Èvariste Galois and the resolution of equations by radicals
|
|
- Moses Mason
- 5 years ago
- Views:
Transcription
1 Èvariste Galois and the resolution of equations by radicals A Math Club Event Department of Mathematical Sciences Florida Atlantic University Boca Raton, FL November 9, 2012
2 Outline 1 The Problem
3 Solving Equations The set of real numbers is a strange, messy, murky and unreal type of world. As the Greeks discovered, and Eudoxus made clear, we can know when two numbers are different. But equality is harder; two numbers are equal if they are not different; TRICHOTOMY RULES! What does it mean to know a number?
4 Solving Equations Let us start with the integers; we all know the integers of course. We can add, subtract, and multiply them, and still have integers. Now construct all numbers you can construct from the integers by division; you get the rational numbers. Next add roots; square roots, cubic roots, fourth roots, all roots. Also roots of numbers you get by adding or subtracting roots. If the real numbers were the ocean, all the numbers obtained by the previous steps are not even a drop of ocean water. The vast majority of numbers is still out there, forever unknown.
5 Solving Equations If we allow calculus methods, all algebraic equations can be solved. In fact, Newton s method, occasionally covered in Calculus 1 courses, is an efficient method for solving algebraic equations. And there are many others. The trick is to solve them using only the operations from the steps mentioned before: Addition, subtraction, multiplication, division, extraction of roots. Come up with a method, an algorithm that works. This is what is meant by solving equations by radicals. To be precise, the solution should be obtained in a finite number of steps by the operations of addition, subtraction, multiplication, division, and extraction of roots, starting with the integers (or the rationals) and the coefficients of the equation.
6 Solving Equations The Babylonians already figured out how to solve the quadratic equation. The solution of the equation ax 2 + bx + c = 0 is given by x = b ± b 2 4ac. 2a Apart from the standard operations of addition, subtraction, multiplication and division, all that is involved is taking square roots.
7 Solving Equations Around 1500, the cubic and quartic equations were solved by Scipio dal Ferro, Tartaglia, Cardano, Ferrari. The search for a formula to solve quintic equations was on. Who would discover it? And when?
8 A Slightly Different Radical Idea Instead of extracting roots, think of adjoining roots. To solve by radicals can also be interpreted as follows: We start with the rational numbers, and the coefficients of the equation, and (at least mentally) close this set with respect to the four basic operations: addition, subtraction, multiplication and division. I ll call this set (for now) our starting field K 0. If we can show that we ll always have the roots of the equations we are considering in K 0, our work is done. Except for first order equations, we can t. Next, we adjoin some roots of (a finite number of) elements of K 0 to K 0, close again by the four basic operations to get a second field of activity, the field K 1. Can we show that the roots are in K 1? Yes, for quadratic equations; we only need to adjoin b 2 ac.
9 Adjoining Radicals Around 1800, three mathematicians, Ruffini, Abel, and Galois, had similar ideas on how to analyze what happens when you keep adjoining radicals. It involved looking at what turned out to be a group of permutations of the roots. When a radical was adjoined to the field of numbers you had so far, the group could get smaller, or remain the same. If it got smaller, it did so in a very specific way: If G was the group before adjunction, and after adjunction it shrank to a subgroup G 1, then in modern day parlance (don t worry to much, I ll hardly use it again), G 1 had to be a normal subgroup of G and G/G 1 had to be commutative.
10 When Was The Equation Solved? As we shall see, if the group G of the equation had a single permutation; in other words just the identity permutation, the equation was solved. And conversely. So, to solve the equation one needed an algorithm by which one can construct a sequence G 0 G 1 G N of subgroups of G such that G 0 = G, G N = {e} (the identity), G i+1 is a normal subgroup of G i and G i /G i+1 is commutative (i = 0,..., N 1). A group for which such a sequence exists is called solvable Is every group of permutations that can appear as group of an equation solvable?
11 The conclusion To solve the general equation of degree n one had to assume that its group was S n, the group of ALL permutations of n elements. S 2, S 3, S 4 are solvable. If n 5, then S n is not solvable. The general equation of order n 5 cannot be solved by radicals.
12 g-rational Numbers From now on we assume given an equation a m x m + a m 1 x m a 1 x + a 0 = 0 (1) to be known as the main equation. The coefficients a 0,..., a m are complex numbers; a m 0. We will assume that all its roots, which Galois denotes a, b, c,..., but I will denote r 1, r 2,..., r m, are distinct. Galois then calls a number rational if it is rational, or if it is a coefficient of the equation, or can be obtained from these first rationals by the basic operations +,,, /. But it is a fluid concept; once a number gets adjoined to the original rationals, it can be used to create more rationals. I will call a Galois rational a g-rational number.
13 g-rational Numbers, a Better Definition The Language of Fields Algebraists consider some very abstract fields; for us a field is just a subset K of the complex numbers that is closed under addition, subtraction, multiplication and division by non-zero elements. The smallest such set is Q, the field of rational numbers. If K is a field, α a complex number, then K(α) is the smallest field containing all the elements of K as well as α. If α is the zero of an irreducible polynomial of degree n, p(x ) = a n X n + + a 1 X + a 0 with coefficients in K, then one can show that K(α) = {c 0 + c 1 α + + c n 1 α n 1 : c 0,..., c n 1 K}. If α, β C, then K(α, β) = K(α)(β) is the smallest field containing all the elements of K, and α, and β. More generally, K(a 1,..., a n ) is the smallest field containing K as well as a 1,..., a n.
14 g-rational Numbers Let K= Q(a 0,..., a m ); Q extended by the coefficients of the main equation; just Q if these coefficients are rational. These are our first g-rational numbers. Every time a number α is adjoined to K, the set of g-rationals expands to include the elements of K(α). Eventually, K(α) becomes the new K.
15 How Galois Defined his Group Let r 1,..., r m be the distinct roots of the main equation. One can prove, it is not terribly hard, that one can find integers A 1,..., A m with the property that if we consider all the different m! linear combinations of the roots using these coefficients, they all take on different values. Precisely, if σ, τ S m, σ τ, then A 1 r σ(1) + A m r σ(m) A 1 r τ(1) + A m r τ(m). Let V = A 1 r A m r m. This is a highly non-unique choice!
16 The Group of Galois, by Galois Let Q(X ) = σ S m X m A j r σ(j). This is a polynomial of degree m!. The coefficients of this polynomial are symmetric in the roots of the main equation; this implies that they are polynomials in the coefficients of the main equation; in other words they are in K. So Q is a polynomial of degree m! and Q(V ) = 0. Q(X ) is not necessarily irreducible over K; because Q(V ) = 0 there is an irreducible polynomial Q 0 with coefficients in K such that Q 0 (V ) = 0. This polynomial divides Q(X ). Let n be the degree of Q 0. j=1
17 The Group is Coming, but not here yet Denote the zeros of Q 0 (X ) by V, V,..., V n 1. Each V k is obtained from V by applying a permutation to the roots, since it also is a zero of Q(X ). Galois now proves that every root of the main equation can be expressed as a polynomial in V with coefficients in K. Denoting as one does the set of polynomials with coefficients in K by K[X ], there exist ϕ(x ), ϕ 1 (X ),..., ϕ m 1 (X ) K[X ] such that r 1 = ϕ(v ), r 2 = ϕ 1 (V ),..., r m = ϕ m 1 (V ). These polynomials can be assumed of degree at most n 1 (n the degree of Q 0 ). In this case they are unique.
18 Almost The Problem Galois now proves that for every root W of Q(X ), the values ϕ(w ), ϕ 1 (W ),..., ϕ m 1 (W ) are a rearrangement (permutation) of the roots r 1,..., r m and he sets up the following table:
19 ! As Galois says: Let ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V be the roots of the proposed equation. Let us write down the following permutations of the roots: (V ) ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V (V ) ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V (V ) ϕv ϕ 1 V ϕ 2 V..... ϕ m 1 V (V (n 1) ) ϕv (n 1) ϕ 1 V (n 1) ϕ 2 V (n 1)..... ϕ m 1 V (n 1)
20 Properties, Independent Characterization Galois calls it a group, but it is likely that he used the word as a synonym for set. But, as he remarked, it is a set of permutations closed under composition, thus a subgroup of S m. The permutations in this group satisfy: If F (X 1,..., X m ) is a polynomial in m-variables with coefficients in K and if F (r 1,..., r m ) = c K, then F (r τ(1),..., r τ(m) ) = c for all permutations in this group. If F (X 1,..., X m ) is a polynomial in m-variables with coefficients in K that satisfies F (r τ(1),..., r τ(m) ) = F (r 1,..., r m ) for all permutations τ of the group, then F (r 1,..., r m ) K As a corollary one obtains: A permutation τ S m is in the group if and only if it satisfies: Whenever F (X 1,..., X m ) is a polynomial in m-variables with coefficients in K such that F (r 1,..., r m ) = 0, then F (r τ(1),..., r τ(m) ) = 0. This last property shows that the definition of the group is independent of the choice of V and that it is, in fact, a subgroup of S m.
21 A modern perspective Ignore if confusing Given abstract up in the cloud fields F, E with F a subfield of E, one considers the set of all automorphisms of E that leave F fixed. One says E is a normal extension of F if it is also true that if an element of E is left fixed by all these automorphisms, then it has to be in F. If E is a normal extension of F, then the order of the group of automorphisms of E leaving F fixed equals the dimension [E : F ] of E as a vector space over F. If F, E are fields, if p(x ) is a polynomial with coefficients in F and E is a minimal field extending F and containing all roots of p(x ), then E is called a splitting field for p(x ) over F. One proves all splitting fields are isomorphic, so one talks of the splitting field of p(x ). One proves: E is a normal extension of F if and only if it is the splitting field of some polynomial p(x ) having no repeated roots. If p(x ) is a polynomial with coefficients in F and E is a splitting field for p(x ) over F, then the group of automorphisms of E leaving F fixed is the Galois group of p(x ). This group can be identified with a subgroup of the group of permutations of the roots and, as such, is the same as the group defined by Galois.
22 In Fact, The Problem The zeros of Q(X ), hence also of Q 0 (X ) are all distinct. K(V ) is the smallest field containing V and K. Because all the roots of the main equation are polynomials in V, K(V ) contains them all. Because all other zeros of Q(X ) and of Q 0 (X ) are linear combinations of the roots of the main equation, all these zeros are also in K(V ) It follows that K(V ) is a splitting field of both the polynomial defined by the main equation and for Q 0 (X ), hence a normal extension of K.
23 Example 1. The Problem Details are left out. Lets look at x 3 2 = 0, with K = Q. With ω = e 2πi/3 = i, its roots are 3 2, 3 2 ω, 3 2 ω 2. It is easy to see that a working choice of A 1, A 2, A 3 is A 1 = 1, A 2 = 2, A 3 = 0. With this choice, V = i.
24 x 3 2 = 0 continued The irreducible polynomial over Q satisfied by V is x = 0. The other roots of this polynomial are V = V = 3 2 3i, V = V (4) = V = ( 3 ) i ( i ) 3 2, V (5) = V (4) = A bit of arithmetic/algebra shows that we have to take 3 2, V = V = ϕ(x) = 1 18 x 4, ϕ 1 (x) = 1 2 x 1 36 x 4, ϕ 2 (x) = 1 2 x 1 36 x 4. ( ) i (
25 x 3 2 = 0 continued Galois table looks as follows: (V ) ϕ(v ) = 3 2 ϕ 1 (V ) = 3 2ω ϕ 2 (V ) = 3 2ω 2 (V ) ϕ(v ) = 3 2 ϕ 1 (V ) = 3 2ω 2 ϕ 2 (V ) = 3 2ω (V ) ϕ(v ) = 3 2ω 2 ϕ 1 (V ) = 3 2ω ϕ 2 (V ) = 3 2 (V ) ϕ(v ) = 3 2ω 2 ϕ 1 (V ) = 3 2 ϕ 2 (V ) = 3 2ω (V (4) ) ϕ(v (4) ) = 3 2ω ϕ 1 (V (4) ) = 3 2ω 2 ϕ 2 (V (4) ) = 3 2 (V (5) ) ϕ(v (5) ) = 3 2ω ϕ 1 (V (5) ) = 3 2 ϕ 2 (V (5) ) = 3 2ω 2 As we can see, all permutations occur. The Galois group is (isomorphic to ) S 3.
26 Example 2 The Problem We consider again x 3 2 = 0, but after adjoining the root 3 2. V = i as before, but now the irreducible polynomial it satisfies is X = 0. V = V is the second root of the equation satisfied by V. The new group reduces to the first two lines of the old one; it consists of the identity and one transposition.
27 Example 3 The Problem We consider the equation x 4 + x 3 + x 2 + x + 1 = 0, with roots r, r 2, r 3, r 4, where r = e 2πi/5 = cos(2π/5) + i sin(2π/5), over Q. Because r, r 2, r 3, r 4 are independent over Q, any four distinct A 1, A 2, A 3, A 4 will work. For example, A 1 = 1, A 2 = 2, A3 = 2, A4 = 1, giving ( V = r + 2r 2 2r 3 r 4 = 2 sin 2π sin π ) i. 5
28 Example 3 contd., x 4 + x 3 + x 2 + x + 1 = 0 With some effort (and help from Maple) one gets that the polynomials ϕ, ϕ 1, ϕ 2, ϕ 3 are ϕ(x ) = 1 25 x x x 3, ϕ(v ) = r, (2) 2 ϕ 1 (X ) = 1 50 x x x + 1, ϕ 1(V ) = r 2, (3) ϕ 2 (X ) = 1 50 x x x + 1, ϕ 2(V ) = r 3,(4) ϕ 3 (X ) = 1 25 x x x 3 2, ϕ(v ) = r 4. (5)
29 Example 3 concluded, x 4 + x 3 + x 2 + x + 1 = 0 Here is how the table of permutations looks like. ϕ(v ) = r ϕ 1 (V ) = r 2 ϕ 2 (V ) = r 3 ϕ 3 (V ) = r 4 ϕ(v ) = r 4 ϕ 1 (V ) = r 3 ϕ 2 (V ) = r 2 ϕ 3 (V ) = r ϕ(v ) = r 3 ϕ 1 (V ) = r ϕ 2 (V ) = r 4 ϕ 3 (V ) = r 2 ϕ(v ) = r 2 ϕ 1 (V ) = r 4 ϕ 2 (V ) = r ϕ 3 (V ) = r 3 If τ denotes the permutation r r 3, r 2 r, r 3 r 4, r 4 r 2, then G is the cyclic group {e, τ, τ 2, τ 3 }.
30 The heart The Problem of the matter Galois divides his Memoir into Preliminaries (Four Lemmas) and eight Propositions. Propositions can be Theorems (Propositions I, II, III, IV, VIII), Problems (Propositions V, VII), or Lemmas (Proposition VI). The construction and basic properties of what we now call the Galois group of an equation is the content of Proposition I (and what precedes it) in the Memoir. Propositions II and III are the main results; very roughly what nowadays is called the Fundamental Theorem of Galois Theory. Proposition IV is a bit of a digression. Proposition V gives the necessary and sufficient condition for an equation to be solvable by radicals. Propositions VI-VIII expand on this matter.
31 Propositions II and III In the notes I am writing on this Memoir, I have it the way Galois stated it. But here I have to be brief, and use modern terms. Suppose the group of the equation, as constructed above, has n permutations. Proposition II states that if we extend the field by adjoining the root of an irreducible polynomial of degree p, two things can happen: Nothing; the group of the equation with respect to the extended field is the same as before, or the new group is a subgroup of size n/p of the previous group. Implicit in this is that p has to divide n for the second option to be possible. Proposition III in modern language says precisely that if we add ALL (Galois emphasis) roots of an irreducible equation to the field, then new group has to be a normal subgroup of the previous group.
32 Solving Equations An equation is considered solved if all its roots have become g-rational, are in the field K in which we are working. In this case V K,the irreducible equation satisfied by V is x V = 0 of degree 1, so that the Galois group is of order (size) 1; reduced to the identity. Can the Galois group be reduced, by adjoining roots of objects already in the field, to size 1?
33 Equations that hate radicals, and don t allow them to do their work Adjoining radicals means adjoining things like k a, where a K. But all roots can be obtained by taking prime roots. For example, 45 a = a. We can also assume that our field contains all (complex) roots of 1; since we can also just adjoin them. So adjoining radicals becomes equivalent to adjoining ALL roots of an irreducible equation of the form x p a, where p is prime. This means that IF the original group G can be reduced, it can also be reduced to a normal subgroup G 0 such that G/G 0 is prime. Groups of prime order are cyclic, in particular commutative.
34 Equations that hate radicals, and don t allow them to do their work, contd. To solve the general equation of order m, because the coefficients could all be transcendental, unrelated to each other, we must assume its Galois group is S m. One thing that happens in S m if m 5 is that you can have two 3-cycles having exactly one element in common. A 3-cycle denoted by (abc), where a, b, c {1,..., m} is the permutation σ such that σ(a) = b, σ(b) = c, σ(c) = a, and σ(k) = k if k a, b, c. Suppose G is a subgroup of S m that contains all 3-cycles. Let σ = (abc) be one of the 3-cycles; a b c a. If and only if m 5, we can find d, e so a, b, c, d, e are 5 distinct elements and form the 3-cycles τ = (dac) and λ = (ceb) Suppose H is a normal subgroup of G and suppose G/H is abelian.
35 Equations that hate radicals, and don t allow them to do their work, contd. Because G/H is abelian (commutative), id G/H = (τh) 1 (λh) 1 (τh)(λh) = (τ 1 λ 1 τλ)h, implying τ 1 λ 1 τλ H. Going from right to left (usual composition order) τ 1 λ 1 τλ = (dca)(cbe)(dac)(ceb) = (abc) = σ. This proves: If a subgroup G of S m, m 5, contains all 3-cycles, every normal subgroup H such that G/H is commutative, must also contain all three cycles. This severely reduces the amount by which G can be reduced. It shows that S 5 is not solvable.
36 Solvable equations But why should an equation whose group G has the property that one can find a sequence of normal subgroups G 0 = G G 1 G 2 {e} such that G i1 /G i is of prime order, be solvable by radicals? Here is Galois argument. We must show that if the group is G, and G has a normal subgroup subgroup H such that G/H is of order p, p a prime, then by adjoining a radical (a p-th root it will turn out to be), the new group of the equation is H. So assume all this, and let α be a primitive p-th root of 1; α = e 2πi/p. Assume the equation has m distinct roots r 1,..., r m.
37 Galois argument continued Let f K[X 1,..., X m ] be a polynomial in m-variables such that if for σ S m we set σ f (X 1,..., X m ) = f (X σ(1),..., X σ(m) ), then σ f (r 1,... r m ) = f (r 1,... r m ) σ H, σ f (r 1,... r m ) f (r 1,... r m ) σ G\H, Let σ G be such that σh is a generator of G/H. Let θ = f (r 1,..., r m ), θ 1 = σ f (r 1,..., r m ), θ 2 = σ 2 f (r 1,..., r m ),..., θ p 1 = σ f (r 1,..., r m ).
38 The argument cont d Let ξ = ( θ + αθ α p 1 θ p 1 ) p. If there is time, I will argue at the board that ξ K, ξ 1/p / K and if we adjoin ξ 1/p to K, then H becomes the new group of the equation.
39 Second degree equations Suppose the equation is ax 2 + bx + c = 0, with roots r, s. The group S 2 = {e, σ} with σ(r) = s, σ(s) = r. We have to reduce it to {e}. In this case p = 2, α = 1 (Actually, this is always the first reduction). We need a polynomial in two variables left invariant by {e} (any polynomial will do) but not by σ. The simplest choice is f (X 1, X 2 ) = X 1 X 2. The quantity Galois says should be adjoined is ξ where ξ = (f (r, s) f (s, r)) 2 = (2(r s)) 2 = 4(r 2 + s 2 2rs) = 4 ( (r + s) 2 4rs ) ( (b ) ) 2 = 4 4c = 4(b2 4ac) a a a 2.
40 The End The Problem THE END Thanks for listening
GALOIS; The first Memoir
GALOIS; The first Memoir Trying to understand it 1 Preliminaries 1.1 Notational Conventions Results directly copied from Galois Memoir, or revisions of such results, are stated using small caps; for example,
More informationGalois Theory and the Insolvability of the Quintic Equation
Galois Theory and the Insolvability of the Quintic Equation Daniel Franz 1. Introduction Polynomial equations and their solutions have long fascinated mathematicians. The solution to the general quadratic
More informationMATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11
MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11 3. Examples I did some examples and explained the theory at the same time. 3.1. roots of unity. Let L = Q(ζ) where ζ = e 2πi/5 is a primitive 5th root of
More informationThe Galois group of a polynomial f(x) K[x] is the Galois group of E over K where E is a splitting field for f(x) over K.
The third exam will be on Monday, April 9, 013. The syllabus for Exam III is sections 1 3 of Chapter 10. Some of the main examples and facts from this material are listed below. If F is an extension field
More informationMath 201C Homework. Edward Burkard. g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) a 2,k u k v a 1,k u k v + k=0. k=0 d
Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise 5.1.7. If v is algebraic over K(u) for some u F and v is transcendental over K, then u is algebraic over K(v).
More informationChapter 11: Galois theory
Chapter 11: Galois theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 410, Spring 014 M. Macauley (Clemson) Chapter 11: Galois theory
More informationSOLVING SOLVABLE QUINTICS. D. S. Dummit
D. S. Dummit Abstract. Let f(x) = x 5 + px 3 + qx + rx + s be an irreducible polynomial of degree 5 with rational coefficients. An explicit resolvent sextic is constructed which has a rational root if
More informationGalois theory and the Abel-Ruffini theorem
Galois theory and the Abel-Ruffini theorem Bas Edixhoven November 4, 2013, Yogyakarta, UGM A lecture of two times 45 minutes. Audience: bachelor, master and PhD students, plus maybe some lecturers. This
More informationAN INTRODUCTION TO GALOIS THEORY
AN INTRODUCTION TO GALOIS THEORY STEVEN DALE CUTKOSKY In these notes we consider the problem of constructing the roots of a polynomial. Suppose that F is a subfield of the complex numbers, and f(x) is
More informationJean-Pierre Escofier. Galois Theory. Translated by Leila Schneps. With 48 Illustrations. Springer
Jean-Pierre Escofier Galois Theory Translated by Leila Schneps With 48 Illustrations Springer Preface v 1 Historical Aspects of the Resolution of Algebraic Equations 1 1.1 Approximating the Roots of an
More informationSchool of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information
MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon
More informationSymmetries and Polynomials
Symmetries and Polynomials Aaron Landesman and Apurva Nakade June 30, 2018 Introduction In this class we ll learn how to solve a cubic. We ll also sketch how to solve a quartic. We ll explore the connections
More informationGALOIS THEORY: LECTURE 20
GALOIS THEORY: LECTURE 0 LEO GOLDMAKHER. REVIEW: THE FUNDAMENTAL LEMMA We begin today s lecture by recalling the Fundamental Lemma introduced at the end of Lecture 9. This will come up in several places
More informationLecture 6.6: The fundamental theorem of Galois theory
Lecture 6.6: The fundamental theorem of Galois theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 410, Modern Algebra M. Macauley
More information1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism
1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials
More informationFields and Galois Theory. Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory.
Fields and Galois Theory Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory. This should be a reasonably logical ordering, so that a result here should
More informationMT5836 Galois Theory MRQ
MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and
More informationGALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2)
GALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2) KEITH CONRAD We will describe a procedure for figuring out the Galois groups of separable irreducible polynomials in degrees 3 and 4 over
More informationCOMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS
COMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS JOHN KOPPER 1. Introduction Given an arbitrary irreducible polynomial f with rational coefficients it is difficult to determine the Galois group
More informationGalois theory - Wikipedia, the free encyclopedia
Page 1 of 8 Galois theory From Wikipedia, the free encyclopedia In mathematics, more specifically in abstract algebra, Galois theory, named after Évariste Galois, provides a connection between field theory
More informationCourse 311: Hilary Term 2006 Part IV: Introduction to Galois Theory
Course 311: Hilary Term 2006 Part IV: Introduction to Galois Theory D. R. Wilkins Copyright c David R. Wilkins 1997 2006 Contents 4 Introduction to Galois Theory 2 4.1 Polynomial Rings.........................
More informationGalois fields/1. (M3) There is an element 1 (not equal to 0) such that a 1 = a for all a.
Galois fields 1 Fields A field is an algebraic structure in which the operations of addition, subtraction, multiplication, and division (except by zero) can be performed, and satisfy the usual rules. More
More informationNotes on Galois Theory
Notes on Galois Theory Math 431 04/28/2009 Radford We outline the foundations of Galois theory. Most proofs are well beyond the scope of the our course and are therefore omitted. The symbols and in the
More informationGalois Theory, summary
Galois Theory, summary Chapter 11 11.1. UFD, definition. Any two elements have gcd 11.2 PID. Every PID is a UFD. There are UFD s which are not PID s (example F [x, y]). 11.3 ED. Every ED is a PID (and
More informationThe following results are from the review sheet for the midterm.
A. Miller M542 Galois Theory Spring 2000 For the material on Galois theory we will be assuming that the fields all have characteristic zero. When we get to solvability by radicals we will assume that all
More informationRings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.
Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary
More informationRUDIMENTARY GALOIS THEORY
RUDIMENTARY GALOIS THEORY JACK LIANG Abstract. This paper introduces basic Galois Theory, primarily over fields with characteristic 0, beginning with polynomials and fields and ultimately relating the
More informationGalois Theory. This material is review from Linear Algebra but we include it for completeness.
Galois Theory Galois Theory has its origins in the study of polynomial equations and their solutions. What is has revealed is a deep connection between the theory of fields and that of groups. We first
More informationFIELD THEORY. Contents
FIELD THEORY MATH 552 Contents 1. Algebraic Extensions 1 1.1. Finite and Algebraic Extensions 1 1.2. Algebraic Closure 5 1.3. Splitting Fields 7 1.4. Separable Extensions 8 1.5. Inseparable Extensions
More informationALGEBRA PH.D. QUALIFYING EXAM September 27, 2008
ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely
More information1 The Galois Group of a Quadratic
Algebra Prelim Notes The Galois Group of a Polynomial Jason B. Hill University of Colorado at Boulder Throughout this set of notes, K will be the desired base field (usually Q or a finite field) and F
More informationdisc f R 3 (X) in K[X] G f in K irreducible S 4 = in K irreducible A 4 in K reducible D 4 or Z/4Z = in K reducible V Table 1
GALOIS GROUPS OF CUBICS AND QUARTICS IN ALL CHARACTERISTICS KEITH CONRAD 1. Introduction Treatments of Galois groups of cubic and quartic polynomials usually avoid fields of characteristic 2. Here we will
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More information1 Finite abelian groups
Last revised: May 16, 2014 A.Miller M542 www.math.wisc.edu/ miller/ Each Problem is due one week from the date it is assigned. Do not hand them in early. Please put them on the desk in front of the room
More informationCourse 311: Abstract Algebra Academic year
Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 3 Introduction to Galois Theory 41 3.1 Field Extensions and the Tower Law..............
More informationMTH 401: Fields and Galois Theory
MTH 401: Fields and Galois Theory Semester 1, 2014-2015 Dr. Prahlad Vaidyanathan Contents Classical Algebra 3 I. Polynomials 6 1. Ring Theory.................................. 6 2. Polynomial Rings...............................
More informationChapter V. Solvability by Radicals
Matematisk Institut Mat 3AL 5.1 Chapter V. Solvability by Radicals One of the oldest problems in algebra was to find roots of an equation. Already in the antiquity solutions of quadratic equations were
More informationNOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22
NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22 RAVI VAKIL Hi Dragos The class is in 381-T, 1:15 2:30. This is the very end of Galois theory; you ll also start commutative ring theory. Tell them: midterm
More informationALGEBRA EXERCISES, PhD EXAMINATION LEVEL
ALGEBRA EXERCISES, PhD EXAMINATION LEVEL 1. Suppose that G is a finite group. (a) Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of its normalizer. (b) Use (a)
More informationGALOIS THEORY. Contents
GALOIS THEORY MARIUS VAN DER PUT & JAAP TOP Contents 1. Basic definitions 1 1.1. Exercises 2 2. Solving polynomial equations 2 2.1. Exercises 4 3. Galois extensions and examples 4 3.1. Exercises. 6 4.
More informationExtension fields II. Sergei Silvestrov. Spring term 2011, Lecture 13
Extension fields II Sergei Silvestrov Spring term 2011, Lecture 13 Abstract Contents of the lecture. Algebraic extensions. Finite fields. Automorphisms of fields. The isomorphism extension theorem. Splitting
More informationFields and Galois Theory
Fields and Galois Theory Rachel Epstein September 12, 2006 All proofs are omitted here. They may be found in Fraleigh s A First Course in Abstract Algebra as well as many other algebra and Galois theory
More informationTheorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.
5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field
More informationMath 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille
Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is
More informationMath 414 Answers for Homework 7
Math 414 Answers for Homework 7 1. Suppose that K is a field of characteristic zero, and p(x) K[x] an irreducible polynomial of degree d over K. Let α 1, α,..., α d be the roots of p(x), and L = K(α 1,...,α
More informationGalois Theory Overview/Example Part 1: Extension Fields. Overview:
Galois Theory Overview/Example Part 1: Extension Fields I ll start by outlining very generally the way Galois theory works. Then, I will work through an example that will illustrate the Fundamental Theorem
More informationCommutative Rings and Fields
Commutative Rings and Fields 1-22-2017 Different algebraic systems are used in linear algebra. The most important are commutative rings with identity and fields. Definition. A ring is a set R with two
More information5 Group theory. 5.1 Binary operations
5 Group theory This section is an introduction to abstract algebra. This is a very useful and important subject for those of you who will continue to study pure mathematics. 5.1 Binary operations 5.1.1
More informationSection 0.2 & 0.3 Worksheet. Types of Functions
MATH 1142 NAME Section 0.2 & 0.3 Worksheet Types of Functions Now that we have discussed what functions are and some of their characteristics, we will explore different types of functions. Section 0.2
More informationGalois theory (Part II)( ) Example Sheet 1
Galois theory (Part II)(2015 2016) Example Sheet 1 c.birkar@dpmms.cam.ac.uk (1) Find the minimal polynomial of 2 + 3 over Q. (2) Let K L be a finite field extension such that [L : K] is prime. Show that
More informationSimple groups and the classification of finite groups
Simple groups and the classification of finite groups 1 Finite groups of small order How can we describe all finite groups? Before we address this question, let s write down a list of all the finite groups
More informationECEN 5682 Theory and Practice of Error Control Codes
ECEN 5682 Theory and Practice of Error Control Codes Introduction to Algebra University of Colorado Spring 2007 Motivation and For convolutional codes it was convenient to express the datawords and the
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More informationList of topics for the preliminary exam in algebra
List of topics for the preliminary exam in algebra 1 Basic concepts 1. Binary relations. Reflexive, symmetric/antisymmetryc, and transitive relations. Order and equivalence relations. Equivalence classes.
More informationCDM. Finite Fields. Klaus Sutner Carnegie Mellon University. Fall 2018
CDM Finite Fields Klaus Sutner Carnegie Mellon University Fall 2018 1 Ideals The Structure theorem Where Are We? 3 We know that every finite field carries two apparently separate structures: additive and
More informationQuasi-reducible Polynomials
Quasi-reducible Polynomials Jacques Willekens 06-Dec-2008 Abstract In this article, we investigate polynomials that are irreducible over Q, but are reducible modulo any prime number. 1 Introduction Let
More informationWhat is the Langlands program all about?
What is the Langlands program all about? Laurent Lafforgue November 13, 2013 Hua Loo-Keng Distinguished Lecture Academy of Mathematics and Systems Science, Chinese Academy of Sciences This talk is mainly
More informationbut no smaller power is equal to one. polynomial is defined to be
13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition 13.1. Let K be a field. An element ω K is said
More information* 8 Groups, with Appendix containing Rings and Fields.
* 8 Groups, with Appendix containing Rings and Fields Binary Operations Definition We say that is a binary operation on a set S if, and only if, a, b, a b S Implicit in this definition is the idea that
More informationOverview: The short answer is no because there are 5 th degree polynomials whose Galois group is isomorphic to S5 which is not a solvable group.
Galois Theory Overview/Example Part 2: Galois Group and Fixed Fields I ll repeat the overview because it explains what I m doing with the example. Then I ll move on the second part of the example where
More informationCSIR - Algebra Problems
CSIR - Algebra Problems N. Annamalai DST - INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli -620024 E-mail: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com
More informationTHE CYCLOTOMIC EQUATION AND ITS SIGNIFICANCE TO SOLVING THE QUINTIC EQUATION
THE CYCLOTOMIC EQUATION AND ITS SIGNIFICANCE TO SOLVING THE QUINTIC EQUATION Jay Villanueva Florida Memorial University Miami, FL 33055 jvillanu@fmuniv.edu ICTCM 2013 I. Introduction A. The cyclotomic
More informationSection 6: Field and Galois theory
Section 6: Field and Galois theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Section
More informationNOVEMBER 22, 2006 K 2
MATH 37 THE FIELD Q(, 3, i) AND 4TH ROOTS OF UNITY NOVEMBER, 006 This note is about the subject of problems 5-8 in 1., the field E = Q(, 3, i). We will see that it is the same as the field Q(ζ) where (1)
More informationGalois theory of fields
1 Galois theory of fields This first chapter is both a concise introduction to Galois theory and a warmup for the more advanced theories to follow. We begin with a brisk but reasonably complete account
More informationGALOIS GROUPS AS PERMUTATION GROUPS
GALOIS GROUPS AS PERMUTATION GROUPS KEITH CONRAD 1. Introduction A Galois group is a group of field automorphisms under composition. By looking at the effect of a Galois group on field generators we can
More informationGalois Theory TCU Graduate Student Seminar George Gilbert October 2015
Galois Theory TCU Graduate Student Seminar George Gilbert October 201 The coefficients of a polynomial are symmetric functions of the roots {α i }: fx) = x n s 1 x n 1 + s 2 x n 2 + + 1) n s n, where s
More informationContents. 1 Solving algebraic equations Field extensions Polynomials and irreducibility Algebraic extensions...
Contents 1 Solving algebraic equations............................................ 1 2 Field extensions....................................................... 9 3 Polynomials and irreducibility.........................................
More informationPublic-key Cryptography: Theory and Practice
Public-key Cryptography Theory and Practice Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Chapter 2: Mathematical Concepts Divisibility Congruence Quadratic Residues
More informationarxiv: v1 [math.gr] 3 Feb 2019
Galois groups of symmetric sextic trinomials arxiv:1902.00965v1 [math.gr] Feb 2019 Alberto Cavallo Max Planck Institute for Mathematics, Bonn 5111, Germany cavallo@mpim-bonn.mpg.de Abstract We compute
More informationPart II Galois Theory
Part II Galois Theory Definitions Based on lectures by C. Birkar Notes taken by Dexter Chua Michaelmas 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly)
More information5 Algebraicially closed fields and The Fundamental Theorem of Algebra
5 Algebraicially closed fields and The Fundamental Theorem of Algebra In this chapter all fields will be of characteristic zero. Definition. A field K is algebraically closed if every polynomial in K[x],
More informationDepartment of Mathematics, University of California, Berkeley
ALGORITHMIC GALOIS THEORY Hendrik W. Lenstra jr. Mathematisch Instituut, Universiteit Leiden Department of Mathematics, University of California, Berkeley K = field of characteristic zero, Ω = algebraically
More informationPart II Galois Theory
Part II Galois Theory Theorems Based on lectures by C. Birkar Notes taken by Dexter Chua Michaelmas 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after
More informationGalois Theory and Some Applications
Galois Theory and Some Applications Aparna Ramesh July 19, 2015 Introduction In this project, we study Galois theory and discuss some applications. The theory of equations and the ancient Greek problems
More informationAbstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications
1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the
More informationFinite Fields: An introduction through exercises Jonathan Buss Spring 2014
Finite Fields: An introduction through exercises Jonathan Buss Spring 2014 A typical course in abstract algebra starts with groups, and then moves on to rings, vector spaces, fields, etc. This sequence
More informationVARIETIES WITHOUT EXTRA AUTOMORPHISMS II: HYPERELLIPTIC CURVES
VARIETIES WITHOUT EXTRA AUTOMORPHISMS II: HYPERELLIPTIC CURVES BJORN POONEN Abstract. For any field k and integer g 2, we construct a hyperelliptic curve X over k of genus g such that #(Aut X) = 2. We
More informationPage Points Possible Points. Total 200
Instructions: 1. The point value of each exercise occurs adjacent to the problem. 2. No books or notes or calculators are allowed. Page Points Possible Points 2 20 3 20 4 18 5 18 6 24 7 18 8 24 9 20 10
More informationGENERAL REDUCIBILITY AND SOLVABILITY OF POLYNOMIAL EQUATIONS
General reducibility solvability of polynomial equations GENERAL REDUCIBILITY AND SOLVABILITY OF POLYNOMIAL EQUATIONS E.N.A. P. O. Box CT 75, Cantonments, Accra ABSTRACT A complete work on general reducibility
More informationChapter 4. Fields and Galois Theory
Chapter 4 Fields and Galois Theory 63 64 CHAPTER 4. FIELDS AND GALOIS THEORY 4.1 Field Extensions 4.1.1 K[u] and K(u) Def. A field F is an extension field of a field K if F K. Obviously, F K = 1 F = 1
More information11-6. Solving All Polynomial Equations. Vocabulary. What Types of Numbers Are Needed to Solve Polynomial Equations? Lesson
Chapter 11 Lesson 11-6 Solving All Polynomial Equations Vocabulary double root, root of multiplicity 2 multiplicity BIG IDEA Every polynomial equation of degree n 1 has exactly n solutions, counting multiplicities.
More informationRings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.
Rings 10-26-2008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over
More informationIUPUI Qualifying Exam Abstract Algebra
IUPUI Qualifying Exam Abstract Algebra January 2017 Daniel Ramras (1) a) Prove that if G is a group of order 2 2 5 2 11, then G contains either a normal subgroup of order 11, or a normal subgroup of order
More informationNotes on Polynomials from Barry Monson, UNB
Notes on Polynomials from Barry Monson, UNB 1. Here are some polynomials and their degrees: polynomial degree note 6x 4 8x 3 +21x 2 +7x 2 4 quartic 2x 3 +0x 2 + 3x + 2 3 cubic 2 2x 3 + 3x + 2 3 the same
More informationSelected exercises from Abstract Algebra by Dummit and Foote (3rd edition).
Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 14.2 Exercise 3. Determine the Galois group of (x 2 2)(x 2 3)(x 2 5). Determine all the subfields
More informationName: Solutions Final Exam
Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] For
More informationGalois theory. Philippe H. Charmoy supervised by Prof Donna M. Testerman
Galois theory Philippe H. Charmoy supervised by Prof Donna M. Testerman Autumn semester 2008 Contents 0 Preliminaries 4 0.1 Soluble groups........................... 4 0.2 Field extensions...........................
More informationTHE REGULAR ELEMENT PROPERTY
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 7, July 1998, Pages 2123 2129 S 0002-9939(98)04257-9 THE REGULAR ELEMENT PROPERTY FRED RICHMAN (Communicated by Wolmer V. Vasconcelos)
More informationPartial Fractions. June 27, In this section, we will learn to integrate another class of functions: the rational functions.
Partial Fractions June 7, 04 In this section, we will learn to integrate another class of functions: the rational functions. Definition. A rational function is a fraction of two polynomials. For example,
More informationA BRIEF INTRODUCTION TO LOCAL FIELDS
A BRIEF INTRODUCTION TO LOCAL FIELDS TOM WESTON The purpose of these notes is to give a survey of the basic Galois theory of local fields and number fields. We cover much of the same material as [2, Chapters
More informationPractice problems for first midterm, Spring 98
Practice problems for first midterm, Spring 98 midterm to be held Wednesday, February 25, 1998, in class Dave Bayer, Modern Algebra All rings are assumed to be commutative with identity, as in our text.
More informationInsolvability of the Quintic (Fraleigh Section 56 Last one in the book!)
Insolvability of the Quintic (Fraleigh Section 56 Last one in the book!) [I m sticking to Freleigh on this one except that this will be a combination of part of the section and part of the exercises because
More informationA Harvard Sampler. Evan Chen. February 23, I crashed a few math classes at Harvard on February 21, Here are notes from the classes.
A Harvard Sampler Evan Chen February 23, 2014 I crashed a few math classes at Harvard on February 21, 2014. Here are notes from the classes. 1 MATH 123: Algebra II In this lecture we will make two assumptions.
More informationMath Introduction to Modern Algebra
Math 343 - Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains
More informationINTRODUCTION TO GALOIS THEORY. 1. Introduction and History. one of the most interesting and dramatic tales in the history of mathematics.
INTRODUCTION TO GALOIS THEORY JASON PRESZLER 1. Introduction and History The life of Évariste Galois and the historical development of polynomial solvability is one of the most interesting and dramatic
More informationLecture Notes on Fields (Fall 1997)
Lecture Notes on Fields (Fall 1997) By George F. Seelinger Last Revised: December 7, 2001 NOTE: All references here are either made to Hungerford or to Beachy/Blair (2nd Edition). The references to Hungerford
More informationSection V.2. The Fundamental Theorem (of Galois Theory)
V.2. The Fundamental Theorem (of Galois Theory) 1 Section V.2. The Fundamental Theorem (of Galois Theory) Note. In this section, we define the Galois group of an arbitrary field extension. We prove (after
More informationRECOGNIZING GALOIS GROUPS S n AND A n
RECOGNIZING GALOIS GROUPS S n AND A n KEITH CONRAD 1. Introduction If f(x) K[X] is a separable irreducible polynomial of degree n and G f is its Galois group over K (the Galois group of the splitting field
More informationA PROOF OF BURNSIDE S p a q b THEOREM
A PROOF OF BURNSIDE S p a q b THEOREM OBOB Abstract. We prove that if p and q are prime, then any group of order p a q b is solvable. Throughout this note, denote by A the set of algebraic numbers. We
More information