DOCSLIB.ORG

Solubility of Polynomials of the Form X6 Ax

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Solubility of Polynomials of the Form X6 Ax School of Mathematics and Systems Engineering Reports from MSI - Rapporter från MSI Solubility of Polynomials of the Form x6 ax + b Jenny Löwerot Oct MSI Report 06150 2006 Växjö University ISSN 1650-2647 SE-351 95 VÄXJÖ ISRN VXU/MSI/MA/E/--06150/--SE Jenny Löwerot Solubility of Polynomials of the Form x6 ax + b − Master´s Thesis Matematik 2006 Växjö University Abstract This thesis is an attempt to find a criterion of when a sixtic equation of the form x6 ax + b = 0, with a; b Z, is soluble. To do this we use Galois theory and a theorem− by Frobenius. We 2state some different criteria of the coefficients, a and b, which make the sixtic soluble. Sammanfattning Det här arbetet är ett försök att finna ett kriterium för när en sjättegradsekvation på formen x6 ax + b = 0, med a; b Z är lösbar. För att göra detta använder − 2 vi oss av Galoisteori och en sats av Frobenius. Vi visar på några olika kriterier på koefficienterna, a och b, vilka ger lösbarhet. Acknowledgments First of all, I would like to thank my supervisor Per-Anders Svensson for introducing me to the subject and for guiding me through the working process. I also want to thank Per-Anders for his help understanding the necessary commands in the programs Mathematica and GP/PARI. Talking about programs, I would like to thank Robert Nyqvist for his help with Mathematica and LATEX. I also want to thank him for his introduction of Chebotarëv's Density Theorem and the theorem by Frobenius. I would like to thank associate professor Alexander Hulpke at Colorado State University for answering my e-mail quickly and for sorting out the which articles I might need. It helped me understand how humble my own knowledge of Galois groups is. I thank my fiancé Tomas Nilsson for helping me with the installation of the above mentioned programs. I also thank him for reading and commenting the thesis during the writing process. I thank my mother Margareta Löwerot for reading the thesis and helping me find and correct all the faults in the language. She did this without quite liking the subject, or even understanding the purpose of it. That is a fine proof of a mother's love and support. iii Contents 1 Introduction 1 2 Basic Group Theory 2 3 Basic Galois Theory 4 4 The Theorem of Frobenius 5 5 Procedure 6 6 Analysis 8 6.1 Types One, Two, and Three . 8 6.2 Types Four and Eight . 9 6.3 Types Five and Seven . 9 6.4 Type Six . 10 6.5 Type Nine . 11 7 Conclusions 11 A Symbol Index 14 B Program functions 15 C Polynomials and their Galois groups 17 D Output from Mathematica 19 E Graphs 23 E.1 Type One . 23 E.2 Type Two . 27 E.3 Type Three . 28 References 30 iv 1 Introduction The problem of solving polynomial equations has interested mathematicians for ages. According to Stewart [12], the Babylonians had methods for solving some quadratic equations in 1600 BC. The ancient Greeks had other methods for solving quadric equations and their geometric approach also gave them a tool for solving some cubic equations. In AD 1500 a formula for solving cubic equations was found, although it is uncertain who was first. The result was published by Cardano 1545 in Ars Magna, which also contained a method for solving the quartic equation. Algebraic notations of equations were introduced by Descartes in the 17th century ([13], page 96). With modern notation the equations of degree 4 have the solutions shown in table 1.1, where a; b; c; d C. ≤ 2 Degree Equation Solution 1 ax + b = 0; a = 0 x = b 6 − a 2 a a2 2 x + ax + b = 0 x = 2 4 b − q − 3 2 3 3 2 3 3 x3 + ax2 + bx + c = 0 y = q + q + p + q q + p r 2 4 27 r 2 4 27 − q 2 3 − − q3 a a −2a +3b 2a −9ab+27c x = y 3 ; p = 3 ; q = 27 4 3 2 − 4 x + ax + bx + cx + d = 0 too complicated for this presentation Table 1.1: Solutions for polynomial equations Since it was now possible to solve all polynomial equations of degree 4 by radicals, the next problem was how to solve the quintic equation. In 1770 Lagrange≤ proved that the tricks used to solve equations of a lower degree do not work for the quintic. This arose the suspicion that the quintic equation may not be soluble by radicals. The first person to publish a proof for this was Ruffini. He made a first attempt in 1799 in his book Teoria Generale delle Equazioni and then tried again with a better, but still not accurate, proof in a journal in 1813. In 1824 Abel filled the gap in Ruffini´s work and came up with a less faulty but long proof. Kronecker published in 1879 a simpler proof that there is no formula for solving all quintic equations by radicals. But he did not prove that all quintics cannot be solved by a special formula for each and every one. This led to a new question: How can we see if a special equation can be solved by radicals? In 1843 Liouville wrote to the Academy of Science in Paris that, among the papers of the late Galois, he had found a proof that the quintic is insoluble by radicals. This was the origin of the Galois theory, which will be further explained in the next chapter. After this, mathematicians have tried to figure out which equations are soluble and which are not. This has resulted in a sufficient, but not necessary, condition for when quintics are insoluble. Theorem 1.1. Assume that f = x5 + ax4 + bx3 + cx2 + dx + e Q[x]. Then f is insoluble if f is irreducible over Q and has exactly three real zeros.2 It has also resulted in theorems like the one on page 376 in Cox [2]. Theorem 1.2. Assume that f = x5 + ax + b F [x], where a = 0, is irreducible and that F has characteristic 0. Then f is soluble2by radicals over6 F if and only if there are λ, µ F such that 2 3125λµ4 3125λµ5 a = ; b = : (λ 1)4(λ2 6λ + 25) (λ 1)4(λ2 6λ + 25) − − − − 1 This thesis is an attempt to find something similar for the sixtic equation. 2 Basic Group Theory To fully understand the beauty of the Galois theory we need some basic knowledge of groups and fields and their different operations. This chapter will therefore be somewhat of an enumeration of definitions. Definition 2.1 (Group). A group (G; ) is a set G together with an operation ∗ ∗ such that (i) (G; ) is closed under the operation , meaning that for each pair g; h of elemen∗ ts of G, g h is an element of G∗ ∗ (ii) is associative, meaning that for all elements g; h; j of G, ∗ (g h) j = g (h j) ∗ ∗ ∗ ∗ (iii) there is an element e in G such that for all g in G e g = g = g e; ∗ ∗ this e is called the identity element (iv) given an element g in G, there is an element g−1 in G such that g g−1 = e = g−1 g; ∗ ∗ g−1 is called the inverse of g. We call a group abelian if the group operation is commutative, meaning that for ∗ all g and h in G, g h = h g: ∗ ∗ Example 2.1. (Z; +) is an abelian group because for all a Z there is an a−1 = a, 2 − such that a + ( a) = 0, the identity element. But (N; +) is not a group, since 5 N but 5 = N. − 2 − 2 From now on we will use the notation G instead of (G; ) for a group. In abstract algebra there is also another interesting object that we need∗ to define. Definition 2.2 (Field). A field (F; +; ) is a set F with two operations + and , · · called addition and multiplication respectively, such that (i) (F; +) is an abelian group with identity element 0F , called zero (ii) is associative · (iii) is commutative · (iv) the distributive laws are valid for all a; b; c F , meaning that 2 a (b + c) = (a b) + (a c) · · · and (a + b) c = (a c) + (b c) · · · 2 (v) there is an identity element 1 = 0 , such that F 6 F 1 f = f 1 = f F · · F for all f F 2 (vi) all elements except 0F have a multiplicative inverse, meaning that if f = 0 then there exists f −1 F such that f f −1 = 1 : 6 F 2 · F From now on we will use the notation F instead of (F; +; ) for a field. · Example 2.2. Q and R are fields. Z is not a field, since 2 Z, but 2−1 = 1 = Z. 2 2 2 Now we need something to make it easier to work with these new objects. Definition 2.3 (Isomorphism). An isomorphism between two groups K and L is a map σ : K L such that ! σ(a b) = σ(a) σ(b) ∗ ∗ for all a; b K, where σ is one-to-one and onto. If K and L are instead two fields, 2 the map needs to fullfil σ(a + b) = σ(a) + σ(b) and σ(a b) = σ(a) σ(b) · · for all a; b K. 2 An isomorphism from a group to itself is called an automorphism.
Load more

Recommended publications
  • Computing Infeasibility Certificates for Combinatorial Problems Through
    1 Computing Infeasibility Certificates for Combinatorial Problems through Hilbert’s Nullstellensatz Jesus´ A. De Loera Department of Mathematics, University of California, Davis, Davis, CA Jon Lee IBM T.J. Watson Research Center, Yorktown Heights, NY Peter N. Malkin Department of Mathematics, University of California, Davis, Davis, CA Susan Margulies Computational and Applied Math Department, Rice University, Houston, TX Abstract Systems of polynomial equations with coefficients over a field K can be used to concisely model combinatorial problems. In this way, a combinatorial problem is feasible (e.g., a graph is 3- colorable, hamiltonian, etc.) if and only if a related system of polynomial equations has a solu- tion over the algebraic closure of the field K. In this paper, we investigate an algorithm aimed at proving combinatorial infeasibility based on the observed low degree of Hilbert’s Nullstellensatz certificates for polynomial systems arising in combinatorics, and based on fast large-scale linear- algebra computations over K. We also describe several mathematical ideas for optimizing our algorithm, such as using alternative forms of the Nullstellensatz for computation, adding care- fully constructed polynomials to our system, branching and exploiting symmetry. We report on experiments based on the problem of proving the non-3-colorability of graphs. We successfully solved graph instances with almost two thousand nodes and tens of thousands of edges. Key words: combinatorics, systems of polynomials, feasibility, Non-linear Optimization, Graph 3-coloring 1. Introduction It is well known that systems of polynomial equations over a field can yield compact models of difficult combinatorial problems. For example, it was first noted by D.
    [Show full text]
  • Gsm073-Endmatter.Pdf
    http://dx.doi.org/10.1090/gsm/073 Graduat e Algebra : Commutativ e Vie w This page intentionally left blank Graduat e Algebra : Commutativ e View Louis Halle Rowen Graduate Studies in Mathematics Volum e 73 KHSS^ K l|y|^| America n Mathematica l Societ y iSyiiU ^ Providence , Rhod e Islan d Contents Introduction xi List of symbols xv Chapter 0. Introduction and Prerequisites 1 Groups 2 Rings 6 Polynomials 9 Structure theories 12 Vector spaces and linear algebra 13 Bilinear forms and inner products 15 Appendix 0A: Quadratic Forms 18 Appendix OB: Ordered Monoids 23 Exercises - Chapter 0 25 Appendix 0A 28 Appendix OB 31 Part I. Modules Chapter 1. Introduction to Modules and their Structure Theory 35 Maps of modules 38 The lattice of submodules of a module 42 Appendix 1A: Categories 44 VI Contents Chapter 2. Finitely Generated Modules 51 Cyclic modules 51 Generating sets 52 Direct sums of two modules 53 The direct sum of any set of modules 54 Bases and free modules 56 Matrices over commutative rings 58 Torsion 61 The structure of finitely generated modules over a PID 62 The theory of a single linear transformation 71 Application to Abelian groups 77 Appendix 2A: Arithmetic Lattices 77 Chapter 3. Simple Modules and Composition Series 81 Simple modules 81 Composition series 82 A group-theoretic version of composition series 87 Exercises — Part I 89 Chapter 1 89 Appendix 1A 90 Chapter 2 94 Chapter 3 96 Part II. AfRne Algebras and Noetherian Rings Introduction to Part II 99 Chapter 4. Galois Theory of Fields 101 Field extensions 102 Adjoining
    [Show full text]
  • Finite Field
    Finite field From Wikipedia, the free encyclopedia In mathematics, a finite field or Galois field (so-named in honor of Évariste Galois) is a field that contains a finite number of elements. As with any field, a finite field is a set on which the operations of multiplication, addition, subtraction and division are defined and satisfy certain basic rules. The most common examples of finite fields are given by the integers mod n when n is a prime number. TheFinite number field - Wikipedia,of elements the of free a finite encyclopedia field is called its order. A finite field of order q exists if and only if the order q is a prime power p18/09/15k (where 12:06p is a amprime number and k is a positive integer). All fields of a given order are isomorphic. In a field of order pk, adding p copies of any element always results in zero; that is, the characteristic of the field is p. In a finite field of order q, the polynomial Xq − X has all q elements of the finite field as roots. The non-zero elements of a finite field form a multiplicative group. This group is cyclic, so all non-zero elements can be expressed as powers of a single element called a primitive element of the field (in general there will be several primitive elements for a given field.) A field has, by definition, a commutative multiplication operation. A more general algebraic structure that satisfies all the other axioms of a field but isn't required to have a commutative multiplication is called a division ring (or sometimes skewfield).
    [Show full text]
  • 494 Lecture: Splitting Fields and the Primitive Element Theorem
    494 Lecture: Splitting fields and the primitive element theorem Ben Gould March 30, 2018 1 Splitting Fields We saw previously that for any field F and (let's say irreducible) polynomial f 2 F [x], that there is some extension K=F such that f splits over K, i.e. all of the roots of f lie in K. We consider such extensions in depth now. Definition 1.1. For F a field and f 2 F [x], we say that an extension K=F is a splitting field for f provided that • f splits completely over K, i.e. f(x) = (x − a1) ··· (x − ar) for ai 2 K, and • K is generated by the roots of f: K = F (a1; :::; ar). The second statement implies that for each β 2 K there is a polynomial p 2 F [x] such that p(a1; :::; ar) = β; in general such a polynomial is not unique (since the ai are algebraic over F ). When F contains Q, it is clear that the splitting field for f is unique (up to isomorphism of fields). In higher characteristic, one needs to construct the splitting field abstractly. A similar uniqueness result can be obtained. Proposition 1.2. Three things about splitting fields. 1. If K=L=F is a tower of fields such that K is the splitting field for some f 2 F [x], K is also the splitting field of f when considered in K[x]. 2. Every polynomial over any field (!) admits a splitting field. 3. A splitting field is a finite extension of the base field, and every finite extension is contained in a splitting field.
    [Show full text]
  • Splitting Fields
    SPLITTING FIELDS KEITH CONRAD 1. Introduction When K is a field and f(T ) 2 K[T ] is nonconstant, there is a field extension K0=K in which f(T ) picks up a root, say α. Then f(T ) = (T − α)g(T ) where g(T ) 2 K0[T ] and deg g = deg f − 1. By applying the same process to g(T ) and continuing in this way finitely many times, we reach an extension L=K in which f(T ) splits into linear factors: in L[T ], f(T ) = c(T − α1) ··· (T − αn): We call the field K(α1; : : : ; αn) that is generated by the roots of f(T ) over K a splitting field of f(T ) over K. The idea is that in a splitting field we can find a full set of roots of f(T ) and no smaller field extension of K has that property. Let's look at some examples. Example 1.1. A splitting field of T 2 + 1 over R is R(i; −i) = R(i) = C. p p Example 1.2. A splitting field of T 2 − 2 over Q is Q( 2), since we pick up two roots ± 2 in the field generated by just one of the roots. A splitting field of T 2 − 2 over R is R since T 2 − 2 splits into linear factors in R[T ]. p p p p Example 1.3. In C[T ], a factorization of T 4 − 2 is (T − 4 2)(T + 4 2)(T − i 4 2)(T + i 4 2).
    [Show full text]
  • An Extension K ⊃ F Is a Splitting Field for F(X)
    What happened in 13.4. Let F be a field and f(x) 2 F [x]. An extension K ⊃ F is a splitting field for f(x) if it is the minimal extension of F over which f(x) splits into linear factors. More precisely, K is a splitting field for f(x) if it splits over K but does not split over any subfield of K. Theorem 1. For any field F and any f(x) 2 F [x], there exists a splitting field K for f(x). Idea of proof. We shall show that there exists a field over which f(x) splits, then take the intersection of all such fields, and call it K. Now, we decompose f(x) into irreducible factors 0 0 p1(x) : : : pk(x), and if deg(pj) > 1 consider extension F = f[x]=(pj(x)). Since F contains a 0 root of pj(x), we can decompose pj(x) over F . Then we continue by induction. Proposition 2. If f(x) 2 F [x], deg(f) = n, and an extension K ⊃ F is the splitting field for f(x), then [K : F ] 6 n! Proof. It is a nice and rather easy exercise. Theorem 3. Any two splitting fields for f(x) 2 F [x] are isomorphic. Idea of proof. This result can be proven by induction, using the fact that if α, β are roots of an irreducible polynomial p(x) 2 F [x] then F (α) ' F (β). A good example of splitting fields are the cyclotomic fields | they are splitting fields for polynomials xn − 1.
    [Show full text]
  • Finite Fields and Function Fields
    Copyrighted Material 1 Finite Fields and Function Fields In the first part of this chapter, we describe the basic results on finite fields, which are our ground fields in the later chapters on applications. The second part is devoted to the study of function fields. Section 1.1 presents some fundamental results on finite fields, such as the existence and uniqueness of finite fields and the fact that the multiplicative group of a finite field is cyclic. The algebraic closure of a finite field and its Galois group are discussed in Section 1.2. In Section 1.3, we study conjugates of an element and roots of irreducible polynomials and determine the number of monic irreducible polynomials of given degree over a finite field. In Section 1.4, we consider traces and norms relative to finite extensions of finite fields. A function field governs the abstract algebraic aspects of an algebraic curve. Before proceeding to the geometric aspects of algebraic curves in the next chapters, we present the basic facts on function fields. In partic- ular, we concentrate on algebraic function fields of one variable and their extensions including constant field extensions. This material is covered in Sections 1.5, 1.6, and 1.7. One of the features in this chapter is that we treat finite fields using the Galois action. This is essential because the Galois action plays a key role in the study of algebraic curves over finite fields. For comprehensive treatments of finite fields, we refer to the books by Lidl and Niederreiter [71, 72]. 1.1 Structure of Finite Fields For a prime number p, the residue class ring Z/pZ of the ring Z of integers forms a field.
    [Show full text]
  • Galois Groups of Mori Trinomials and Hyperelliptic Curves with Big Monodromy
    European Journal of Mathematics (2016) 2:360–381 DOI 10.1007/s40879-015-0048-2 RESEARCH ARTICLE Galois groups of Mori trinomials and hyperelliptic curves with big monodromy Yuri G. Zarhin1 Received: 22 December 2014 / Accepted: 12 April 2015 / Published online: 5 May 2015 © Springer International Publishing AG 2015 Abstract We compute the Galois groups for a certain class of polynomials over the the field of rational numbers that was introduced by Shigefumi Mori and study the monodromy of corresponding hyperelliptic jacobians. Keywords Abelian varieties · Hyperelliptic curves · Tate modules · Galois groups Mathematics Subject Classification 14H40 · 14K05 · 11G30 · 11G10 1 Mori polynomials, their reductions and Galois groups We write Z, Q and C for the ring of integers, the field of rational numbers and the field of complex numbers respectively. If a and b are nonzero integers then we write (a, b) for its (positive) greatest common divisor. If is a prime then F, Z and Q stand for the prime finite field of characteristic , the ring of -adic integers and the field of -adic numbers respectively. This work was partially supported by a grant from the Simons Foundation (# 246625 to Yuri Zarkhin). This work was started during author’s stay at the Max-Planck-Institut für Mathematik (Bonn, Germany) in September of 2013 and finished during the academic year 2013/2014 when the author was Erna and Jakob Michael Visiting Professor in the Department of Mathematics at the Weizmann Institute of Science (Rehovot, Israel): the hospitality and support of both Institutes are gratefully acknowledged. B Yuri G. Zarhin [email protected] 1 Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA 123 Galois groups of Mori trinomials and hyperelliptic curves..
    [Show full text]
  • 11. Splitting Field, Algebraic Closure 11.1. Definition. Let F(X)
    11. splitting field, Algebraic closure 11.1. Definition. Let f(x) F [x]. Say that f(x) splits in F [x] if it can be decomposed into linear factors in F [x]. An∈ extension K/F is called a splitting field of some non-constant polynomial f(x) if f(x) splits in K[x]butf(x) does not split in M[x]foranyanyK ! M F . If an extension K of F is the splitting field of a collection of polynomials in F [x], then⊇ we say that K/F is a normal extension. 11.2. Lemma. Let p(x) be a non-constant polynomial in F [x].Thenthereisafiniteextension M of F such that p(x) splits in M. Proof. Induct on deg(p); the case deg(p) = 1 is trivial. If p is already split over F ,thenF is already a splitting field. Otherwise, choose an irreducible factor p1(x)ofp(x)suchthat deg(p1(x)) 1. By, there is a finite extension M1/F such that p1 has a root α in M1.Sowe can write p≥(x)=(x α)q(x) in M [x]. Since deg(q) < deg(p), by induction, there is a finite − 1 extension M/M1 such that q splits in M.ThenM/F is also finite and p splits in M. ! 11.3. Lemma. Let σ : F L be an embedding of a field F into a field L.Letp(x) F [x] be an irreducible polynomial→ and let K = F [x]/(p(x)) be the extension of of F obtained∈ by adjoining a root of p.Theembeddingσ extends to K if and only if σ(p)(x) has a root in L.
    [Show full text]
  • Section V.3. Splitting Fields, Algebraic Closure, and Normality (Partial)
    V.3. Splitting Fields, Algebraic Closure, and Normality 1 Section V.3. Splitting Fields, Algebraic Closure, and Normality (Partial) Note. The topic of this section is the identification and construction of Galois extensions. Our attention is turned to factoring polynomials and finding their roots. We restrict our attention to finite collections of polynomials and omit the part of this section concerning roots of infinite collections of polynomials. After this section, we have the equipment to give a mostly-algebriac proof of the Fundamental Theorem of Algebra. Definition. Let F be a field and f F [x] a polynomial of positive degree. f is ∈ said to split over F if f can be written as a product of linear factors in F [x]; that is, f = u0(x u1)(x u2) (x un) with ui F . (So f splits over F if field F − − ··· − ∈ contains all roots of f.) Definition V.3.1. Let K be a field and f K[x] a polynomial of positive degree. ∈ An extension field F of K is a splitting field over K of polynomial f if f splits in F [x] where F = K(u1, u2,...,un) with u1, u2,...,un the roots of f in F . Let S be a set of polynomials of positive degree in K[x]. An extension field F of K is a splitting field over K of the set S of polynomials if every polynomial in S splits in F [x] and F is generated over K by the roots of all the polynomials in S. V.3.
    [Show full text]
  • Constructing Algebraic Closures, I
    CONSTRUCTING ALGEBRAIC CLOSURES KEITH CONRAD Let K be a field. We want to construct an algebraic closure of K, i.e., an algebraic extension of K which is algebraically closed. It will be built out of the quotient of a polynomial ring in a very large number of variables. Let P be the set of all nonconstant monic polynomials in K[X] and let A = K[tf ]f2P be the polynomial ring over K generated by a set of indeterminates indexed by P . This is a huge ring. For each f 2 K[X] and a 2 A, f(a) is an element of A. Let I be the ideal in A generated by the elements f(tf ) as f runs over P . Lemma 1. The ideal I is proper: 1 62 I. Pn Proof. Every element of I has the form i=1 aifi(tfi ) for a finite set of f1; : : : ; fn in P and a1; : : : ; an in A. We want to show 1 can't be expressed as such a sum. Construct a finite extension L=K in which f1; : : : ; fn all have roots. There is a substitution homomorphism A = K[tf ]f2P ! L sending each polynomial in A to its value when tfi is replaced by a root of fi in L for i = 1; : : : ; n and tf is replaced by 0 for those f 2 P not equal to an fi. Under Pn this substitution homomorphism, the sum i=1 aifi(tfi ) goes to 0 in L so this sum could not have been 1. Since I is a proper ideal, Zorn's lemma guarantees that I is contained in some maximal ideal m in A.
    [Show full text]
  • Mid-Semestral Exam 2014-2015
    Mid-Semestral Exam 2014-2015 January 31, 2016 Problem 1.(a). State true or false with justification. For fields F ⊆ K, and α 2 K, if [F (α): F ] is odd then F (α) = F (α2). Proof. Suppose F (α) 6= F (α2). Clearly this implies that α2 = F (α2). We can also conclude that the minimal polynomial of α over F (α2) is x2 − α2. Hence [F (α): F (α2)] = 2. But we know that [F (α): F ] = [F (α): F (α2)][F (α2): F ]. Hence we must have 2j[F (α): F ] and thus we arrive at a contradiction. So F (α) = F (α2) and the given statement is true. Problem 1.(b). State true or false with justification. The regular 5-gon is not constructible by straightedge and compass. Proof. The problem of constructing the regular n-gon is equivalent to the problem of constructing the angle 2π=n which in turn is equivalent to the problem of construct- ing cos(2π=n). In our problem we need to check the constructibility of cos(2π=5). Now cos(2π=5) = (exp2πi=5 + exp−2πi=5)=2. So we have Q ⊂ Q(cos(2π=5)) ⊂ Q(cos(exp2πi=5)). Now Q(cos(exp2πi=5))=Q is a cyclotomic extension of degree 5−1 = 4. Also [Q(cos(exp2πi=5)) : Q(cos(2π=5))] = 2 because cos(exp2πi=5) satisfies the polynomial x2 − 2 cos(2π=5)x + 1 and these two fields can not be equal. Hence [Q(cos(2π=5)) : Q] = 2. By the fundamental theorem of Galois theory this extension is Galois.
    [Show full text]