CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Frobenius compatibility in finite towers

Statement

Let M/L/K have M/K and L/K finite Galois, and let $Q ∣ P ∣ p$ be nonzero primes with Q unramified over p. Then $Frob (Q / p) ∣_{L} = Frob (P / p), Frob (Q / P) = Frob (Q / p)^{f (P / p)} .$ If $M = L_{1} L_{2}$ with both $L_{i} / K$ finite Galois and Q unramified over p, the Frobenius elements at its two contractions determine $Frob (Q / p)$ uniquely by restriction.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Decomposition and inertia in towers: Let M/L/K be a tower of number fields with M/K and L/K finite Galois, and fix nonzero primes $Q ∣ P ∣ p$ . With $H = Gal (M / L)$ , $D (Q / P) = D (Q / p) \cap H, I (Q / P) = I (Q / p) \cap H .$ Restriction gives exact sequences $1 \to D (Q / P) \to D (Q / p) \to D (P / p) \to 1,$ $1 \to I (Q / P) \to I (Q / p) \to I (P / p) \to 1.$ The intersection identities also hold without L/K Galois; the displayed quotient assertions use that hypothesis.

[F2]

Unramified frobenius element exists uniquely: For finite Galois L/K and a nonzero prime $P ∣ p$ with $e (P / p) = 1$ , there is a unique ${Frob}_{P} \in D (P / p)$ satisfying ${Frob}_{P} (a) \equiv a^{N p} (m o d P) (a \in O_{L}) .$ It is the arithmetic Frobenius element, the unique lift of the arithmetic Frobenius coset.

[F3]

Ramification and residue degrees in towers: For $M / L / K$ and $Q ∣ P ∣ p$ , $e (Q / p) = e (Q / P) e (P / p), f (Q / p) = f (Q / P) f (P / p) .$

Proof

1.1F1F2F3

Multiplicativity of e and positivity give e=1 for both steps. Restriction of $σ = Frob (Q / p)$ belongs to D(P/p), and its congruences on $O_{L}$ reduce to the $q = N p$ power map at P. Uniqueness gives the first formula.

2.1F2step 1.1

Set $f = f (P / p)$ . The residue power map on $κ (P)$ has order f: this follows directly since $κ (P)$ has $q^{f}$ elements, while for $0 < j < f$ the polynomial $T^{q^{j}} - T$ has fewer roots than that field. Since the unramified residue action identifies D(P/p) with its image, the restriction of $σ^{f}$ is identity. Thus $σ^{f} \in D (Q / P)$ and acts on $κ (Q)$ as the $q^{f} = N P$ power map. The relative unramified uniqueness gives the second formula.

3.1step 1.1∎

An automorphism of a compositum is determined by its restrictions to the generating fields: if both restrictions are identity it fixes every field expression in those generators. Apply the first formula to each $L_{i}$ to obtain the stated determining pair. This assertion assumes the compositum prime is unramified.

Depends on

Used by

Decomposition groups in a tower Example

Dependency tree · two levels

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Sources

Chapter 8, Propositions 8.15–8.17, p.142 (standard reference, not scraped)