Cyclotomic polynomials

Jordan Bell
April 12, 2017

1 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 multiplicative when f(nm)=f(n)f(m) for all n,m1.

Write

Un={e2πik/n:1kn}={e2πik/n:0kn-1},

the nth roots of unity. For n>1, there is an element ζ of Un with ζ1. Because ξζξ is a bijection UnUn we have ζξUnξ=ξUnξ, hence (1-ζ)ξUnξ=0. But ζ1, which means that

k=0n-1e2πik/n=ξUnξ=0,n>1.

Write

Δn={e2πik/n:1kn,gcd(k,n)=1},

the primitive nth roots of unity. Let ϕ be the Euler phi function:

ϕ(n)=|{k:1kn,gcd(k,n)=1}|=|Δn|.

ϕ is multiplicative, and for prime p and for r1, ϕ(pr)=pr-1(p-1).

Let μ be the Möbius function:

μ(n)=1kn,gcd(k,n)=1e2πik/n=ξΔnξ.

For p prime, as Δp=Up{1},

μ(p)=-1+ξUpξ=0-1=-1.

For r2, as Δpr=UprUpr-1,

μ(pr)=-ξUpr-1ξ+ξUprξ=-0+0=0.

Furthermore, one proves that μ is multiplicative. Thus

μ(n)={1n is a square-free integer with an even number of prime factors-1n is a square-free integer with an odd number of prime factors0otherwise.

The Möbius inversion formula states that if f and g are arithmetic functions satisfying

g(n)=dnf(d),n1,

then

f(n)=dnμ(n/d)g(d),n1.

We can write

Un=dnΔd,

and ΔdΔe= for de. So

n=dnϕ(d).

Therefore by the Möbius inversion formula,

ϕ(n)=dndμ(n/d).

Also, for n>1,

dnμ(d)=dnξΔdξ=ξUnξ=0. (1)

Let

d(n)=dn1,

the number of divisors of n, for example, d(6)=4. Let

ω(n)=pn1,

the number of prime divisors of n: for n=p1α1prαr, α1,,αr1, we have ω(n)=r, for example ω(12)=ω(223)=2.

2 Definition and basic properties of cyclotomic polynomials

For n1, let

Φn(x)=1kn,gcd(k,n)=1(x-e2πik/n)=ξΔn(x-ξ),

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 n1,

xn-1=dnΦd(x),

and for xUn,

Φn(x)=dn(xd-1)μ(n/d).
Proof.

For Fn(x)=xn-1, each of e2πik/n, 1kn, is a distinct root of Fn(x), so

xn-1 =1kn(x-e2πik/n)
=dn1kn,gcd(k,n)=d(x-e2πik/n)
=dn1jn/d,gcd(j,n/d)=1(x-e2πijd/n)
=dnΦn/d(x)
=dnΦd(x).

That is, logFn=dnlogΦd. Therefore applying the Möbius inversion formula yields logΦn=dnμ(n/d)logFd and so Φn=dnFdμ(n/d). ∎

Lemma 2.

When p is a prime,

Φp(x)=xp-1++x+1.

When p is an odd prime,

Φ2p(x)=xp-1-xp-2+xp-3-+x2-x+1.
Proof.

When p is a prime, xp-1=Φ1(x)Φp(x), i.e.

Φp(x)=xp-1Φ1(x)=xp-1x-1=xp-1++x+1.

When p is an odd prime,

Φ2p(x)=x2p-1Φ1(x)Φ2(x)Φp(x)=x2p-1(xp-1)Φ2(x)=(xp-1)(xp+1)(xp-1)(x+1)=xp+1x+1,

and because (x+1)(xp-1-xp-2+xp-3-+x2-x+1)=xp+1,

Φ2p(x)=xp-1-xp-2+xp-3-+x2-x+1.

Lemma 3.

If p is a prime and m1,

Φpm(x)={Φm(xp)p|mΦm(xp)/Φm(x)pm.

For k1,

Φpkm(x)={Φm(xpk)p|mΦm(xpk)/Φm(xpk-1)pm,
Proof.

Using Lemma 1,

Φpm(x) =d(pm)(xd-1)μ(pm/d)
=d(pm),p|d(xd-1)μ(pm/d)d(pm),pd(xd-1)μ(pm/d)
=em(xpe-1)μ(m/e)d(pm),pd(xd-1)μ(pm/d)
=Φm(xp)d(pm),pd(xd-1)μ(pm/d).

If m=ap and d|(pm) and pd, then μ(pm/d)=μ(ap2/d)=0 and

Φpm(x)=Φm(xp)da(xd-1)μ(ap2/d)=Φm(xp).

If pm and d(pm) and pd, then μ(pm/d)=μ(p)μ(m/d)=-μ(m/d) and

Φpm(x)=Φm(xp)d(pm),pd(xd-1)μ(pm/d)=Φm(xp)dm(xd-1)-μ(m/d).

For k2,

Φpkm(x)=Φppk-1m(x)=Φpk-1m(xp)==Φpm(xpk-1),

and using the expression we obtained for Φpm(x) we get the expression stated for Φpkm(x). ∎

Lemma 4.

For n=p1α1prαr, where pi are prime and αi1, and N=p1pr,

Φn(x)=ΦN(xn/N).
Proof.

If d|n and dN then μ(d)=0, hence

Φn(x) =dn(xn/d-1)μ(d)
=dN(xn/d-1)μ(d)
=dN((xn/N)N/d-1)μ(d)
=ΦN(xn/N).

Lemma 5.

If n>1 then

Φn(x-1)=x-ϕ(n)Φn(x).
Proof.
Φn(x-1)=dn(x-d-1)μ(n/d)=dn(1-xd)μ(n/d)(x-d)μ(n/d),

hence

Φn(x-1)=dn(-x-d)μ(n/d)dn(xd-1)μ(n/d).

Because n>1 it holds that dnμ(n/d)=0, and using this and dndμ(n/d)=ϕ(n) yields

Φn(x-1)=x-ϕ(n)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

Φ2r(x) =d(2r)(xd-1)μ(2r/d)
=dr(xd-1)μ(2r/d)d|r(x2d-1)μ(2r/(2d))
=dr(xd-1)μ(2r/d)(x2d-1)μ(r/d)
=dr(xd-1)μ(2)μ(r/d)+μ(r/d)(xd+1)μ(r/d)
=dr(xd+1)μ(r/d).

Because r is odd, any divisor d of r is odd and then xd+1=-((-x)d-1), so

Φ2r(x)=dr(-1)μ(r/d)((-x)d-1)μ(r/d)=(-1)ϕ(r)dr((-x)d-1)μ(r/d).

Because r is odd and >1, ϕ(r) is even, so we have obtained the claim. ∎

Theorem 7.

Φn[x].

Proof.

It is a fact that if R is a unital commutative ring, fR[x] is a monic polynomial and gR[x] is a polynomial, then there are q,rR[x] with

g=qf+r,

r=0 or degr<degf.

First, Φ1(x)=x-1[x]. For n>1, assume that Φd(x)[x] for 1d<n. Then let

f=dn,d<nΦd,

which by hypothesis belongs to [x]. Since each Φd is monic, so is f. On the one hand, since g(x)=xn-1[x], there are q,r[x] with g=qf+r and r=0 or degr<degf. On the other hand, by Lemma 1 we have g=Φnf[x]. Thus Φnf=qf+r[x], so r=f(Φn-q)[x]. If Φnq then degr=degf+deg(Φn-q)degf, contradicting that r=0 or degr<degf. Therefore Φn=q[x], and because q[x] this means that Φn[x]. ∎

In fact, it can be proved that Φn is irreducible in [x]. Gauss states in entry 40 of his mathematical diary, dated October 9, 1796, that Φp is irreducible in [x] when p is prime, and he proves this in Disqisitiones Arithmeticae, Art. 341. Gauss further states in entry 136 of his mathematical diary, dated June 12, 1808, that for any n, Φn is irreducible in [x], and Kronecker proves this in his 1854 Mémoire sur les facteurs irréductibles de l’expression xn-1. Gauss’s work on cyclotomic polynomials is surveyed by Neumann [40]. For any ξΔn, Φn(ξ)=0, and since Φn is irreducible in [x] and is monic, Φ is the minimal polynomial of ξ over , which implies that [(ξ):]=degΦn=ϕ(n).

There is a group isomorphism Gal((ξ)/)(/n)* [16, p. 596, Theorem 26].

The discriminant [44, p. 12, Proposition 2.7]:

d((e2πi/n))=(-1)ϕ(n)/2nϕ(n)pnpϕ(n)/(p-1).

It can be proved that 𝒪(e2πi/n)=[e2πi/n] [39, p. 60, Proposition 10.2].

Let p be prime, let q=pr for r1, let 𝔽q be a finite field with q elements, and for n1 with gcd(n,q)=1, let ν be the multiplicative order of q modulo n: ν is the minimum positive integer satisfying qν1(modn). It can be proved that there are monic, degree ν, irreducible polynomials P1,,Pϕ(n)/ν𝔽q[x] such that Φn=P1Pϕ(n)/ν𝔽q[x] [28, p. 65, Theorem 2.47]; cf. Bourbaki [7, p. 581] on Kummer. In particular, q is a generator of the multiplicative group (/n)* if and only if ν=ϕ(n) if and only if Φn is irreducible in 𝔽q[x]. We remark that (/n)* is cyclic if and only if n is 2, 4, some power of an odd prime, or twice some power of an odd prime (Gauss, Disquisitiones Arithmeticae, Art. 89–92). This follows from (i) the multiplicative group (/n)* is isomorphic with the direct product (/p1α1)*××(/prαr)* for n=p1α1prαr, (ii) (/2α)* is isomorphic with /2×/2α-2, α2, and (iii) (/pα)* is a cyclic group with pα-1(p-1) elements when p is an odd prime, α1 [16, p. 314, Corollary 20].

3 Special values

Lemma 8.

Φ1(0)=-1, and for n2, Φn(0)=1.

Proof.

Φ1(x)=x-1, so Φ1(0)=-1. For n2, using (1),

Φn(0)=dn(-1)μ(n/d)=(-1)dnμ(n/d)=(-1)dnμ(d)=(-1)0=1.

Let Λ be the von Mangoldt function: Λ(n)=logp if n=pα for some prime p and some integer α1, and is Λ(n)=0 otherwise. Thus Λ(2)=log2, Λ(8)=log2, Λ(3)=log3, and Λ(6)=0. One sees that

logn=dnΛ(d).

Therefore by the Möbius inversion formula,

Λ(n)=dnμ(n/d)logd.
Theorem 9.

For n>1,

Φn(1)=eΛ(n)

and

Φn(1)=12eΛ(n)ϕ(n).
Proof.

For n>1,

xn-1++x+1=dn,d>1Φd(x),

hence

logn=dn,d>1logΦd(1).

Therefore by the Möbius inversion formula,

logΦn(1)=dn,d>1μ(n/d)logd=dnμ(n/d)logd=Λ(n).

Because xn-1=dnΦd(x), taking the logarithm and then taking the derivative yields

nxn-1xn-1=dnΦd(x)Φd(x).

Φ1(x)=x-1 and so Φ1(x)Φ1(x)=1x-1, hence

nxn-1xn-1-1x-1=dn,d>1Φd(x)Φd(x),

i.e.

nxn-1-(xn-1+xn-2++x+1)xn-1=dn,d>1Φd(x)Φd(x).

Doing polynomial long division we find

(n-1)xn-1-xn-2--x-1x-1=(n-1)xn-2+(n-2)xn-3++2x+1.

Hence

(n-1)xn-2+(n-2)xn-3++2x+1xn-1+xn-2++x+1=dn,d>1Φd(x)Φd(x),

and for x=1 this is

n-12=dn,d>1Φd(1)Φd(1).

By the Möbius inversion formula,

Φn(1)Φn(1)=dn,d>1μ(n/d)d-12,

and using (i) Φn(1)=eΛ(n) for n>1, (ii) dnμ(n/d)=0 for n>1, and (iii) dndμ(n/d)=ϕ(n), we have

Φn(1)=eΛ(n)12dnμ(n/d)d-eΛ(n)12dnμ(n/d)=12eΛ(n)ϕ(n).

Because Φn[x], it is the case that Φn(-i)=Φn(i)¯.

Theorem 10.

Φ1(i)=i-1, Φ2(i)=i+1, Φ4(i)=0, and otherwise we have the following.

  • If n is odd and has a prime factor p1(mod4), then Φn(i)=1.

  • If p3(mod4) is prime and k1 is odd, then Φpk(i)=i.

  • If p3(mod4) is prime and k1 is even, then Φpk(i)=-i.

  • If p3(mod4) is prime and k1 is odd, then Φ2pk(i)=-i.

  • If p3(mod4) is prime and k1 is even, then Φ2pk(i)=i.

  • If p,q3(mod4) are distinct primes and k,l1, then Φpkql(i)=-1.

  • If p,q3(mod4) are distinct primes and k,l1, then Φ2pkql(i)=-1.

  • If p is an odd prime and k1, then Φ4pk(i)=p.

  • If ω(n)3 then Φn(i)=1.

Proof.

Φ1(x)=x-1, Φ2(x)=x+1, so Φ1(i)=i-1 and Φ2(i)=i+1. As iΔ4, Φ4(i)=0.

Suppose that n is odd, that p1(mod4) is a prime factor of n, and write n=pkm with gcd(m,p)=1. Lemma 3 tells us

Φn(x)=Φpkm(x)=Φm(xpk)Φm(xpk-1),

and as pk-11(mod4) and i4=1, this yields

Φn(i)=Φm(i)Φm(i)=1.

Suppose that n is odd, that p3(mod4) is a prime factor of n, and write n=pkm with gcd(m,p)=1. If k is odd then pk3(mod4), so

Φn(i)=Φm(ipk)Φm(ipk-1)=Φm(i3)Φm(i)=Φm(-i)Φm(i),

and if m=1 then

Φn(i)=Φ1(-i)Φ1(i)=-i-1i-1=i.

If k is even then pk1(mod4), so

Φn(i)=Φm(i)Φm(-i),

and if m=1 then Φn(i)=-i.

Suppose that n=2k, k3. Lemma 4 tells us that

Φn(x)=Φ2(xn/2)=Φ2(x2k-1)=x2k-1+1,

thus

Φn(i)=i2k-1+1=1+1=2.

Suppose that n=2m with m>1 odd. Lemma 6 tells us Φn(x)=Φ2m(x)=Φm(-x), so Φn(i)=Φm(-i).

Suppose that n=2km with k2 and m>1 odd. Lemma 3 tells us

Φ2km(x)=Φ2k-12m(x)=Φ2m(x2k-1),

and then Lemma 6 tells us Φ2m(x2k-1)=Φm(-x2k-1). For k=2 this yields

Φ4m(i)=Φm(1),

and for k>2,

Φn(i)=Φm(-i2k-1)=Φm(-1).

Kurshan and Odlyzko [25]

Montgomery and Vaughan [36, pp. 131–132, Exercise 9].

Theorem 11.

If n=py,p2,3(mod5)p with ω(n) odd, then

|Φn(e2πi/5)|=(1+52)d(n)/2.
Proof.

Write e(x)=e2πix, let dn, d>1, and write d=p1pkq1ql where p1,,pk2(mod5) and q1,,ql3(mod5) are prime. Then ω(d)=k+l and, as 233(mod5),

d2k3l2k23l2k+l(-1)l(mod5).

If ω(d)0(mod4) then 2k+l1(mod5) and if ω(d)2(mod4) then 2k+l-1(mod5), and therefore if ω(d) is even then d1(mod5) or d-1(mod5). Since |e(-1/5)-1|=|e(1/5)-1|, we have |e(d/5)-1|=|e(1/5)-1|.

If ω(d)1(mod4) then 2k+l2(mod5) and if ω(d)3(mod4) then 2k+l-2(mod5), and therefore if ω(d) is odd then d2(mod5) or d-2(mod5). Since |e(-2/5)-1|=|e(2/5)-1|, we have |e(d/5)-1|=|e(2/5)-1|.

Now using Lemma 1 and |e(1/5)-1|-1=|e(2/5)-1|,

|Φn(e(1/5))| =dn|e(d/5)-1|μ(n/d)
=dn,ω(d) even|e(1/5)-1|-1dn,ω(d) odd|e(2/5)-1|.

Hence, for ω(n)=2ν+1 and for A=|e(1/5)-1|-1 and B=|e(2/5)-1|,

log|Φn(e(1/5))| =r=0ν(2ν+12r)logA+r=0ν(2ν+12r+1)logB
=22νlogA+22νlogB
=log((AB)2ω(n)/2),

and using d(n)=r=0ω(n)(ω(n)r)=2ω(n) this is |Φn(e(1/5))|=(AB)d(n)/2. Finally,

AB=|e(2/5)-1||e(1/5)-1|=|e(1/5)+1|=1+52.

4 Primes in arithmetic progressions

For prime p, pn, the following theorem relates the order of an element of the multiplicative group (/p)* with Φn [44, p. 13, Lemma 2.9]. We remind ourselves that Φn[x] (Theorem 7), and so Φn(a) for a.

Lemma 12.

Let p be prime, pn, and a. Then pΦn(a) if and only if n is the multiplicative order of a modulo p.

Proof.

Suppose that pΦn(a). Now, let b with pΦn(b). By Lemma 1, bn-1=dnΦd(b), and because Φn(b)0(modp) this yields bn-10(modp), i.e. bn1(modp); in particular, pb. Let ν=min{k>0:ak1(modp)}, the multiplicative order of a modulo p, so νn, and suppose by contradiction that ν<n. Using xν-1=dνΦd(x) we have bν-1=dνΦd(b). Using this with b=a, as aν1(modp) and because p is prime it follows that for some d0ν<n, Φd0(a)0(modp). As νn,

bn-1=Φn(b)Φd0(b)dn,dd0,nΦd(b).

Applying the above with b=a yields an-10(modp2). Moreover, by the binomial theorem, Φn(a+p)Φn(a)0(modp) and Φd0(a+p)Φd0(a)0(modp), so applying the above with b=a+p yields (a+p)n-10(modp2). But by the binomial theorem, (a+p)n-1=j=0n(nj)an-jpj-1, whence (a+p)n-1an+nan-1p-1(modp2), hence an+nan-1p-10(modp2). Together with an-10(modp2) this yields nan-1p0(modp2), i.e. nan-10(modp), contradicting that pn,a. Therefore ν=n.

Suppose that an1(modp) and that aν1(modp) for 0<ν<n. As dnΦd(a)=an-10(modp), there is some d0n for which Φd0(a)0(modp). Suppose by contradiction that d0<n. As d0n,

ad0-1=dd0Φd(a)=Φd0(a)dd0,d<d0Φd(a)0(modp),

contradicting that aν1(modp) for 0<ν<n. Therefore Φn(a)0(modp), i.e. pΦn(a). ∎

Lemma 13.

Let p be prime, pn. There is some a such that pΦn(a) if and only if p1(modn).

Proof.

Suppose that a and pΦn(a). Then by Lemma 12, n is the multiplicative order of a modulo p. As the multiplicative group (/p)* has p-1 elements, this implies that n(p-1), i.e. p-10(modn).

Suppose that p1(modn), i.e. n(p-1). Because (/p)* is a cyclic group with p-1 elements, it is a fact that there is some a, a+p(/p)*, whose multiplicative order modulo p is n. (Generally, if G is a cyclic group with m elements and n divides m then there is some gG with order n.) Then by Lemma 12, pΦn(a). ∎

We now use Lemma 13 to prove an instance of Dirichlet’s theorem on primes in arithmetic progressions [44, p. 13, Lemma 2.9].

Theorem 14.

For any n1, there are infinitely many primes p with p1(modn).

Proof.

The claim for n=1 follows from the claim for n=2. For n2, by Lemma 8, Φn(0)=1, namely the constant coefficient of Φn(x) is 1. Suppose by contradiction that there are at most finitely many such primes p1,,pt and let M=np1pt. For N, Φn(NM)1(modM) and from M(Φn(NM)-1) it follows that pi(Φn(NM)-1), 1it, and n(Φn(NM)-1). Hence if p is a prime factor of Φn(NM) then ppi, 1it, and pn. As Φn is a monic polynomial that is not a constant, for all sufficiently large N, Φn(NM) is an integer 2 and thus has a prime factor p, amd we have established that pn. Therefore Lemma 13 tells us that p1(modn). But we have also established that ppi, 1ir, a contradiction. Therefore there are infinitely many primes p with p1(modn). ∎

One can prove that for any integers n,b2 it holds that

12bϕ(n)Φn(b)2bϕ(n).

Using this, Thangadurai and Vatwani [42] prove that for n2, the least prime p1(modn) satisfies

p2ϕ(n)+1-1.

5 Zsigmondy’s theorem

[20, pp. 167–169, §8.3.1]

6 Newton’s identities and Ramanujan sums

For positive integers n and n, let

cn(k)=1jn,gcd(n,j)=1e2πijk/n=ξΔnξk,

called a Ramanujan sum.

Lemma 15.
cn(k)=dgcd(n,k)dμ(n/d).
Proof.

Let

ηn(k)=j=1ne2πijk/n={0nkmnk.

We can write ηn(k) as

ηn(k)=dncd(k),

so by the Möbius inversion formula,

cn(k)=dnμ(n/d)ηd(k).

Theorem 16.

For n>1 and for |x|<1,

Φn(x)=exp(-m=1cn(m)mxm).
Proof.

Using that ξξ-1 is a bijection ΔnΔn,

ddxlogΦn(x) =ddxξΔnlog(x-ξ)
=ξΔn1x-ξ
=ξΔn-1ξ11-xξ
=-ξΔn1ξm=0(xξ)m
=-m=0xmξΔnξm+1.

Because n>1, Φn(0)=1, and integrating,

Φn(x)=exp(-m=0xm+1m+1ξΔnξm+1)=exp(-m=1xmmcn(m)).

A formula due to Hölder [36, p. 110, Theorem 4.1] is that

cn(k)=μ(n/gcd(n,k))ϕ(n)ϕ(n/gcd(n,k)). (2)

This identity is used to prove the following lemma that we use later.

Lemma 17.

If n is square-free then kμ(n)cn(k) is multiplicative.

Lemma 18.

For n1 and Res>1,

k=1cn(k)k-s=ζ(s)dnμ(n/d)d1-s.
Proof.

By Lemma 15,

k=1cn(k)k-s =k=1k-sdn,dkμ(n/d)d
=dnm=1(md)-sμ(n/d)d
=dnm=1m-sd-sμ(n/d)d
=m=1m-sdnd-sμ(n/d)d
=ζ(s)dnμ(n/d)d1-s.

Write

j=1n(x-αj)=k=0n(-1)kskxn-k,

and put, for k1,

pk=j=1nαjk.

Newton’s identities [19, p. 32, Proposition 3.4] state that for k1,

pk=j=1k-1(-1)j-1sjpk-j+(-1)k-1ksk. (3)

Write

Φn(x)=k=0ϕ(n)an(k)xk.

Let n>1, and for integer j define

χ1(j)={1gcd(n,j)=10gcd(n,j)>1,

namely the principal Dirichlet character modulo n. We can then write

Φn(x)=1kn,gcd(n,k)=1(x-e2πik/n)=x-n+ϕ(n)j=1n(x-αj)

for αj=χ1(j)e2πij/n, and thus

xn-ϕ(n)Φn(x)=j=1n(x-αj).

Because χ1(j)k=χ1(j) for k1,

pk=j=1nαjk=j=1nχ1(j)e2πijk/n=1jn,gcd(n,j)=1e2πijk/n=cn(k).

Now, from

xn-ϕ(n)k=1ϕ(n)an(k)xk=k=0n(-1)kskxn-k

we have, for 0kn,

(-1)ksk=an(ϕ(n)-k).

In fact by Lemma 20, an(ϕ(n)-k)=an(k), so an(k)=(-1)ksk. Thus (3) yields the following, and in particular

an(1)=-cn(1)=-μ(n).
Theorem 19.

For n1 and k1,

kan(k)=-cn(k)-j=1k-1an(j)cn(k-j).

Let n be a product of distinct odd primes and for a let χ(a)=(an) be the Jacobi symbol. Dedekind, in Supplement I to Dirichlet’s Vorlesungen über Zahlentheorie [15, pp. 208–210], §116, proves that

1jnχ(j)e2πijh/n=χ(h)i(n-1)2/4n; (4)

this is proved earlier by Gauss in his Summatio quarumdam serierum singularium [22, pp. 9–45], dated 1808. The expression G(h,χ)=1jnχ(j)e2πijh/n is called a Gauss sum. Dedekind, in Supplement VII to Dirichlet’s Vorlesungen, says what amounts to the following. Define

An(x)=1an,χ(a)=1(x-e2πia/n)=jαn(j)xj

and

Bn(x)=1bn,χ(b)=-1(x-e2πib/n)=jβn(j)xj,

and write

Sn(k)=1an,χ(a)=1e2πika/n,Tn(k)=1bn,χ(b)=-1e2πikb/n.

Then

Φn(x)=An(x)Bn(x),cn(k)=Sn(k)+Tn(k),

and by (4), writing

n*=(-1)(n-1)/2n,

we have

Sn(k)-Tn(k)=1jnχ(j)e2πikj/n=χ(k)n*,

hence

2Sn(k)=cn(k)+χ(k)n*,2Tn(k)=cn(k)-χ(k)n*.

We have established in Lemma 15 that cn(k), so this shows that Sn(k),Tn(k)(n*). Newton’s identities yield for k1,

Sn(k)=-j=1k-1αn(n-j)Sn(k-j)-kαn(n-k)

and

Tn(k)=-j=1k-1βn(n-j)Tn(k-j)-kβn(n-k),

and it follows that αn(k),βn(k)(n*). Furthermore, αn(k),βn(k) are algebraic integers, so αn(k),βn(k)𝒪(n*). If D is a square-free, it is a fact [16, p. 698, §15.3] that 𝒪(D)=[ω] for

ω={DD2,3(mod4)1+D2D1(mod4),

and n*=(-1)(n-1)/2n1(mod4), we have 𝒪(n*)=[(1+n*)/2]. Thus αn(k),βn(k)[(1+n*)/2].

It is a fact that (n*)(e2πi/n) [23, p. 19, Proposition 5.13]

Gauss, Disquisitiones Arithmeticae, Art. 357

7 Algebraic theorems about coefficients of cyclotomic polynomials

For n1, we write

Φn(x)=k=0ϕ(n)an(k)xk.

Let

A(n)=max0kϕ(n)|an(k)|

and

S(n)=k=0ϕ(n)|an(k)|.

It is immediate that A(n)S(n).

Lemma 20.

For n>1 and for 0kϕ(n),

an(ϕ(n)-k)=an(k).
Proof.

For P(x)=j=0na(j)xj, check that a(j)=a(n-j) for each 0jn is equivalent to xnP(x-1)=P(x). But because n>1, by Lemma 5 we have Φn(x-1)=x-ϕ(n)Φn(x), so we obtain the claim. ∎

Migotti [35] proves the following, and also calculates a105(7)=-2. The following is also proved by Bang [2]; cf. Beiter [4].

Theorem 21 (Bang).

For odd primes p<q,

apq(k){0,-1,1}.
Proof.

By Lemma 1,

Φpq(x) =(xpq-1)(x-1)(xp-1)(xq-1)
=(1-x)α=0p-1xαq1-xp
=(1-x)0αp-1xαqβ0xβp
=0αp-1,β0xαq+βp-0αp-1,β0xαq+βp+1
=0αp-1,β0,0δ1(-1)δxαq+βp+δ.

Suppose by contradiction that α1q+β1p+δ1=α2q+β2p+δ2 with δ1=δ2. Then q(α1-α2)=p(β2-β1), which implies that p divides α1-α2. But 0α1,α2p-1 means 0|α1-α2|p-1, so α1-α2=0 and thence β2-β1=0, which means that (α1,β1,δ1)=(α2,β2,δ2). Therefore, for 0kϕ(pq) there are zero, one, or two triples (α,β,δ) such that k=αq+βp+δ; if there are two such triples, then one has δ=0 and one has δ=1. If there are no such triples, then an(k)=0. If there is one such triple (α,β,δ), then an(k)=(-1)δ. If there are two such triples, then an(k)=(-1)0+(-1)1=0. ∎

Lam and Leung [26] determine the following explicit formula.

Theorem 22 (Lam and Leung).

Suppose that p<q are primes. Then there are nonnegative integers r,s such (p-1)(q-1)=rp+sq, and for 0kϕ(pq)=(p-1)(q-1),

apq(k)={10ir0js with k=ip+jq-1r+1iq-1s+1jp-1 with k+pq=ip+jq0otherwise

Furthermore,

|{k:0kϕ(pq),apq(k)=1}|=(r+1)(s+1)

and

|{k:0kϕ(pq),apq(k)=-1}|=(p-s-1)(q-r-1).
Proof.

Because gcd(p,q)=1, there is some 0rq-1 such that

rp-p+1(modq).

If r=q-1 then we get from the above that 10(modq), which is false because q1, so in fact 0rq-2. Now,

s=(p-1)(q-1)-rpq=pq-p-q+1-rpq

is an integer and

s=p(q-r-1)-q+1q-q+1q>-1,

hence s0. Also, s(p-1)(q-1)q<p-1, so sp-2. We then have

rp+sq=rp+(p-1)(q-1)-rp=(p-1)(q-1).

For ξΔpq, because Φq(ξp)=0 and Φp(ξq)=0,

i=0r(ξp)i=-i=r+1q-1(ξp)i,j=0s(ξq)j=-j=s+1p-1(ξq)j.

(Because 0rq-2 and 0sp-2, each of the above four sums has a nonempty index set.) From this we have

(i=0r(ξp)i)(j=0s(ξq)j)-(i=r+1q-1(ξp)i)(j=s+1p-1(ξq)j)=0.

Because ξ-pq=1, this implies that each ξΔpq is a zero of the polynomial

f(x)=(i=0rxip)(j=0sxjq)-(i=r+1q-1xip)(j=s+1p-1xjq)x-pq;

that this is indeed a polynomial follows from

(r+1)p+(s+1)q-pq=rp+sq+p+q-pq=1.

The first product is a monic polynomial of degree rp+sq=ϕ(pq). The second product is a polynomial of degree

(q-1)p+(p-1)q-pq=-p-q+pq=ϕ(pq)-1.

Therefore f(x) is a monic polynomial of degree ϕ(pq). Because each ξΔpq is a zero of f(x) and f(x) is monic, f(x)=Φpq(x). ∎

Carlitz [10] proves the following.

Theorem 23.

Let p<q be primes, let

qu-1(modp),0<u<p,

let θ(pq) be the number of terms of Φpq with nonzero coefficients, and let θ0(pq) be the number of terms of Φpq with positive coefficients. Then

θ(pq)=2θ0(pq)-1

and

θ0(pq)=(p-u)(uq+1)/p.

Cobeli, Gallot, Moree and Zaharescu [13] give an exposition of apqr(k) where p<q<r are primes, p is fixed, and q,r are free.

Bang [2] proves the following.

Theorem 24 (Bang).

For odd primes p<q<r,

A(pqr)p-1.

Beiter [5] proves the following improvement for a case of the above theorem. If p,q,r, 3<p<q<r, are odd primes for which either q±1(modp) or r±1(modp), then

A(pqr)12(p+1).

Bloom [6] proves the following.

Theorem 25 (Bloom).

For odd primes p<q<r<s,

A(pqrs)p(p-1)(pq-1).

Gallot and Moree [21]

The following is from Lehmer [27], who says that it appears in an unpublished letter of Schur to Landau; cf. Bourbaki [8, V. 165, §11, Exercise 19].

Theorem 26 (Schur).

For any odd m3 there are primes p1<p2<<pm, with p1+p2>pm. For such primes,

ap1p2pm(pm)=-m+1.
Proof.

Write

π(x)=|{p:p is prime and px}|.

For m3, suppose by contradiction that if p1<p2<<pm are primes then p1+p2pm, and thus 2p1<pm. For k1, as there are infinitely many primes, let p1 be the least prime >k, and let kp1<p2<<pm. Then

π(2k)-π(k)=π(2k)-π(p1)+1π(2p1)-π(p1)+1(m-1)+1=m.

This yields, for j1,

π(2j)m+π(2j-1)m+m+π(2j-2)jm.

But the prime number theorem tells us

π(2j)2jjlog2,j,

with which we get a contradiction.

Let m3 be odd and let p1<p2<<pm be primes satisfying p1+p2>pm, and let n=p1p2pm. Since p1+p2>pm, for 1j,km we have pj+pkpm+1. It follows that if d is a divisor of n aside from 1 and p1,,pm, and μ(n/d)0, then

(xd-1)μ(n/d)xpm+1[x].

Therefore

Φn(x)+xpm+1[x] =d|n(xd-1)μ(n/d)+xpm+1[x]
=d|n,μ(d/n)0(xd-1)μ(n/d)+xpm+1[x]
=(x-1)-1j=1m(xpj-1)μ(n/pj)+xpm+1[x]
=(x-1)-1j=1m(xpj-1)+xpm+1[x]
=(x-1)-1(-1+xp1++xpm)+xpm+1[x].

Now,

(x-1)-1(-1+xp1+-xpm)+xpm+1[x]=(1+x+x2++xpm)(1-xp1--xpm)+xpm+1[x].

For 1im, there is one and only one 0jpm such that pi+j=pm. This implies that the coefficient of xpm in the above expression is -m+1. ∎

Lehmer also states that in Rolf Bungers’ 1934 dissertation, Über die Koeffizienten von Kreisteilungspolynomen (University of Göttingen), it is proved that if there exist infinitely many twin primes then for any M there are primes p<q<r such that A(pqr)M. Lehmer proves this without the hypothesis that there are infinitely many twin primes.

For power series A(x)=k=0akxk and B(x)=k=0bkxk, write

AB

if |ak|bk for all k. For power series A,B,P,Q with AP and BQ,

|ak+bk||ak|+|bk|pk+qk,

so A+BP+Q, and

|i+j=kaibj|i+j=k|aibj|i+j=kpiqj,

so ABPQ.

Now,

xd-1k=0xkd,1k=0xkd,(xd-1)-1k=0xkd,

and since μ(n/d){0,1,-1},

Φn(x)=dn(xd-1)μ(n/d)dn(k=0xkd)=dn11-xd. (5)

Hence, because 111-xj,

Φn(x)j=111-xj.

Let np(n) be the partition function, the number of ways of writing n as a sum of positive integers, where the order does not matter. p(0)=1 and p(n)=0 for n<0, and for example, p(4)=5 because 4=4,3+1,2+2,2+1+1,1+1+1+1. It is a fact that for |x|<1,

j=111-xj=k=0p(k)xk,

found by Euler.

Theorem 27.
|an(k)|p(k),

and so

A(n)=max0kϕ(n)|an(k)|max0kϕ(n)p(k)p(ϕ(n))p(n).

It is proved by Hardy and Ramanujan [12, p. 166, Chapter VII] that for K=π23 and λn=n-124,

p(n)=eKλn43λn2+O(eKλnλn3),n.

This implies

p(n)eKn43n,n.

Therefore,

A(n)=O(eKnn),S(n)=O(eKn),n

Now let

Qn(x)=d|n(1+xd+x2d++xn-d).

It is straightforward that for 0k<n, the coefficient of xk in Qn(x) is equal to the coefficient of xk in dn11-xd. For n>1, because the degree of Φn(x) is ϕ(n)<n, using (5) we get

Φn(x)Qn(x).

Let

d(n)=dn1,

the number of positive integer divisors of n. It is straightforward that

dnd=nd(n)/2,

so

Qn(1)=dnnd=dnd=nd(n)/2.

But from Φn(x)Qn(x) we have that S(n) is the sum of the coefficients of the polynomial Qn(x), i.e.

S(n)Qn(1)=nd(n)/2.

This is found by Bateman [3]; cf. [36, p. 64, Exercise 7].

Theorem 28 (Bateman).
S(n)exp(12d(n)logn).

A result due to Wigert [12, p. 19, Theorem 6], proved using the prime number theorem, is that

lim supnlogd(n)loglognlogn=log2.

Thus, for each ϵ>0, there is some nϵ such that when nnϵ,

logd(n)loglognlognlog2+ϵ,

so

logd(n)lognloglogn(ϵ+log2).

Then

logS(n)d(n)2lognlogn2exp(lognloglogn(ϵ+log2)).

Wirsing [46]

Konyagin, Maier and Wirsing [24]

Maier [29], [30], [31], [32]

Bachman [1]

Bzdęga [9]

Nicolas and Terjanian [41]

Let Ψn(x)=xn-1Φn(x), i.e. Ψn(x)=dn,d<nΦd(x), which belongs to [x] and is monic. Moree [37] proves the following.

8 Analytic theorems about coefficients of cyclotomic polynomials

Erdős [18]

Erdős and Vaughan [17] prove the following.

Vaughan [43] proves the next theorem. Vaughan’s original proof is complicated and delightful, and we first outline it and then give a radically simplified proof using Theorem 11, attributed to Saffari by Montgomery and Vaughan [36, pp. 131–132, Exercise 9].

For n=py,p2,3(mod5)p with ω(n) odd, let cm=-cn(m)m. Because n is square-free and μ(n)=-1, it follows from Lemma 17 that mcm is multiplicative. Because cm=O(m-1), the following Euler product expansions hold [36, p. 20, Theorem 1.9]:

m=1cmm-s=pk=0cpkp-ks,Res>0

and

m=1χ(m)cmm-s=pk=0χ(pk)cpkp-ks,Res>0,

where χ is the quadratic Dirichlet character modulo 5. Using Hölder’s formula (2) one works out that for pn,

k=0cpkp-ks=1-p-s1-p-(s+1)

and for pn,

k=0cpkp-ks=11-p-(s+1),

thus

m=1cmm-s=ζ(1+s)pn(1-p-s),Res>0.

Using Hölder’s formula and that χ is completely multiplicative, one works out that for pn,

k=0χ(pk)cpkp-ks=1+p-s1-χ(p)p-(s+1)

and for pn,

k=0χ(pk)cpkp-ks=11-χ(p)p-(s+1),

thus

m=1χ(m)cmm-s=L(1+s,χ)pn(1+p-s),Res>0.

Using (i) the fact that the Gauss sum r=14χ(r)e2πira/5 is equal to χ(a)5, (ii) the fact that c5m=cm5, and (iii) e2πim/5+e2πi4m/5=2Ree2πim/5, one works out that for x>0,

4Rem=1cme2πim/5e-m/x=m=1cm(5χ(m)e-m/x+e-5m/x-e-m/x).

Using this and the above Euler product expansions we get for s>0,

0(Rem=1cme(m/5)e-m/x)x-s-1𝑑x=Γ(s)4(5L(1+s,χ)pn(1+p-s)-(1-5-s)ζ(1+s)pn(1-p-s)).

For x>0, writing f(x)=Rem=1cme2πim/5e-m/x, one has for 0<σ<1,

0f(x)x-σ-1𝑑x11-σ+1σsupx1f(x),

so

supx1f(x)σ0f(x)x-σ-1𝑑x-σ1-σ=σΓ(σ)4(5L(1+σ,χ)pn(1+p-σ)-(1-5-σ)ζ(1+σ)pn(1-p-σ))-σ1-σ.

As σ0 we have σΓ(σ)=1+O(σ), (1-5-σ)ζ(1+σ)=log5+O(σ), and 1-p-σ=σlogp+O(σ2), thus

supx1f(x)145L(1,χ)2ω(n)=145L(1,χ)d(n).

But Theorem 16 tells us that for |z|<1,

|Φn(z)|=exp(Rem=1cmzm),

so |Φn(e2πi/5e-1/x)|=ef(x) and thus

sup|z|<1|Φn(z)|exp(145L(1,χ)d(n)).

As χ is the quadratic Dirichlet character modulo 5, it is a fact that L(1,χ) can be explicitly evaluated (this is an instance of Dirichlet’s class number formula), and using this one checks that exp(125L(1,χ))=1+52. Therefore

sup|z|<1|Φn(z)|(1+52)d(n)/2.
Theorem 29 (Vaughan).

If n=py,p2,3(mod5)p with ω(n) odd, then

|Φn(e2πi/5)|=(1+52)d(n)/2.

There are infinitely many n such that

logA(n)>exp((log2)(logn)loglogn).
Proof.

Vaughan further proves the following.

Theorem 30 (Vaughan).

There is some C such that for infinitely many k,

logmaxn1|an(k)|Ck1/2(logk)-1/4.

9 Fourier analysis

Let 𝕋=/. For p1, define

fLp=(01|f(x)|p𝑑x)1/p

and fL=supx[0,1]|f(x)|. By Jensen’s inequality, if 1pq then

fLpfLq.

For fL1(𝕋), define f^: by

f^(k)=01e-2πikxf(x)𝑑x.

Define

f^p=(k|f^(k)|p)1/p

and f^=supk|f^(k)|. For 1pq,

f^qf^p.

Plancherel’s theorem tells us that

fL2=f^2.

The Hausorff-Young inequality states that for 1p2 and 1p+1q=1,

f^qfLp.

Nikolsky’s inequality [14, p. 102, Theorem 2.6] says that if f^(k)=0 for |k|>n, namely f is a trigonometric polynomial of degree n, then for 0<pq and for rp2 an integer,

fLq(2nr+1)1p-1qfLp.

On the other hand, using Jensen’s inequality for sums one proves that if f is a trigonometric polynomial of degree n, then for 1pq,

f^p(2n+1)1p-1qf^q.

For f:𝕋, define

f^0=|suppf^|=|{n:f^(n)0}|.

McGehee, Pigno and Smith [33] prove that there is some K such that for all N, if n1,,nN are distinct integers and c1,,cN satisfy |ck|1, then

k=1Ncke2πinktL1KlogN.

That is, if f:𝕋 is a trigonometric polynomial with |f^(n)|1 when f^(n)0, then

fL1Klogf^0.

For F:/N, define F^:/N by

F^(k)=1Nj=0N-1F(j)e-2πijk/N,0kN-1.

One checks that [36, pp. 109–110, §4.1]

F(j)=k=0N-1F^(k)e2πijk/N,0jN-1

and

k=0N-1|F^(k)|2=1Nj=0N-1|F(j)|2.

For a0,,aN-1, define f:𝕋 by

f(x)=k=0N-1ake2πikx

and define F:/N by

F(j)=f(j/N)=k=0N-1ake2πikj/N,0jN-1,

for which we calculate F^(k)=ak, for 0kN-1. Then

k=0N-1|ak|2=k=0N-1|F^(k)|2=1Nj=0N-1|F(j)|2=1Nj=0N-1|f(j/N)|2.

Carlitz [11]

10 Algebraic topology

Musiker and Reiner [38]

Meshulam [34]

References

  • [1] G. Bachman (1993) On the coefficients of cyclotomic polynomials. Memoirs of the American Mathematical Society 106 (510), pp. 1–80. Cited by: §7.
  • [2] A. S. Bang (1895) Om Ligningen Φm(X)=0. Nyt Tidsskrift for Mathematik, Afdeling B 6, pp. 6–12. Cited by: §7, §7.
  • [3] P. T. Bateman (1949) Note on the coefficient of the cyclotomic polynomial. Bull. Amer. Math. Soc. 55 (12), pp. 1180–1181. Cited by: §7.
  • [4] M. Beiter (1964) The midterm coefficient of the cyclotomic polynomial Fpq(x). Amer. Math. Monthly 71 (7), pp. 769–770. Cited by: §7.
  • [5] M. Beiter (1968) Magnitude of the coefficients of the cyclotomic polynomial Fpqr(x). Amer. Math. Monthly 75 (4), pp. 370–372. Cited by: §7.
  • [6] D. M. Bloom (1968) On the coefficients of the cyclotomic polynomials. Amer. Math. Monthly 75, pp. 372–377. Cited by: §7.
  • [7] N. Bourbaki (1972) Elements of mathematics. Commutative algebra. Addison-Wesley. Cited by: §2.
  • [8] N. Bourbaki (1990) Elements of mathematics. Algebra II, chapters 4–7. Springer. Note: Translated by P. M. Cohn and J. Howie Cited by: §7.
  • [9] B. Bzd\kega (2012) On the height of cyclotomic polynomials. Acta Arith. 152 (4), pp. 349–359. Cited by: §7.
  • [10] L. Carlitz (1966) The number of terms in the cyclotomic polynomial Fpq(x). Amer. Math. Monthly 73 (9), pp. 979–981. Cited by: §7.
  • [11] L. Carlitz (1967) The sum of the squares of the coefficients of the cyclotomic polynomial. Acta Math. Acad. Sci. Hungar. 18, pp. 295–302. Cited by: §9.
  • [12] K. Chandrasekharan (1970) Arithmetical functions. Die Grundlehren der mathematischen Wissenschaften, Vol. 167, Springer. Cited by: §7, §7.
  • [13] C. Cobeli, Y. Gallot, P. Moree, and A. Zaharescu (2013) Sister Beiter and Kloosterman: a tale of cyclotomic coefficients and modular inverses. Indagationes Mathematicae 24, pp. 915–929. Cited by: §7.
  • [14] R. A. DeVore and G. G. Lorentz (1993) Constructive approximation. Die Grundlehren der mathematischen Wissenschaften, Vol. 303, Springer. Cited by: §9.
  • [15] P. G. L. Dirichlet (1999) Lectures on number theory. History of Mathematics, Vol. 16, American Mathematical Society, Providence, RI. Note: Supplements by R. Dedekind, translated from the German by John Stillwell Cited by: §6.
  • [16] D. S. Dummit and R. M. Foote (2004) Abstract algebra. third edition, John Wiley & Sons. Cited by: §2, §2, §6.
  • [17] P. Erdős and R. C. Vaughan (1974) Bounds for the r-th coefficients of cyclotomic polynomials. J. London Math. Soc. (2) 8 (3), pp. 393–400. Cited by: §8.
  • [18] P. Erdős (1957) On the growth of the cyclotomic polynomial in the interval (0,1). Proceedings of the Glasgow Mathematical Association 3 (2), pp. 102–104. Cited by: §8.
  • [19] J. Escofier (2001) Galois theory. Graduate Texts in Mathematics, Vol. 204, Springer. Note: Translated by Leila Schneps Cited by: §6.
  • [20] G. Everest and T. Ward (2005) An introduction to number theory. Graduate Texts in Mathematics, Vol. 232, Springer. Cited by: §5.
  • [21] Y. Gallot and P. Moree (2009) Ternary cyclotomic polynomials having a large coefficient. J. reine angew. Math. 632, pp. 105–125. Cited by: §7.
  • [22] C. F. Gauss (1876) Carl Friedrich Gauss. Werke. Zweiter Band. Königlichen Gesellschaft der Wissenschaften zu Göttingen. Cited by: §6.
  • [23] K. Kato, N. Kurokawa, and T. Saito (2011) Number theory 2: introduction to class field theory. Translations of Mathematical Monographs, Vol. 240, American Mathematical Society, Providence, RI. Note: Translated by Masato Kuwata and Katsumi Nomizu Cited by: §6.
  • [24] S. Konyagin, H. Maier, and E. Wirsing (2004) Cyclotomic polynomials with many primes dividing their orders. Period. Math. Hungar. 49 (2), pp. 99–106. Cited by: §7.
  • [25] R. P. Kurshan and A. M. Odlyzko (1981) Values of cyclotomic polynomials at roots of unity. Math. Scand. 49, pp. 15–35. Cited by: §3.
  • [26] T. Y. Lam and K. H. Leung (1996) On the cyclotomic polynomial Φpq(X). Amer. Math. Monthly 103 (7), pp. 562–564. Cited by: §7.
  • [27] E. Lehmer (1936) On the magnitude of the coefficients of the cyclotomic polynomial. Bull. Amer. Math. Soc. 42 (6), pp. 389–392. Cited by: §7.
  • [28] R. Lidl and H. Niederreiter (1997) Finite fields. Encyclopedia of Mathematics and Its Applications, Vol. 20, Cambridge University Press. Cited by: §2.
  • [29] H. Maier (1990) The coefficients of cyclotomic polynomials. In Analytic Number Theory. Proceedings of a Conference in Honor of Paul T. Bateman, B. C. Berndt, H. G. Diamond, H. Halberstam, and A. Hildebrand (Eds.), Progress in Mathematics, Vol. 85, pp. 349–366. Cited by: §7.
  • [30] H. Maier (1996) The size of the coefficients of cyclotomic polynomials. In Analytic Number Theory. Proceedings of a Conference in Honor of Heini Halberstam, Volume 2, B. C. Berndt, H. G. Diamond, and A. J. Hildebrand (Eds.), Progress in Mathematics, Vol. 139, pp. 633–639. Cited by: §7.
  • [31] H. Maier (2006) The distribution of the L2-norm of cyclotomic polynomials on the unit circle. In Elementare und analytische Zahlentheorie, W. Schwarz and J. Steuding (Eds.), Schriften der Wissenschaftlichen Gesellschaft an der Johann Wolfgang Goethe-Universität Frankfurt am Main, Vol. 20, pp. 164–179. Cited by: §7.
  • [32] H. Maier (2008) Anatomy of integers and cyclotomic polynomials. In Anatomy of Integers, J. De Koninck, A. Granville, and F. Luca (Eds.), CRM Proceedings & Lecture Notes, Vol. 46, pp. 89–95. Cited by: §7.
  • [33] O. C. McGehee, L. Pigno, and B. Smith (1981) Hardy’s inequality and the L1 norm of exponential sums. Ann. of Math. (2) 113 (3), pp. 613–618. Cited by: §9.
  • [34] R. Meshulam (2012) Homology of balanaced complexes via the Fourier transform. J. Algebraic Combin. 35, pp. 565–571. Cited by: §10.
  • [35] A. Migotti (1883) Zur Theorie der Kreistheilungsgleichung. Sitzungsberichte der Mathematisch-Naturwissenschaftlichen Classe der Kaiserlichen Akademie der Wissenschaften 87, pp. 7–14. Note: Heft I, Abt. II Cited by: §7.
  • [36] H. L. Montgomery and R. C. Vaughan (2006) Multiplicative number theory I: classical theory. Cambridge Studies in Advanced Mathematics, Vol. 97, Cambridge University Press. Cited by: §3, §6, §7, §8, §8, §9.
  • [37] P. Moree (2009) Inverse cyclotomic polynomials. J. Number Theory 129, pp. 667–680. Cited by: §7.
  • [38] G. Musiker and V. Reiner (2014) The cyclotomic polynomial topologically. J. reine angew. Math. 687, pp. 113–132. Cited by: §10.
  • [39] J. Neukirch (1999) Algebraic number theory. Grundlehren der mathematischen Wissenschaften, Vol. 322, Springer. Note: Translated from the German by Norbert Schappacher Cited by: §2.
  • [40] O. Neumann (2007) The Disquisitiones Arithmeticae and the theory of equations. In The Shaping of Arithmetic after C. F. Gauss’s Disquisitiones Arithmeticae, C. Goldstein, N. Schappacher, and J. Schwermer (Eds.), pp. 107–127. Cited by: §2.
  • [41] J. Nicolas and G. Terjanian (1999) Une majoration de la longueur des polynômes cyclotomiques. Enseign. Math. (2) 45 (3-4), pp. 301–309. Cited by: §7.
  • [42] R. Thangadurai and A. Vatwani (2011) The least prime congruent to one modulo n. Amer. Math. Monthly 118 (8), pp. 737–742. Cited by: §4.
  • [43] R. C. Vaughan (1974) Bounds for the coefficients of cyclotomic polynomials. Michigan Math. J. 21, pp. 289–295 (1975). Cited by: §8.
  • [44] L. C. Washington (1997) Introduction to cyclotomic fields. second edition, Graduate Texts in Mathematics, Vol. 83, Springer. Cited by: §2, §4, §4.
  • [45] A. Weil (1984) Number theory: an approach through history from Hammurapi to Legendre. Birkhäuser. Cited by: §2.
  • [46] E. Wirsing (2006) The third logarithmic momentum of the cyclotomic polynomial on the unit circle and factorizations with a linear side condition. In Elementare und analytische Zahlentheorie, W. Schwarz and J. Steuding (Eds.), Schriften der Wissenschaftlichen Gesellschaft an der Johann Wolfgang Goethe-Universität Frankfurt am Main, Vol. 20, pp. 297–312. Cited by: §7.