A Project Example 1: Topics in Galois Theory Preamble For this project, the student explored an area of mathematics that was outside the course, and was difficult. The information was available and well embedded in the literature: however the concepts were advanced. In style therefore, this project most closely resembles the Hypergeometric Functions project outlined in Section 3.7. Abstract Galois Theory was originally formulated to determine whether the roots of a rational polynomial could be expressed in radicals. The theory is based on the association of a group to each polynomial and the analysis of the group to determine solubility in terms of radicals. We first study Galois Theory with the approach of the author Evariste Galois, then in its modern form based on abstract algebra, particularly field theory, which allows a more general interpre- tation. We then outline the theory of soluble groups and give some examples. Finally, we consider the application of methods in Galois Theory to the three classical construction problems and the construction of n-gons. 151 152 Managing Mathematical Projects - With Success! Introduction Evariste Galois (1811–1832) was a French mathematician who was born near Paris. Although he died in tragic circumstances at 20 years old he had already produced work that would make him famous. Unfortunately during his lifetime Galois suffered many setbacks and was considered to be a troublemaker by the government. He also found himself unaccepted by the mathematical establish- ment. It was not until 1846 that Joseph Liouville published some of Galois’ work in his Journal de Mathematiques [ED, p1]. Since then the importance of Galois’ results has been realised and his work has provided the impetus for many developments in algebra. In his Memoir on the Conditions for Solvability of Equations by Radicals Galois was solving a problem that interested many mathematicians including Lagrange. The solution to the general quadratic equation has been known since Babylonian times. In the 16th century formulae were discovered for the cubic and quartic. The solution for the cubic equation x3 + px = q found by Niccolo Fontana is 3 q p3 q2 3 q p3 q2 x = + + + − + . 2 27 4 2 27 4 This is an expression built up from the coefficients of the equation using the usual arithmetical operators with the nth roots, or radicals [ST, p xiv]. If the roots of an equation can be expressed in this way it is said to be “soluble by radicals”. Mathematicians hoped that they would be able to find formulas for general equations of degree n ≥ 5. Lagrange worked for many years on the solution of general equations, finding improved solutions for both the cubic and quartic [ED, pp18–22]. He was unable to find a solution for the quintic and expressed doubts that a solution existed. His work was almost certainly a source of inspiration for Galois. In 1824 Abel proved that the general quintic was not soluble by radicals so that a general formula did not exist but was unable to find a method for identifying which specific are soluble. This question was answered by Galois comprehensively in a piece of brilliant and original mathematics known as Galois Theory. This is the subject of this project and we now give an outline of each section. In Section A.1 we closely follow the English translation of Galois’ memoir by Harold M. Edwards in his book Galois Theory [ED]. The style, both of the translation and accompanying text, is traditional and often difficult to follow so much of the material in this section is a modified interpretation. The book takes a concrete rather than abstract approach to Galois Theory, and Edwards has attempted to present the theory with as little modern algebra as possible, for the most part drawing only on the techniques available to Galois. Some A. Project Example 1: Topics in Galois Theory 153 basic field theory is used in the text, particularly in the proofs, but we delay the introduction of fields until Section A.2. In Section A.2 we study Galois Theory in its modern form which makes use of the developments in abstract algebra over the last 150 years. Field theory has allowed mathematicians to interpret Galois Theory in a formal, abstract sense, suited to the more rigorous standards of definition and proof introduced at the turn of the century. The main source for this chapter is Ian Stewart’s Galois Theory [ST] which has a more accessible style than Edwards’ book. In separating the two approaches to Galois Theory we are able to appreciate the simplicity of Galois’ approach which the more structured methods of abstract algebra do not share although they allow Galois Theory to be applied to more general areas than rational polynomials. Groups play an essential part in Galois Theory and Galois’ work in this area provided a foundation for the study of group theory. This is expanded on in Section A.3, where we give an outline of the relevant theory and some examples. Sources for this chapter include Stewart’s book and Allenby’s Rings, Fields and Groups [AL] which was useful for a more detailed approach and worked examples. Finally, Section A.4 explores the applications of field theory to classical geometry, using the methods developed in our study of Galois Theory. Here we combine elements from Stewart, Allenby and also John R. Durbin’s Modern Algebra: An Introduction. This project is intended to be an introduction to Galois Theory although, due to the depth of the subject and restrictions upon the project, it has not been possible to give more than an outline of many of the proofs. It is hoped however that it will provide the reader with the impetus for a more detailed study of the subject. A.1 Galois’ Approach A.1.1 Preparation We begin this section by reviewing some facts from basic algebra. A.1.1.1 Polynomials. We consider polynomials with rational coefficients un- less specifically declared otherwise. A polynomial of degree n is irreducible in Q if it cannot be expressed as a product of two or more polynomials whose degrees are less than n but greater than 0. The polynomial of degree 0 is the constant 154 Managing Mathematical Projects - With Success! polynomial. The factorisation of a polynomial into irreducible polynomials is unique up to constant factors and the order in which the factors are written. n If f(x)=anx + ...+ a1x + a0 and an =1thenf is a monic polynomial. Since we are interested in the roots of a polynomial, when two polynomials differ only by a constant factor we consider them to be essentially the same. This is demonstrated by the example 1 1 1 f(x)= x2 − x − 2= (x +2)(x − 3) 3 3 3 g(x)=x2 − x − 6=(x +2)(x − 3). 1 So f = 3 g but f and g havethesameroots. A.1.1.2 Some Methods of Checking Irreducibility. If f is a polynomial that is irreducible in Z then f is also irreducible in Q. For example 1 1 let g(x)= x2 + x +6 3 3 and f(x)=x2 + x +18=3g(x). Now f is irreducible in Z so f is irreducible in Q. Eisensteins’s Irreducibility Criterion [ST, p20] Let n f(x)==anx + ...+ a1x + a0,ai ∈ Z,i=0, 1,...,n. Suppose there is a prime q such that an,q|ai for i =0, 1, 2,...,n− 1, and 2 q a0 then f is irreducible in Q. For example, if f(x)=x5 +4x4 +6x3 +2x2 +8x + 2 and we take q =2 then f is irreducible in Q. Unfortunately not every polynomial can be treated with this method. How- ever, we can sometimes rewrite a polynomial so that the method will be suit- able. For example, f(x)=g(x)h(x)ifandonlyiff(x +1)=g(x +1)h(x + 1). Hence f(x) is irreducible if and only if f(x + 1) is irreducible [ST,p21]. An example of this is f(x)=x4 +x3 +x2 +x+1 where Eisenstein’s criterion does not apply. But f(x +1)=x4 +5x3 +10x2 +10x + 5 and is irreducible by Eisenstein’s criterion with q = 5. Hence f(x) is irreducible in Q. A.1.1.3 Roots of Polynomials. Each root of a polynomial is either simple or multiple (repeated). We will want to work with polynomials that have only simple roots, but it is quite straightforward to detect simple roots. If f(x) has roots α1,α2,...,αn which are roots of multiplicity m1,m2,...,mn respectively then m1 m2 mn f(x)=(x − α1) (x − α2) ...(x − αn) . A. Project Example 1: Topics in Galois Theory 155 If we differentiate formally then $ m1−1 m2−1 mn−1 Df =(x − α1) (x − α2) ...(x − αn) mj (x − αi) j i= j and it is clear that if, for any i, mi > 1thenf and Df have a common factor. Thus we can say Lemma 1 [ST,p85] A polynomial has a multiple root if and only if f and Df have a common factor of degree ≥ 1. If f has no multiple roots then f and Df will be coprime, i.e. the highest common factor will be a polynomial of degree 0. Any irreducible polynomial with rational coefficients must have only simple roots since if it had a multiple root then f and Df would have a common factor of degree≥ 1 which can only be a multiple of f.Butf is irreducible and Df has degree less than n so we must have Df = 0 which implies that f is a constant.