[15] Eric W. Weisstein: Polynomial Discriminant, MathWorld Wolfram, 2005, http://mathworld.wolfram.com/PolynomialDiscriminant.html was valid on October 20, 2006

30

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