ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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The reduced conductor of Q(zeta_6)

Example

$Q (ζ_{6}) = Q (ζ_{3})$ , the conductor of this field is $3$ , and the rational prime $2$ is unramified in it: $2 O_{Q (ζ_{3})}$ is prime, with residue degree $2$ .

Facts & Assumptions

Given: A primitive sixth root of unity $ζ_{6}$ and a primitive third root of unity $ζ_{3}$ , with $Q (ζ_{6})$ and $Q (ζ_{3})$ the corresponding cyclotomic fields (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

[F1]

For odd $m$ , $- ζ_{m}$ is a primitive $2 m$ -th root of unity, so $Q (ζ_{2 m}) = Q (ζ_{m})$ (Conductor of a full cyclotomic field).

[F2]

The conductor of $Q (ζ_{n})$ is $n$ when $n$ is odd or $4 ∣ n$ and is $n / 2$ when $n \equiv 2 (m o d 4)$ (Conductor of a full cyclotomic field, Cyclotomic conductor of a full cyclotomic field).

[F3]

Ramification criterion: for a cyclotomic field presented by its reduced index $f$ , a rational prime $ℓ$ ramifies if and only if $ℓ ∣ f$ (Ramification primes of a reduced cyclotomic conductor).

[F4]

Unramified decomposition: if $ℓ ∤ f$ for the reduced index $f$ of $Q (ζ_{f})$ , then every prime above $ℓ$ has residue degree ${ord}_{f} (ℓ)$ and there are $φ (f) / {ord}_{f} (ℓ)$ of them (Decomposition of an unramified prime in a cyclotomic field).

Verification

technique · direct

1.1F1

Taking $m = 3$ in [F1], $- ζ_{3}$ is a primitive sixth root of unity, so $Q (ζ_{6}) = Q (ζ_{3})$ .

1.2F2

By [F2] with $n = 6 \equiv 2 (m o d 4)$ , the conductor of $Q (ζ_{6})$ is $6 / 2 = 3$ .

2.1F3F4step 1.1step 1.2∎

The reduced index of this field is $3$ , and $2 ∤ 3$ , so [F3] shows that $2$ is unramified in $Q (ζ_{6}) = Q (ζ_{3})$ ; by [F4] with $f = 3$ , $ℓ = 2$ , the residue degree is ${ord}_{3} (2) = 2$ and the number of primes above $2$ is $φ (3) / 2 = 1$ , so $2 O$ is prime of degree $2$ .

Remarks

Why the displayed index $6$ is a trap. The index $6$ is not reduced: an inference of the form " $ℓ ∣ (displayed index) \Rightarrow ℓ$ ramifies" would wrongly make $2$ ramified, since $2 ∣ 6$ ; the actual conductor is $3$ , and $2 ∣ 3$ fails. This is the exceptional shape excluded in the ramification criterion.
Frobenius viewpoint. Since ${ord}_{3} (2) = 2$ , the arithmetic Frobenius at $2$ acts by $ζ_{3} \mapsto ζ_{3}^{2}$ and has order $2$ , matching the single degree-two prime above $2$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Number Theory, Ch. 6, Remark 6.6 (standard reference, not scraped)
Conrad-Landesman, Math 154 Algebraic Number Theory, Ch. 11, Remark 11.7 (standard reference, not scraped)