Galois Theory Master's Dissertation Victor Summers 9th May 2013 Contents Introduction 2 1 Rings, Irreducible Polynomials and Field Extensions 5 2 Separable Polynomials and the Cyclotomic Extensions 26 3 Automorphisms, Galois Groups and the Fundamental Theorem 34 4 General Polynomials and the Discriminant 50 5 Algebraic Closures and Composite Extensions 57 6 Solvability by Radicals and Galois’ Great Theorem 63 7 Solvability of the Symmetric Groups 68 Bibliography 72 1 Introduction Galois theory is generally considered to be the birth of modern algebra. Unlike some areas of mathematics, it is motivated by a simple clear cut question. Most secondary school children are familiar with the so-called ”quadratic formula” which expresses the two complex roots of b pb2 4ac a polynomial of degree 2; the formula λ = − ± − is instilled in us by the end of 1,2 2a high school. So why isn’t a similar looking formula for cubic polynomials also pushed into our brains as well? Does such a thing even exist? The answers to these questions are 1. yes, there does exist an analogous cubic formula and even a quartic formula, and 2. the cubic and quartic formulas are wildly complicated in comparison. Nevertheless, such formulas do exist; cubics and quartics are ”solvable by radicals”. The question arises, are there also such formulas for expressing the roots of quintics? or more generally of polynomials of degree higher than 4? The quadratic formula has been known in one form or another for thousands of years but it was as late as circa 1550 when the cubic formula was first discovered, with the quartic formula coming to light shortly after that. The methods for arriving at such general formulas seemed very much like they might generalise to polynomials of degree 5 and above, yet for almost 200 years mathematicians tried and tried but got precisely nowhere. It was only when a select few mathematicians in the 18th and 19th centuries, most notably Joseph-Louis Lagrange, Niels Henrik Abel and E´variste Galois (pictured below), began to investigate the true underlying structure of the problem in its full abstraction that significant developments in algebra were made. Lagrange introduced entirely new ways of approaching the problem. For instance, instead of trying to find clever tricks for producing explicit formulas for the roots of polynomials, he would ask what kind of structure a solution must have if it were to exist. Also, he would refer to a solution as being “known” if he could prove it to exist. This type of existential thinking is, in my opinion, the true birth of modern algebra. Galois built greatly upon the ground work put in place by Lagrange and is generally credited as the inventor of the notion of a group which, although they can be seen lurking in the background in earlier works, were not brought 2 into the light and recognised for their profound nature until then. As it turns out, there can exist no such analogous formulas for the general quintic and above! This fact is known as the Abel-Ruffini theorem, named after Paolo Ruffini and Niels Henrik Abel. Ruffini produced many proofs between 1799 and 1813 all of which were rejected by the mathematical community due to his use of overly awkward and unsatisfying personal notations and conventions. Abel produced a proof of the unsolvability of the general quintic and polynomials of higher degree in 1823, which in fact turned out to be slightly incomplete, but nevertheless was deemed satisfactory. The problem with the Abel-Ruffini theorem is that it says very little about why this is the case and gives no information about when a particular polynomial may or may not be solvable by radicals. Galois’ brilliance was in shedding light on the answer to these questions, giving a nec- essary and sufficient condition under which a given polynomial is solvable by radicals! Galois’ beautiful method can be summarised as follows: ”Give me any polynomial and I will assign a group to that polynomial in such a way that the possible forms of its roots are reflected in the structure of the group. This then enables us to transform the problem to a group theoretical question which is simpler to answer”. In this project I aim to display the foundations upon which modern Galois theory stands, illustrating the core concepts with worked examples and culminating in a rigorous proof of what has been dubbed Galois great theorem. Throughout the project I assume a knowledge of the basic terminology and concepts from ring theory. For example, I will not provide formal definitions of a ring, Euclidean rings, prin- cipal ideal domains and polynomial rings, and I will take for granted some minor theorems and lemmas such as Euclid’s lemma and ”zero divisors if and only if the cancellation law holds”. However, in the construction of objects which are of crucial importance to the end goal I will give full formal definitions, theorems and proofs. The only rings considered in this project are assumed to be commutative rings with a multiplicative identity. 3 Chapter 1 Rings, Irreducible Polynomials and Field Extensions Our first discussion is motivated by the following question. The polynomial x2 +1 Q[x] has 2 no roots in Q, so is there some larger field E containing Q in which it does have a root? More generally we may ask, given an arbitrary polynomial f(x) F [x] which has no roots in the 2 field F , is there some field E containing F in which f(x) has a root? The answer is yes and we begin here by drawing upon some tools from ring theory which enable us to construct such objects, as well as other tools which will be of great use to us throughout. Let R be a ring and I R be an ideal. Define a relation on R by ⇢ a b a b I. ⇠ () − 2 We see that is an equivalence relation. Indeed, ⇠ i) Reflexivity: a a =0 I for each a R so a a for each a R; − 2 2 ⇠ 2 ii) Symmetry: a b a b I b a = (a b) I b a; ⇠ () − 2 () − − − 2 () ⇠ iii) Transitivity: a b and b c a b, b c I = a c =(a b)+(b c) I ⇠ ⇠ () − − 2 ) − − − 2 () a c. ⇠ Let us denote the equivalence class of a R by [a]; note that b [a] b = a + i for some 2 2 () i I so that [a]= a + i i I .LetR/I denote the set of all equivalence classes of , i.e 2 { | 2 } ⇠ R/I = [a] a R , equipped with the operations of addition, multiplication and negation { | 2 } defined for each a, b R by 2 [a]+[b]=[a + b] 5 [a][b]=[ab] [a]=[ a]. − − We call R/I the quotient ring of I in R. Of course we need to validate the use of the word ”ring” in this definition: Since the operations are defined on equivalence classes, in order for them to be well-defined we require the definitions to be independent of representatives. Suppose a0 a and b0 b, then ⇠ ⇠ ( ) a0 = a + i and b0 = b + j for some i, j I. Therefore, a0 + b0 = a + b +(i + j) and i + j I ⊕ 2 2 so that a0 + b0 a + b = [a0]+[b0]=[a + b]. ⇠ ) ( ) a0b0 =(a + i)(b + j)=ab +(aj + ib + ij) and aj + ib + ij I. Therefore, a0b0 ab = 2 ⇠ ) [a0][b0]=[ab]. ( ) a0 = (a + i)= a i and i I. Therefore, a0 a = [a0]=[ a]. − − − − − 2 − ⇠ ) − All the axioms of a ring follow immediately from the corresponding properties of the ring R itself. In particular, observe that the zero element is 0 := [0] = 0+i i I = I. { | 2 } Important pieces of machinery are those maps between rings which preserve certain algebraic relations. Definition 1.1. Let R and S be rings. A function σ : S is called a ring homomorphism ! (or ring map for short) if it preserves the ring operations. That is, for each a, b R 2 i) σ(1R)=1S ii) σ(ab)=σ(a)σ(b) iii) σ(a + b)=σ(a)+σ(b), where 1R and 1S denote the multiplicative identities of R and S respectively. In general, a ring map needn’t be injective or surjective. In the case where a ring map is both injective and surjective, we call it an isomorphism and say that the rings R and S are isomorphic; write R ⇠= S. In this sense we identify rings which are isomorphic to one another; although the two rings may be very di↵erent in appearance and their respective operations may even appear dissimilar, they share all ring theoretic properites and are thus the same up to relabelling of elements. We will use this mode of identification throughout this project. These isomorphisms also provide a way of assigning the afore mentioned group to a polynomial! 6 Theorem 1.2 (The first ring isomorphism theorem). Let R and S be rings amd let σ : R S ! be a ring map. Then, R ker(σ) ⇠= Im(σ) . and an isomorphism is given by the map h : R ker(σ) Im(σ) with ! . h([a]) := σ(a). Proof. We first need to show that the definition of h is independent of representatives as it is defined on equivalence classes. To this end, suppose a0 a so that a0 = a + i for some ⇠ i ker(σ)=: I. 2 Then, h([a0]) = σ(a0) = σ(a + i) = σ(a)+σ(i) = σ(a)(asi ker(σ)) 2 = h([a]) as required. h is a ring map: h([a]+[b]) = h([a + b]) = σ(a + b)=σ(a)+σ(b)=h([a]) + h([b]) • h([a][b]) = h([ab]) = σ(ab)=σ(a)σ(b)=h([a])h([b]) • h(1 = [1]) = σ(1) = 1.