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Total ramification at a prime-power cyclotomic level

Statement

For a prime p and an integer a≥1, put K=Q(ζpa) and λ=1−ζpa. Then pOK=(λ)φ(pa), and the principal ideal λOK is the unique prime ideal of OK lying above p, with residue field OK/λOK≅Fp.

Facts & Assumptions

Given: A prime p, an integer a≥1, e:=φ(pa), a primitive pa-th root of unity ζ=ζpa, K:=Q(ζ), λ:=1−ζ and R:=Z[ζ]=OK.

[F1]

OK=R=Z[ζ] and pR=(λ)e; moreover 1,ζ,…,ζe−1 is an integral basis of OK (Prime-power cyclotomic ring, discriminant support and p factor, Ring of integers of every cyclotomic field).

[F2]

Φpa∈Z[t] is monic of degree e and Φpa(1)=p (Φpr(t)=∑k<ptkpr−1, and Φpr(t+1) is Eisenstein at p).

[F3]

Division by the monic polynomial t−1 in Z[t] writes every f∈Z[t] uniquely as f=q⋅(t−1)+f(1) with q∈Z[t] and constant remainder f(1) (Division by a monic polynomial over a commutative ring); hence the evaluation homomorphism Z[t]→Z, f↦f(1), is a surjective ring homomorphism with kernel (t−1), so Z[t]/(t−1)≅Z (First isomorphism theorem for rings: R/ker⁡f≅im⁡f).

[F4]

An ideal M of a commutative ring R is maximal if and only if R/M is a field (R/M is a field if and only if M is a maximal ideal), and every maximal ideal is prime (Every maximal ideal of a commutative ring is prime). Also Z/p is a field (For every prime p, the two operations on Z/p make it a field).

[F5]

A nonzero prime P of OK lies above p when P∩Z=pZ, and its residue degree is [OK/P:Fp] (Primes above and residue degree).

Proof

technique · direct
1.1F1F2F3F4

Under the presentation R=Z[ζ]=Z[t]/(Φpa) with t↦ζ, the element λ=1−ζ corresponds to the class of 1−t, so R/λR≅Z[t]/(Φpa, t−1); sending t to 1 via [F3] identifies this quotient with Z/(Φpa(1))=Z/pZ=Fp.

2.1F1F4F5step 1.1

Since R/λR≅Fp is a field, λR is a maximal and hence prime ideal of R, and it is proper; consequently λR∩Z is a proper ideal of Z containing pZ (as pR=(λ)e⊆λR by [F1]) and therefore equals pZ, so λR lies above p with residue field OK/λOK≅Fp, of residue degree 1.

3.1F1F4step 2.1

If P is any prime ideal of R with p∈P, then λe∈λeR=pR⊆P, so λ∈P because P is prime, hence λR⊆P; as λR is maximal and P is proper, P=λR. Thus λR is the unique prime above p.

4.1F1step 2.1step 3.1∎

Combining [F1] with steps 2.1 and 3.1, pOK=(λ)φ(pa) and λOK is the unique prime of OK above p, with residue field Fp. For p=2, a=1 this reads K=Q, λ=2, e=1, and 2Z is the unique prime above 2 with residue field F2.

Remarks

  • Total ramification. The exponent equals the degree [Q(ζpa):Q]=φ(pa), and the residue degree is 1, so p is totally ramified; the equality pOK=(λ)e exhibits the ramification index without invoking any general ramification theory beyond the definitions.
  • The quotient computation is the only place where Φpa(1)=p is used, and it also shows that no prime other than λOK can contain p.

Depends on

Used by

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Sources