SOME SPECIAL VALUES OF COSINE
|
|
- Bridget Bradford
- 5 years ago
- Views:
Transcription
1 SOME SPECIAL VALUES OF COSINE JAKE LEVINSON. Introduction We all learn a few specific values of cos(x) (and sin(x)) in high school such as those in the following table: x 0 6 π 4 π π π π cos(x) sin(x) Surely there are other nice values of cos(mπ), where m? In fact, it turns out there are at least a couple that are more or less as nice as those in the above table, such as (.) cos( π 5 ) = 4 ( 5 ), cos( π ) = ( + 6), cos( π 8 ) = +. We ll find these and a few others, using basic Galois theory and algebraic number theory. In particular, we ll classify the values of m Z such that cos( π m ) is, respectively, rational, quadratic, biquadratic and quartic (i.e., a nested quadratic). Our goal is to proceed algebraically as much as possible, so we avoid embedding into C. In fact, the only truly analytic fact we need is (.) cos( π m ) 0 if m 4. We also observe that complex conjugation (always) restricts to the involution ζ ζ of Gal((ζ)/).. Setup (and rational values of cos( π m ) for m Z) Our approach is based on the following fact (pointed out by Adam Kaye): let ζ = ζ m = e πi/m C be a primitive m-th root of unity. Then (.) α = ζ + ζ = Re(ζ) = cos( π m ). In particular, α is real, and ζ satisfies a quadratic polynomial over (α), (.) αζ = ζ +. Assuming m,, so that [(ζ) : ] = φ(m), we therefore have the tower of fields (ζ) φ(m) (α) = (ζ) R φ(m) In particular, the degree of cos( π m ) over is φ(m)/. We ll consider the cases where it is degree 4, since those are, for example, root extensions of, hence reasonably tractable to work with.
2 .. Rational values. First of all, we see that α if and only if m =, or φ(m) =, which is to say m =,,, 4, 6. We note that when m is odd, the identities (.) ζ m = ζ (m+)/ m and ζ m = ζ m show that the cyclotomic extension is the same and that α m is a Galois conjugate of α m (the map ζ m ζ m (m+)/ is an automorphism of (ζ m )). In particular, we can skip m = 6: we ll get it easily once we have done m =. For the others, we have the cyclotomic polynomials (.4) f (x) = x, f (x) = x x = x +, f (x) = x x = x + x +, f 4 (x) = x4 x = x +. For the first two, we obtain ζ =,, respectively, and so (.5) α = cos( π m ) =, for m =,, as we learned in high school. For the other two, we observe that α = (ζ + ζ ) is the linear term of the minimal polynomial of ζ! We conclude that (.6) α =, 0 for m =, 4, respectively, as we had hoped. Finally, for m = 6, we compute using our result from m =, (.7) α 6 = ζ 6 + ζ 6 = ζ ζ = α, so α 6 =. We have shown: Corollary. The only values of m for which cos( π m ) are m =,,, 4, 6. We have, respectively, cos( π m ) =,,, 0,.. uadratic, Biquadratic and uartic values The quadratic values are straightforward, but will nicely illustrate the tools we ll use for the degree-4 case. We make use of the following standard facts: Fact. The cyclotomic field (e πi/p ), where p is an odd prime, contains p if p mod 4 and p if p mod 4. The field (e πi/8 ) contains. Fact. The Galois group of the m-th cyclotomic field is isomorphic to the unit group (Z/mZ). The unit u corresponds to the automorphism ζ ζ u. In particular, complex conjugation is always represented by the element Z/mZ. Also, given a field extension L/K, we denote by NK L and TrL K the norm and trace maps L K, respectively. Our approach is first to identify the extension (α) = R (ζ) as a root extension (i.e. a quadratic, biquadratic or nested quadratic), so that α can be expressed in a familiar basis. We then use the trace and norm of certain intermediate fields to finish the computations.
3 .. uadratic values. For this case, we need φ(m) = 4, which yields m = 5, 8, 0,. As before, we save the case m = odd for the end. For m = 5, the cyclotomic polynomial is (.) f 5 (x) = + x + x + x + x 4. We have (Z/5Z) = Z/4Z, so there is exactly one quadratic subfield, which must be ( 5) by Fact. Hence (α) = ( 5), so α = a + b 5 for some rational a, b. We will compute the trace and norm of α (which are, respectively, a and a 5b ). The nontrivial automorphism σ Gal((α)/)must come from ζ ζ (or, equivalently, ζ ), since the other map restricts to the identity. Hence the Galois conjugate of α is ζ + ζ = ζ + ζ, so the trace and norm are (.) Tr (α) (α) = ζ + ζ + ζ + ζ 4 =, N (α) (α) = (ζ + ζ )(ζ + ζ ) = ζ + ζ 4 + ζ + ζ =. Hence a = and a 5b =, which gives (.) α 5 = cos( π 5 ) = ± 5. Finally, we know (analytically) that cos( π 4 ) should be positive, so we conclude (.4) cos( π 5 ) = 4 ( 5 ). We also get the m = 0 case as a Galois conjugate, namely (.5) α 0 = ζ 0 + ζ 0 = ζ 5 ζ 5 σ(α), the nontrivial Galois conjugate. We conclude (.6) cos( π 0 ) = 4 ( 5 + ). For m = 8,, we can proceed similarly. The cyclotomic polynomials are (.7) f 8 (x) = x 4 +, f (x) = x 4 x +. We observe that (ζ 8 ) contains (ζ 4 ) = (i), and by Fact, hence must be the biquadratic field (i, ). The real subfield is then ( ). Likewise, (ζ ) contains i and, so must be (i, i ), with real subfield ( ). As before, we compute the trace and norm, writing (.8) α 8 = a + b, α = c + d, a, b, c, d. Note that the traces are a, c and the norms are a b and c d. By abuse of notation, we can write the nontrivial automorphism in both cases as the map ζ ζ 5, so the remaining Galois conjugate is (.9) σ(α) = ζ 5 + ζ 5. Hence the trace and norm are (.0) Tr (α) (α) = ζ + ζ + ζ 5 + ζ 5. N (α) (α) = ζ 6 + ζ 4 + ζ 4 + ζ 6. For m = 8, we simplify using ζ 4 8 =, which gives Tr(α 8) = 0 and N(α 8 ) =. For m =, we instead simplify using ζ 6 =, to get Tr(α ) = 0, and ζ 4 = ζ, to get N(α ) =. After solving for a, b, c, d, we obtain: (.) α 8 = cos( π 8 ) = ±, α = cos( π ) = ±.
4 Finally, we know (analytically) that cos( π 4 ) and cos( π 6 ) should be positive. Putting these results together, we have the following: Corollary. The only m Z for which cos( π m ) is quadratic over are m = 5, 8, 0,. We have: m cos( π m ) 4 ( 5 ) 4 ( 5 + )... The Only Biquadratic Value. It may come as a surprise that there is only one value of m making cos( π m ) biquadratic over, namely m = 4. We expected (correctly) that these would be the easiest non-quadratic values, but all the others are all nested quartics. The following facts explain why: Fact (Units mod m). The following facts determine the group structure of (Z/mZ) : (a) If p k is an odd prime power, then (Z/p k ) = Z/(p ) Z/p k. Moreover, any such isomorphism sends (Z/p k ) to the ordered pair ( p, 0) having order. (b) For any k, (Z/ k ) = Z/ Z/ k. Any such isomorphism sends (Z/ k Z) to an element of order. (c) If a, b are coprime, then (Z/ab) = (Z/a) (Z/b), and this isomorphism maps to (, ). Fact 4. Consider the group G = Z/ k Z/ kr and the element g = ( k,..., kr ), which has order two. Then the quotient G/ g has the same decomposition into cyclic groups, except with the smallest k i decremented by. In particular, if G/ g is isomorphic to (Z/) n, then every k i = and n = r. Thus Fact 4 shows that the only way for (α m ) to be biquadratic (with Galois group Z/ Z/) is if (ζ m ) is triquadratic, with Galois group (Z/mZ) = (Z/). By Fact, this can only occur for m = = 4. Note that the cyclotomic polynomial is then (.) f 4 (x) = x 8 x 4 +, and that ζ 4 also satisfies ζ 4 =. For this case, we see that (ζ 4 ) contains (ζ ) and (ζ 8 ), hence is precisely (i,, ). The real subfield is thus (α 4 ) = (, ), so we can write (.) α 4 = ζ 4 + ζ 4 = a + b + c + d 6. A direct calculation shows that the Galois conjugates of α 4 are (.4) β = ζ ζ 9 4, γ = ζ ζ 7 4, δ = ζ 4 + ζ 4. We observe that α 4 = δ. The corresponding automorphism ζ 4 ζ 4 sends ζ ζ, hence fixes α =. Thus it must be the map that negates and 6. We equate coefficients, concluding that (.5) a = c = 0. By similar reasoning, the map ζ 4 ζ 7 4 fixes α 8 =. Thus it negates 6, so (.6) (α + γ) = (b ) = 8b 4
5 and, using ζ 4 =, (.7) (ζ 4 + ζ 4 + ζ7 4 + ζ 7 4) = ζ 8 4 ζ =, where the last equality is from the cyclotomic polynomial. A similar computation with α + β shows that d = ±, and so (.8) α 4 = ± ( ± 6). Both signs must be the same (and both give positive real numbers), but we have chosen our ζ s so that α 4 = α =, which forces the inner sign to be +. Corollary. The only m Z for which cos( π m ) is biquadratic is m = 4, and the resulting value is cos( π 4 ) = 4 ( + 6)... uartic Values. Lastly, we consider the other values of m for which φ(m) = 8, namely m = 5, 6, 0, 0. As usual, we will get m = 0 for free as a Galois conjugate. In all three of these cases, the Galois groups are (.9) Gal((ζ)/) = Z/ Z/4, Gal((α)/) = Z/4. In particular, (α) contains a unique quadratic subfield. In each case, we will identify the subfield as ( r) for some r, then express (α) as ( r, a + b r). In other words, we will ultimately write (.0) cos( π m ) = c + c r + c a + b r + c 4 ar + br r. We will work through the case m = 0; the cases m = 5, 6 are similar. We make the following observations: (ζ 5 ), ( 5), (i) are all subfields of (ζ 0 ); Since Gal((ζ 0 )/) = Z/4 Z/, there is a unique biquadratic subfield, which must therefore be (i, 5). The other two degree-4 subfields have Galois groups isomorphic to Z/4; these must be (α) and (ζ 5 ) (to distinguish them, note that (α) is real and (ζ 5 ) is not.) Putting these thoughts together, we see that the lattice of subfields of (ζ 0 ) is the following: (ζ 0 ) (i, 5) (ζ 5 ) (α) = (ζ) R ( 5) (i) ( 5) In particular, we conclude that the quadratic subfield of (α 0 ) is ( 5). By a similar analysis (identifying the biquadratic subfield and its real quadratic degree- subfield) we determine that the quadratic subfield of (α 5 ) is ( 5) and the quadratic subfield of (α 6 ) is ( ). 5
6 Now we proceed as follows: we know that α 0 is quadratic over ( 5). We will compute its (relative) norm and trace, (.) Tr (α) ( 5) (α) = T = a + b 5, N (α) ( 5) (α) = N = c + d 5. These determine the minimal polynomial of α over ( 5), namely (.) α T α + N = 0, so we can solve for α using the quadratic formula. The Galois conjugates of α over are (.) β = ζ 0 + ζ 7 0, γ = ζ ζ 0, δ = ζ ζ 0. We note that ζ ζ 9 fixes ζ ζ 4 0 = α 5. We proved earlier that (α 5 ) = ( 5), so in fact this is the identity automorphism of ( 5). Thus the nontrivial conjugate of α over ( 5) is δ. Consequently, (.4) (.5) T = α + δ = ζ 0 + ζ ζ 0 + ζ 9 0 = 0, where we have used ζ0 0 =. In particular, N = α δ = ζ ζ 0 + ζ ζ 0 0 = (ζ 0 + ζ 0 ), (.6) α 0 = ± As always, we take the positive root. We apply this same method to the other values m = 5, 6, 0. Our results are summarized as follows: Corollary 4. The only m Z for which cos( π m ) is a nested quartic are m = 5, 6, 0, 0. We have: m cos( π m ) 8 ( ) + 4 Note: the field α 5 and α 0 generate the same field extension over ( ). 6
Galois 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 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 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 informationALGEBRA PH.D. QUALIFYING EXAM SOLUTIONS October 20, 2011
ALGEBRA PH.D. QUALIFYING EXAM SOLUTIONS October 20, 2011 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved
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 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 informationGALOIS THEORY AT WORK: CONCRETE EXAMPLES
GALOIS THEORY AT WORK: CONCRETE EXAMPLES KEITH CONRAD 1. Examples Example 1.1. The field extension Q(, 3)/Q is Galois of degree 4, so its Galois group has order 4. The elements of the Galois group are
More informationMath 121 Homework 6 Solutions
Math 11 Homework 6 Solutions Problem 14. # 17. Let K/F be any finite extension and let α K. Let L be a Galois extension of F containing K and let H Gal(L/F ) be the subgroup corresponding to K. Define
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 informationSOME EXAMPLES OF THE GALOIS CORRESPONDENCE
SOME EXAMPLES OF THE GALOIS CORRESPONDENCE KEITH CONRAD Example 1. The field extension (, ω)/, where ω is a nontrivial cube root of unity, is Galois: it is a splitting field over for X, which is separable
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 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 informationAlgebraic number theory Revision exercises
Algebraic number theory Revision exercises Nicolas Mascot (n.a.v.mascot@warwick.ac.uk) Aurel Page (a.r.page@warwick.ac.uk) TA: Pedro Lemos (lemos.pj@gmail.com) Version: March 2, 20 Exercise. What is the
More informationSome algebraic number theory and the reciprocity map
Some algebraic number theory and the reciprocity map Ervin Thiagalingam September 28, 2015 Motivation In Weinstein s paper, the main problem is to find a rule (reciprocity law) for when an irreducible
More informationKeywords and phrases: Fundamental theorem of algebra, constructible
Lecture 16 : Applications and Illustrations of the FTGT Objectives (1) Fundamental theorem of algebra via FTGT. (2) Gauss criterion for constructible regular polygons. (3) Symmetric rational functions.
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 informationSection V.7. Cyclic Extensions
V.7. Cyclic Extensions 1 Section V.7. Cyclic Extensions Note. In the last three sections of this chapter we consider specific types of Galois groups of Galois extensions and then study the properties of
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 informationORAL QUALIFYING EXAM QUESTIONS. 1. Algebra
ORAL QUALIFYING EXAM QUESTIONS JOHN VOIGHT Below are some questions that I have asked on oral qualifying exams (starting in fall 2015). 1.1. Core questions. 1. Algebra (1) Let R be a noetherian (commutative)
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 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 informationMath 121 Homework 5 Solutions
Math 2 Homework Solutions Problem 2, Section 4.. Let τ : C C be the complex conjugation, defined by τa + bi = a bi. Prove that τ is an automorphism of C. First Solution. Every element of C may be written
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 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 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 informationClassification of Finite Fields
Classification of Finite Fields In these notes we use the properties of the polynomial x pd x to classify finite fields. The importance of this polynomial is explained by the following basic proposition.
More informationLesson 2 The Unit Circle: A Rich Example for Gaining Perspective
Lesson 2 The Unit Circle: A Rich Example for Gaining Perspective Recall the definition of an affine variety, presented last lesson: Definition Let be a field, and let,. Then the affine variety, denoted
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 121. Fundamental Theorem and an example
Math 121. Fundamental Theorem and an example Let K/k be a finite Galois extension and G = Gal(K/k), so #G = [K : k] by the counting criterion for separability discussed in class. In this handout we will
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 AT WORK
GALOIS THEORY AT WORK KEITH CONRAD 1. Examples Example 1.1. The field extension Q(, 3)/Q is Galois of degree 4, so its Galois group has order 4. The elements of the Galois group are determined by their
More informationThus, the integral closure A i of A in F i is a finitely generated (and torsion-free) A-module. It is not a priori clear if the A i s are locally
Math 248A. Discriminants and étale algebras Let A be a noetherian domain with fraction field F. Let B be an A-algebra that is finitely generated and torsion-free as an A-module with B also locally free
More information3 Galois Theory. 3.1 Definitions and Examples
3 Galois Theory 3.1 Definitions and Examples This section of notes roughly follows Section 14.1 in Dummit and Foote. Let F be a field and let f (x) 2 F[x]. In the previous chapter, we proved that there
More informationField Theory Qual Review
Field Theory Qual Review Robert Won Prof. Rogalski 1 (Some) qual problems ˆ (Fall 2007, 5) Let F be a field of characteristic p and f F [x] a polynomial f(x) = i f ix i. Give necessary and sufficient conditions
More informationA SIMPLE PROOF OF KRONECKER-WEBER THEOREM. 1. Introduction. The main theorem that we are going to prove in this paper is the following: Q ab = Q(ζ n )
A SIMPLE PROOF OF KRONECKER-WEBER THEOREM NIZAMEDDIN H. ORDULU 1. Introduction The main theorem that we are going to prove in this paper is the following: Theorem 1.1. Kronecker-Weber Theorem Let K/Q be
More informationLemma 1.1. The field K embeds as a subfield of Q(ζ D ).
Math 248A. Quadratic characters associated to quadratic fields The aim of this handout is to describe the quadratic Dirichlet character naturally associated to a quadratic field, and to express it in terms
More informationAlgebraic number theory
Algebraic number theory F.Beukers February 2011 1 Algebraic Number Theory, a crash course 1.1 Number fields Let K be a field which contains Q. Then K is a Q-vector space. We call K a number field if dim
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 informationGraduate Preliminary Examination
Graduate Preliminary Examination Algebra II 18.2.2005: 3 hours Problem 1. Prove or give a counter-example to the following statement: If M/L and L/K are algebraic extensions of fields, then M/K is algebraic.
More informationp-divisible Groups and the Chromatic Filtration
p-divisible Groups and the Chromatic Filtration January 20, 2010 1 Chromatic Homotopy Theory Some problems in homotopy theory involve studying the interaction between generalized cohomology theories. This
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 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 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 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 informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
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 informationAlgebra Qualifying Exam Solutions. Thomas Goller
Algebra Qualifying Exam Solutions Thomas Goller September 4, 2 Contents Spring 2 2 2 Fall 2 8 3 Spring 2 3 4 Fall 29 7 5 Spring 29 2 6 Fall 28 25 Chapter Spring 2. The claim as stated is false. The identity
More informationSection X.55. Cyclotomic Extensions
X.55 Cyclotomic Extensions 1 Section X.55. Cyclotomic Extensions Note. In this section we return to a consideration of roots of unity and consider again the cyclic group of roots of unity as encountered
More informationHomework 4 Algebra. Joshua Ruiter. February 21, 2018
Homework 4 Algebra Joshua Ruiter February 21, 2018 Chapter V Proposition 0.1 (Exercise 20a). Let F L be a field extension and let x L be transcendental over F. Let K F be an intermediate field satisfying
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 informationField Theory Problems
Field Theory Problems I. Degrees, etc. 1. Answer the following: (a Find u R such that Q(u = Q( 2, 3 5. (b Describe how you would find all w Q( 2, 3 5 such that Q(w = Q( 2, 3 5. 2. If a, b K are algebraic
More informationIntroduction to Algebraic Geometry. Franz Lemmermeyer
Introduction to Algebraic Geometry Franz Lemmermeyer February 6, 2005 Chapter 1 The Unit Circle We will start our journey to the land of algebraic geometry by discussing the simplest algebraic varieties,
More informationAutomorphisms and bases
Chapter 5 Automorphisms and bases 10 Automorphisms In this chapter, we will once again adopt the viewpoint that a finite extension F = F q m of a finite field K = F q is a vector space of dimension m over
More informationCLASS FIELD THEORY WEEK Motivation
CLASS FIELD THEORY WEEK 1 JAVIER FRESÁN 1. Motivation In a 1640 letter to Mersenne, Fermat proved the following: Theorem 1.1 (Fermat). A prime number p distinct from 2 is a sum of two squares if and only
More informationALGEBRAIC NUMBER THEORY - COURSE NOTES
ALGEBRAIC NUMBER THEORY - COURSE NOTES STEVE DONNELLY Housekeeping (1) There ll be a final exam (3 hours or so) on the whole semester (representation theory and algebraic number theory) weighted towards
More informationSolutions 13 Paul Garrett garrett/
(March 1, 005) Solutions 13 Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [13.1] Determine the degree of Q( 65 + 56i) over Q, where i = 1. We show that 65 + 56i is not a suare in
More information(January 14, 2009) q n 1 q d 1. D = q n = q + d
(January 14, 2009) [10.1] Prove that a finite division ring D (a not-necessarily commutative ring with 1 in which any non-zero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.)
More informationSolutions of exercise sheet 11
D-MATH Algebra I HS 14 Prof Emmanuel Kowalski Solutions of exercise sheet 11 The content of the marked exercises (*) should be known for the exam 1 For the following values of α C, find the minimal polynomial
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 informationDirichlet Characters. Chapter 4
Chapter 4 Dirichlet Characters In this chapter we develop a systematic theory for computing with Dirichlet characters, which are extremely important to computations with modular forms for (at least) two
More information9. Finite fields. 1. Uniqueness
9. Finite fields 9.1 Uniqueness 9.2 Frobenius automorphisms 9.3 Counting irreducibles 1. Uniqueness Among other things, the following result justifies speaking of the field with p n elements (for prime
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 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 informationFebruary 1, 2005 INTRODUCTION TO p-adic NUMBERS. 1. p-adic Expansions
February 1, 2005 INTRODUCTION TO p-adic NUMBERS JASON PRESZLER 1. p-adic Expansions The study of p-adic numbers originated in the work of Kummer, but Hensel was the first to truly begin developing the
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 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 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 informationTHE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - SPRING SESSION ADVANCED ALGEBRA II.
THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - SPRING SESSION 2006 110.402 - ADVANCED ALGEBRA II. Examiner: Professor C. Consani Duration: 3 HOURS (9am-12:00pm), May 15, 2006. No
More informationDirichlet Series Associated with Cubic and Quartic Fields
Dirichlet Series Associated with Cubic and Quartic Fields Henri Cohen, Frank Thorne Institut de Mathématiques de Bordeaux October 23, 2012, Bordeaux 1 Introduction I Number fields will always be considered
More information22. Galois theory. G = Gal(L/k) = Aut(L/k) [L : K] = H. Gal(K/k) G/H
22. Galois theory 22.1 Field extensions, imbeddings, automorphisms 22.2 Separable field extensions 22.3 Primitive elements 22.4 Normal field extensions 22.5 The main theorem 22.6 Conjugates, trace, norm
More informationMAT 535 Problem Set 5 Solutions
Final Exam, Tues 5/11, :15pm-4:45pm Spring 010 MAT 535 Problem Set 5 Solutions Selected Problems (1) Exercise 9, p 617 Determine the Galois group of the splitting field E over F = Q of the polynomial f(x)
More informationDirichlet s Theorem and Algebraic Number Fields. Pedro Sousa Vieira
Dirichlet s Theorem and Algebraic Number Fields Pedro Sousa Vieira February 6, 202 Abstract In this paper we look at two different fields of Modern Number Theory: Analytic Number Theory and Algebraic Number
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 informationThe Kronecker-Weber Theorem
The Kronecker-Weber Theorem Lucas Culler Introduction The Kronecker-Weber theorem is one of the earliest known results in class field theory. It says: Theorem. (Kronecker-Weber-Hilbert) Every abelian extension
More informationTHE ARTIN-SCHREIER THEOREM KEITH CONRAD
THE ARTIN-SCHREIER THEOREM KEITH CONRAD 1. Introduction The algebraic closure of R is C, which is a finite extension. Are there other fields which are not algebraically closed but have an algebraic closure
More informationOhio State University Department of Mathematics Algebra Qualifier Exam Solutions. Timothy All Michael Belfanti
Ohio State University Department of Mathematics Algebra Qualifier Exam Solutions Timothy All Michael Belfanti July 22, 2013 Contents Spring 2012 1 1. Let G be a finite group and H a non-normal subgroup
More information18. Cyclotomic polynomials II
18. Cyclotomic polynomials II 18.1 Cyclotomic polynomials over Z 18.2 Worked examples Now that we have Gauss lemma in hand we can look at cyclotomic polynomials again, not as polynomials with coefficients
More informationNumber Theory Fall 2016 Problem Set #3
18.785 Number Theory Fall 2016 Problem Set #3 Description These problems are related to the material covered in Lectures 5-7. Your solutions are to be written up in latex and submitted as a pdf-file via
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 informationD-MATH Algebra II FS18 Prof. Marc Burger. Solution 26. Cyclotomic extensions.
D-MAH Algebra II FS18 Prof. Marc Burger Solution 26 Cyclotomic extensions. In the following, ϕ : Z 1 Z 0 is the Euler function ϕ(n = card ((Z/nZ. For each integer n 1, we consider the n-th cyclotomic polynomial
More informationCOUNTING MOD l SOLUTIONS VIA MODULAR FORMS
COUNTING MOD l SOLUTIONS VIA MODULAR FORMS EDRAY GOINS AND L. J. P. KILFORD Abstract. [Something here] Contents 1. Introduction 1. Galois Representations as Generating Functions 1.1. Permutation Representation
More informationALGEBRA 11: Galois theory
Galois extensions Exercise 11.1 (!). Consider a polynomial P (t) K[t] of degree n with coefficients in a field K that has n distinct roots in K. Prove that the ring K[t]/P of residues modulo P is isomorphic
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 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 informationThese warmup exercises do not need to be written up or turned in.
18.785 Number Theory Fall 2017 Problem Set #3 Description These problems are related to the material in Lectures 5 7. Your solutions should be written up in latex and submitted as a pdf-file via e-mail
More informationNotes on graduate algebra. Robert Harron
Notes on graduate algebra Robert Harron Department of Mathematics, Keller Hall, University of Hawai i at Mānoa, Honolulu, HI 96822, USA E-mail address: rharron@math.hawaii.edu Abstract. Graduate algebra
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 informationMath 553 Qualifying Exam. In this test, you may assume all theorems proved in the lectures. All other claims must be proved.
Math 553 Qualifying Exam January, 2019 Ron Ji In this test, you may assume all theorems proved in the lectures. All other claims must be proved. 1. Let G be a group of order 3825 = 5 2 3 2 17. Show that
More informationRamification Theory. 3.1 Discriminant. Chapter 3
Chapter 3 Ramification Theory This chapter introduces ramification theory, which roughly speaking asks the following question: if one takes a prime (ideal) p in the ring of integers O K of a number field
More informationProfinite Groups. Hendrik Lenstra. 1. Introduction
Profinite Groups Hendrik Lenstra 1. Introduction We begin informally with a motivation, relating profinite groups to the p-adic numbers. Let p be a prime number, and let Z p denote the ring of p-adic integers,
More informationAre ζ-functions able to solve Diophantine equations?
Are ζ-functions able to solve Diophantine equations? An introduction to (non-commutative) Iwasawa theory Mathematical Institute University of Heidelberg CMS Winter 2007 Meeting Leibniz (1673) L-functions
More informationA short proof of Klyachko s theorem about rational algebraic tori
A short proof of Klyachko s theorem about rational algebraic tori Mathieu Florence Abstract In this paper, we give another proof of a theorem by Klyachko ([?]), which asserts that Zariski s conjecture
More informationHONDA-TATE THEOREM FOR ELLIPTIC CURVES
HONDA-TATE THEOREM FOR ELLIPTIC CURVES MIHRAN PAPIKIAN 1. Introduction These are the notes from a reading seminar for graduate students that I organised at Penn State during the 2011-12 academic year.
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 informationCP Algebra 2. Unit 3B: Polynomials. Name: Period:
CP Algebra 2 Unit 3B: Polynomials Name: Period: Learning Targets 10. I can use the fundamental theorem of algebra to find the expected number of roots. Solving Polynomials 11. I can solve polynomials by
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 informationTHE HASSE DAVENPORT RELATION. F = F q. is an automorphism of F, and the group of automorphisms of F is the cyclic group of order r generated by σ p,
THE HASSE DAVENPORT RELATION 1. Environment: Field, Traces, Norms Let p be prime and let our ground field be F o = F p. Let q = p r for some r 1, and let the smaller of our two main fields be The map F
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 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 informationComputing the Galois group of a polynomial
Computing the Galois group of a polynomial Curtis Bright April 15, 2013 Abstract This article outlines techniques for computing the Galois group of a polynomial over the rationals, an important operation
More information