CYCLOTOMIC POLYNOMIALS JORDAN BELL Contents 1. Introduction 1 2. Preliminaries 1 3. Definition and basic properties of cyclotomic polynomials 3 4. Special values 6 5. Primes in arithmetic progressions 10 6. Zsigmondy’s theorem 11 7. Newton’s identities and Ramanujan sums 11 8. Algebraic theorems about coefficients of cyclotomic polynomials 15 9. Analytic theorems about coefficients of cyclotomic polynomials 21 10. Fourier analysis 23 11. Algebraic topology 24 References 25 1. Introduction 2. Preliminaries By an arithmetical function we mean a function whose domain contains the positive integers. We say that an arithmetical function f is multiplicative when gcd(n; m) = 1 implies f(nm) = f(n)f(m), and that it is completely multiplica- tive when f(nm) = f(n)f(m) for all n; m ≥ 1. Write 2πik=n 2πik=n Un = fe : 1 ≤ k ≤ ng = fe : 0 ≤ k ≤ n − 1g; the nth roots of unity. For n > 1, there is an element ζ of Un with ζ 6= 1. Because ξ 7! ζξ is a bijection U ! U we have ζ P ξ = P ξ, hence n n ξ2Un ξ2Un (1 − ζ) P ξ = 0. But ζ 6= 1, which means that ξ2Un n−1 X X e2πik=n = ξ = 0; n > 1: k=0 ξ2Un Write 2πik=n ∆n = fe : 1 ≤ k ≤ n; gcd(k; n) = 1g; the primitive nth roots of unity. Let φ be the Euler phi function: φ(n) = jfk : 1 ≤ k ≤ n; gcd(k; n) = 1gj = j∆nj: φ is multiplicative, and for prime p and for r ≥ 1, φ(pr) = pr−1(p − 1). Date: April 12, 2017. 1 2 JORDAN BELL Let µ be the Möbius function: X X µ(n) = e2πik=n = ξ: 1≤k≤n;gcd(k;n)=1 ξ2∆n For p prime, as ∆p = Up n f1g, X µ(p) = −1 + ξ = 0 − 1 = −1: ξ2Up For r ≥ 2, as ∆pr = Upr n Upr−1 , X X µ(pr) = − ξ + ξ = −0 + 0 = 0: r ξ2Upr−1 ξ2Up Furthermore, one proves that µ is multiplicative. Thus 8 1 n is a square-free integer with an even number of prime factors <> µ(n) = −1 n is a square-free integer with an odd number of prime factors :>0 otherwise: The Möbius inversion formula states that if f and g are arithmetic functions satisfying X g(n) = f(d); n ≥ 1; djn then X f(n) = µ(n=d)g(d); n ≥ 1: djn We can write [ Un = ∆d; djn and ∆d \ ∆e = ; for d 6= e. So X n = φ(d): djn Therefore by the Möbius inversion formula, X φ(n) = d · µ(n=d): djn Also, for n > 1, X X X X (1) µ(d) = ξ = ξ = 0: djn djn ξ2∆d ξ2Un Let X d(n) = 1; djn the number of divisors of n, for example, d(6) = 4. Let X !(n) = 1; pjn α1 αr the number of prime divisors of n: for n = p1 ··· pr , α1; : : : ; αr ≥ 1, we have !(n) = r, for example !(12) = !(22 · 3) = 2. CYCLOTOMIC POLYNOMIALS 3 3. Definition and basic properties of cyclotomic polynomials For n ≥ 1, let Y 2πik=n Y Φn(x) = (x − e ) = (x − ξ); 1≤k≤n;gcd(k;n)=1 ξ2∆n the nth cyclotomic polynomial. The first of the following two identities was found by Euler [45, pp. 199–200, Chap. III, §VI]. Lemma 1. For n ≥ 1, n Y x − 1 = Φd(x); djn and for x 62 Un, Y d µ(n=d) Φn(x) = (x − 1) : djn n 2πik=n Proof. For Fn(x) = x − 1, each of e , 1 ≤ k ≤ n, is a distinct root of Fn(x), so Y xn − 1 = (x − e2πik=n) 1≤k≤n Y Y = (x − e2πik=n) djn 1≤k≤n;gcd(k;n)=d Y Y = (x − e2πijd=n) djn 1≤j≤n=d;gcd(j;n=d)=1 Y = Φn=d(x) djn Y = Φd(x): djn P That is, log Fn = djn log Φd. Therefore applying the Möbius inversion formula P Q µ(n=d) yields log Φn = djn µ(n=d) log Fd and so Φn = djn Fd . Lemma 2. When p is a prime, p−1 Φp(x) = x + ··· + x + 1: When p is an odd prime, p−1 p−2 p−3 2 Φ2p(x) = x − x + x − · · · + x − x + 1: p Proof. When p is a prime, x − 1 = Φ1(x) · Φp(x), i.e. p p x − 1 x − 1 p−1 Φp(x) = = = x + ··· + x + 1: Φ1(x) x − 1 When p is an odd prime, x2p − 1 x2p − 1 (xp − 1)(xp + 1) xp + 1 Φ2p(x) = = p = p = ; Φ1(x)Φ2(x)Φp(x) (x − 1)Φ2(x) (x − 1)(x + 1) x + 1 and because (x + 1)(xp−1 − xp−2 + xp−3 − · · · + x2 − x + 1) = xp + 1, p−1 p−2 p−3 2 Φ2p(x) = x − x + x − · · · + x − x + 1: 4 JORDAN BELL Lemma 3. If p is a prime and m ≥ 1, ( p Φm(x ) pjm Φpm(x) = p Φm(x )=Φm(x) p - m: For k ≥ 1, ( pk Φm(x ) pjm Φpkm(x) = pk pk−1 Φm(x )=Φm(x ) p - m; Proof. Using Lemma 1, Y d µ(pm=d) Φpm(x) = (x − 1) dj(pm) Y Y = (xd − 1)µ(pm=d) · (xd − 1)µ(pm=d) dj(pm);pjd dj(pm);p-d Y Y = (xpe − 1)µ(m=e) · (xd − 1)µ(pm=d) ejm dj(pm);p-d p Y d µ(pm=d) = Φm(x ) · (x − 1) : dj(pm);p-d If m = ap and dj(pm) and p - d, then µ(pm=d) = µ(ap2=d) = 0 and p Y d µ(ap2=d) p Φpm(x) = Φm(x ) · (x − 1) = Φm(x ): dja If p - m and d j (pm) and p - d, then µ(pm=d) = µ(p)µ(m=d) = −µ(m=d) and p Y d µ(pm=d) p Y d −µ(m=d) Φpm(x) = Φm(x ) · (x − 1) = Φm(x ) · (x − 1) : dj(pm);p-d djm For k ≥ 2, p pk−1 Φpkm(x) = Φp·pk−1m(x) = Φpk−1m(x ) = ··· = Φpm(x ); and using the expression we obtained for Φpm(x) we get the expression stated for Φpkm(x). α1 αr Lemma 4. For n = p1 ··· pr , where pi are prime and αi ≥ 1, and N = p1 ··· pr, n=N Φn(x) = ΦN (x ): Proof. If djn and d - N then µ(d) = 0, hence Y n=d µ(d) Φn(x) = (x − 1) djn Y = (xn=d − 1)µ(d) djN Y = ((xn=N )N=d − 1)µ(d) djN n=N = ΦN (x ): CYCLOTOMIC POLYNOMIALS 5 Lemma 5. If n > 1 then −1 −φ(n) Φn(x ) = x Φn(x): Proof. −1 Y −d µ(n=d) Y d µ(n=d) −d µ(n=d) Φn(x ) = (x − 1) = (1 − x ) (x ) ; djn djn hence −1 Y −d µ(n=d) Y d µ(n=d) Φn(x ) = (−x ) · (x − 1) : djn djn P P Because n > 1 it holds that djn µ(n=d) = 0, and using this and djn d · µ(n=d) = φ(n) yields −1 −φ(n) Φn(x ) = x Φn(x): Lemma 6. If r > 1 is odd then Φ2r(x) = Φr(−x): Proof. Because r is odd, if d1; : : : ; dl are the divisors of r then d1; : : : ; dl; 2d1;:::; 2dl are the divisors of 2r, so Y d µ(2r=d) Φ2r(x) = (x − 1) dj(2r) Y Y = (xd − 1)µ(2r=d) · (x2d − 1)µ(2r=(2d)) djr djr Y = (xd − 1)µ(2r=d)(x2d − 1)µ(r=d) djr Y = (xd − 1)µ(2)µ(r=d)+µ(r=d)(xd + 1)µ(r=d) djr Y = (xd + 1)µ(r=d): djr Because r is odd, any divisor d of r is odd and then xd + 1 = −((−x)d − 1), so Y µ(r=d) d µ(r=d) φ(r) Y d µ(r=d) Φ2r(x) = (−1) ((−x) − 1) = (−1) · ((−x) − 1) : djr djr Because r is odd and > 1, φ(r) is even, so we have obtained the claim. Theorem 7. Φn 2 Z[x]. Proof. It is a fact that if R is a unital commutative ring, f 2 R[x] is a monic polynomial and g 2 R[x] is a polynomial, then there are q; r 2 R[x] with g = qf + r; r = 0 or deg r < deg f. 6 JORDAN BELL First, Φ1(x) = x − 1 2 Z[x]. For n > 1, assume that Φd(x) 2 Z[x] for 1 ≤ d < n. Then let Y f = Φd; djn;d<n which by hypothesis belongs to Z[x]. Since each Φd is monic, so is f. On the one hand, since g(x) = xn − 1 2 Z[x], there are q; r 2 Z[x] with g = qf + r and r = 0 or deg r < deg f. On the other hand, by Lemma 1 we have g = Φnf 2 C[x]. Thus Φnf = qf + r 2 C[x], so r = f · (Φn − q) 2 C[x]. If Φn 6= q then deg r = deg f + deg(Φn − q) ≥ deg f, contradicting that r = 0 or deg r < deg f. Therefore Φn = q 2 C[x], and because q 2 Z[x] this means that Φn 2 Z[x]. In fact, it can be proved that Φn is irreducible in Q[x]. Gauss states in entry 40 of his mathematical diary, dated October 9, 1796, that Φp is irreducible in Q[x] when p is prime, and he proves this in Disqisitiones Arithmeticae, Art.