While developing any theory, it is always helpful to have explicit examples at hand. We have previously encountered the family of quadratic fields, for which it is possible to work out many of their properties (eg. generators of the number ring). In this post, we introduce the cyclotomic fields, which form another such family of fields where explicit computations are viable.
Cyclotomic fields
For an integer m≥1, let ω=e2πi/m∈C be a primitive m-th root of unity. Then Q[ω] is called the m-th cyclotomic field.
We have already encountered some examples of cyclotomic fields. For instance, the m-th cyclotomic field is Q for m=1,2, Q[21+−3] for m=3,6, and Q[i] for m=4.
Proposition: Let m≥1 be odd. Then the m-th cyclotomic field is equal to the 2m-th cyclotomic field, ie. Q[e2πi/m]=Q[e2πi/2m].
Proof: ⊆ follows from e2πi/m=(e2πi/2m)2, and ⊇ follows from e2πi/2m=−(e2πi/m)(m+1)/2. □
The above result essentially follows from the fact that −1∈Q is a primitive square root of unity.
Our first task is to determine the degree of the extension [Q[ω]:Q], which is equal to the number of conjugates of ω in C, or the degree of the minimal polynomial of ω over Q. Note that this minimal polynomial divides Xm−1, but does not divide Xn−1 for any 1≤n<m (since ωm=1 but ωn=1). Hence the only possible conjugates of ω are ωk for k coprime to m. (Note that this immediately implies that Q[ω] is Galois over Q.) We prove the converse statement below.
Proposition: For any k with gcd(k,m)=1, we have ωk is conjugate to ω.
Proof: We will prove the following statement: for any θ=ωl with gcd(l,m)=1, and any prime p not dividing m, we have θp is conjugate to θ. Then since conjugacy is transitive, by induction we have ω is conjugate to ωk for any k which is a product of primes not dividing m, ie. for any k≥1 coprime to m, and we are done.
Let f be the minimal polynomial of θ over Q, so Xm−1=f(X)g(X) for some monic g∈Q. By Gauss’s lemma, we have f,g∈Z[X].
Assume for sake of contradiction that θp is not conjugate to θ. Then f(θp)=0, so g(θp)=0. Thus θ is a root of g(Xp), so f(X)∣g(Xp) in Q[X], and by Gauss’s lemma again we have f(X)∣g(Xp) in Z[X].
Therefore, under the mod p homomorphism ⋅:Z[X]→Fp[X], we have f(X)∣g(Xp)=g(X)p in Fp[X]. Now Fp[X] is a UFD, so f and g have a nontrivial common factor, say h.
Thus h2∣fg=Xm−1, so h divides the derivative mXm−1. By unique factorisation, we have h is a multiple of some Xb, b≥1, but this contradicts h∣Xm−1, and we are done. □
Recall that the Euler totient functionφ:Z>0→Z>0 is defined by φ(n)=∣(Z/nZ)×∣; equivalently, φ(n) is the number of integers in {1,2,…,n} which are coprime to n.
Corollary: [Q[ω]:Q]=φ(m).
Proof: By the above result, the conjugates of ω are precisely
{ωk:1≤k≤m,gcd(k,m)=1},
which is a set with φ(m) many elements. Hence the minimal polynomial of ω has degree φ(m). □
Corollary: The Galois group of Q[ω] over Q is isomorphic to (Z/mZ)×. More precisely, each k∈(Z/mZ)× is mapped to the automorphism sending ω↦ωk.
Proof: It suffices to check that the composition law for automorphisms corresponds to multiplication mod m: performing ω↦ωk then ω↦ωl yields
ω↦ωk↦(ωl)k=ωkl,
and we are done. □
Corollary: The only roots of unity in Q[ω] are the m-th roots of unity for m even, and the 2m-th roots of unity for m odd.
Proof: Since the m-th and 2m-th cyclotomic fields are equal for m odd, it suffices to prove the claim for m even. Suppose that Q[ω] contains a primitive k-th root of unity θ=e2πih/k, with gcd(h,k)=1. Let gcd(k,m)=d, so that gcd(k,hm)=d. Hence there exists a,b∈Z with ak+bhm=d. Then Q[ω] contains
ωaθb=e2πid/km,
which is a primitive cm-th root of unity (where c=dk). Hence Q[ω] contains the cm-th cyclotomic field, so φ(m)≥φ(cm). By well-known properties of the Euler totient function, this implies that c=1, so d=k divides m; hence θ is an m-th root of unity, as desired. □
Corollary: The m-th cyclotomic fields, for even m, are pairwise non-isomorphic.
Proof: Since all such fields are Galois over Q, and any such isomorphism is an embedding in C, it suffices to show that these fields are pairwise distinct.
Now if Q[e2πi/m]=Q[e2πi/m′], then the above result implies that e2πi/m is an m′-th root of unity, so m′∣m; similarly, we have m∣m′, so m=m′. □
Discriminant of an element
We want to investigate the ring of algebraic integers in cyclotomic fields. However, we first need some properties of the discriminant of n-tuples with a special form.
For an algebraic number α of degree n over Q, we will write disc(α) to denote disc(1,α,…,αn−1).
Lemma: Let K=Q[α], and let α1,…,αn be the conjugates of α over Q. Then
disc(α)=r<s∏(αr−αs)2=(−1)2n(n−1)NK(f′(α)),
where f is the minimal (monic irreducible) polynomial of α over Q.
Proof: Let σ1,…,σn be the n embeddings of K into C such that σk(α)=αk. Then
where the last product is over all n(n−1) ordered pairs (r,s) with r=s. Then note that
r<s∏(ar−as)2=(−1)n(n−1)/2r=s∏(ar−as),
which yields the second equality. □
Algebraic integers in cyclotomic fields
We now consider the ring of algebraic integers A∩Q[ω] in Q[ω]. (Recall that ω=e2πi/m is a primitive m-th root of unity.) We first observe that ω is a root of Xm−1, and is hence an algebraic integer. This implies Z[ω]⊆A∩Q[ω]. In this section, we show that this is in fact an equality.
Lemma: disc(ω) divides mφ(m).
Proof: Let f be the minimal polynomial for ω over Q. As before, we have Xm−1=f(X)g(X) with f,g∈Z[X] monic. Differentiating both sides yields
mXm−1=f′(X)g(X)+f(X)g′(X).
Now multiply both sides by X, and substitute X=ω to get
m=ωf′(ω)g(ω).
Taking norms yields
mφ(m)=±disc(ω)N(ωg(ω)),
and we finish by noting that N(ωg(ω))∈Z (since ω and g(ω) are algebraic integers). □
Lemma: Let m≥3. Then Z[1−ω]=Z[ω], and disc(1−ω)=disc(ω).
Proof: The first statement is clear. For the second statement, note that if αi are the conjugates of ω, then 1−αi are the conjugates of 1−α. Hence
We will first determine the ring of integers in the m-th cyclotomic field Q[ω] in the case where m is a power of a prime.
Lemma: Let p be a prime, and let m=pr. Then
1≤k≤m,p∤k∏(1−ωk)=p.
Proof: Let
f(x)=xpr−1−1xpr−1=1+xpr−1+x2pr−1+⋯+x(p−1)pr−1.
Then for all p∤k, we have ωk is a root of Xpr−1 but not Xpr−1−1. Thus all the ωk are roots of f. Furthermore, we have
#{1≤k≤m:p∤k}=φ(m)=pm−pm−1=degf,
so the ωk are all the roots of f. Thus
f(X)=1≤k≤m,p∤k∏(X−ωk),
and the result follows by substituting X=1. □
Theorem: Let ω=e2πi/m, where m=pr for some prime p. Then A∩Q[ω]=Z[ω].
Proof: We write K=Q[ω], R=A∩K, and n=φ(pr)=[K:Q]. Also, let d:=disc(1−ω)=disc(ω). Note that d=0, since
{1,ω,…,ωn−1}
is a basis for K over Q. Hence
{1,1−ω,…,(1−ω)n−1}
is also a basis of K over Q. Thus by a theorem in the previous post, every α∈R can be written as
α=d1j=0∑n−1mj(1−ω)j,
where mj∈Z. Also, by a previous lemma, we have d∣mφ(m), so d is a power of p.
Assume for the sake of contradiction that R=Z[ω]=Z[1−ω]. Then there exists α∈R such that not all mj above are divisible by d. By multiplying a suitable power of p to all the mj, we may assume that pd∣mj; by subtracting some initial terms, we see that R contains an element of the form
For the case of general m, we need the following corollary of the last theorem in the previous post:
Corollary: Let K,L be number fields, and let R,S,T be the number rings of K,L,KL respectively. If [KL:Q]=[K:Q][L:Q] and gcd(discR,discS)=1, then T=RS.
We can now prove the main theorem for this section.
Theorem: Let ω=e2πi/m. Then A∩Q[ω]=Z[ω].
Proof: We proceed by strong induction on m. We have already solved the case when m is a prime power. If m is not a prime power, then we can write m=m1m2 for coprime integers m1,m2>1.
Let K=Q[ω], R=A∩K. For k=1,2, write ωk=e2πi/mk, Kk=Q[ωk], and Rk=A∩Kk=Z[ωk] by inductive hypothesis.
Let a,b∈Z satisfy am1+bm2=gcd(m1,m2)=1. Then ω=ω1aω2b implies K⊆K1K2 and Z[ω]⊆R1R2. Note that the reverse inclusions are clear, so K=K1K2 and Z[ω]=R1R2.
Hence it suffices to show that R=R1R2, so we are left to check the conditions of the previous corollary. The degree condition follows from φ(m)=φ(m1)φ(m2), since m1,m2 are coprime. The discriminant condition follows since discR1 divides a power of m1, and similarly for discR2, so m1,m2 coprime implies discR1,discR2 coprime. □
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