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Galois Theory Galois Theory Willem A. de Graaf Contents Preface 1 1 Galois Theory 3 1.1 How it all began . .3 1.1.1 Cardano and the others . .3 1.1.2 Lagrange . .4 1.1.3 Ruffini and Abel . .5 1.1.4 Galois . .6 1.2 Separable polynomials . .6 1.3 Separable, normal and Galois extensions . .9 1.4 The Galois correspondence . 12 1.5 Cyclotomic fields . 17 1.5.1 Cyclotomic polynomials . 18 1.5.2 The Galois group of a cyclotomic extension . 20 1.6 Solvability of polynomial equations by radicals . 23 1.6.1 Solvable groups . 23 1.6.2 Radical extensions . 26 1.6.3 A polynomial not solvable by radicals . 31 1.6.4 The discriminant of a polynomial . 32 1.6.5 Casus irreducibilis . 33 2 Applications of Galois theory 35 2.1 The fundamental theorem of algebra . 35 2.2 Proving that certain primitive functions are not elementary . 35 2.3 Galois descent . 41 2.4 Galois cohomology . 45 Index 53 Bibliography 54 Preface Galois theory started, now about two hundred years ago, with the work of Galois (1811-1832) on root of polynomials. His goal was to give a criterion for deciding whether a given polynomial was solvable by radicals or not. The criterion he came up with involved a (for his time) new algebraic structure, namely groups. Galois introduced them to study the symmetries that exist between the roots of a polynomial. For this reason Galois theory was a very important step in the development of algebra as we know it today. Moreover the theory of Galois provided a striking example of the power of group actions. After the publication of Galois' work (in 1846) his theory has seeen a tremendous development and is now key to understanding various areas of algebra and geometry. The theory found in Galois' work is rather different from the way it is done nowadays. One of the reasons for this is the development that mathematics and in particular algebra has undergone since the times of Galois. When Galois was alive there was no group theory (indeed, the origin of this concept lies in his work) and almost no field or ring theory. To mention just one main difference between Galois' and the modern approach: Galois used permutations of the roots of a polynomial, whereas nowadays we use automorphisms of a splitting field. For a thorough account of the history of the study of the roots of a polynomial we refer to the book by Tignol, [Tig01]. In these notes we describe some of the basic concepts and constructions of modern Galois theory, focusing on field extensions of finite degree. (The infinite case is very important and interesting as well, but requires a lot more technical tools.) We will need many basic results from algebra, and for these I will refer to my course notes on algebra, available at http://www.science.unitn.it/~degraaf/algnotes/algebranotes.pdf A reference like 3(Algebra refers to Chapter 3 of the algebra notes. The main concepts that we need from basic algebra are groups (3(Algebra) and field extensions (Chapter 4(Algebra)). From the latter especially the concept of splitting field (Section 4.6(Algebra)) will play a major role. There are a great many great books on Galois theory (most of them with the same title as these notes). We cannot list all of them, so we just mention [Bos18], [Cox12]. 2 CONTENTS Chapter 1 Galois Theory We start with a very brief description of the historical roots of Galois theory; here we see how the concept and use of symmetry has been developed. 1.1 How it all began Galois theory is concerned with roots of polynomials. The main problem that led Galois to invent his theory was the one of finding radical expressions for the roots of a polynomial (or proving that no such expressions can be found). This problem has a rather long history as demonstrated by certain clay tablet dating back to at least 1500 BC, and found in current Iraq: This shows that the ancient Babylonians knew how to solve an equation of degree two, in modern notation: x2 + bx + c = 0. Given this knowledge the question comes up whether we can find formulas for the roots of polynomials of higher degree, for example of degree three: x3 + ax2 + bx + c = 0. The history regarding those equations starts rather spectacularly in the italian renaissance. 1.1.1 Cardano and the others In the sixteenth century mathematicians from northern Italy found magnificent formulas for finding the roots of polynomial equations of degree three and four. The main players in this story are Scipione del Ferro (1465-1526, from Bologna), Nicol`oFontana, called il Tartaglia (1499-1557, from Brescia) and Girolamo Cardano (1501-1576, from Milano). Scipione del Ferro Nicol`oTartaglia Girolamo Cardano 4 Galois Theory Scipione del Ferro was the first to find a formula for the roots of polynomials of degree 3. However, he did not publish it in any form, preferring to keep his discovery to himself, written down in a little notebook. Apparently the mathematicians of those days had a bit of a habit to keep their discoveries to themselves. The reason for this tendency to secrecy was that they often tried to impress other people with their mathematical skill, and even fought mathematical duels between each other. It is clear that the person with more secrets was more likely to excell at such events. However, in the centuries after del Ferro's death his notebook was lost, so we do not know exactly what he discovered. Before he passed away del Ferro revealed the secret of his formula for the cubic equation to a student of his, Antonio Maria del Fiore. Antionio del Fiore went round in mathematical circles bragging about his knowledge. Tartaglia, who was interested in the problem of solving cubic equations himself, heard about it, and accepted a cartello di matematica disfida from Antonio del Fiore. This was a mathematical duel where each contestant gave thirty mathematical problems to his opponent. The winner would be the one who managed to solve more problems. Tartaglia proposed thirty problems of various nature, whereas del Fiore wrote down thirty cubic equations. However, in the meantime Tartaglia had on his own discovered the formula for solving the cubic, so he easily won the contest. At around that time Girolamo Cardano was writing a book, called the Ars Magma (or in full Artis Magnae, Sive de Regulis Algebraicis Liber Unus), on algebra. He learned about the contest won by Tartaglia and was very keen on including the formula for the cubic in his book. However, Tartaglia, like Scipione del Ferro, wanted to keep the secret to himself, and only revealed the formula to Cardano when the latter promised not to publish it. Lodovico Ferrari (1522-1565), who was a collaborator of Cardano, then also found a way to solve equations of degree four. One step of his method consisted in solving an equation of degree 3. So without Tartaglia's formula Cardano and Ferrari could not publish their new discovery either. At this point Cardano and Ferrari recalled the contest with Antionio del Fiore, who had obtained the solution to the cubic from his bolgnese master Scipione del Ferro. They went to Bologna where they found del Ferro's notebook that had been kept by Annibale della Nave, who was his successor at the university of Bologna. Because he had now found the formula also from a different source, Cardano found that he was no longer bound by the promise made to Tartaglia and published the formulas for the cubic and quartic equations in his Ars Magna. Cardano did write that the formulas for the cubic were due to del Ferro and Tartaglia. But the latter was extremely annoyed just the same. 1.1.2 Lagrange After the publication of the Ars Magna the obvious question arose whether a formula could be found to solve equations of degree 5. The first to systematically investigate this question was Joseph-Louis Lagrange (1736-1813) Joseph-Louis Lagrange who in his great book R´eflexionssur la r´esolutionalg´ebriquedes ´equations devised a method for solving polynomial equations of arbitrary degree. For degrees 2,3,4 his method yielded the known formulas, but for degree 5 the method failed. Here we have a short look at Lagrange's method. Consider, for example, an equation of degree 3, 3 2 x + ax + bx + c = 0. Let α1; α2; α3 be its solutions. Lagrange considered symmetric expressions in these αi; these are polynomials in the αi that remain invariant when we permute the αi. For example 2 2 α1α2α3 + α1α2α3 1.1 How it all began 5 is not symmetric because if we interchange α1; α2 the expression becomes 2 2 2 2 α2α1α3 + α2α1α3 = α1α2α3 + α1α2α3 which is not equal to the original expression. On the other hand, 4 4 4 α1α2α3 + α1α2α3 + α1α2α3 is symmetric. A fundamental fact concerning these symmetric expressions (known before Lagrange) is the following: a symmetric expression in α1; α2; α3 can be computed as a polynomial in a; b; c. For example, 4 4 4 3 2 α1α2α3 + α1α2α3 + α1α2α3 = a c − 3abc + 3c : p −1+ −3 3 Now let ! = 2 , then ! 2 C has the property ! 6= 1 but ! = 1. Set 1 2 z1 = 3 (α1 + !α2 + ! α3) 1 2 z2 = 3 (α1 + !α3 + ! α2) 1 2 z3 = 3 (α2 + !α1 + ! α3) 1 2 z4 = 3 (α2 + !α3 + ! α3) 1 2 z5 = 3 (α3 + !α1 + ! α2) 1 2 z6 = 3 (α3 + !α2 + ! α1); and define the so called resolvent polynomial r = (x − z1)(x − z2)(x − z3)(x − z4)(x − z5)(x − z6).
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