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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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Q(μm)∩Q(μn)=Q(μgcd⁡(m,n))

Statement

Let m,n≥1, put d:=gcd⁡(m,n) and ℓ:=lcm⁡(m,n) (Common divisor, and the greatest common divisor gcd⁡(a,b), with the convention gcd⁡(0,0):=0, Common multiple, and the least common multiple lcm⁡(a,b), taken to be 0 when a=0 or b=0), and let Ω be a splitting field of tℓ−1 over Q (Every nonzero polynomial over a field has a splitting field), inside which the cyclotomic extensions Q(μk) for k∣ℓ are taken (The cyclotomic extension K(μn) as a splitting field of tn−1). Then

Q(μm)∩Q(μn)=Q(μd).

Facts & Assumptions

Given: Integers m,n≥1 with d=gcd⁡(m,n) and ℓ=lcm⁡(m,n); Q is an ordered field (The rationals form a totally ordered field), so char⁡Q=0 (The characteristic of a ring: the least n≥1 with n⋅1R=0 when one exists, and 0 otherwise) and divides no positive integer (Divisibility in Z: d∣a when a=dq for some integer q); a splitting field Ω of tℓ−1 over Q; and, for each positive divisor k of ℓ, the subfield Q(μk):=Q(μk(Ω)) of Ω. Write I:=Q(μm)∩Q(μn).

[L1]

For a positive divisor k of ℓ, the subfield Q(μk(Ω)) generated by the k-th roots of unity in Ω is a cyclotomic extension of Q of order k (The cyclotomic extension K(μn) as a splitting field of tn−1, The group μn(K) of n-th roots of unity in a field, and primitive n-th roots of unity); because char⁡Q=0 divides no positive integer, tn−1 is separable over K exactly when the characteristic does not divide n, and then a splitting field carries n distinct n-th roots of unity gives μk(Ω) cyclic of order k, with exactly φ(k) primitive k-th roots of unity.

[L4]

For E/F finite Galois and L/F finite inside a common field, [EL:F]=[E:F][L:F]/[E∩L:F] (For E/F finite Galois and L/F finite inside a common field, [EL:F]=[E:F][L:F]/[E∩L:F]).

[L5]

Q(μm)Q(μn)=Q(μℓ) inside Ω (K(μm)K(μn)=K(μlcm⁡(m,n))).

[L6]
[L7]

For fields F⊆K⊆M with K/F and M/K finite, [M:F]=[M:K][K:F] (Tower law for finite extensions: [L:F]=[L:K][K:F]).

Proof

technique · direct
1.1L1givenalgebra

d divides m and n, and m,n,d all divide ℓ. For each positive divisor k of ℓ, write ℓ=kq; then tℓ−1=(tk)q−1=(tk−1)(tk(q−1)+⋯+tk+1), so tk−1 splits over the splitting field Ω of tℓ−1, and [L1] applies to k. In particular all four cyclotomic extensions for k=m,n,d,ℓ sit inside Ω.

2.1step 1.1givenalgebra

For the inclusion Q(μd)⊆I: since d∣m, every x with xd=1 satisfies xm=1, so μd(Ω)⊆μm(Ω) and hence Q(μd)⊆Q(μm); the same argument with n gives Q(μd)⊆Q(μn).

2.2step 1.1L2L3L4L5

By [L3] the extension Q(μm)/Q is finite Galois and Q(μn)/Q is finite, both inside Ω, so [L4] and [L5] give φ(ℓ)=[Q(μℓ):Q]=[Q(μm)Q(μn):Q]=φ(m)φ(n)/[I:Q], using [L2] twice.

3.1step 2.2L6

Hence [I:Q]=φ(m)φ(n)/φ(ℓ)=φ(d) by [L6].

4.1step 2.1step 3.1L2L7∎

By step 2.1 the tower Q⊆Q(μd)⊆I is defined, and [L7] with [L2] gives φ(d)=[I:Q]=[I:Q(μd)] φ(d), so [I:Q(μd)]=1 and I=Q(μd).

Remarks

  • Where the base field is used. Only through [L2]: the equality [Q(μk):Q]=φ(k) for every k, which is irreducibility of Φk over Q. Over a base field where some Φk becomes reducible the degrees drop unevenly and the degree count in step 2.2 no longer forces the intersection down to Q(μd).

  • The base field really matters. Over a general base field the same formula can fail; the companion page gives a finite-field witness in F3(μ5)∩F3(μ7) is larger than F3 although five and seven are coprime ↗.

Depends on

Used by

Dependency tree · two levels

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Sources