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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02
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Decomposition of an unramified prime in a cyclotomic field

Statement

Let f be the conductor of the cyclotomic field K=Q(ζf), and let ℓ be a rational prime with ℓ∤f. Then every prime P of OK above ℓ has residue degree deg⁡(P/ℓ)=ord⁡f(ℓ), the multiplicative order of ℓ modulo f, and there are exactly φ(f)/ord⁡f(ℓ) such primes.

Facts & Assumptions

Given: A cyclotomic field K=Q(ζf) presented by its conductor f, so that f is the least admissible index for K (Cyclotomic conductor of a full cyclotomic field), a rational prime ℓ with ℓ∤f, and the factorisation f=ℓam with gcd⁡(ℓ,m)=1 (so a=0 and m=f under the hypothesis ℓ∤f).

[F1]

The conductor f of a full cyclotomic field is a reduced index: it is odd or divisible by 4 (Conductor of a full cyclotomic field, Cyclotomic conductor of a full cyclotomic field).

[F2]

Prime factorisation in a reduced cyclotomic field: for the reduced index f, a rational prime ℓ, and f=ℓam with gcd⁡(ℓ,m)=1, ℓOK=(P1⋯Pg)e,e=φ(ℓa), d=ord⁡m(ℓ), g=φ(m)/d, with the Pi pairwise distinct primes of residue degree d; here ord⁡1(ℓ)=1 (Prime factorisation in a cyclotomic field).

[F3]

For ℓ∤f the arithmetic Frobenius at every prime above ℓ is the automorphism σℓ with σℓ(ζf)=ζf ℓ (Arithmetic Frobenius is the power map in an unramified cyclotomic field).

Proof

technique · direct
1.1F1

By [F1], f is a reduced index; as ℓ∤f the ℓ-adic valuation is a=0, so m=f and e=φ(1)=1.

2.1F2step 1.1

Applying [F2] with a=0, m=f, the ideal ℓOK factors as P1⋯Pg with g=φ(f)/d pairwise distinct primes of residue degree d=ord⁡f(ℓ), so ℓ is unramified and these are exactly the primes above ℓ.

3.1F2F3step 2.1given∎

Therefore every prime above ℓ has residue degree ord⁡f(ℓ) and their number is φ(f)/ord⁡f(ℓ). This degree also equals the order of the arithmetic Frobenius in [F3]: for r≥1, σℓr(ζf)=ζfℓr, so, since ζf has order f and generates K, σℓr=id exactly when ℓr≡1(modf). The least such r is ord⁡f(ℓ), including r=1 for f=1.

Remarks

  • Conductor versus displayed index. The statement is about the conductor f; for an unreduced displayed index such as 6 the count and degrees are those of the reduced index 3, as recorded in The reduced conductor of Q(zeta_6) ↗.
  • Order-one convention. For f=1 one has ord⁡1(ℓ)=1 and the formula gives the single prime ℓZ=ℓOQ, consistent with φ(1)=1.

Depends on

Used by

Dependency tree · two levels

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Sources