Roots of Unity, Roots of Complex Numbers, and Cyclotomic Polynomials
Gary D. Knott Civilized Software Inc. 12109 Heritage Park Circle Silver Spring MD 20906 email:[email protected] URL:www.civilized.com
November 14, 2017
0.1 Complex Numbers
The set of complex numbers is the set α + iβ α,β where i denotes the “imaginary” value C { | ∈ R} √ 1; α + iβ is a “complex” of a real part α and an imaginary part iβ; the imaginary part is 0 in − the case where β = 0. The real numbers are a subset of the complex numbers, and we generally R leave out the real or imaginary part in writing a complex number when that part is 0. Arithmetic with complex numbers follows the usual rules, including i−1 = i and i2 = 1, and is a field. − − C There are some special features that make complex number arithmetic different than computation with real numbers. Given a complex number z = α+iβ, we define Re(z)= α and Im(z)= β. Let z and w be two complex numbers. The sum z + w is the complex number Re(z)+Re(w)+ i(Im(z)+ Im(w)) and the product zw is the complex number Re(z)Re(w) Im(z)Im(w)+ i(Re(z)Im(w)+ − Im(z)Re(w)). This is consistent with i2 = 1. The conjugate of z is the complex number − z∗ = α iβ = Re(z) i Im(z). (Often the conjugate of z is writtenz ¯.) The magnitude of z is the − − non-negative real number (zz∗)1/2 denoted by z . (The magnitude of a complex number z is also | | called the modulus of z.) Note Re(z)=(z + z∗)/2 and Im(z)=(z z∗)/(2i). − Exercise 0.1: Let z . Show that z2 = z 2. ∈C | | | | Exercise 0.2: Let z,w . Show that z∗∗ = z, (z + w)∗ = z∗ + w∗, and (zw)∗ = z∗w∗. This ∈C means the self-inverse mapping z z∗ is an automorphism of ; when we adjoin i to (and → C R “stir”) to form , we can’t tell the difference between choosing i = √ 1 versus i = √ 1 from C − − − “inside” . C Exercise 0.3: Let z = a + ib with a, b and z = 0. Show that 1/z = z∗/(z∗z). ∈R 6 Exercise 0.4: Show that for z with z = 1, z + z−1 = 2Re(z). ∈C | | The field of complex numbers has the wonderful property that any polynomial p(z)= p + p z + C 0 1 + p zn with coefficients p ,p ,...,p where p = 0 can be factored into n linear factors so ··· n 0 1 n ∈C n 6
1 2 that p(z)= p (z r )(z r ) (z r ) where r ,r ,...,r . (The values r ,...,r need not be n − 1 − 2 ··· − n 1 2 n ∈C 1 n distinct.) The values r1,...,rn are called the roots or zeros of the polynomial p, since p(rj)=0for j = 1,...,n. This fact is called the Fundamental Theorem of Algebra: every degree-n polynomial p(z) with its coefficients in has n roots in , i.e., n linear factors z r with r such that C C − j j ∈ C p(r ) = 0; it is what we mean by calling the complex numbers algebraically complete. j C
The distinct roots of p are determined by the coefficients of p, and when pn = 1, the remain- ing coefficients of p are determined by the n roots of p. The coefficients of p(x)/pn are given pj in terms of elementary symmetric functions of the roots r1,...,rn; for j = 0, 1, 2,...,n, = pn ( 1)n−je (r ,r ,...,r ) where − n−j,n 1 2 n e (x ,x ,...,x ) := x x x k,n 1 2 n j1 j2 ··· jk 1≤j1
Exercise 0.5: Let p(x) = p + p x + + p xn = p (x r )(x r ) (x r ) with 0 1 ··· n n − 1 − 2 ··· − n p ,p ,...,p ,r , ,r . Define p∗(x) = p∗ + p∗x + + p∗ xn. Show that the polyno- 0 1 n 1 ··· n ∈ C 0 1 ··· n mial p(x)p∗(x) has real coefficients. Also show that the roots of p∗ are the conjugates of the roots of p. Thus if the polynomial p has only real coefficients, then if r is a root of p, r∗ is also a root of p. Hint: the polynomials p and p∗ are functions defined on . C Euler’s formula eiθ = cos(θ)+ i sin(θ) allows us to write any non-zero complex number z in polar form z = z ei arg(z) where arg(z) := atan2(Im(z)/ z , Re(z)/ z ) for z = 0. Treating z as a point | | · | | | | 6 (Re(z), Im(z)) in 2, arg(z) is the angle in ( π,π] between the x-axis and the vector (Re(z), Im(z)); R − this angle is negative if Im(z) < 0 and non-negative if Im(z) 0. We define arg(z) := π/2 for ≥ z = 0. When we look at the components of points in 2 as being the real values comprising R the real and imaginary parts of complex numbers, we see that the points of 2 are in one-to-one R correspondence with , and in this context, we call 2 the complex plane. C R Exercise 0.6: Let z . Show that z = z cos(arg(z))+i z sin(arg(z)). Hint: draw a vector ∈C | | | | in the complex plane corresponding to z.
Exercise 0.7: Show that eiπ = e−iπ = 1. Here again, we can’t tell the difference be- − tween √ 1 and √ 1, or equivalently, between clockwise rotations in righthanded 2-space and − − − counter-clockwise rotations in lefthanded 2-space.
Let the complex numbers z and w be given in polar form as z = z eiθ and w = w eiφ. Then | | | | the product zw in polar form is zw = z eiθ w eiφ = z w ei(θ + φ). This exhibits the fact that, | | | | | || | looking at z, w, and zw as vectors in the right-handed complex plane, zw is the vector z rotated counter-clockwise arg(w) radians and scaled in length by w . | | Exercise 0.8: Look-up the development of Euler’s formula based on the power-series for ex, sin(θ), and cos(θ). 3
Because angle(x,y) (without restricting qualification), is multi-valued, (i.e., the angle between two non-zero 2-vectors is not a unique value but instead can be any value in the set of real numbers R arccos((x,y)/( x y )) + k2π k ), arg is more properly taken to be a multi-valued function; { | || | | ∈ Z} for z = 0, we may take arg(z) to be any of the values atan2(Im(z)/ z , Re(z)/ z ) + k2π with 6 | | | | k . The definition we gave before is the principal value arg function. We may have multi-valued ∈Z expressions involving complex numbers when Euler’s formula is involved in the definition of the expression. (Note cos(arg(z)) is not a multi-valued function, but arg(z) cos(arg(z)) is a multi- · valued function.) Generally however, in the sequel we shall take arg(z) as denoting its principal value.
Exercise 0.9: Show that the principal value of arg(1) is 0.
Note ei2π = 1. This is consistent with the fact that for r +, really, reiθ = r(cos(θ + k2π)+ ∈ R i sin(θ + k2π)) for k . We generally take k = 0 to obtain the principal value of arg(reiθ) ∈ Z in ( π,π]. Since ew + i2π = 1, ew = ew + i2π, and we see that the complex function ew is a − periodic function with the period i2π. This far-reaching fact leads to a large body of intricate relationships and requires us to deal with multi-valued functions. Of course, multi-valued functions are not functions – at least not functions producing single complex numbers as output – they are specifications of sets. But by suitably restricting the range of such “functions” to so-called principal values, we can reduce a multi-valued function to a proper function, generally with the loss of continuity along a boundary. (Even in real-number computation, multi-valued functions arise when square-roots of positive values occur. When k-th roots of positive values occur with i(2π/k) i2(2π/k) i(k 1)(2π/k) k > 2, the complex roots of unity e , e , ..., e − , 1 slip-in.) (Generally computations with real numbers are implicitly restricted to principal value computations, so that, for example, when we write √3 2 we mean the positive real cube-root of 2.)
Powers and logarithms may be defined for complex numbers. First, for z 0 with z = aeiθ ∈C−{ } where a = z and π<θ π, we define log(z) to be the complex number log(a)+ iθ. This is | | − ≤ consistent with log(aeiθ)= log(a)+ log(eiθ) and log(eiθ)= iθ.
Now let w 0 be given as w = beiφ where b = w and 0 φ < 2π. Then for z = 0, zw is ∈C−{ } | | ≤ 6 defined as ew log(z). In particular, ew = eRe(w) ei Im(w). · · Because we really have z = aei(θ + k2π) and w = bei(φ + m2π) for k, m , both log(z) and zw ∈Z are multi-valued. Fixing k = 0 defines the principal values of log(z) and zw defined just above. (Why don’t we have to fix m = 0?)
Exercise 0.10: w log(a)b cos(θ) θb sin(φ) i(log(a)b sin(θ) θb cos(φ)) Show that z = [e − ][e − ]. By writing (θ + k2π) for θ in this identity, we make the multi-valued nature of zw explicit.
Exercise 0.11: What is the set of complex numbers defined by 1i? Hint: 1= e2πin for n . ∈Z Exercise 0.12: Let z = x + iy with x,y . Show that ez = ex. ∈R | | x y Exercise 0.13: Let C be the set of 2 2 matrices x,y . Note the × y x | ∈R − matrices in C are scalar multiples of orthogonal matrices with (M row 1,M row 2) = 0 and 4
((M col 1)T, (M col 2)T) = 0 for M C. Show that C is an algebra of dimension 2 over the ∈ real numbers with respect to matrix sums and products, and give a basis for C.
Also show that C is a field and show that C is isomorphic to the field of complex numbers . C
0.2 Roots of Unity
m Let us consider what taking a rational power a n of a complex number a means, where m , and m 1 m m ∈Z n +. Since a n =(a n ) , and we understand what b is for any complex number b: namely for ∈Z 1 m< 0 and b = 0, bm = , and for m 0, bm = b b b = b where an empty product 6 b−m ≥ · ······ 1≤i≤m 0 m Y 1 is taken to be 1, i.e., b = 1, we will understand what a n means, when we understand what a n means.
1 n The expression a n denotes an n-th root of a, i.e., a value v such that v = a. Except for a = 1 0, there are n distinct n-th roots of a, (for a = 0, a n = 0). For a non-zero complex number 1 a, the notation √n a or a n is ambiguous; either denotes one, or perhaps all, of the n distinct n values x such that x = a. These n solutions are distinct complex values r1, r2, ..., rn such that xn a = (x r )(x r ) (x r ), i.e., r , r , ..., r are the n roots of xn a. Thus − − 1 − 2 ··· − n 1 2 n − ( 1)nr r r =( 1)ne (r ,...,r )= a and in general, ( 1)n−je (r ,r ,...,r )=0 for − 1 2 ··· n − n,n 1 n − − n−j,n 1 2 n j = 2, 3,...,n 1; in particular, e (r ,r ,...,r )= (r + r + + r ) = 0 when n> 1, (and − − 1,n 1 2 n − 1 2 ··· n we have e0,n(r1,r2,...,rn) = 1). (For a = 0, the nn-th roots of a are all the value 0, i.e., the value 0 with multiplicity n.)
For a = 1, The values r1, r2, ..., rn are known as the n-th roots of unity; in this case, we may write 1 in polar form: 1 = e2πik for some arbitrary integer k, and then 11/n = e2πik/n. More generally, we take 11/n to be a multi-valued function taking values in e2πi1/n, e2πi2/n, ..., e2πi(n 1)/n, { − e2πin/n , or more expansively, 11/n is the set e2πi1/n,e2πi2/n,...,e2πi(n 1)/n,e2πin/n . } { − } Note eπin/n = eπi0/n = e0 = 1. In general, e2πik/n = e2πi(k mod )n/n for k . ∈Z The principal value of 11/n is e2πi0/n = e2πin/n =1= √n 1; in general, we shall denote the principal n-th root of a complex number a by underlining the corresponding expression, e.g., √n a.
Exercise 0.14: Show that for k , e2πik/n e2πi1/n,e2πi2/n,...,e2πi(n 1)/n,e2πin/n . ∈Z ∈ { − } Hint: e2πik = 1 and e2πik/n = e2πi(k mod n)/n for k . ∈Z The set of n-th roots of unity e2πi1/n,e2πi2/n,...,e2πi(n 1)/n,e2πin/n =: R is an { − } n ⊂ C order-n cyclic multiplicative group. The identity is e2πin/n = 1 R , and e2πi1/n generates R , ∈ n n i.e., e2πik/n =(e2πi/n)k for k . ∈Z Exercise 0.15: 2πik/n 2πik/n Show that the inverse of e in the multiplicative group Rn is e− = 2πi(n k)/n (e − ). 5
Exercise 0.16: Let n 1. Show that for ζ R , ζ = 1. Also show that for ζ R , ≥ ∈ n | | ∈C− n ζ = 1. | | 6 Exercise 0.17: Let ζ R with n 1. Show that ζ−1 = 1/ζ = ζ∗. And show that, for ζ = 0, ∈ n ≥ 6 if ζ = 1, then 1/ζ = ζ∗. | | 6 6 2πij/n Denoting the roots of unity by r1, r2, ..., rn where rj = e , we have that r1, r2, ..., rn satisfy rn = 1, and r r r = ( 1)n, and r r r r is an integer for 0 k n. Not j 1 2 ··· n − j1 j1 j2 ··· jk ≤ ≤ 1≤j1 The generators of the cyclic group Rn are called the primitive n-th roots of unity; these primitive n-th roots of unity are the values in gen(R ) = e2πik/n R 1 k n, gcd(k,n) = 1 . For n { ∈ n | ≤ ≤ } n = 1, 1 is the sole primitive n-th root of unity; for n> 1, 1 gen(R ) We have #gen(R )= φ(n) 6∈ n n where φ(n):=# k 1 k n, gcd(k,n) = 1 is Euler’s totient function (see section ??). { | ≤ ≤ } The set of generators gen(Rd) for the order-d cyclic subgroup Rd consists of the order-d elements of Rn, thus the primitive n-th roots of unity are the φ(n) order-n elements of Rn and the Finite Cyclic Group Theorem tells us that Rn has τ(n) (cyclic) subgroups, Rd, for d ranging over the positive divisors of n (τ(n) is the number of divisors of n in +). The first proof of the Finite Cyclic Group Z Theorem (page ??) tells us that R = gen(R ) where gen(R ) and gen(R ) are disjoint sets n ∪d|n d d1 d2 when d = d . Since #gen(R )= φ(d), we have again that n = φ(d). 1 6 2 d Xd|n Exercise 0.18: Show that for n +, r = δ where δ = 1 for n = 1 and δ = 0 ∈ Z n,1 n,1 n,1 r∈Rn otherwise. Hint: see page ??. X The n-th roots of unity in are the roots of the polynomial xn 1. This polynomial factors over C − the real numbers into a product of linear and quadratic factors with real coefficients. When n is even, the linear factors are x 1 and x + 1, and the quadratic factors are − 2πk n 1 (x e2πik/n) (x e2πi(n k)/n)= x2 2 cos( )x +1 for k = 1, 2,..., − . − · − − − n ⌊ 2 ⌋ When n is odd, we have only the single linear factor x 1, and the quadratic factors are again − 2πk n 1 (x e2πik/n) (x e2πi(n k)/n)= x2 2 cos( )x +1 for k = 1, 2,..., − . − · − − − n ⌊ 2 ⌋ 6 2πk Exercise 0.19: Show that e2πik/n + e2πi(n k)/n = cos( ). Hint: see exercise 0.4. − n The multiplicative group composed of the n-th roots of unity Rn is isomorphic to the additive group = 1, 2,...,n 1, 0 with the operation a + b =(a + b) mod n. The associated multiplicative Zn { − } n group of units, U = k gcd(k,n) = 1 with the operation a b =(ab) mod n, corresponds n { ∈Zn | } ×n to the set of primitive n-th roots of unity, gen(R ); we have gen(R )= e2πik/n k U . (Note n n { | ∈ n} however, gen(Rn) is not a multiplicative group unless n = 1.) We may use the correspondence between gen(Rn) and Un to show that r = 1 for n > 2. r∈genY(Rn) This is because r = e2πik/n = e2πiS/n, where S = k = nφ(n)/2 for n 2 (see ≥ r∈genY(Rn) kY∈Un kX∈Un exercise ??), and thus for n 2, r = e2πi(nφ(n)/2)/n = eπiφ(n). ≥ r∈genY(Rn) We have φ(2) = 1 and φ(n) = 2h for some h + when n > 2 (page ??); thus eπiφ(n) = 1 for ∈ Z − n = 2 and eπiφ(n) = 1 for n> 2, i.e., eπiφ(n) = 1 2δ = r for n 2. And for n = 1, − n,2 ≥ r∈genY(Rn) r = 1, so altogether, for n 1, r = 1 2δ . [QED] ≥ − n,2 r∈genY(Rn) r∈genY(Rn) Exercise 0.20: Show that r =( 1)δn,2 . − r∈genY(Rn) Exercise 0.21: Compute r =( 1)δn,2 . − r∈Rn−Ygen(Rn) Exercise 0.22: Show that for n prime, gen(R ) 1 = R , so that gen(R ) is “almost” a n ∪ { } n n group when n is prime. 2πi1/n 2πik/n k We will denote the primitive root value e by ωn so that e = ωn; the n n-th roots 2 3 n n 0 of unity are then ωn, ωn, ωn, ..., ωn, where ωn = ωn = 1. We can think of ωn as the primary primitive n-th root of unity. Note the principal n-th root of unity is 1 which is not a primitive n-th 2 n root of unity unless n = 1. (Often the n n-th roots of unity are denoted by ζn, ζn, ..., ζn since z is often used to denote a complex number.) The single 1-th root of unity is 1, the two square-roots of unity are 1 and 1, and the three cube- 1+ i√3 1 i√3 − roots of unity are 1, − , and − − , (and the latter two are primitive cube roots of 2 2 unity). The value 1 is the principal n-th root of unity for n 1; when n is even, both 1 and 1 ≥ − comprise the real n-th roots of unity; when n> 2 is odd, there is only the single real n-th root of unity, 1 and all the other n-th roots of unity are non-real complex numbers. Since the n-th roots of unity e2πik/n = cos(2πk/n)+ i sin(2πk/n) with 0 k spaced on the unit circle C1 with (1, 0) being one of these points; the unit circle C1 is the radius-1 circle centered at (0, 0). The cube roots of unity are the roots of the polynomial x3 1. In this case, as for every polynomial − p(x) with real coefficients, if r is a root of p, then the conjugate r∗ is a root of p, (and r = r∗ only when r is real). Exercise 0.23: Show that every non-constant polynomial with real coefficients factors into a product of [x]-irreducible linear and/or quadratic polynomials with real coefficients; the R quadratic polynomial factors have non-real roots. We have e2πi/5 = ω = cos(2π/5) + i sin(2π/5) = [ 1+ √5+ i 10 + 2√5]/4 where √5 is the 5 − positive real square-root of 5, and 10 + 2√5 is the positive realp square-root of 10 + 2√5. The 2πi/5 2πi2/5 four combinations of choices of thep signs for √5 and 10 + 2√5 yield the values e , e , 2πi3/5 2πi4/5 e , and e . p We may obtain the above expression in radicals for the primitive 5-th roots of unity as follows. We 5 4 4 2 4 3 2 2 3 4 have x 1=(x 1)(x + x 3+ x + x +1) and the roots of x + x + x + x +1 are ω5, ω5, ω5, ω5 − 2πi/5− 2 3 4 5 5 2 3 4 5 with ω5 = e . But ω5 +ω5 +ω5 +ω5 +ω5 = 0 with ω5 = 1, (and indeed, ζ +ζ +ζ +ζ +ζ = 0 for ζ ω ,ω2,ω3,ω4 since each such choice is a primitive 5-th root of unity.) See exercise 0.18. ∈ { 5 5 5 5} Let ζ ω ,ω2,ω3,ω4 = gen(R ), and let α = ζ4 +ζ. Then α2 +α 1= ζ3 +2+ζ2 +ζ4 +ζ 1=0, ∈ { 5 5 5 5} 5 − − so, using the quadratic formula, 1 √1 + 4 1 √5 α = − ± = − ± . 2 2 Now ζ2 ζα +1 = 0 since ζ(ζ4 + ζ)=1+ ζ2. Thus again using the quadratic formula, − α α2 4 α √1 α 4 ζ = ± − = ± − − p2 2 1 1 1 √5 1 √5 2 = − ± ∓ 3 2 2 ± " 2 − # 1 1 = ( 1 √5 [2 2√5 12] 2 ) 4 − ± ± ∓ − 1 1 = ( 1 √5 [ 2√5 10] 2 ) 4 − ± ± ∓ − 1 = ( 1 √5 i 10 2√5) as given above. 4 − ± ± ± q Here the sign of both instances of √5 must be the same. Exercise 0.24: Explain why the sign of √5 in the formula for the primitive 5-th roots of unity must be the same at each occurrence. 8 Gauss gave an algorithm for expressing any n-th root of unity using the operations of negation, taking reciprocals, addition, multiplication, and radicals (i.e., root-extraction) [Gau86] [Web96], although in general p-th roots, not just square-roots, may be required with p prime; and as seen in the above formula for the primitive 5-th roots of unity, iterated (i.e., nested) radicals may occur. Moreover, radicals of radically-expressible complex numbers may occur, i.e., a p-th root of a number a + ib where a and b are radically-expressible real values (which themselves may involve radically-expressible complex numbers). 2π 2π The primary primitive 17-th root of unity is ω = e2πi/17 = cos + i sin , and we 17 17 17 2 3 17 have the 17-th roots of unity ω17, ω17, ω17, ..., ω17, These 17 complex values are the vertices of a regular 17-gon inscribed in the unit circle of the complex plane. The “initial” edge of this 17-gon is the line-segment 2π 2π segment[(1, 0), (cos , sin )] 17 17 17 connecting ω17 and ω17. Note if we know this line-segment, we can “mark” it with a compass, and then, given the unit-circle graph, we can draw the remaining 16 edges of the inscribed 17-gon. Gauss obtained 2π 1 1 1 cos = + √17 + 34 2√17 17 −16 16 16 − q 1 + 17 + 3√17 34 2√17 2 34 + 2√17. 8 − − − r q q 2π 2π where ω = e2πi/17 = cos + i sin . 17 17 17 2π If we can construct the line-segment of length cos on the x-axis, we can then draw a per- 17 2π pendicular line through (cos , 0) parallel to the y-axis and mark its intersection with the 17 2π 2π unit-circle above the x-axis as the point A; the point A is just (cos , sin ). Now with 17 17 this initial edge, we can succesively mark-off the remaining edges of the inscribed 17-gon. 2π We can construct a line-segment of length cos graphically since Gauss’s formula requires 17 only constructing multiples of the unit distance, repeated bisections to effect division by powers of two, and constructing similar right triangles to compute square-roots. Thus the regular 17-gon is constructable by compass. Exercise 0.25: Explain exactly how to construct a line-segment whose length is the square- root of the length of a given line-segment. Gauss also showed that a regular n-sided polygon is constructable with a compass if and only if mj n is an integer of the form 2k (22 + 1), where k 0, M 0, and m , m , ..., m are · ≥ ≥ 1 2 M 1≤Yj≤M 9 mj positive integers such that the Fermat numbers (22 + 1) are all prime. (This ensures that the 2π edge-length cos is constructable using only the equivalent of repeated division by 2, integer- n multiplication, addition, and subtraction, and computation of square-roots of positive values.) Exercise 0.26: Show that for d n, ωd = ω . | n n/d Exercise 0.27: Show that every element ζ of Rn is a primitive d-th root of unity for some positive integer d. Exercise 0.28: Show that ω ω2 ω3 ωn =( 1)n+1. Hint: 1+2+ + n = n(n + 1)/2 n · n · n ····· n − ··· and eiπ = 1. − Exercise 0.29: Show that for n odd, gen(R )= gen(R ). 2n − n Solution 0.29: Let g = e2πik/n gen(R ); necessarily gcd(k,n) = 1. Let s = e2πih/2 ∈ n ∈ gen(R ); necessarily gcd(h, 2) = 1 and hence s = 1. Note s,g R and o(s) = 2 and 2 − ∈ 2n o(g)= n, and since n is odd, we have gcd(2,n)=1. Also, for 1 j But now from exercise ??, we may conclude that o(sg)= lcm(2,n) = 2n, so sg = g gen(R ). − ∈ 2n Thus gen(R ) R . There are φ(n) choices for g and, with n odd, o(R )) = φ(2n)= φ(n)= − n ⊆ 2n 2n o(R ), and hence #gen(R )=#gen(R ). Therefore gen(R )= gen(R ). [QED] n n 2n − n 2n Exercise 0.30: Show that when n is even with n 2, R = R . Hint: eiπ = 1. ≥ − n n − Exercise 0.31: + + 1 n−k Show that for n and k 0 , k = ωn . Thus 1/Rn = ∈ Z ∈ Z ∪ { } ωn 1/ω , 1/ω2, ..., 1/ωn = ωn−1, ωn−2,...,ω0 = R . { n n n} { n n n} n Exercise 0.32: Show that if w is an n-th root of unity of order k then the conjugate w∗ is also an n-th root of unity of order k. This is an instance of the fact that if p(x) is a polynomial with real coefficients and r is a root of p, then r∗ is also a root of p (see exercise 0.5). Then ∗ conclude that Rn = Rn. Exercise 0.33: Let ζ be a primitive n-th root of unity with n + and let a be a rational ∈ Z number. Show that ζa is an m-th root of unity for some integer m +. ∈Z Exercise 0.34: Let R + = + R ; R + is the set of all roots of unity on the unit circle in Z ∪n∈Z n Z the complex plane. Show that RZ+ is a countably-infinite Abelian group. Are the points in the complex plane corresponding to the complex numbers in RZ+ rational points with Re(r) and Im(r) rational for r R + ? ∈ Z The set of complex numbers RZ+ defined in the previous exercise are roots of polynomials with integer coefficients, (or equivalently, monic polynomials with rational coefficients); such complex numbers are called algebraic numbers. Let A denote the set of algebraic numbers. We write Z ⊂C 10 A because A = ; the set A is known as the set of transcendental numbers which has Z ⊂C C− Z 6 ∅ C− Z been shown to include π and e. In fact, the set A is countable, so the set A is uncountable. Z C− Z ′ The AZ -subset of complex numbers AZ that are roots of monic polynomials with integer coefficients n ′ ′ ′ are called algebraic integers. Since x 1 is monic, RZ+ AZ ; in fact, RZ+ AZ since AZ ′ − ⊆ ⊂ Z ⊂ while RZ+ . However, i AZ as well as RZ+ . Indeed, letting C1 denote the set of points on Z 6⊆ ∈ ′ the unit circle in the complex plane, we have R + A C A C C and C (A C ) Z ⊆ Z ∩ 1 ⊂ Z ∩ 1 ⊂ 1 1 − Z ∩ 1 is uncountable. Exercise 0.35: Show that the set of rational numbers is a subset of the algebraic numbers Q A , and show that the algebraic integers A′ satisfy A′ = , so no non-integer rational Z Z Z ∩Q Z number is an algebraic integer. a Solution 0.35: First we note that if a rational number r = with b = 0 and gcd(a, b)=1is b 6 the root of a monic polynomial with integer coefficients then, in fact, r , i.e., b = 1. This ∈ Z is because, for f(x)= xk + c xk−1 + + c x + c with k 1 and c ,c ,...,c , if we k−1 ··· 1 0 ≥ o 1 k−1 ∈Z a ak ak−1 a a have f( ) = 0 then + c + + c + c = 0, and thus bkf( )= ak + c bak−1 + b bk k−1 bk−1 ··· 1 b 0 b k−1 + c bk−1a + c bk = 0, so c bk = ak + c bak−1 + + c bk−1a. And since c bk is divisible ··· 1 0 − 0 k−1 ··· 1 − 0 by b, and c bak−1 + + c bk−1a is divisible by b, ak must be divisible by b which is contrary k−1 ··· 1 to the assumption gcd(a, b) = 1 unless b = 1. Hence we conclude that b = 1 and r = a . ∈Z c Now A since s with s = and c,d with d = 0 and gcd(c,d) = 1 satisfies Q ⊂ Z ∈ Q d ∈ Z 6 dx c = 0. And only when d = 1 does dx c = 0 have an integer root in the case where − − gcd(c,d) = 1. Therefore we see that A′ Q = . [QED] Z ∩ Z Exercise 0.34 shows that there are an uncountable number of complex numbers z with z = 1 that | | are not n-th roots of unity for any n +. Conrad [?] discusses polynomials with roots on the unit ∈Z circle in the complex plane. Among others, he considers the polynomial x4 2z3 2z + 1 which − − has two real roots and two complex roots on the unit circle that are not roots of unity. The real roots are 2.29663 ... and .43542 .... Are there polynomials with integer coefficients that have both roots of unity and non-roots of unity in C as roots? The answer is yes: (x4 2z3 2z +1)(x3 1) 1 − − − is an example. Are there polynomials with integer coefficients that have all their roots in C1, some or all of which are not roots of unity? Exercise 0.36: Let z C . Given the real value ǫ > 0, does there exist an integer n such ∈ 1 that zn 1 <ǫ? | − | 0.3 Roots of Complex Numbers For a 0 , we may write a in polar form as a ei(arg(a) + 2πk) for k , and then ∈C−{ } | | ∈Z a1/n = n a ei(arg(a) + 2πk)/n for k | | ∈Z p n where here arg(a) denotes the principle arg(a) value in ( π,π] and a denotes the principal − | | positive real n-th root of a with a = √aa∗ +. The n distinct n-th roots of a are | | | | ∈R p 11 n a ei(arg(a) + 2π1)/n, n a ei(arg(a) + 2π2)/n, | | | | n i(arg(a) + 2π(n 1))/n n i(arg(a) + 2πn)/n p p ..., a e , a e . | | − | | These are just the values p p ω n a ei arg(a)/n,ω2 n a ei arg(a)/n,...,ωn−1 n a ei arg(a)/n,ωn n a ei arg(a)/n n | | · n | | · n | | · n | | · p i2π/n pn 0 p p where ωn = e . (Recall ωn = ωn = 1.) 1/n n i arg(a)/n n 1/n n The principal value of a is a e · which we denote by √a or by a . Note √a is | | n a non-negative real value whenparg(a) = 0, i.e., when a is a non-negative real value, and √a is a complex number otherwise. Since the principal value of a1/n is n a ei arg(a)/n = √n a, we can determine the n n-th roots of | | · a as the principal n-th root of a times each of the n n-th roots of unity, i.e., a1/n ω , ω2, ω3, p · { n n n ..., ωn . n} Exercise 0.37: Use Euler’s formula eiθ = cos(θ)+ i sin(θ) to show that the 3 cube roots 1+ i√3 1 i√3 of unity are: 1, − , and − − , and that the latter two are primitive cube roots of 2 2 unity. Exercise 0.38: Show that, for a , h √k a = hk√a. ∈C q For a, b 0 , the laws of exponents yield the identity (ab)1/n = a1/nb1/n. Now taking a1/n ∈C−{n } 1/n n 1/n n n n as the set √a Rn and b as the set √b Rn, we see that (ab) = √a Rn √b Rn = √ab Rn since i j R R = ω ωn 1 i,j n = R because R is a multiplicative Abelian group (!) n n { n | ≤ ≤ } n n For a 0 , we have a1/(hk) = (a1/h)1/k. This is reflected in the fact that R = R R ∈ C−{ } pq p q when gcd(p,q) = 1. We have R = e2πi0/p), e2πi1/p), ..., e2πi(p 1)/p) and R = e2πi0/q), p { − } q { e2πi1/q), ..., e2πi(q 1)/q) and − } R R = e2πih/p e2πik/q h 0, 1, 2,...,p 1 and k 0, 1, 2,...,q 1 p q { · | ∈ { − } ∈ { − }} = e2πi(hq + kp)/(pq)) h 0, 1, 2,...,p 1 and k 0, 1, 2,...,q 1 { | ∈ { − } ∈ { − }} = e2πi((hq + kp) mod (pq))/(pq)) h 0, 1, 2,...,p 1 and k 0, 1, 2,...,q 1 . { | ∈ { − } ∈ { − }} And, when gcd(p,q) = 1, as h runs through 0, 1, 2,...,p 1 and k runs through 0, 1, 2,...,q 1, − − (hq + kp) mod pq runs through 0, 1, 2,...,pq 1. Therefore R = R R when gcd(p,q)=1. − pq p q Exercise 0.39: Show that when gcd(p,q) = 1, as h runs through 0, 1, 2,...,p 1 and k runs − through 0, 1, 2,...,q 1, (hq + kp) mod pq runs through 0, 1, 2,...,pq 1. Hint: see exercise − − ??. Exercise 0.40: Show that gen(R ) = gen(R ) gen(R ) when gcd(p,q) = 1. Hint: x pq p · q ∈ gen(R ) implies o(x)= p and y gen(R ) implies o(y)= q; now consider o(xy). p ∈ q 12 Exercise 0.41: Show that R = R R when gcd(p,q) > 1. pq 6 p q Exercise 0.42: Let a be a negative real number. Show that the n values of a1/n are ω1 n a , 2n | | ω3 n a , ω5 n a , ..., ω2n−1 n a , where ω1 , ω2 , ..., ω2n−1, ω2n are the 2n-th roots of 2n | | 2n | | 2n | | 2n 2n 2n 2n p 1 unity.p (Here ωp, ω3 , ..., ω2n−p1 are the n n-th roots of 1. The principal n-th root of 1 is 2n 2n 2n − − √n 1 = eiπ/n which is the primary primitive 2n-th root of unity. − For example, the polar form of 1 is ei(π + 2kπ) for k , so − ∈Z ( 1)1/3 = ei(π + 2 0π)/3, ei(π + 2 1π)/3, ei(π + 2 2π)/3 = eiπ/3, eiπ, eiπ 5/3 − { · · · } { · } where the principal value √3 1 = ei arg( 1)/3 = eiπ/3.) − · − Exercise 0.43: iπ 1/3 iπ 3/3 iπ 5/3 iπ 1/3 iπ 3/3 iπ 5/3 What is the set e · , e · , e · e · , e · , e · ? iπ 1/3 iπ 3{/3 iπ 5/3}·{ iπ 2 } Note e · has order 6, e · has order 2, and e · has order 6; we have (e ) = 1 corresponding to ( 1)2/3. − 0.4 Cyclotomic Polynomials For d n, the d-th-roots-of-unity elements of the order-d R -subgroup R are the roots of the | n d polynomial xd 1, and in particular, the elements of the order-n R -group R are the roots of the − n n polynomial xn 1; all these polynomials share the root 1, and various subsets of these polynomials − share other subsets of the n-th roots of unity as roots. We have xn 1=(x 1)(xn−1 + xn−2 + + x1 + 1), and depending on n, the polynomial − − ··· xn−1 + xn−2 + + x1 +1 may factor into lower-degree monic polynomials with integer coefficients, ··· or may itself be irreducible over the integers. For n = 1, 2,..., the irreducible monic factors of the monic polynomials xn 1 are called cyclotomic polynomials since their roots lie on the unit circle − in the complex plane. xhk 1 yk 1 Note that when n is composite with n = hk, − = − = yk−1 + yk−2 + + y1 +1, where xh 1 y 1 ··· xhk 1 − − y = xh, so − =(xh(k−1) + xh(k−2) + + xh + 1), and xh 1 ··· − xhk 1 = (xh 1)(xh(k−1) + xh(k−2) + + xh + 1) − − ··· = (x 1)(xh−1 + xh−2 + + x1 + 1)(xh(k−1) + xh(k−2) + + xh + 1). − ··· ··· xh(k−1) 1 Also − =(xh(k−2) + xh(k−3) + + xh + 1) so xh(k−1) =(xh 1)(xh(k−2) + xh(k−3) + + xh 1 ··· − ··· xh +1)+1;− we may substitute to attempt to derive more elaborate factorizations of xhk 1. Many − such factorizations have been determined, for example: xn 1=(x 1)[(x 1)(xn−2 + 2xn−3 + 3xn−4 + +(n 2)x +(n 1)) + n]. − − − ··· − − 13 The set of polynomials with integer coefficients, [x], is a unique-factorization domain; thus there Z will be “prime” (irreducible) polynomials in [x] and every polynomial in [x] will be uniquely Z Z expressible as a product of irreducible polynomials. Seeking the “prime” irreducible factors of xn 1 − leads to the cyclotomic polynomials. Exercise 0.44: Show that when n is prime, the polynomial xn−1 + xn−2 + + x1 +1 is ··· irreducible in [x]. Z Solution 0.44: There is a famous irreducibility criterion known as Eisenstein’s criterion which states that for n 1, if c(x) = c xn + c xn−1 + + c x + c with c ,c ,...,c ,c ≥ n n−1 ··· 1 0 0 1 n−1 n ∈ Z and cn = 1 (so that c(x) is a monic polynomial in [x]), then if p is a prime such that p divides Z2 each of the coefficients cn−1,cn−2,...,c1,c0, but p does not divide c0, then c(x) is irreducible in [x] (and also irreducible in [x], where denotes the rational numbers and [x] is the set Z Q Q Q of polynomials with rational coefficients). We can prove this as follows. Let c(x) be a degree n 1 polynomial with integer coefficients. ≥ Suppose the conditions of Eisenstein’s criterion are met, but c(x)= a(x) b(x) with degree(a)= · s 0 and degree(b)= t 0 such that s + t = n where a(x)= a xs + a xs−1 + + a x + a , ≥ ≥ s s−1 ··· 1 0 b(x) = b xt + b xt−1 + + b x + b , and a = 1 and b = 1 with a , b for i . Then t t−1 ··· 1 0 s t i i ∈ Z ∈ Z c = a b with a = 0 for j 0, 1,...,s and b = 0 for j 0, 1,...,t . k j k−j j 6∈ { } j 6∈ { } 0≤Xj≤k Now c = a b , and p2 ∤ c , so not both a and b are divisible by p. Let us arbitrarily assume 0 0 · 0 0 0 0 that p a and p ∤ b . Further suppose that p also divides a , a , ..., a and p ∤ a where | 0 0 1 2 i−1 i 0 Consider c = a b = a b + a b + a b . We have p c , and p divides a , a , a , i j i−j 0 i 1 i−1 ··· i 0 | i 0 1 2 0X≤j≤i ..., ai−1, but p ∤ ai and, as chosen above, p ∤ b0; thus p does not divide the sum ajbi−j, 0X≤j≤i and hence p ∤ ci contrary to assumption, unless t = 0, and in that case b(x) must be 1, and c(x) does not factor, and thus is irreducible in [x]. If we have p b and p ∤ a , we obtain the same Z | 0 0 conclusion. [qed] Now, we may employ Eisenstein’s criterion to show that xn−1 + xn−2 + + x1 +1 is irreducible ··· in [x] when n is prime. Let c(x)= c xn−1 + + c x + c = xn−1 + xn−2 + + x1 + 1, so Z n−1 ··· 1 0 ··· that c = 1 for 0 j But (x + 1)n 1 c(x +1) = − (x + 1) 1 − n n n n n xn + xn−1 + xn−2 + + x2 + x1 0 1 2 ··· n 2 n 1 = − − x n n n n n = xn−1 + xn−2 + xn−3 + + x + . 0 1 2 ··· n 2 n 1 − − n n (n 1) (n j + 1) And thus c(x + 1) is a polynomial whose coefficients are = · − ··· − for j j (j 1) 2 1 j = 0, 1, ,n 1. And when n is prime, all these coefficients, except the· leading− ··· coefficent· , are ··· − n divisible by n, (exercise ??), and for n prime, the constant term coefficient = n is not n 1 divisible by n2. Thus Eisenstein’s criterion applies and we see that when n is prime,− c(x +1) is irreducible in [x], and then c(x) is also irreducible in [x]. [QED] Z Z Recall that xn 1 = (x r) with R = ω ,ω2,...,ωn−1, 1 where ω = e2πi/n; R is − − n { n n n } n n r∈Rn the order-n multiplicativeY group consisting of the n-th roots of unity. The roots of cyclotomic polynomials will thus be roots of unity as observed above. The cyclotomic polynomials can be numbered; the n-th cyclotomic polynomial is the monic poly- nomial Φ (x) = (x r) where gen(R ) is the set of primitive n-th roots in R n − n n r gen(R ) ∈ Y n that comprise the generators of Rn, i.e., gen(Rn) is the set of order-n elements in Rn. Recall #gen(Rn) = φ(n), where φ is Euler’s totient function; thus Φn(x) is the degree-φ(n) monic poly- nomial whose roots are the φ(n) primitive n-th roots of unity: Φ (x) = (x e2πik/n). n − 1≤k≤n gcd(Yk,n)=1 We shall take this as the definition of the cyclotomic polynomials Φ1, Φ2, .... In order for this definition to conform to our prior notion of cyclotomic polynomials, we must show that the Φn- polynomials have integer coefficients, and are irreducible in [x], and that xk 1 factors into a Z − product of such polynomials for k 1. Note Φ (x) = (x e2πik/n) since the group of units ≥ n − k∈Un U = k 1 k n, gcd(k,n) = 1 ; thus we again seeY the correspondence between gen(R ) and n { | ≤ ≤ } n the order-φ(n) group Un. The cyclotomic polynomials Φ1(x), Φ2(x), ... do not satisfy the entire chain of inequalitites degree(Φ ) degree(Φ ) ; this is because φ(n) is not a monotonically-nondecreasing function 1 ≤ 2 ≤··· of n. Exercise 0.45: Can you characterize the integers n such that φ(n) φ(n + 1)? ≤ φ(n)−1 Note the coefficient of the x -term of the cyclotomic polynomial Φn is the negative of the sum 15 of the roots of Φ , this is the value r = c (1), where c (m) = e2πimh/n is n − − n n r∈gen(Rn) 1≤h≤n X gcd(Xh,n)=1 known as Ramanujan’s sum. See page ??, there we derived cn(1) = µ(n) where µ is the M¨obius function (see page ??). Thus the coefficient of the xφ(n)−1-term of Φ (x) is µ(n) and the sum of n − the roots of Φn(x) is µ(n). 0 The constant term (the coefficient of the x -term) of the cyclotomic polynomial Φn(x) is the product of the negative roots of Φ , this is ( 1)φ(n) r; we have computed r = 1 2δ n − − n,2 r∈genY(Rn) r∈genY(Rn) above. Thus the constant term of Φ (x) is ( 1)φ(n) (1 2δ ). Since φ(1) = φ(2) = 1 and φ(n) is n − − n,2 even for n 3, (exercise ??), with δ = 0 unless n = 2. we see that the constant term of Φ (x) ≥ n,2 n is 1 for n> 1, and the constant term of Φ (x) is 1, so 1 2δ is the constant term of Φ (x) for 1 − − n,1 n n 1. ≥ j Exercise 0.46: Show that the cyclotomic polynomial Φp(x)= x when p is prime. Hint: 0≤j Here is a table of the first 6 cyclotomic polynomials. n primitive roots Φn(x) 1 1 x 1 { } − 2 1 x + 1 {− } 3 ω ,ω2 x2 + x + 1 { 3 3} 4 ω ,ω3 x2 + 1 { 4 4} 5 ω ,ω2,ω3,ω4 x4 + x3 + x2 + x + 1 { 5 5 5 5} 6 ω ,ω5 x2 x + 1 { 6 6} − We have xn 1= (x r), and thus the set-identity R = gen(R ) expressing R as a union − − n d n r R d|n ∈Y n [ of disjoint sets implies xn 1 = Φ (x). For example, x6 1 = Φ (x)Φ (x)Φ (x)Φ (x). (See − d − 6 3 2 1 Yd|n page 5.) This identity, in turn, allows us to see that the coefficients of the cyclotomic polynomials are integers, i.e., Φ (x) [x] for n = 1, 2,.... This follows by induction. We have Φ (x)= x 1, n ∈Z 1 − and for n > 1, we may write xn 1 = Φ (x)Φ (x) Φ (x)Φ (x) where 1 = d < d < − d1 d2 ··· dτ(n)−1 dτ(n) 1 2 Now, we may take the assertion that the cyclotomic polynomials Φk(x) are monic and have in- teger coefficients for 1 k Φ (x)Φ (x) Φ (x); then f(x) is a monic polynomial with integer coefficients. Then d1 d2 ··· dτ(n)−1 xn 1 = f(x) Φ (x); xn 1 is a monic polynomial with integer coefficients as is f(x), so − · dτ(n) − Φdτ(n) (x) must be monic with integer coefficients as well, and hence all the cyclotomic polynomials have integer coefficients. [QED] Although we might think that the coefficients of cyclotomic polynomials are always among the values 1, 0, 1, this is not the case. As we shall see later, for n equal to a product of certain − sufficiently-many and sufficiently-large distinct prime factors, a coefficient of arbitrarily large mag- nitude can be seen in Φn [Leh36] [Erd46]. Exercise 0.47: Use the identity xn 1= Φ (x) to construct an algorithm to compute the − d Yd|n cyclotomic polynomial Φn+1(x) given all the earlier cyclotomic polynomials Φn(x), Φn−1(x), ..., Φ2(x), Φ1(x). Exercise 0.48: Show that xn−1 + xn−2 + + x +1= Φ (x). Thus we again see that ··· d d|n,d>Y 1 for n prime, Φ (x)= xn−1 + xn−2 + + x + 1. n ··· xn 1 Exercise 0.49: Show that − = Φ (x). xm 1 d − d | n,Y d ∤ m xn 1 Exercise 0.50: Show that Φn(x)= − . gcd(xn 1, (xj 1)) − − 1≤Yj Exercise 0.52: Show that for n 3, Φ (x) > 0 for x . Hint: Φ (x) has no real roots for ≥ n ∈R n n 3. ≥ Exercise 0.53: (Jameson) Show that for n 1, Φ (x) is strictly-increasing on the interval ≥ n (1, ). ∞ Solution 0.53: Recall U = v gcd(v,n) = 1 where = 1, 2,...,n 1, 0 . We n { ∈ Zn | } Zn { − } have Φ (x) = x 1 and Φ (x) = x + 1 both strictly-increasing as x increases. Now for n 3, 1 − 2 ≥ we have Φ (x) = (x e2πik/n) (x e2πi(n k)/n) n − · − − k∈UnY,k n n This is because, for n 3, U = k U 1 k < n k U 1 k < where ≥ n { ∈ n | ≤ 2 } ∪ { − ∈ n | ≤ 2 } n n φ(n) # k U 1 k < =# n k U 1 k < = . (See exercise ??.) { ∈ n | ≤ 2 } { − ∈ n | ≤ 2 } 2 2πk n Now let α = cos( ) and q (x) = x2 2α x + 1. Note since when k U with k < , k n k − k ∈ n 2 2πk gcd(k,n)=1 and 1 < cos( ) < 1, so 1 <α < 1. Then q (x)=(x α )2 + (1 α2) > 0 − n − k k − k − k for x 0, and q (x) is strictly-increasing for x 1. And we have Φ (x)= q (x), so ≥ k ≥ n k k∈UnY,k Exercise 0.55: Show that for n> 1, Φ (v) > (v 1)φ(n) for v . | n | − ∈R Solution 0.55: Since Φ (x)= (x ωk), for v real, we have n − n kY∈Un Φ (v) 2 = v ωk 2 | n | | − n| kY∈Un = [(v Re(ωk))2 + Im(ωk)2] − n n kY∈Un = [v2 2v Re(ωk)+Re(ωk)2 + Im(ωk)2] for n> 1. − n n n kY∈Un And thus, for n> 1, Φ (v) 2 = [v2 2v Re(ωk)+1] > [v2 2v +1] = (v 1)2 =(v 1)2φ(n) | n | − n − − − kY∈Un kY∈Un kY∈Un so Φ (v) > (v 1)φ(n). [QED] | n | − Exercise 0.56: Show that for n > 2, Φ ( x ) as x . Hint: degree(Φ ) = φ(n) is n | | → ∞ | | → ∞ n even for n> 2. 18 Note that, since we have xn 1= Φ (x), we may apply the product M¨obius inversion formula: − d d|n n Y h(n) = H( )µ(d) for n + where H(m) = h(d) with H(m) = 0 for m +. (See page d ∈ Z 6 ∈ Z Yd|n dY|m ??.) Here µ is the M¨obius function discussed in chapter ??; we have µ(1) = 1, µ(p p p )=( 1)s 1 2 ··· s − where p1, p2, ..., ps are distinct primes, and µ(n) = 0 when n is not squarefree. We obtain Φn(x)= (xd 1)µ(n/d) = (xn/d 1)µ(d) for n + and x = 1. Note that (xd 1)µ(n/d) expresses − − ∈Z 6 ± − Yd|n Yd|n Yd|n the polynomial Φ (x) as a rational expression when n> 1. For example, Φ (x)=(x 1)−1(x5 1)1. n 5 − − Now invoking L’Hˆopital’s rule and continuity, we may assert Φ (x) = (xd 1)µ(n/d) for x . n − ∈ C Yd|n (Alternately, Jameson [Jam15] notes that we may tranform Φ (x)= (xn/d 1)µ(d) to n − Yd|n Φ (x) (xn/d 1) = (xn/d 1)µ(d) = (xn/d 1) n · − − − d|n,µY(d)=−1 d|n,µY(d)6=−1 d|n,µY(d)=1 since (xn/d 1)0 = 1; this necessarily holds for all x . − ∈C Also, for n = pe1 pe2 pek where p , p , ..., p are distinct primes and e 1, e 1, ..., 1 2 ··· k 1 2 k 1 ≥ 2 ≥ e 1, # d d n and µ(d) = 1 = 2k−1 and # d d n and µ(d) = 1 = 2k−1 and k ≥ { | | − } { | | } # d d n and µ(d) = 0 = τ(n) 2k. (See exercise ??.) { | | } − We have Φ (x) (xn/d 1)/(x 1) = (xn/d 1)/(x 1) n · − − − − d|n,µY(d)=−1 d|n,µY(d)=1 where both products have 2k−1 factors. And thus Φ (x) s (x) = s (x) n · n/d n/d d|n,µY(d)=−1 d|n,µY(d)=1 where s (x)= xh−1 + xh−2 + + x + 1. h ··· Now for n> 1, we have −1 Φ (x) = s (x)µ(d) s (x)µ(d) s (x) n n/d · n/d · n/d d|n,µY(d)=1 d|n,µY(d)=0 d|n,µY(d)=−1 = s (x)µ(d) s (x)µ(d) s (x)µ(d) n/d · n/d · n/d d|n,µY(d)=1 d|n,µY(d)=0 d|n,µY(d)=−1 µ(d) = sn/d(x) . Yd|n Exercise 0.57: Show that for n> 1, Φ (x)= (1 xn/d)µ(d) = (1 xd)µ(n/d). n − − Yd|n Yd|n 19 Solution 0.57: We have Φ (x) = (xn/d 1)µ(d) so ( 1)S(n)Φ (x) = ( 1)µ(d)(xn/d n − − n − − Yd|n Yd|n 1)µ(d) where S(n)= µ(d). But then ( 1)µ(d)(xn/d 1)µ(d) = (1 xn/d)µ(d). − − − Xd|n Yd|n Yd|n And S(n) = µ(d) = δ (see page ??), so ( 1)S(n) = ( 1)0 = 1 for n > 1 and ( 1)S(n) = 1,n − − − Xd|n ( 1)1 = 1. Thus for n > 1, ( 1)S(n)Φ (x) = Φ (x) = (1 xn/d)µ(d) = (1 xd)µ(n/d). − − − n n − − Yd|n Yd|n [QED] We can use the identity Φ (x)= (xn/d 1)µ(d) to show that, for p prime, Φ (x) = Φ (xp) when n − pn n Yd|n p n and Φ (x) = Φ (xp)/Φ (x) when p ∤ n. | pn n n In the first case where p n, we have | Φ (x) = (xpn/d 1)µ(d) pn − dY|pn = (xpn/d 1)µ(d) (xpn/d 1)µ(d) − · − Yd|n d|pn,dY ∤n = Φ (xp) (xpn/d 1)µ(d) . n · − d|pn,dY ∤n And when the condition d pn and d ∤ n holds, we have p2 d (since p n); but then d is not | | | squarefree and hence µ(d)=0. Thus (xpn/d 1)µ(d) = (xpn/d 1)0 = 1 and then Φ (x) = Φ (xp) when p n. − − pn n | d|pn,dY ∤n d|pn,dY ∤n In the alternative case where p ∤ n, we have Φ (x) = (xpn/d 1)µ(d) pn − dY|pn = (xpn/d 1)µ(d) (xpn/(pd) 1)µ(pd) − · − Yd|n Yd|n = Φ (xp) (xn/d 1)−µ(d) n · − Yd|n p = Φn(x )/Φn(x). 20 d Note with p ∤ n, the condition d pn splits into two distinct cases: d n and n with p d. Also, | | p | | with gcd(d,p) = 1, we have µ(pd)= µ(p)µ(d)=( 1)µ(d) since µ is a multiplicative function (page p − ??). Thus Φpn(x) = Φn(x )/Φn(x) when p ∤ n. [QED] Note that iterating the above pair of identities yields p pk−1 pk Φ k (x) = Φ k−1 (x )= = Φ (x ) = Φ (x ) p n p n ··· pn n p when p n with p prime. And for k > 1, Φ k (x) = Φ k−1 (x) = Φ k−1 (x ), so | p n p(p n) p n p pk−1 pk pk−1 Φ k (x) = Φ k−1 (x )= = Φ (x ) = Φ (x )/Φ (x ) p n p n ··· pn n n when p ∤ n with p prime. p + pk−1 y 1 p−1 Exercise 0.58: Show that for p prime and k , Φ k (x) = Φ (x ) = − = y + ∈ Z p p y 1 k−1 − k yp−2 + ... + y + 1 with y = xp . Thus for x = 2, we obtain the Fermat numbers 22 +1= Φ2k+1 (2). k Exercise 0.59: (Jameson) Show that for n> 1, Φn(1) = p if n = p with k > 1 and p prime, and Φn(1) = 1 otherwise. k Solution 0.59: Take p prime and write n = mp with k 1 and p ∤ m. Then Φn(x) = pk pk−1 ≥ k Φm(x )/Φm(x ), and then with m > 1, Φn(1) = Φm(1)/Φm(1) = 1. And for n = p , k−1 k−1 k−1 exercise 0.58 yields Φ (1) = ((xp )p−1 +(xp )p−2 + +(xp )+1) = p. [QED] n ··· |[x=1] Exercise 0.60: Generalize the above result that Φ (xp) = Φ (xp) when p n by showing pn n | that Φ (x) = Φ (xn/m) when m is a non-deficient divisor of n, i.e., m n and if p n then n m | | p m for every prime divisor p of n. Hint: if d n but d ∤ m then d is not squarefree, and then | | µ(d)=0. Exercise 0.61: Restate the result of exercise 0.60 to show that for n,k + with k n, we k ∈ Z | have Φnk(x) = Φn(x ). Solution 0.61: Note with k n, ζ gen(R ) if and only if ζk gen(R ) since when | ∈ nk ∈ n o(ζ) = nk, o(ζk) = nk/gcd(nk,k) = n and when o(ζk) = n, we have n = o(ζ)/gcd(o(ζ),k), or equivalently, o(ζ)= n gcd(o(ζ),k), and so o(ζ)= nk when k n. (exercise ??). · | k But now ζ is a root of Φnk(x) if and only if ζ is a root of Φn(x ); thus the monic polynomials k Φn(x ) and Φnk(x) have the same roots, so they are identical. [QED] (You may wish to verify that both Φ (xk) and Φ (x) have degree kφ(n) when k n.) n nk | These formulas allow us to compute cyclotomic polynomials with certain composite indices more easily; for example Φ (x) = Φ (x3)=(x3)2 (x3)+1= x6 x3 +1. Indeed, if we know how to 18 6 − − compute Φn(x) for n squarefree, then the prior identities allow us to compute Φn(x) for arbitrary n. Exercise 0.62: Show that for n odd and greater than 1, Φ (x) = Φ ( x). 2n n − 21 Solution 0.62: Since for n 3 odd, gen(R ) = gen(R ) from exercise 0.29, we have ≥ 2n − n Φ (x)= (x r ) with gen(R )= r ,r ,...,r . and Φ (x)= (x + r ). n − j n { 1 2 φ(n)} 2n j 1≤jY≤φ(n) 1≤jY≤φ(n) And thus Φ ( x) = ( x r ) n − − − j 1≤jY≤φ(n) = ( 1)φ(n)(x + r ) − j 1≤jY≤φ(n) = (x + rj) 1≤jY≤φ(n) = Φ2n(x). (We have ( 1)φ(n) = 1 because n 3 and hence φ(n) is even (exercise ??). [QED] − ≥ (xpq 1)(x 1) Exercise 0.63: Show that, for p and q distinct primes, Φ (x)= − − . pq (xp 1)(xq 1) − − Exercise 0.64: Show that, for p, q, and r distinct primes, (xpqr 1)(xp 1)(xq 1)(xr 1) Φ (x)= − − − − . pqr (x 1)(xpq 1)(xpr 1)(xqr 1) − − − − Exercise 0.65: (A. Migotti) Let p and q distinct primes. Show that Φpq(x) has only 1 and 1 ( and 0) as coefficients. Hint: look at Φ (x) (xp 1)(xq 1)=(xpq 1)(x 1). Also see − pq · − − − − [Bro16]. Exercise 0.66: Let n = p p p where p , p , ..., p are distinct primes; n is squarefree. 1 2 ··· k 1 2 k Generalize exercise 0.63 and exercise 0.64 to show that, for k odd we have Φn(x)= An(x)/Bn(x), and for k even we have Φn(x)= Bn(x)/An(x), where p p ···p p p ···p A (x)= (x j1 j2 jt 1) and B (x)= (x j1 j2 jt 1). n − n − 0≤t≤k 1≤j1<··· Hint: let a = p p p t 0, 2,..., 2 k/2 and 1 j < Then A (x) = (xd 1) and B (x) = (xd 1). See exercise ??. Now note that µ(d) = 1 n − n − dY∈a Yd∈b for d a and µ(d)= 1 for d b, and use the identity Φ (x)= (xn/d 1)µ(d). ∈ − ∈ n − Yd|n 22 Exercise 0.67: Show that Φ (x)=(xn 1)/lcm xd 1 1 d (1) Is every polynomial with integer coefficents whose roots all have magnitude 1 a product of cyclotomic polynomials? (2) Is every even-degree-k polynomial with integer coefficents whose roots r1, r2, ..., rk satisfy r + r + + r = k a product of cyclotomic polynomials? | 1| | 2| ··· | k| Looking at the first few cyclotomic polynomials, we may note that, except for Φ1(x), they have R R a symmetry: Φn (x) = Φn(x) where Φn (x) is the degree-φ(n) polynomial obtained from Φn(x) by reversing the sequence of coefficients. Specifically, for p(x)= c +c x+c x2 + +c xk−1 +c xk, 0 1 2 ··· k−1 k pR(x) = c + c x + c x2 + + c xk−1 + c xk. Conrad [?] calls a polynomial p such that k k−1 k−2 ··· 1 0 p(x) = pR(x) palindromic. Note a degree-k palindromic polynomial p(x) has no roots equal to 0, i.e., is not divisible by x. 1 Consider g(x) = xkp( ); g(x) = xk(c + c x−1 + c x−2 + + c x−k) = pR(x). Now taking x 0 1 2 ··· k r1,r2,...,rk to be the roots of p, we have p(x) = ck(x r1)(x r2) (x rk), and thus g(x) = 1 1 1 − − ··· − xkc ( r )( r ) ( r ) = c (1 r x)(1 r x) (1 r x). k x − 1 x − 2 ··· x − k k − 1 − 2 ··· − k The constant term c of p is c ( 1)kr r r . Assume that none of c , r , r , ..., r are 0, then 0 k − 1 2 ··· k k 1 2 k c ( 1)k c = 0 and k = − . And then 0 6 c r r r 0 1 2 ··· k c ( 1)k k g(x) = − g(x) c r r r 0 1 2 ··· k ( 1)k = − c (1 r x)(1 r x) (1 r x) r r r k − 1 − 2 ··· − k 1 2 ··· k 1 1 1 = ck(x )(x ) (x ). − r1 − r2 ··· − rk 1 1 1 Thus cancelling ck on both sides, we have g(x) = c0(x )(x ) (x ). Note if p(x) = − r1 − r2 ··· − rk 1 1 xkp( )= g(x) then p is palindromic and pR(x)= g(x). In this case, if r =0 is a root of p then x 6 r is also a root of p; when p is palindromic, the roots of p come in mutually-reciprocal pairs; except for 1 and 1, these pairs are composed of two distinct values. − Now for p(x) = Φ (x) with the φ(n) distinct non-zero roots r ,r ,...,r gen(R ), we have the n 1 2 φ(n) ∈ n constant term c =( 1)φ(n)r r r . But we have previously seen that r r r = 1 2δ 0 − 1 2 ··· φ(n) 1 2 ··· φ(n) − n,2 where r1, r2,...,rφ(n) are the φ(n) primitive n-th roots of unity and that the constant term c0 of Φ (x)is( 1)φ(n)(1 2δ ) = 1 2δ . Also Φ (x) is a degree-φ(n) monic polynomial, so c = 1. n − − n,2 − n,1 n φ(n) 23 Thus, with p(x) = Φn(x), we have R g(x) = Φn (x) 1 = xφ(n)Φ ( ) n x 1 1 1 = c0(x )(x ) (x ) − r1 − r2 ··· − rφ(n) 1 1 1 = (1 2δn,1)(x )(x ) (x ) − − r1 − r2 ··· − rφ(n) = (1 2δ )(x r )(x r ) (x r ) − n,1 − 1 − 2 ··· − φ(n) = (1 2δ )Φ (x). − n,1 n R Thus we have the symmetry Φn (x) = Φn(x) for n> 1. 1 1 1 1 1 We have (x )(x ) (x )=(x r1)(x r2) (x rφ(n)) used above since , , ..., − r1 − r2 ··· − rφ(n) − − ··· − {r1 r2 1 2πik/n 2πi(n k)/n = r1, r2,...,rφ(n) ; this is because, if e gen(Rn) then e − gen(Rn), rφ(n) } { } ∈ ∈ 2πik/n 2πi(n k)/n i.e., e and e − have the same order n, since gcd(n k,n) = 1. (Of course, 2πi(n k)/n 2πik/n − e − in Rn is just the inverse of e .) [QED] 1 The symmetry relationship ΦR(x) = Φ (x) for n > 1 is equivalent to the relationship xnΦ ( ) = n n n x Φn(x) for n> 1; this symmetry relation is not a characterization of cyclotomic polynomials however. There are palindromic polynomials in [x] with roots in C R + , see page 10 and 26, and Z 1 − Z the example Conrad gives: f(x) = x4 2x3 2x + 1 with the palindromic coefficient sequence − − 1, 2, 0, 2, 1 , satisfies f R(x) = f(x). Note, however, although the constant term of f(x) is 1, h − − i the coefficient of x3 is not µ(n) for any value of n. Exercise 0.69: Can you give a useful characterization of cyclotomic polynomials? That is, given a polynomial b(x) [x], can you give a means to relatively-easily determine whether ∈ Z b(x) is a cyclotomic polynomial? Conrad [?] also shows that, for any degree-n polynomial in [x] with n > 1 and n even, the 1 Z 1 identity f(x)= xnf( ) is equivalent to the identity f(x)= xn/2g(x + ) for a unique degree-n/2 x x polynomial g in [x]. Note the transformation f g “converts” each pair of complex-conjugate Z → 1 x∗ roots of f lying on the unit circle into a real root of g since = = x∗ when xx∗ = 1, and then x xx∗ 1 x + x∗ = 2Re(x), as we have observed above. Also note that with degree(f)= n, f(x)= xnf( ) x 1 implies f(0) = 0; the constant term of f is not zero if f(x)= xnf( ). 6 x 24 The proof uses the prior result that the degree-n polynomial f(x) [x] is palindromic if and only 1 ∈Z if f(x)= xnf( ). Suppose f(x)= c xn + c xn−1 + + c x + c is a even degree-n palindromic x n n−1 ··· 1 0 1 polynomial satisfying f(x)= xnf( ). Take m = n/2 1. Since n = 2m, we have x ≥ 1 1 1 1 ( )mf(x)= c xm + c xm−1 + + c + c + + c + c . x 2m 2m−1 ··· m m−1 x ··· 1 xm−1 0 xm R 1 m But f(x)= f (x), so c2m−k = ck for k = 0, 1,..., 2m and in particular c2m = c0. Now c2m(x+ ) x 1 1 has its highest-power term equal to c xm and its least-power term equal to c ( )m = c ( )m. 2m 2m x 0 x 1 (Note if a polynomial g such that f(x)= xmg(x + ) exists then its leading term must be c xm x 2m and its constant term must be c2m since f(x) is palindromic.) 1 1 1 Thus ( )mf(x) c (x+ )m is a polynomial in x and of degree less than m. (This circumlocution x − 2m x x 1 1 just means that xk[( )mf(x) c (x + )m] is a polynomial in x of degree less than m + k, with x − 2m x 1 k equal to the exponent of in the trailing term which is the same as the exponent of x in the x leading term in this case.) 1 1 m m The polynomial in x and , c (x + )m, is palindromic (since = ), and the x 2m x j m j 1 − difference of two palindromic degree-m polynomials in x and with the same leading coefficient x and the same trailing coefficient is a palindromic polynomial of degree less than m. 1 Exercise 0.70: With f, m, and c as defined above, show that the polynomial xm[( )mf(x) 2m x − 1 c (x + )m] is palindromic. 2m x Thus, introducing an induction argument with the induction hypothesis that for all palindromic polynomials h [x] of even degree k less than n, there exists a unique polynomial g [x] of ∈ Z 1 ∈ Z degree k such that h(x) = xk/2g(x + ), we can verify this holds for n = 2, and then, since we x 1 1 have that, for some k 0, 1,...,m 1 , xk[( )mf(x) c (x + )m] =: h(x) is a even degree-2k ∈ { − } x − 2m x palindromic polynomial in [x] with k < m, by our induction hypothesis there exists a polynomial Z 1 g˜ of degree k/2 less than m such that h(x) = xkg˜(x + ). But then g(x) :=g ˜(x)+ c xm is a x 2m 1 polynomial of degree m such that f(x)= xmg(x + ). x 1 Exercise 0.71: Show that ax2 + bx + a = xg(x + ) where g(x)= ax + b. x 1 1 1 The palindromic polynomial in x and , ( )mf(x) c (x + )m of degree less than m, has the x x − 2m x 25 form: 1 1 1 0 xm + +0 xm−j+1 +c ˜ xm−j + +c ˜ x0 + +c ˜ ( )m−j +0 ( )m−j+1 + +0 ( )m · ··· · 2m−j · ··· 2m · ··· j · x · x ··· · x 1 1 withc ˜ =c ˜ for some j 1, 2,...,m and j t 2m j. If j = m, ( )mf(x) c (x + )m 2m−t t ∈ { } ≤ ≤ − x − 2m x 1 is identically 0 and we may take g(x) = c xm to obtain f(x) = xmg(x + ). Otherwise, j < m 2m x andc ˜ = 0. 2m−j 6 Suppose j < m. Then the length-(2m +1) sequence of exponents is: m, m 1,..., 1, 0, 1,...,m h − − 1, m and the length-(2m+1) sequence of coefficients is: 0, 0,..., 0, c˜ ,..., c˜ ,..., c˜ , 0, 0,..., 0 i h 2m−j m j i with the same positive number of leading and trailing 0 coefficients. And this sequence of coefficients 1 1 1 is palindromic. Thus ( )mf(x) c (x + )m is a palindromic polynomial in x and , of degree x − 2m x x 1 1 m j with 1 j m, and letting k = m j, xk[( )mf(x) c (x+ )m] is a degree-2k palindromic − ≤ ≤ − x − 2m x polynomial in x with 2k Now, by our induction hypothesis, there is a degree-k polynomialg ˜ in [x] such that Z 1 1 1 xk[( )mf(x) c (x + )m]= xkg˜(x + ). x − 2m x x 1 1 1 Hence, ( )mf(x) c (x + )m =g ˜(x + ), so x − 2m x x 1 1 1 1 f(x) c xm(x + )m = xmg˜(x + ), or equivalently, f(x)= xm[˜g(x + )+ c (x + )m]. − 2m x x x 2m x 1 Thus, taking g(x) =g ˜(x)+ c xm where degree(g)= m, we have f(x)= xmg(x + ). 2m x 1 1 1 1 Finally, if f(x) = xmg(x + ) with necessarily degree(g) = m, then f( ) = g( + x), so x x xm x 1 1 1 1 1 xmf( )= g( + x), and hence x2mf( )= xmg( + x)= f(x), i.e., f(x)= xnf( ). x x x x x 1 Altogether, for f a polynomial of degree n with n > 0 even, f(x) = xnf( ) if and only if f(x) = x 1 xmg(x + ) for some (unique) degree-n/2 polynomial g [x]. [QED] x ∈Z 1 1 The result that for n even, f(x) = xnf( ) if and only if f(x) = xn/2g(x + ) for some (unique) x x degree-n/2 polynomial g [x], together with the fact that f(x) is necessarily palindromic when 1 ∈ Z f(x) = xnf( ), establishes a one-to-one correspondence between the palindromic polynomials of x degree 2m in [x] and the degree-m polynomials in [x]. Z Z If f is -irreducible, then g is -irreducible, since if g(x) = a(x)b(x) then f(x) = xdegree(a)a(x + Z Z 1 1 )xdegree(b)b(x + ) where degree(a)+ degree(b) = m, and if degree(a) 1 and degree(b) 1, x x ≥ ≥ then f is not irreducible. The converse is false, however. 26 One reason this result is of special interest is that if f(x) is a palindromic polynomial of even degree n then each pair of conjugate roots of f lying in the unit circle C1 correspond to a real root of g 1 in the interval [ 2, 2]. If f(r) = 0 then g(r + ) = 0 and conversely. Note that f(0) = 0 since f is − r 6 palindromic, so the issue of dividing by zero does not arise. 1 Now if r = 1 then r + = r + r∗ = 2 cos(arg(x)) [ 2, 2]. And if g(t) = 0 with t [ 2, 2], then | | r ∈ − ∈ − 1 t = 2 cos(θ) for some θ ( π,π], and then t = r+ for r = cos(θ)+i sin(θ) or r = cos(θ) i sin(θ), ∈ − r − 1 1 and with g(t) = 0, we have g(r + ) = 0, and hence f(r)=0 and f( )=0. r r For example, using the recursive construction producing the polynomial g above, we can express Conrad’s polynomial, given before, as: 1 1 f(x)= x4 2x3 2x +1= x2 (x + )2 2(x + ) 2 . − − x − x − 1 2+ √12 Thus f(x) = x2g(x + ) where g(x) = x2 2x 2. The roots of g are s := = x − − 1 2 2 √12 2.73205 ... and s := − = .73205 .... The g-root s lies in ( 2, 2), and hence corresponds 2 2 − 2 − 1 1 to a conjugate pair of distinct complex roots r and of f on the unit circle. Since r + = r r 1 1 s s = 2 cos(arg(r)) = 2 cos(arg( )), we have arg(r) = arg( ) = cos−1( 2 ) = 1.94553 ..., and 2 r r 2 cos(arg(r)) = .36602 ... and sin(arg(r)) = .93060 .... Our pair of complex conjugate roots of f − on the unit circle are ( .36602 ...)+ i(.93060 ...) and ( .36602 ...) i(.93060 ...). These complex − − − 2πk numbers are not roots of unity since if ζ is a non-real n-th root of unity, we have Re(ζ)= cos( ) n 2πk and arg(ζ)= for some integer k with k 0 (mod n), and s /2 is an algebraic number while n 6≡ 2 2πk is transcendental; more directly, we would have an absurd value for π. n 1 The pair of real roots t1 and t2 of f corresponding to the root s1 of g outside [ 2, 2] satisfy t2 = − t1 1 2 and t1 + t2 = s1. Thus t1 = = s1, so t1 s1t1 + 1 = 0. This quadratic equation has the roots t1 − s + s2 4 s s2 4 1 1 − = 2.29660 ... and 1 − 1 − = .43542 ... which are real and reciprocal. (Note 2 2 p 1 p the symmetries t2 = and t1 + t2 = s1 ensure that t2 satisfies the same quadratic equation as t1, t1 so the two roots of t2 s t +1 = 0 obtained above are the real roots of f(x). − 1 Now, turning back to cyclotomic polynomials, we may show that the integer coefficients of cyclo- tomic polynomials may be arbitrarily large in absolute value; i.e., there exists an integer n such that Φ (x) has a coefficient of magnitude greater than α for any fixed α +. n ∈Z If we compute the cyclotomic polynomials Φn(x) for n = 1, 2, 3,..., we discover that Φ105(x) is the first cyclotomic polynomial with a coefficient greater than 1 in absolute value; the coefficient of x7 27 in the degree-φ(105) polynomial Φ (x) is 2. The special property of 105 that we may note is 105 − that 105 = 3 5 7; this is the first integer that is the product of three distinct odd primes. · · Emma Lehmer gives a proof due to Issac Schur that, for n = p1p2 pt, where t 3 is odd and p1, ··· pt ≥ p2, ..., pt are distinct odd primes such that p1 + p2 > pt, the coefficient of x in the cyclotomic polynomial Φ (x) is greater than t 2 in absolute value [Leh36]. n − Let M(n) denote the maximum of the absolute values of the coefficients of Φn(x). Then Schur’s result shows that the arithmetic function M is unbounded. Schur’s proof is as follows. Let t denote an odd integer greater than 2, and let p1, p2, ..., pt be distinct odd primes such that p1 + p2 > pt. For example, for t = 5, 11, 13, 17, 19, 23 is such a sequence of primes. Let n = p p p . Now recall that Φ (x)= (1 xd)µ(n/d) (exercise 0.57). 1 2 ··· t n − Yd|n Then Φ (x)=(1 x)µ(p1···pt) (1 xd)µ(n/d) (1 xd)µ(n/d). n − · − · − d|n d|n d primeY d compositeY And t is odd, so µ(p p )= 1 and µ(n/p ) = 1, and thus 1 ··· t − j Φ (x) = (1 x)−1 (1 xpj )1 (1 xd)µ(n/d) n − · − · − 1≤j≤t d|n Y d compositeY 1 xpt = − (1 xpj ) (1 xd)µ(n/d) 1 x · − · − − 1≤j = (xpt−1 + xpt−2 + + x + 1) (1 xpj ) (1 xd)µ(n/d). ··· · − · − 1≤j Now we consider Φ (x) pmod xpt+1. (Recall for monic polynomials f(x) and g(x) in [x], f(x) pmod g(x) n Z is the unique polynomial r(x) such that f(x) = q(x)g(x) + r(x) where q(x),r(x) [x] with m m−1 ∈ Z pt+1 degree(r) < degree(g). For f(x)= fmx +fm−1x + +f1x+f0 with m>pt, f(x) pmod x = pt ··· fpt x + + f1x + f0; this degree-pt polynomial is just the polynomial f truncated to include only ··· pt+1 pt its low-order pt + 1 terms. By reducing Φn modulo x we thus obtain the coefficient of x in pt+1 Φn as the coefficient of the leading term of Φn modulo x . Now, we have Φ (x) (xpt−1 + xpt−2 + + x + 1) (1 xpj ) (1 0)µ(n/d) (pmod xpt+1) n ≡ ··· · − · − 1≤j p p ···p p +1 since x i1 i2 ik 0 (pmod x t ) for k > 1. ≡ But then Φ (x) (xpt−1 + xpt−2 + + x + 1) (1 xp1 ) (1 xp2 ) (1 xpt−1 ) 1 (pmod xpt+1). n ≡ ··· · − · − ··· − · 28 And p p p p p p k p p ···p (1 x 1 ) (1 x 2 ) (1 x t−1 ) = 1 x 1 x 2 x t−1 + ( 1) x i1 i2 ik , − · − ··· − − − −···− − 1 Now xpi xpt−pi = xpt for i = 1, 2,...,t 1, so the coefficient of xpt in (xpt−1 + xpt−2 + + x + − ··· 1) (1 xp1 xp2 xpt−1 ) is (t 1). (Each term xpi multiplies xpt−pi to contribute 1 to · − − −···−pt − − − pt+1 p−t−1 the final coefficent of x .) And hence, coefpt (Φn(x)) = coefpt (Φn(x) pmod x )= coefpt ((x + xpt−2 + + x + 1) (1 xp1 xp2 xpt−1 )) = (t 1) where coef (f(x)) is the coefficient ··· · − − −···− − − pt of xpt in f(x). We have found a cyclotomic polynomial Φ (x) with M(n) t 1. n ≥ − Also xpt−2 = xpi xpt−2−pi for i = 1, 2,...,t 1, and 1 xpt−2 = xpt−2, so the terms xp1 , xp2 , − · − − ..., xpt−1 multiply the terms +xpt−2−p1 , +xpt−2−p2 , ..., +xpt−2−pt−1 to contribute (t 1) to − − − the final coefficient of xpt−2, and 1 multiplies xpt−2 to contribute 1 to the final coefficient of xpt−2, which is thus seen to be (t 2). − − Now to conclude this proof, we must show that the condition [p , p , ..., p with p p ] is not vacuous for any t 3 1 2 t ≥ with t odd, i.e., such primes p1, p2, ..., pt exist for every admissible value of t. This argument is given in [Suz87]. Suppose to the contrary that we have an odd integer t 3 such that for any t distinct primes ≥ p 1, we ∈Z would have 2k k2t log(2) But then lim 1, and taking logarithms of the numerator and denominator yields k→∞ 2k ≥ 2 log(k)+ log(t log(2)) lim 1, and then differentiating the numerator and denominator yields k→∞ k log(2) ≥ 2 lim k which clearly has the limit 0, not a value 1. Thus to avoid this contradiction, we k→∞ log(2) ≥ must deny that for any t distinct primes p 3, there is always an increasing sequence of primes p p . 1 2 ··· t 1 2 t And with this conclusion we conclude Schur’s Theorem that the integer coefficients of cyclotomic polynomials may be arbitrarily large in absolute value. [QED] Note we have shown that for t 3 odd, there is a cyclotomic polynomial Φ (x) with both the ≥ n negative even coefficient 1 t and the negative odd coefficient 2 t. The index n is an odd number − − formed by the product of t odd primes p Exercise 0.72: What does the fact that for every integer t> 3, there is always an increasing sequence of primes p p say about the distribution of primes in 1 2 ··· t 1 2 t +? Z The use of the Prime Number Theorem is avoidable; Emma Lehmer invokes an equally-analytic theorem of Dirichlet: [given n> 1 and a + with gcd(a,n) = 1, there are infinitely many primes ∈Z p that satisfy p a (mod n)] to show that for α +, there are always some three primes p, q, and ≡ ∈Z r such that the maximum of the absolute values of the coefficients of Φpqr, M(pqr), is greater than α [Leh36]. Paul Erd¨os used the Prime Number Theorem to show that, for any k +, M(n) >nk ∈Z for infinitely many values of n [Erd46]. Much work continues on understanding the nature of the coefficients of the cyclotomic polynomials; some of this work is reviewed in [Tha00], [Mor09] and [Jam15]. Now we may return to the topic of the irreduciblity of cyclotomic polynomials in [x]. Since Z Φ (x) and Φ (x) are irreducible monic polynomials in [x], as we shall see below, we have n m Z gcd(Φ (x), Φ (x)) = 1 in [x] when n = m. We can see this directly as follows. To begin, m n Z 6 suppose that m Now the divisibility relation Φ (x) (xn 1)/(xd 1) shows that, with m Given the monic cyclotomic polynomial Φ (x) [x] there must be monic polynomials f(x) and n ∈ Z g(x) in [x] such that f(x) is irreducible in [x] and Φ (x) = f(x)g(x) with degree(f) 1. (If Z Z n ≥ Φ (x) is irreducible in [x], f(x) = Φ (x) and g(x) = 1.) n Z n Note degree(f) 1, and if Φ (x) is not irreducible in [x], degree(g) 1. Also, since Φ (x) ≥ n Z ≥ n | (xn 1), f(x) (xn 1) and g(x) (xn 1). − | − | − The roots of Φn(x) are the φ(n) distinct order-n roots of unity, gen(Rn) so we have the roots of f lying in gen(R ) (and indeed, roots(f) roots(g) = gen(R )). Thus, letting ζ denote a root of n ∪ n f, if the elements of ζj 2 j n and gcd(j,n) = 1 are also roots of f, then the roots of f { | ≤ ≤ } coincide with the roots of Φ (x), and f(x) = Φ (x) (and g(x)=1), so Φ (x) is irreducible in [x], n n n Z (since f is irreducible in [x]). Thus to show that Φ (x) is irreducible in [x], it suffices to show Z n Z that f(ζj)=0 for j U = j gcd(j,n) = 1 . ∈ n { ∈Zn | } p j If f(ζ ) = 0 for every prime p that does not divide n, then f(ζ ) = 0 for j Un, since with α1 α2 αk p ∈ j = p1 p2 pk , where p1 ∤ n, p2 ∤ n, ..., pk ∤ n, [f(ζ) = 0 implies f(ζ ) = 0] can be iterated to ··· 2 a yield f(ζp) = 0 implies f(ζp ) = 0, ..., and in general, we obtain f(ζ) = 0 implies f(ζp )=0 for a b a> 0. And also f(ζp) = 0 implies f((ζp )q ) = 0 where q is a prime that does not divide n, so that α α α p 1 p 2 ···p k j we can eventually obtain f(ζ) = 0 implies f(ζ 1 2 k )= f(ζ ) = 0 with gcd(j,n)=1. Thus, since Φn(x) and f share the order-n n-th root of unity ζ as a root, when f(ζ) = 0 implies p f(ζ ) = 0, Φn(x) and f share all the φ(n) elements of gen(Rn) as roots, so Φn(x)= f(x), and then Φ (x) is irreducible in [x]. Hence, Φ (x) is irreducible in [x] if we can establish that f(ζp) = 0 n Z n Z given f(ζ)=0 for p prime with p ∤ n. Now, take p prime with p ∤ n, and suppose that f(ζ) = 0, but f(ζp) = 0. This assumption will lead 6 to a contradiction. p p Since Φn(x)= f(x)g(x), and ζ is a root of Φn(x), but not a root of f(x), we must have g(ζ )=0. Let h(x) [x] be the monic [x]-irreducible polynomial dividing the monic polynomial g(x) such ∈Z Z that h(ζp) = 0. Then ζ is a root of the polynomial h(xp). Note h(x) is irreducible in [x], but Z 31 h(xp) is not necessarily irreducible in [x]. Since the monic polynomials f(x) and h(xp) share the Z root ζ, and f(x) is irreducible in [x], and hence the minimal degree monic polynomial in [x] Z Z having the root ζ, f(x), must divide h(xp), so that h(xp)= f(x) s(x) for some monic polynomial · s(x) [x]. ∈Z Now we will consider the relations f(x) g(x) = Φ (x) m(x) and f(x) s(x)= h(xp) in [x], i.e., · n | · Zp with respect to modulo-p addition and multiplication, where m(x)=(xn 1). Recall is a finite − Zp field with the associated multiplicative group U = 1, 2,...,p 1 where each element in U 1 p { − } p − { } has order p 1, i.e., U 1 = gen(U ); this is the content of Fermat’s Theorem. − p − { } p Let fˆ(x)= f(x) cmod p,s ˆ(x)= s(x) cmod p, hˆ(x)= h(x) cmod p, andm ˆ (x)= m(x) cmod p where a(x) cmod p is computed by replacing each coefficient in a(x) by its value modulo p in ; (this is Zp done by replacing each negative coefficient c by its “canonical” value in obtained by adding the Zp multiple of p, p c /p and replacing each non-negative coefficient c by its “canonical” value in · ⌈| | ⌉ obtained by subtracting the multiple of p, p c /p ). Recall that a(x) cmod p = b(x) cmod p Zp · ⌊| | ⌋ implies a(x) b(x) (mod p) for all x . (Note reducing a polynomial f(x) to f(x) cmod p ≡ ∈ Z might be “incomplete”, since degree(F (x) cmod p) may be greater than p 1, and hence reducible − modulo-p via Fermat’s Theorem to a lower-degree polynomial; to do a complete reduction, we would use Fermat’s Theorem recursively as in exercise ??.) Arithmetic with the coefficients entailed in any algebraic expressions involving such cmod -reduced polynomials is to be understood to be done in , i.e., using the addition and multiplication Zp operations of ; we denote this by using the annotation “[in [x]]” since such algebraic expressions Zp Zp specify polynomials in [x]. In particular, hˆ(x)p denotes hˆ(x) hˆ(x) hˆ(x) (with p factors) Zp · ··· computed in [x] and fˆ(x) gˆ(x) computed in [x] is fˆ(x) gˆ(x) cmod p. Remember, however, Zp · Zp · we may compute modulo-p expressions with ordinary arithmetic, reducing modulo-p only when we wish. (Reduction modulo-p is just invoking the ring homomorphism (page ??).) Z→Zp In general, if a(x)= b(x) where a(x) and b(x) are polynomials with integer coefficients, then a(x) ≡ b(x) (mod p) for x . And then fora ˆ(x) = a(x) cmod p and ˆb(x) = b(x) cmod p,a ˆ(x) = ˆb(x) ∈ Z for x when all integers ina ˆ(x) and ˆb(x) are reduced to their canonical representations in . ∈Zp Zp The ring-homomorphism between and ensures that equalities and divibility relations in Z Zp Z are preserved in . Zp Note fˆ(x),s ˆ(x), hˆ(x), andm ˆ (x) are monic polynomials in [x] with the same degrees as the monic Zp polynomials f, s, h, and m in [x]. Also f(x) s(x) = h(xp) so fˆ(x) sˆ(x) = hˆ(xp) [in [x]]. Z · · Zp Furthermore, fˆ(x) gˆ(x) mˆ (x) in [X] since f(x) g(x) m(x) in [X], and similarly, hˆ(x) gˆ(x) · | Zp · | Z | in [X] since h(x) g(x) in [X]. Zp | Z We recall that for x , h(xp) h(x)p (mod p) (see exercise ??) and hˆ(xp) = hˆ(x)p [in [x]] ∈ Zp ≡ Zp (exercise ??). Now, although fˆ(x) is not necessarily irreducible in [x], since fˆ(x) sˆ(x)= hˆ(xp) Zp · [in [x]] and hˆ(xp) = hˆ(x)p [in [x]], we have fˆ(x) sˆ(x) = hˆ(x)p [in [x]]; but then we see Zp Zp · Zp that fˆ and hˆ have a common [x]-irreducible factor t(x). And fˆ(x) gˆ(x) mˆ (x) [in [x]] and Zp · | Zp hˆ(x) gˆ(x) in [x], so fˆ(x) gˆ(x) mˆ (x) in [x], and thus t(x)2 mˆ (x) in [ [X]]. But this is | Zp · | Zp | Zp impossible unless t(x) = 1 sincem ˆ (x) has no repeated roots [in ], whereas t(x)2 certainly has Zp repeated roots unless t(x)=1. REFERENCES 32 To see that (xn 1) cmod p =m ˆ (x) = xn +(p 1) has no repeated modulo-p roots, recall that − − gcd(m ˆ (x), mˆ ′(x))=1in [X] exactly whenm ˆ (x) has no repeated roots in , i.e., does not factor Zp Zp as (x r)2a(x) in [x]. Butm ˆ ′(x) = αxn−1 where α = n mod p = 0 since p ∤ n, and αxn−1 has − Zp 6 just the root 0 with multiplicity n 1 (since nxn−1 = n(x 0)n−1). And none of the n modulo-p − − roots of xn +(p 1) are zero, so gcd(m ˆ (x), mˆ ′(x)) = 1 and hencem ˆ (x) has no repeated roots. (We − can also carry-out the computation of gcd(m ˆ (x), mˆ ′(x)) in [x] with the Euclidean Algorithm to Zp obtain gcd(m ˆ (x), mˆ ′(x)) = 1.) But t(x) = 1 since t(x) has all the roots shared by fˆ and hˆ in , and fˆ and hˆ are not constant 6 C polynomials (!) Thus we have a contradiction. p Therefore f(x) must have ζ as a root after all, and hence f(x) = Φn(x) and hence Φn(x) is irreducible for n 1. [QED] ≥ References [Bro16] Gary Brookfield. The coefficients of cyclotomic polynomials. Math. Mag., 89(3):179–188, June 2016. [Erd46] Paul Erd¨os. On the coefficients of the cyclotomic polynomial. Bull. Amer. Math. Soc., 52(2):179–184, 1946. [Gau86] Carl Friedrich Gauss. Disquisitiones Arithmeticae (In Latin). G. Fleischer, Leipzig, 1801 [English Translation: Springer-Verlag, 1986]. [Ge14] Yimin Ge. Elementary properties of cyclotomic polynomials. http://www.yimin- ge.com/doc/cyclotomic polynomials.pdf, 2014. [Jam15]G. J. O. Jameson. The cyclotomic polynomials. http://www.maths.lancs.ac.uk/jameson/cyp.pdf, 2015. [Leh36] Emma Lehmer. On the magnitude of the coefficients of the cyclotomic polynomial. Bull. Amer. Math. Soc., 42:389–392, June 1936. [Mor09] Peter Moree. Inverse cyclotomic polynomials. Journal of Number Theory, 129:667–680, 2009. [Suz87] Jiro Suzuki. On the coefficients of cyclotomic polynomials. Proc. Japan Acad., 63, Series A(7):279–280, 1987. [Tha00] R. Thangadurai. On the coefficients of cyclotomic polynomials. In Cyclotomic Fields and Related Topics, pages 311–322, Pune, 2000. Bhaskaracharya Pratishthana. [Web96] Andreas Weber. Computing radical expressions for roots of unity. SIGSAM Bulletin, 30(3):11–20, 1996. [Wei13] Steven H. Weintraub. Several proofs of the irreducibility of the cyclotomic polynomials. Amer. Math. Monthly, 120(6):537–545, June 2013. 2k+1 and q < 2k+1, while by assumption, taking p ,p ,...,p Z i i+t h 1 2 ti = q ,q ,...,q , we have 2q < q . And q < 2k+1 so 2k+1 < 2q < q < 2k+1 which is h i i+1 i+ti i i+t i+t i i+t impossible. Thus the number of primes in the interval [2k, 2k+1] must be fewer than t for k +. ∈Z Now recall we denote the number of primes less than or equal to n by π(n). The preceeding argument shows that π(22) < t, π(23) < 2t, ..., π(2k+1) < kt, ..., and hence π(2k) < kt for k +. ∈Z π(n) However, we have the famous Prime Number Theorem: lim = 1. (More informatively, n→∞ n log(n) + v! n k! n for k fixed in , we have π(n)= v+1 + O k+1 . ) Z log(n) log(n) ! 0≤Xv