Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02
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Arithmetic of Q(zeta_5)

Example

For K=Q(ζ5) one has OK=Z[ζ5] and dK=53=125. The prime 5 is totally ramified: (5)=(1−ζ5)4 is a fourth power of the unique prime above 5, whose residue field is F5. The prime 2 is unramified with a single prime above it of residue degree 4 (residue field F16) and trivial inertia group; and 11 splits completely into four degree-one primes.

Facts & Assumptions

Given: A primitive fifth root of unity ζ=ζ5 and K=Q(ζ), of degree φ(5)=4 (The cyclotomic extension K(μn) as a splitting field of tn−1).

[F1]

OK=Z[ζ], with integral basis 1,ζ,ζ2,ζ3 (Ring of integers of every cyclotomic field).

[F2]

Discriminant: for the reduced index f=5, dK=(−1)φ(5)/25φ(5)/∏p∣5pφ(5)/(p−1) (Signed discriminant of a cyclotomic field).

[F3]

Total ramification at a prime-power level: with λ=1−ζ, 5OK=(λ)φ(5)=(λ)4, and λOK is the unique prime of OK above 5, with residue field F5 (Total ramification at a prime-power cyclotomic level).

[F4]

Unramified decomposition: for the reduced index 5 and a prime ℓ∤5, every prime above ℓ has residue degree ord⁡5(ℓ) and there are φ(5)/ord⁡5(ℓ)=4/ord⁡5(ℓ) of them (Decomposition of an unramified prime in a cyclotomic field).

[F5]

Complete splitting: for ℓ∤5, the prime ℓ splits completely in K if and only if ℓ≡1(mod5) (Complete splitting criterion for a cyclotomic field).

[F6]

For finite Galois extensions the inertia group at a prime has order equal to the ramification exponent, ∣I(P/p)∣=e(P/p), and P is unramified over p if and only if its inertia group is trivial (Inertia group of a prime, Orders of decomposition and inertia groups).

Verification

technique · direct
1.1F2

Since φ(5)=4, the discriminant formula gives dK=(−1)2⋅54/54/4=54/5=53=125.

1.2F4

The multiplicative order of 2 modulo 5 is 4: the powers of 2 modulo 5 are 2,4,3,1, so ord⁡5(2)=4.

1.3F5

The class 11≡1(mod5) is the identity of (Z/5)×.

1.4F3F6

By [F3], 5OK=(λ)4 with λOK the unique prime above 5 and residue field F5; its ramification exponent is 4=[K:Q], so 5 is totally ramified and its inertia group at λ has order 4, the full Galois group.

2.1F4F6step 1.2

By [F4] and step 1.2 applied with ℓ=2, there is exactly 4/4=1 prime above 2, of residue degree 4; its residue field has 24=16 elements, and since 2 is unramified its inertia group is trivial by [F6], so e=1,f=4,g=1 with efg=4=[K:Q].

2.2F5step 1.3

By [F5] and step 1.3, 11 splits completely in K: there are φ(5)=4 distinct primes above 11, each of ramification exponent 1 and residue degree 1.

3.1F1step 1.1step 1.4step 2.1step 2.2∎

Collecting the results: OK=Z[ζ5] with 1,ζ,ζ2,ζ3 an integral basis, dK=125, the prime 5 is totally ramified with (5)=(1−ζ)4, the prime 2 has one prime above it of residue degree 4 and trivial inertia, and 11 splits into four degree-one primes.

Remarks

  • Unramified decomposition has three cases. For a prime ℓ≠5, the orders modulo 5 are 1,2,4 for the classes 1,4,{2,3}, respectively. Thus [F4] gives four degree-one primes, two degree-two primes, or one degree-four prime. In particular 2 is inert, while 19≡4(mod5) gives two primes, each of residue degree 2.
  • Frobenius orders. The arithmetic Frobenius at 2 has order 4 and generates the full group (Z/5)×, while the Frobenius at 11 is trivial, which is complete splitting in the sense of Complete splitting criterion for a cyclotomic field.

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