On the Irreducibility of the Cauchy-Mirimanoff Polynomials

Transcription

1 University of Tennessee, Knoxville Trace: Tennessee Research and Creative Exchange Doctoral Dissertations Graduate School 5-00 On the Irreducibility of the Cauchy-Mirimanoff Polynomials Brian C. Irick University of Tennessee - Knoxville, birick@utk.edu Recommended Citation Irick, Brian C., "On the Irreducibility of the Cauchy-Mirimanoff Polynomials. " PhD diss., University of Tennessee, This Dissertation is brought to you for free and open access by the Graduate School at Trace: Tennessee Research and Creative Exchange. It has been accepted for inclusion in Doctoral Dissertations by an authorized administrator of Trace: Tennessee Research and Creative Exchange. For more information, please contact trace@utk.edu.

2 To the Graduate Council: I am submitting herewith a dissertation written by Brian C. Irick entitled "On the Irreducibility of the Cauchy-Mirimanoff Polynomials." I have examined the final electronic copy of this dissertation for form and content and recommend that it be accepted in partial fulfillment of the requirements for the degree of Doctor of Philosophy, with a major in Mathematics. We have read this dissertation and recommend its acceptance: David Dobbs, Shashikant Mulay, Soren Sorensen (Original signatures are on file with official student records.) Pavlos Tzermias, Major Professor Accepted for the Council: Dixie L. Thompson Vice Provost and Dean of the Graduate School

3 To the Graduate Council: I am submitting herewith a dissertation written by Brian Christopher Irick entitled On the Irreducibility of the Cauchy-Mirimanoff Polynomials. I have examined the final electronic copy of this dissertation for form and content and recommend that it be accepted in partial fulfillment of the requirements for the degree of Doctor of Philosophy, with a major in Mathematics. Pavlos Tzermias Major Professor We have read this dissertation and recommend its acceptance: David Dobbs Shashikant Mulay Soren Sorensen Accepted for the Council: Carolyn Hodges Vice Provost and Dean of the Graduate School (Original signatures are on file with offical student records.)

4 On the Irreducibility of the Cauchy-Mirimanoff Polynomials A Dissertation Presented for the Doctor of Philosophy Degree The University of Tennessee, Knoxville Brian Christopher Irick May 00

5 Copyright c 00 by Brian Irick. All rights reserved. ii

6 Acknowledgments First and foremost, I would like to thank my advisor, Pavlos Tzermias, for suggesting the topic of this dissertation and for all his help during my research. I am particularly grateful for the freedom I had to pursue this in my own way. His thoughtful suggestions and personal encouragement were always what I needed to move forward. I very much appreciate his efforts, and this dissertation is better because of it. I would also like to thank my doctoral committee members David Dobbs, Shashikant Mulay, and Soren Sorensen for their time and interest in this topic. I am particularly indebted to each for my education, and I am grateful to be able to call each a friend. iii

7 Abstract The Cauchy-Mirimanoff Polynomials are a class of polynomials that naturally arise in various classical studies of Fermat s Last Theorem. Originally conjectured to be irreducible over 00 years ago, the irreducibility of the Cauchy-Mirimanoff polynomials is still an open conjecture. Recently, there has been renewed interest in this conjecture including Helou (997), Beukers (997), Tzermias (007) and (009), and Nanninga (009). This dissertation takes a new approach to the study of the Cauchy-Mirimanoff Polynomials. The reciprocal transform of a self-reciprocal polynomial is defined, and the reciprocal transforms of the Cauchy- Mirimanoff Polynomials are found and studied. Particular attention is given to the Cauchy-Mirimanoff Polynomials with index three times a power of a prime, and it is shown that the Cauchy-Mirimanoff Polynomials of index three times a prime are irreducible. iv

8 Contents Introduction. Background Contributions Arrangement The Cauchy-Mirimanoff Polynomials 3. Definition of the Cauchy-Mirimanoff Polynomials The Roots of P n (X) and E n (X) Elementary Facts Reciprocal and Self-Reciprocal Polynomials Root Orbits of the Cauchy-Mirimanoff Polynomials Factors of E n (X) Self-Reciprocal Polynomials Revisited Reduction Modulo a Prime Bounds on the Degrees of Factors of the Cauchy-Mirimanoff Polynomials Other Results The Cauchy-Mirimanoff polynomials are relatively prime E p (X) is irreducible for all primes p A Generalization of the Irreducibility of E p (X) Chebyshev Polynomials 3 3. The Fractional, Half-Fractional, and Modified Half-Fractional Chebyshev Polynomials Properties of the Half-Fractional and Modified Half-Fractional Chebyshev Polynomials The Reciprocal Transform of the Cauchy-Mirimanoff Polynomials 8 4. The Reciprocal Transform of a Self-Reciprocal Polynomial Definition of the Reciprocal Transform of a Self-Reciprocal Polynomial Dickson s Theorem The Reciprocal Transforms of the Cauchy-Mirimanoff Polynomials An Equation for the Reciprocal Transform of the Cauchy-Mirimanoff Polynomials Roots of En(x) and its Translates Dickson s Theorem for the Cauchy-Mirimanoff Polynomials v

9 5 The Irreducibility of E p (x) and E 3p (x) 5 5. The Irreducibility of E p (x) - A New Proof The Irreducibility of E 3p (x) - First Proof The Newton Polygon of E3p (x ) Root Orbits of E 3p (x) and E3p (x) The Irreducibility of E3p (x) and E 3p(x) A Study of E 3p i(x) The Newton Polygon of E (x ) p i 6. The Root Orbits of E 3p i(x) Technical Results T z = 6 for all Roots z of E 3p i(x) Additional Results Bibliography 4 Appendices 43 A Numerical Evidence Supporting Conjectures.3.0 and Vita 48 vi

10 List of Tables A. For odd n, the first ten primes p, not dividing Disc(E n (X)), such that E n factors into exactly two distinct irreducible polynomials over F p A. For even n, the first ten primes p, not dividing Disc(E n (X)), such that E n is irreducible over F p 46 vii

11 Chapter Introduction In this dissertation, the factorization of a class of polynomials that originally arose in the first proof of Fermat s Last Theorem for the special case n = 7 is studied. This class of polynomials has come to be known as the Cauchy-Mirimanoff Polynomials, and it is a long standing conjecture that every Cauchy-Mirimanoff polynomial is irreducible.. Background In 839, Gabriel Lamé proved the special case of Fermat s Last Theorem for exponent n = 7. In the proof, Lamé used the polynomial identity: (X + Y) 7 X 7 Y 7 = 7XY(X + Y)(X + XY + Y ) In examining Lamé s proof, Augustin-Louis Cauchy and Joseph Liouville indicated a more general identity of which the following is a special case (set Y = ): P n (X) (X + ) n X n = X(X + ) ɛ n (X + X + ) e n E n (X) (.) where for even n, ɛ n = e n = 0; for odd n, ɛ n = and e n = 0,, or according as n 0,, or (mod 3). The remaining factor E n (X) of P n (X) is called the n th Cauchy-Mirimanoff Polynomial. Dimitry Mirimanoff studied E n (X) extensively for prime n, and conjectured its irreducibility over Q[X] (Mirimanoff, 903). Indeed, this conjecture seems to hold for all n. The interested reader may find a more extensive account of the historical connection of the Cauchy-Mirimanoff Polynomials to Fermat s Last Theorem in Ribenboim (979). Little progress was made on the conjecture until the mid 990 s. Helou (997) studied the Galois group of the Cauchy-Mirimanoff Polynomials and included a proof, attributed to Michael Filaseta, that E p (x) is irreducible for all primes p. Many of the results from Helou (997) will be of use in this dissertation and are stated in Sections..3,.3., and.3.. Independently, in an application to the Korteweg-de Vries equation, Beukers (997) proved that the Cauchy-Mirimanoff Polynomials are relatively prime to each other. Tzermias (007) studied the Cauchy-Mirimanoff polynomials of prime index and obtained lower bounds for the degrees of some factors of the Cauchy-Mirimanoff polynomials of prime index p with p (mod 3). Recently, Tzermias (009) obtained bounds for the degrees of some factors of the Cauchy-Mirimanoff polynomials of prime index p with p (mod 3). The main results of both papers are stated in Section.3.3.

12 In a submitted Ph.D. thesis, Nanninga (009) has announced a proof that E n (X) is irreducible when n = k m where m is an odd integer and k {,, 3, 4, 5}. The thesis is under revision, but the results would seem to indicate a generalization of Michael Filaseta s proof of the irreducibility of E p (X).. Contributions This dissertation begins with a broad overview of the known results regarding the Cauchy-Mirimanoff Polynomials. Then the reciprocal transform of a self-reciprocal polynomial is defined, and the reciprocal transforms of the Cauchy-Mirimanoff Polynomials are found (Theorem 4..). It is worth noting that the reciprocal transforms of the Cauchy-Mirimanoff polynomials are closely related to a problem of minimization in Approximation Theory (Chebyshev Approximation), and are also related to Dickson Polynomials. In other words, the Cauchy-Mirimanoff Polynomials could have applications to cryptography, coding theory, and applied mathematics (in addition to the Korteweg-de Vries equation). After applying Dickson s Theorem to the Cauchy-Mirimanoff Polynomials (Theorem 4.3.), we specialize to the polynomials E 3p i(x) for primes p and i N. In particular, it is shown that E 3p (X) is irreducible over Q, and in fact, three distinct proofs are provided (see Section 5..3, Corollary 6..9, and Corollary 6.3.5). It is also shown that for any i N, E 3p i(x) is a product of no more than i irreducible polynomials (Corollary 6..8), every irreducible factor of E 3p i(x) has degree d 3(p ) (Corollary 6.3.7), and E 3p i(x) has an irreducible factor of degree d 3(p )p i (Corollary 6.3.6). It is also shown that E 3p (X) is either irreducible, or a product of two irreducible factors of degree 3(p ) and 3(p )p (Corollary 6.3.8)..3 Arrangement Chapter defines the Cauchy-Mirimanoff Polynomials, and contains many of the known results regarding these polynomials including results from Mirimanoff (903), Helou (997), Beukers (997), Tzermias (007, 009), and Nanninga (009). Reciprocal and self-reciprocal polynomials are defined and studied. Chapter 3 defines the Fractional, Half-Fractional, and Modified Half-Fractional Chebyshev Polynomials. These functions, which are not necessarily polynomials, are closely related to the classic Chebyshev Polynomials and have a number of analogous properties. These properties will be useful in applications to the Cauchy-Mirimanoff Polynomials, and do not seem to appear in the literature. Chapter 4 defines the concept of the reciprocal transform of a self-reciprocal polynomial. This is not a new idea, but there is no universal agreement to name or notation. Dickson s Theorem, which relates factorization of a self-reciprocal polynomial to the factorization of the reciprocal transform, is stated and applied to the Cauchy-Mirimanoff Polynomials. An explicit formula for the reciprocal transform of the Cauchy-Mirimanoff Polynomials is found and the roots are studied. Chapter 5 begins with a new proof that E p (X) is irreducible over Q. Then, the Newton Polygon of the reciprocal transform of E 3p (X) is studied. Subsequently combining many of the results from Chapters and 4, a proof that E 3p (X) is irreducible is given. Chapter 6 generalizes the results of Chapter 5 to the Cauchy-Mirimanoff Polynomials E 3p i(x). Two additional proofs of the irreducibility of E 3p (X) are obtained as corollaries to more general theorems. It is shown that for any i N, E 3p i(x) is a product of no more than i irreducible polynomials, every irreducible factor of E 3p i(x) has degree d 3(p ), and E 3p i(x) has an irreducible factor of degree d 3(p )p i. It is also shown that E 3p (X) is either irreducible, or a product of two irreducible factors of degree 3(p ) and 3(p )p.

13 Chapter The Cauchy-Mirimanoff Polynomials In this chapter, the Cauchy-Mirimanoff Polynomials are defined and many of the known facts regarding these polynomials are presented.. Definition of the Cauchy-Mirimanoff Polynomials Proposition... Let n be an integer, ω a primitive third root of unity, and define P n (X) Z[X] as Then. 0 is a simple root of P n (X). P n (X) (X + ) n X n. If n is even, then is not a root of P n (X); if n is odd, then is a simple root of P n (X). 3. If n is even, then ω is not a root of P n (X); if n is odd, then ω is a root of P n (X) of multiplicity 0,, or according as n 0,, or (mod 3). Proof. Proceed by examining cases. For all n : X P n (X) : P n (0) = 0. X P n (X) : If X P n (X), then 0 is a root of dp n dx. However dp n(0) dx = n 0. Suppose n is even: (X + ) P n (X) : P n ( ) = 0. (X + X + ) P n (X) : P n (ω) = i sin( πn 3 ). So either P n(ω) has a nonzero imaginary part or equals. In either case P n (ω) 0. Suppose n is odd: (X + ) P n (X) : P n ( ) = 0. (X + ) P n (X) : If (X + ) P n (X), then is a root of dp n dx. However, dp n( ) dx = n 0. 3

14 Suppose n is odd and n 0 (mod 3): (X + X + ) P n (X) : P n (ω) = 3 0. Suppose n is odd and n (mod 3): (X + X + ) P n (X) : It is enough to check that ω is a root of both P n (X) and dp n dx. Both are easily verified. (X + X + ) 3 P n (X) : If (X + X + ) 3 P n (X), then ω is a root of d P n dx. However, in this case, d P n (ω) dx = n(n ) 0. Finally, suppose n is odd and n (mod 3): (X + X + ) P n (X) : P n (ω) = 0. (X + X +) P n (X) : If (X + X +) P n (X), then ω is a root of dp n dx. However, dp n(ω) dx = 3in 0. Often Proposition.. is stated more succinctly in terms of the factorization of P n (X). This leads to the definition of the Cauchy-Mirimanoff Polynomials. Definition.. (Cauchy-Mirimanoff Polynomials). Let n be an integer. The n th Cauchy-Mirimanoff Polynomial is the remaining factor E n (X) of P n (X), in Q[X], after removing X and the cyclotomic factors. Specifically, P n (X) = X(X + ) ɛ n (X + X + ) e n E n (X) (.) where for even n, ɛ n = e n = 0; for odd n, ɛ n = and e n = 0,, or according as n 0,, or (mod 3). Corollary..3. If n is even, then deg(e n ) = n. If n is odd, then deg(e n ) = n 3 e n. Proof. By inspection, deg(p n ) = n. By Proposition.., if n is even, then X is a factor of P n (X) deg(e n ) = n. Similarly, if n is odd, then X and X + are factors of P n (X) of degree, and X + X + is a factor of degree e n deg(e n ) = n 3 e n.. The Roots of P n (X) and E n (X) This section collects many of the known facts about the roots of both P n (X) and E n (X)... Elementary Facts Theorem... Let n. Then. P n (X) has no roots that are roots of unity beside and ω in the cases already handled.. P n (X) has no real roots except 0 and (n odd). 3. E n (X) has no roots that are roots of unity. 4. E n (X) has no real roots. 5. E n (X) has no repeated roots in any splitting field over Q. 4

15 Proof.. Let ζ m = e πi m, m 4, and n 3. Then P n (ζ m ) = (ζ m + ) n ζm n (ζ m + ) n ζm n + = ζ n m (ζ m + ζ m ) n ζm n + = ( cos( π m ))n ζm n +. As ( cos( π m ))n > ζm n +, it follows ( cos( π m ))n ζm n + ( cos( π m ))n ζm n + 0 ( cos( π m ))n ζm n + > 0 P n (ζ m ) > 0.. Observe dp n dx = n(x + )n nx n. If α R such that dp n(α) dx = 0 then n(α + ) n = nα n. If n is even, no solution exists; if n is odd, then α = is the only solution. Consequently, if n is even, then P n (X) can have at most one real root; if n is odd, then P n (X) can have at most two roots. From Proposition.., P n (X) has 0 as a root if n is even, and 0 and as roots if n is odd, and so these are the only real roots of P n (X). 3. Consequence of above. 4. Consequence of above. 5. Suppose α C is a multiple root of P n (X). Then both P n (α) and dp n(α) dx equal 0. This implies that (α + ) n α n = 0 and (α + ) n = α n (α + ) n α(α + ) n = (α + ) n = α n =. Therefore, α must be a root of unity. The only roots of unity of P n (X) are and ω, from above, and the multiplicity of those roots was handled in Proposition... Therefore, ω and ω ( complex conjugate of z) are the only possible multiple roots of P n (X) (which do occur in certain cases) E n (X) has no multiple roots... Reciprocal and Self-Reciprocal Polynomials Definition.. (Reciprocal and Self-Reciprocal Polynomials). Let D be an integral domain, 0 p(x) D[x], and let d deg(p). The reciprocal polynomial of p(x) is defined to be the polynomial p (x) x d p( x ). A polynomial is said to be self-reciprocal if it equals its reciprocal, that is, p(x) = p (x). There is no universally agreed upon definition or notation for reciprocal and self-reciprocal polynomials in the literature. The reciprocal of a polynomial is sometimes called the reversal ; self-reciprocal polynomials are sometimes simply called reciprocals or palindromes. Proposition..3. Let D be an integral domain, p(x) D[x], and d deg(p). Proof.. If x p(x), then d = deg(p ).. If x p(x), then d > deg(p ) 3. If p(x) is self-reciprocal, then x p(x).. Let p(x) = c d x d + + c 0. Then p (x) = x d ( c d x d + + c 0 ) = cd + + c 0 x d.. Let p(x) = c d x d + + c α x α with d α > 0. Then p (x) = x d ( c d x d + + c α x α ) = cd + + c α x d α. 3. Assume x p(x). Then, by, d > deg(p ), yet d = deg(p ) p(x) = p (x), a contradiction. 5

16 An alternative way to characterize reciprocal and self-reciprocal polynomials is by their roots. Theorem..4. Let D be an integral domain, F the quotient field of D, and p(x) D[x] with nonzero constant coefficient c 0. Let f (x) F[x] be the monic polynomial whose roots are precisely the multiplicative inverses of the roots of p(x) (counting multiplicity). Then c 0 f (x) = p (x). Proof. Let F be the splitting field of p(x) over F. In F[x], suppose p(x) = c n (x z ) α (x z n ) α n with the z i s unique roots of p(x), and α i the multiplicity of the root z i for i = n. Let d deg(p) = α + + α n. By the definition of f, ( f (x) = x ) α ) (x zn αn z = x d z α zα n n = ( )d x d = z α zα n n x d ( ) d z α zα n n = p (x) c 0 ( z α ( z n x) ) αn x ( ) α ( x z ) p ( x c n x z n Theorem..5. Let D be an integral domain and p(x) D[x] such that ± are not roots of p(x). Then p(x) is self-reciprocal if and only if when z is a root of p(x) of multiplicity α, then z is also a root of p(x) of multiplicity α. Proof. Let F be the quotient field of D, and let F be the splitting field of p(x) over F. ( ) Suppose p(x) is self-reciprocal with d deg(p). By Proposition..3, all roots of p(x) are nonzero, so the multiplicative inverse of each root of p(x) exists in F. Factoring over F[x], say p(x) = c(x z ) α (x z n ) α n with the z i s distinct roots of p(x) each with multiplicity α i. Consequently, p(x) = p (x) ( ) = x d p x = x d c ( x z = ( ) d cz α zα n n ) α ( ) αn x z n ( x ) α z ) αn (x zn ) αn So for each root z i with multiplicity α i of p(x), it is clear that z i is also a root of p(x) with multiplicity α i. ( ) Suppose that when z is a root of p(x) of multiplicity α then z is also a root p(x) of multiplicity α, and note z and z are distinct since z ±. Let c d be the leading coefficient of p(x), c 0 the constant coeffient of p(x), and let f (x) be the monic polynomial whose roots are precisely the multiplicative inverses of the roots of p(x) (counting multiplicity). Consequently, c d f (x) = p(x). By Theorem..4, f (x) = p (x) c 0 so 6

17 c d c 0 p (x) = p(x). The fraction c d c 0 is up to sign the product of the roots of p(x). Therefore, by hypothesis, c d c 0 = ( ) deg(p). Also by hypothesis, it is clear that the number of roots, counting multiplicity, of p(x) is even, so deg(p) is even c d c 0 = p (x) = p(x) p(x) is self-reciprocal. Theorem..6 (Cauchy-Mirimanoff Polynomials are self-reciprocal). Let n. Then. Proof. P n (X) X is a self-reciprocal polynomial.. E n (X) is a self-reciprocal polynomial.. Fix n. Then ( P n(x) X ) = X P ( ) n n X X (( ) n = X n X + ) X n = X ((X + )n X n ) = P n(x) X. Fix n, and let z be a root of E n (X). By Theorem.., z is a root of multiplicity one, so by Theorem..5 it is enough to show z is also a root of E n(x). As z is a root of E n (X), it is also a root of P n (X). Clearly, z 0 E n (X) has no real roots, so z is a root of P n(x) X. Since P n(x) X is a self-reciprocal polynomial, z is a root of P n(x) X. As z, ω, or ω it follows z, ω or ω z is not a root of X + or X + X + z must be a root of E n(x)...3 Root Orbits of the Cauchy-Mirimanoff Polynomials Since the Cauchy-Mirimanoff polynomials are self-reciprocal, if z is a root of E n (X) then z E n (X). While this is true for all n, even more can be said when n 9 is odd. is also a root of Definition..7. Let n 9 be odd, and let z be a root of E n (X). The orbit of z, denoted Orb(z), is the set { Orb(z) z,, z, z z +, z, z } z + Theorem..8. Let n 9 be odd, and let z be a root of E n (X). Then the elements of Orb(z) are distinct roots of E n (X). Proof. Observe that P n ( X ) = ( X) n ( X ) n = P n (X) if P n (z) = 0 then P n ( z ) = 0. Consequently, if E n (z) = 0 then z is a root of P n (X). Since z is not real, z is not real z is not a root of X or X +. If z were a root of X + X +, then ( z ) + ( z ) + = z + z + = 0 z = ω 7

18 or ω, which is impossible. Therefore, z must be a root of E n (X). Since E n (X) is self-reciprocal, z+ is also a root of E n (X). Similarly, since z is a root of E n(x), the same reasoning shows z and z z+ are roots of E n (X). As z,,, ω or ω, a brute force comparison of the elements of Orb(z) verifies that the elements are distinct. The set Orb(z) is the orbit of z under the action of the group of unimodular transformations T = {X, X, X, X+, X+ X, X X+ } on C {0, }. The group T is isomorphic to G 3, the symmetric group on three elements. By Corollary..3, for odd n 9, the roots of E n (X) are partitioned into r n n 3 e n 6 orbits. This leads to an important observation in Helou (997). Theorem..9 (Helou (997) - Lemma ). For odd n 9, the roots of E n (X) in C are partitioned into r n orbits; and E n (X) has exactly r n roots of absolute value, two conjugates in each orbit. It is natural to consider whether the elements of a root orbit of E n (X) are Q-conjugates. So for every root z of E n (X) in C, let g z (X) be the monic polynomial with roots the elements of Orb(z). Definition..0. Let J(X) Q(X) be the rational function J(X) (X + X + ) 3 X (X + ) Proposition.. (Helou (997)). Let n 9 be odd. For z a root of E n (X) in C, g z (X) = X 6 + 3X 5 + (6 J(z))X 4 + (7 J(z))X 3 + (6 J(z))X + 3X + All elements of Orb(z) have the same image under J. In particular, if θ π (mod π) in R, then J(e iθ ( cos θ+)3 ) = (cos θ+). So, if z is a root of E n(x), then J(z) R and g z R[X]. In fact, more can be said of J(z). Proposition.. (Helou (997) - Lemma 3). For odd n 9 and z a root of E n (X), J(z) is a real algebraic number. If n is a prime, then J(z) is a real algebraic integer. Lastly, Helou (997) gives a lemma which will be used to give a partial answer to whether the elements of Orb(z), for a root z of E n (X), are Q-conjugates. Theorem..3 (Helou (997) - Lemma 4). Let n 9 be odd, z a root of E n (X) in C, and K Q(J(z ),, J(z rn )) where z j = e iθ j ( j r n ) are representatives of the root orbits of E n (X) in C.. We have Q(z) K = Q(J(z)) and Gal(K(z) K) Gal(Q(z) Q(J(z))) T z, where T z is a subgroup of T of order T z {, 6}.. For a Q-conjugate z of z, T z = T z. The minimal polynomial of z over Q is the product of [Q(J(z)) : Q] minimal polynomials over K of such z in different orbits. 3. If T z = then, for any z Orb(z), T z = ; and, if z = then all the roots of the minimal polynomial of z over Q have absolute value. It is not immediately obvious to see how Theorem..3 gives a partial answer to whether elements of Orb(z) are Q-conjugates of each other. This connection will be made clear through the next results. Lemma..4. Let n 9 be odd, and z C a root of E n (X). Then for any z Orb(z), we have z Q(z). Proof. Straightforward observation. 8

19 Lemma..5. Let n 9 be odd, and z C a root of E n (X). Then ( ) ( Q(z) = Q = Q( z ) = Q ) ( = Q ) z z + z ( = Q z ) z + Proof. Consequence of Lemma..4. Lemma..6. Let n 9 be odd, and z C a root of E n (X). Let z Orb(z) be such that there exists σ Aut(Q(z)) with σ(z) = z. Then σ Gal(Q(z)/Q(J(z))). Proof. It is enough to show σ(j(z)) = J(z). Observe σ(j(z)) = J(σ(z)) = J(z ). As z Orb(z), both z and z have the same image under J, i.e. J(z ) = J(z) σ(j(z)) = J(z). Corollary..7. Let n 9 be odd, z C a root of E n (X), and z Orb(z) such that z z, z. If z and z are Q-conjugates, then T z = 6. Proof. Let f (x) be the minimal polynomial of z over Q, and let K be the splitting field of f. Since z and z are Q-conjugates, there exists σ Gal(K/Q) such that σ(z) = z. Define φ : Q(z) Q(z ) = Q(z) by φ σ Q(z), and note φ Aut(Q(z)). It follows from Lemma..6 that φ Gal(Q(z)/Q(J(z))). The complex conjugate root theorem guarantees that z and z are Q-conjugates, so as in the previous paragraph, there also exists γ Gal(Q(z)/Q(J(z))) such that γ(z) = z. As id, σ, and γ are three distinct elements of Gal(Q(z)/Q(J(z))), it follows that T z 3. Consequently, by Theorem..3 (), T z = 6. Finally, Theorem..8 and Corollary..9 make explicit how to use Theorem..3 to determine if the elements of Orb(z) are Q-conjugates. Theorem..8. Let n 9 be odd, and z e iθ be a root of E n (X). Then the following are equivalent:. T z =. The only Q-conjugate of z in Orb(z) is z. 3. The only Q-conjugate of z in Orb(z) is z. 4. The only Q-conjugate of z z+ in Orb(z) is z+. Proof. (, 3, 4) Suppose T z =. By the complex conjugate root theorem, z and z are Q-conjugates, z and z are Q-conjugates, and z+ and z z+ are Q-conjugates. By..3 (3), all Q-conjugates of z have absolute value, so the only possible Q-conjugate of z in Orb(z) is z, which establishes (). To establish (3), suppose z had a Q-conjugate other than itself or z in Orb(z). Then by Corollary..7, T z = 6, which contradicts Theorem..3 (3). Similar reasoning establishes (4). (, 3, 4) By Theorem..3 (), if T z, then T z = 6. So suppose T z = 6. As Gal(Q(z)/Q(J(z))) Aut(Q(z)), it follows by the Isomorphism Extension Theorem that the elements of Orb(z) are Q-conjugates of each other. As this is prohibited by (), (3), or (4), it follows T z 6 T z =. Corollary..9. Let n 9 be odd, and z C be a root of E n (X). Then T z = 6 if and only if the elements of Orb(z) are Q-conjugates. Proof. ( ) Established in the proof of Theorem..8 for the ( )-direction. ( ) Consequence of Corollary..7. 9

20 .3 Factors of E n (X).3. Self-Reciprocal Polynomials Revisited Theorem.3.. Suppose ± z C such that z =. If z is algebraic over Q, then the minimal polynomial of z over Q is a self-reciprocal polynomial of even degree. Proof. Since z ± and z =, it follows that z z and z = z. Let f (x) be the minimal polynomial of z over Q, and define d deg( f ). Observe f (z) = z d f ( z ) = zd f (z) = z d f (z) = 0 z is a root of f (x). If f (0) = 0, then x f (x) f (x) = x because f (x) is monic and irreducible. The only root of f (x) = x is 0, which contradicts the hypothesis that f (x) is the minimal polynomial of z with z =. So f (0) 0 deg( f ) = deg( f ) by Proposition..3. Because f (x) is the minimal polynomial of z, it follows f (x) f (x) f (x)h(x) = f (x) for some h(x) Q[x]. As deg( f ) = deg( f ), it follows deg(h) = 0 h(x) = q for some 0 q Q q f (x) = f (x). Let z be any root of f (x). Then z C ( z 0), z z ( z ±), and z is a root of f (x) ( f ( z ) = q ( z ) d f (z ) = 0). Therefore, the roots of f (x) may be partitioned into sets of order consisting of a complex number and its multiplicative inverse. In particular, the degree, d, of f (x) is even and the product of the roots is. Consequently, f (0) =. The leading coefficient of q f (x) is q because f (x) is monic. As f (0) 0, the leading coefficient of f (x) equals the constant coefficient of f (x) the leading coefficient of f (x) is. Because q f (x) = f (x) and the leading coefficients must be equal, q = f (x) = f (x) f (x) is self-reciprocal. This theorem provides information about any possible factors of the Cauchy-Mirimanoff Polynomials. Indeed, a necessary condition that the Cauchy-Mirimanoff polynomials be irreducible is that all the elements in Orb(z) for any root z of E n (X) are Q-conjugates of each other. As the next two corollaries show, this condition is equivalent to checking that factors of E n (X) are self-reciprocal. Corollary.3.. Let n 9 be odd. Then at least one irreducible factor of E n (X) over Q (or Z) is selfreciprocal. Proof. By Theorem..9, there exists at least one root z of E n (X) such that z =. Let f (x) be the minimal polynomial of z, and note that f (x) is an irreducible factor of E n (X). By Theorem.3., f (x) is self-reciprocal. Corollary.3.3. Let n 9 be odd. If T z = 6 for all roots z of E n (X), then all irreducible factors of E n (X) over Q (or Z) are self-reciprocal polynomials. Proof. Suppose T z = 6 for all roots of E n (X). Let z be a root of E n (X). By Corollary..9, every element of Orb(z) is a Q-conjugate of z. Consequently, by Theorem..9, there are at least two Q-conjugates of z with absolute value. Therefore, the minimal polynomial of z, which is an irreducible factor of E n (X) over Q, is a self-reciprocal polynomial by Theorem.3.. This shows that techniques used to study self-reciprocal polynomials could be helpful in determining the irrreducibility or reducibility of the Cauchy-Mirimanoff polynomials. Theorem.3.4 (Helou (997) - Proposition ). For prime n and any root z of E n (X) in C, g z (x) is irreducible over Q(J(z)) and Gal(Q(z) Q(J(z))) G 3. Any irreducible factor of E n (X) over Q is a product of some of the g z s. 0

21 Corollary.3.5. Let p be prime. All irreducible factors of E p (X) are self-reciprocal. Proof. From Theorem.3.4, for any root z of E n (X) it follows that T z = 6. The result is immediate from Corollary Reduction Modulo a Prime Often, to test the irreducibility of a polynomial, it is useful to consider the polynomial over F p for a prime p. It is well-known that if a primitive polynomial ( a polynomial such that the greatest common denominator of the coefficients is ) over Z[X] is irreducible over F p [X] for some prime p, then the polynomial is irreducible over Z[X]. Of course, the converse isn t true. In other words, there exist primitive polynomials that are irreducible over Z[X], yet are reducible modulo every prime (the cyclotomic polynomials are such a class of polynomials). Helou (997) considered the Cauchy-Mirimanoff polynomials modulo a prime. Theorem.3.6 (Helou (997) - Lemma 6). Let f Z[X] be a self-reciprocal polynomial such that there exists z in C {0, } for which Orb(z) consists of six distinct roots of f. Then f is reducible modulo every prime p. Definition.3.7. Let f (x) Z[x] and p a prime. The reduction of f (x) modulo p, denoted f (x), is the polynomial in Z/pZ[x] whose coefficients equal the coefficients of f (x) reduced modulo p. Helou applied Theorem.3.6 to obtain the next two results. Corollary.3.8 (Helou (997) - Lemma 6). For odd n 9 and any prime number p, E n is reducible modulo p. Corollary.3.9 (Helou (997) - Proposition 3). Let n be a prime. If, for some prime p, E n has at most 3 irreducible factors in F p [X], then E n is irreducible in Q[X]. One might wonder how relevant Corollary.3.9 is to testing the irreducibility of the Cauchy-Mirimanoff polynomials. If the hypothesis of Corollary.3.9 is satisfied frequently, then Corollary.3.9 provides a compelling way to try to prove that the Cauchy-Mirimanoff polynomials are irreducible. After performing a series of numerical calculations on the Cauchy-Mirimanoff polynomials, summarized in Table A. in Appendix A, the following conjecture seems reasonable. Conjecture.3.0. Let n 9 be odd. Then there exists a prime p such that E n is a product of exactly two irreducible polynomials over F p. Of course, there is no reason to expect a theorem such as this to hold when n is even. In fact, the numerical evidence, summarized in Table A. also in Appendix A, would suggest the following conjecture when n is even. Conjecture.3.. Let n 8 be even. Then there exists a prime p such that E n is irreducible over F p. The above, if true, would also provide structure information on the Galois groups of the Cauchy- Mirimanoff polynomials via Chebotarev s Density Theorem.

22 .3.3 Bounds on the Degrees of Factors of the Cauchy-Mirimanoff Polynomials If n is a prime, then Theorem.3.4 implies that every irreducible factor of E n (X) has degree 6. Tzermias (007) improves this result for certain primes. Theorem.3. (Tzermias (007) - Theorem.). Let p be a prime such that p (mod 3) and p 7.. Every irreducible factor of E p (X) over Q is of degree at least.. For p 3, E p (X) has an irreducible factor of degree d 8. In a recent paper, Tzermias (009), similar results are found for primes p (mod 3). Theorem.3.3 (Tzermias (009)). Let S be the set of primes greater than or equal to 9 and congruent to (mod 3). There exists an effectively computable subset S 0 of S with S 0 having at most 6 elements and such that, for any p in S \S 0, the polynomial E p (X) has no irreducible factor of degree d over Q. Theorem.3.4 (Tzermias (009)). Let p be a prime congruent to (mod 3). Suppose that there exists a prime q such that p (mod q) and p (mod q ). Then E p (X) has an irreducible factor of degree d 6 q 3 over Q..4 Other Results A couple of results remain that are certainly worth note. While these results will not be directly used in this dissertation, they are nonetheless important in the study of the Cauchy-Mirimanoff polynomials..4. The Cauchy-Mirimanoff polynomials are relatively prime In an application to the Korteweg-de Vries equation, Beukers (997) proved the following theorem: Theorem.4. (Beukers (997) - Theorem 4.). For all < m < n, (E n (X), E m (X)) =. An additional result of interest from Beukers (997) relates to the location of the zeros of P n (X). Theorem.4. (Beukers (997) - Lemma.). The number of distinct zeros z of P k (X) on the unit circle such that z + z + 0 is at least [ k 3 ] [ k 3 ]. In particular, if k, 3, 5, 7 there exists a zero z on the unit circle such that z + < E p (X) is irreducible for all primes p While numerical tests have shown that the Cauchy-Mirimanoff polynomials E n (X) are irreducible for at least all n 00, almost no general irreducibility results are known. The first exception, attributed to Michael Filaseta, was proved in Helou (997). Theorem.4.3 (Helou (997) - Proposition 4). For any odd prime p, E p is irreducible over Q. The proof of this theorem used Newton Polygons to establish the form of any possible factorization of E p, and then a number-theoretic calculation to show that no factorization was possible..4.3 A Generalization of the Irreducibility of E p (X) In a thesis under revision, Nanninga (009) announced a proof that E n (X) is irreducible when n = k m where m is an odd integer and k {,, 3, 4, 5}. The proof has not yet been made available.

23 Chapter 3 Chebyshev Polynomials While there are many excellent resources and even entire books devoted to the Chebyshev Polynomials, the results needed do not appear in the literature. All standard results on Chebyshev polynomials will simply be quoted and can be found in either of the excellent books Mason and Handscomb (003) or Rivlin (990). 3. The Fractional, Half-Fractional, and Modified Half-Fractional Chebyshev Polynomials Recall the classic definition of the Chebyshev Polynomials (of the first kind): Definition 3.. (Chebyshev Polynomial). The n th -Chebyshev Polynomial, T n (X), is defined as T n (x) cos nθ where n is a nonnegative integer, x = cos θ, and 0 θ π. The definition is often abbreviated by T n (X) = cos (n arccos(x)). By de Moivre s Theorem, cos(nθ) is a polynomial of degree n in cos(θ), and making the substitution x = cos(θ) gives the polynomial form of T n (X). The first six Chebyshev Polynomials are as follows: T 0 (x) = T (x) = x T (x) = x T 3 (x) = 4x 3 3x T 4 (x) = 8x 4 8x + T 5 (x) = 6x 5 0x 3 + 5x The choice of n as a nonnegative integer guarantees T n (X) is a polynomial in X. However, if one is willing to forgo the guarantee that T n (X) be a polynomial, then it is reasonable to consider any n R. This leads to the definition of Fractional Chebyshev Polynomials. 3

24 Definition 3.. (Fractional Chebyshev Polynomial). The q th -Fractional Chebyshev Polynomial, T q (X), is defined as T q (x) cos (q arccos x) where q Q, and x [, ]. It is unusual to call functions polynomials when they are not actually polynomials. This strange naming convention is meant to highlight the close relationship of these functions to the Chebyshev Polynomials, rather than indicate a specific form of the function. Additionally, for the sake of clarity, the standard Chebyshev Polynomials will sometimes be called the Integral Chebyshev Polynomials. For the purposes at hand, it will be enough to consider a particular subclass of the Fractional Chebyshev Polynomials, called the Half-Fractional Chebyshev Polynomials. Definition 3..3 (Half-Fractional Chebyshev Polynomial). The n th Half-Fractional Chebyshev Polynomial, T n (X), is defined as ( n ) T n (x) cos arccos x where n is a nonnegative integer, and x [, ]. It should be clear that if n is even, then the Half-Fractional Chebyshev Polynomial is an Integral Chebyshev Polynomial; and if n is odd, then the Half-Fractional Chebyshev Polynomial is not a polynomial. There is one last modification to consider. On the surface, Definition 3..4 appears to be a trivial modification to Half-Fractional Chebyshev Polynomials, but the change will be important later. Definition 3..4 (Modified Half-Fractional Chebyshev Polynomial). The n th Modified Half-Fractional Chebyshev Polynomial, C n (X), is defined as ( x ) C n (x) T n where n is a nonnegative integer, and x [, ]. Of course, the Modified Integral Chebyshev Polynomials, and the Modified Fractional Chebyshev Polynomials are similarly defined. It also worth mention that C n (x) T n ( x ) is the Dickson polynomial Dn (x, ) for nonnegative integers n. 3. Properties of the Half-Fractional and Modified Half-Fractional Chebyshev Polynomials The modification to the definition of the Integral Chebyshev Polynomials yielding the Half-Fractional Chebyshev Polynomials makes it clear that the two collections of functions are related, yet the relationship is even closer than apparent at first blush. Theorem 3... For any nonnegative integer n, and x [, ] x + T n (x) = T n 4

25 Proof. Let θ arccos x+. Then θ = arccos x + cos θ = x + cos θ = x cos (θ) = x θ = arccos x Consequently, x + arccos x + cos n arccos x + T n = arccos x ( n ) arccos x = cos = T n (x) Some ambiguity in Theorem 3.. may be perceived( if n is even. However, let n = k. Using the Nesting Property of Chebyshev Polynomials, T k (x) = T k x+ ) ( ( x+ )) = T k T = T k (x) as expected. Corollary 3... For any nonnegative integer n, and x [, ] x + C n (x) = T n Proof. Immediate from Definition 3..4 and Theorem 3... In the case of odd n, Theorem 3.. gives an effective way to generate the Half-Fractional Chebyshev Polynomials. The first five odd Half-Fractional Chebyshev Polynomials are T (x) = T 3 (x) = T 5 (x) = T 7 (x) = T 9 (x) = x + x + (x ) x + (4x x ) x + (8x 3 4x 4x + ) x + (6x 4 8x 3 x + 4x + ) This suggests that the Half-Fractional Chebyshev Polynomials are always of the form polynomial. This is, in fact, true. x+ times a 5

26 Theorem Let n be an odd nonnegative integer. Then x + T n (x) = f n (x) with f n (x) Z[x]. ( x+ ) Proof. By Theorem 3.., T n (x) = T n. By Definition 3..4, T n (x) = C ( ) n x +. It is known that C n (X) Z[X] (see Rivlin (990) - Theorem 5.5 and Lemma 5..), and C n (X) is an odd function (see Rivlin (990) - Equation.3). Consequently, C n (X) = Xτ(X ) for some τ Z[X]. Therefore, T n (x) = C ( ) n x + = x+ τ(x + ), where clearly τ(x + ) Z[x]. Define f (x) τ(x + ) to obtain the desired result. Once the polynomial form is known for the Integral Chebyshev Polynomials, the domain of the Integral Chebyshev Polynomials is extended to all of R or even all of C, as these are valid for polynomials. The same now will be done for the Half-Fractional Chebyshev Polynomials. With this consideration, the domain for the Integral and Half-Fractional Chebyshev Polynomials will no longer be explicitly mentioned. A known result for nonnegative integers n, is C n (X + X ) = X n + X n (see Rivlin (990) - Exercise 5..). This is now generalized to Modified Half-Fractional Chebyshev Polynomials. Theorem For any nonnegative integer n, Proof. By Corollary 3.., C n (x + x ) = x n + x n x + x C n (x + x ) = T n + For all nonnegative integers n, a well-known identity for Integral Chebyshev Polynomials is (see Rivlin (990) - Exercise..) T n (x) = (x x ) n + (x + x ) n Consequently, x + x C n (x + x ) = T n + n x + x = + x + x + x + x + + x + x + + x + x = + x + x n + x + x x + x + n + 4 x + x = + x + x n x + x x + x + n 4 n 6

27 = = = x + x + x + x n (x ) + x (x ) x x + x x x n + x + x x + x + + x + x + n n (x ) + x (x x x x n ) n = x n + x n 7

28 Chapter 4 The Reciprocal Transform of the Cauchy-Mirimanoff Polynomials Over the last two hundred years, there have been many papers and results giving tests to apply to a polynomial to determine irreducibility. The difficulty in testing the irreducibility of the Cauchy-Mirimanoff Polynomials is that these polynomials, and their translates, seem to satisfy none of the hypotheses of any known tests. So instead of studying the Cauchy-Mirimanoff Polynomials directly, the reciprocal transform is defined and studied. 4. The Reciprocal Transform of a Self-Reciprocal Polynomial 4.. Definition of the Reciprocal Transform of a Self-Reciprocal Polynomial Theorem 4... Let D be an integral domain and f (x) D[x] a self-reciprocal polynomial with deg( f ) = k for some k Z. Then there exists a unique polynomial f (x) D[x] such that f (x) = x k f (x + x ) Proof. Uniqueness is straightforward. For existence, suppose f (x) = a k x k + a k x k + + a x + a 0 (4.) with each a i D. Since a k j = a j for each j = 0,,, k, Equation 4. may be rearranged as f (x) = k a j (x k j + x j ) + a k x k (4.) j=0 k = x k a j (x k j + x k j ) + a k j=0 k ( = x k a j C k j x + ) + a k x j=0 (4.3) (4.4) 8

29 As mentioned in the proof of Theorem 3..3, it is known that C n (x) Z[x], so the coefficients of C n in Equation 4.4 may be interpreted appropriately in D (i.e. as a sum of ± s). Thus, define k f (x) a j C k j (x) + a k j=0 where deg( f ) = k. The existence of f has been known for at least a century, but there is no universal agreement to notation or name. Definition 4.. (Reciprocal Transform of a Self-Reciprocal Polynomial). Let D be an integral domain, f (x) D[x] a self-reciprocal polynomial of even degree, and d deg( f ). The reciprocal transform of f is the unique polynomial f (x) D[x] with deg( f ) = d such that 4.. Dickson s Theorem f (x) = x d f (x + x ) It is natural to consider how the factorization of a polynomial is related to the factorization of its reciprocal transform. Theorem 4..3 (Dickson (908)). Let F be a field, and f (x) F[x] a self-reciprocal polynomial such that f (±) 0 and deg( f ) = k for some k Z. Then f(x) is irreducible over F if and only if. f (x) is irreducible over F, and. f (x) is not a product of two distinct irreducible polynomials, each of degree k. Further commenting on Condition, if f (x) is irreducible over F and f (x) = g(x)h(x) is a product of two distinct irreducible polynomials, then the roots of h are the reciprocals of the roots of g. In other words, there exists a λ F and g(x) F[x] such that f (x) = λg(x)g (x). Moreover, g(x) and g (x) are not self-reciprocal polynomials. On the contrary, if g(x) is a self-reciprocal polynomial, then g (x) is also a self-reciprocal polynomial and both deg(g) and deg(g ) are even integers (consequently the reciprocal transforms exist giving a factorization to f (x)). That the degrees are even follows from the hypothesis that g(±) 0 and g (±) 0, and the fact that every irreducible self-reciprocal polynomial over F, except X +, has even degree. For if z is a root of a self-reciprocal polynomial, then z is also a root, so the roots of a self-reciprocal polynomial come in pairs unless z = z. This condition occurs if and only if z = ± is a root of the polynomial; the minimal polynomial of is X and is not self-reciprocal unless char(f) =, and the minimal polynomial of is X +. Thus, the only irreducible self-reciprocal polynomial of odd degree is X +. If the hypothesis that f (±) 0 is dropped in Theorem 4..3, then the statement remains true if the word distinct is dropped in Condition. Moreover, in this case, there is no guarantee that the two polynomials of Condition are non-self-reciprocal. For an example, consider f (x) = x + x + = (x + ) over Q with f (x) = x +. After Dickson (908), a number of papers have appeared on the topic including Kleiman (974) and Meyn (990). 9

30 4. The Reciprocal Transforms of the Cauchy-Mirimanoff Polynomials By Theorem..6, the Cauchy-Mirimanoff Polynomials are self-reciprocal, so their reciprocal transform exists. In this section, an equation for this reciprocal transform is found and studied. 4.. An Equation for the Reciprocal Transform of the Cauchy-Mirimanoff Polynomials Theorem 4.. (The Reciprocal Transform of the Cauchy-Mirimanoff Polynomials). For any integer n, the reciprocal transform of E n (x) is given by ( x+ ) (x + ) n T n E n(x) = ( x + ) ɛn (x + ) e n (4.5) where for even n, ɛ n = e n = 0; for odd n, ɛ n = and e n = 0,, or according as n 0,, or (mod 3). Proof. Consider the function f n defined by f n (x) (x + ) n C n (x) ( x + ) ɛn (x + ) e n Now using Corollary..3, Definition 4.., and Theorem 3..4, if n is even, then ( x n fn x + ) x = x n = x n (x + x ) n + ( C n x + x) ( x n (x + x + ) n n ) x x n = x ((x + ) n x n ) = E n (x) On the other hand, if n is odd, then ( x n 3 en fn x + ) x = x n 3 en = x n 3 en ( x + x + ) n ( ) C n x + x x + x + ( x + x + ) e n ( x + x + ) n x n x n x n x e n x + x + ( x + x + ) e n ( x + x + ) n x n = x x + x + ( x + x + ) e n ( = x (x + ) n x n ) (x + ) ( x + x + ) e n = E n (x) In either case, f n is a reciprocal transform of E n E n is the reciprocal transform of E n. 0

31 Corollary 4... Let n be an integer.. If n is even, then deg(e n) = n. If n is odd, then deg(e n) = n 3 e n Proof. Immediate from Corollary Roots of E n(x) and its Translates Everything known about the roots of the Cauchy-Mirimanoff Polynomials may be translated to information about the roots of the reciprocal transform. Proposition Let n 9 be an odd integer. If e iθ is a root of E n (x), then r cos θ is a real root of E n(x) such that r. Proof. Suppose e iθ is a root of E n (x). This implies, by Defintion 4.., that r cos θ = e iθ + e iθ is a root of En(x). In Helou (997), it was shown θ ( π 3, ( ) π. Consequently, θ π 3, π) cos θ (, ) cos θ (, ). In Section..3, it is shown for odd n that the roots of E n (x) can be partitioned into orbits, each of which contains two roots of E n (x) of absolute value. The partitioning of the roots of E n (x) induces a partition of roots of E n(x). The parts of the induced partition of the roots of E n(x) will be called the root orbits of E n(x) to emphasize the relationship with the orbits of roots of E n (x). Definition Let n 9 be an odd integer, and z e iθ be a root of E n (x). Then Orb (z) is the set { } Orb (z) cos θ, e iθ + e iθ, e iθ + e iθ Theorem Let n 9 be an odd integer, and z e iθ be a root of E n (x). Then the root orbit of E n(x) induced by Orb(z) is Orb (z), and Orb (z) consists of three distinct roots, one real and two complex conjugates, of E n(x). Proof. Considering each root of Orb(z) separately, we find the corresponding root of E n(x).. e iθ Orb(z): Then e iθ + e iθ = cos θ is a root of E n(x).. e iθ Orb(z): Then e iθ + e iθ = cos θ is a root of E n(x). 3. e iθ Orb(z): Then e iθ + e iθ is a root of E n(x). 4. e iθ Orb(z): Then e iθ eiθ is a root of E n(x). 5. e iθ Orb(z): Then e iθ + e iθ = e iθ + e iθ is a root of E n(x). 6. eiθ eiθ Orb(z): Then e iθ + e iθ + + eiθ e iθ + = e iθ + e iθ is a root of E n(x).

32 So Orb(z) induces a partition of the roots of En(x) with parts { cos θ, e iθ + e iθ, } e iθ + e iθ. It remains to show that these roots are distinct. As cos θ is real, it is enough to show e iθ + e iθ and e iθ + are two distinct nonreal roots. e iθ A complex number is real if and only if it equals its complex conjugate. Clearly, the complex conjugate of e iθ + e iθ is e iθ +, and vice versa, so suppose they are equal: e iθ Let x e iθ, and solve for x: e iθ + e iθ = e iθ + e iθ x + x = x + x (x )(x + x + ) = 0 So x must be a root of unity, a contradiction since E n (x) has no roots that are roots of unity by Theorem... Thus, e iθ + e iθ and e iθ + are two distinct nonreal numbers. e iθ In Section..3, conditions are given to show when the elements of Orb(z) are Q-conjugates. That work can be used to give conditions for when the elements of Orb (z) are Q-conjugates. Lemma Let n 9 be an odd integer, and z e iθ be a root of E n (x). Proof.. T z = 6 if and only if the elements of Orb (z) are Q-conjugates.. T z = if and only if the only Q-conjugate of cos θ in Orb (z) is itself.. ( ) Suppose T z = 6. Let f (x) be the minimal polynomial of z over Q. Then f (x) is a self-reciprocal polynomial of even degree by Theorem.3., so the reciprocal transform f (x) Q[x] exists. Every element of Orb(z) is a root of f (x) by Corollary..9, so every element of Orb (z) is a root of f (x). By Dickson s Theorem (Theorem 4..3), f (x) is irreducible over Q the elements of Orb (z) are Q-conjugates. ( ) Suppose the elements of Orb (z) are Q-conjugates, but T z = (the only possibility if T z 6 by Theorem..3). Let f (x) be the minimal polynomial of z over Q. Then f (x) is a self-reciprocal polynomial by Theorem.3., so the reciprocal transform f (x) Q[x] exists. By Dickson s Theorem (Theorem 4..3), f (x) is irreducible over Q. So, by hypothesis, the elements of Orb (z) are all roots of f (x). In particular, e iθ + e iθ is a root of f (x). Consequently, f ( e iθ ) = x deg( f ) f ( e iθ + e iθ ) = 0 Therefore, e iθ is a root of f (x) e iθ Orb(z) is a Q-conjugate of z. This contradicts Theorem..8 T z = 6.. ( ) Suppose T z =. Then by, the elements of Orb (z) cannot all be Q-conjugates. Since e iθ + e iθ Orb (z) and e iθ + e iθ Orb (z) are complex conjugates, they are necessarily Q-conjugates by the complex conjugate root theorem. So if cos θ were a Q-conjugate to either