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

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Lifting residue frobenius by galois conjugates

Statement

Let L/K be finite Galois and $P ∣ p$ nonzero primes. Put $q = ∣ κ (p) ∣$ . Some $σ \in D (P / p)$ induces the arithmetic power map $x \mapsto x^{q}$ on $κ (P)$ .

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Inertia group of a prime: For finite Galois L/K and a chosen nonzero prime $P ∣ p$ , set $κ (P) = O_{L} / P$ and $κ (p) = O_{K} / p$ . Each $σ \in D (P / p)$ induces a $κ (p)$ -automorphism $\overset{ˉ}{σ}$ of $κ (P)$ : the rule $\overset{ˉ}{a} \mapsto \overline{σ a}$ is independent of the representative because $σ P = P$ . The inertia group is $I (P / p) = \ker ⁣ (D (P / p) ⟶ Gal (κ (P) / κ (p))) .$ Equivalently, $σ \in I (P / p)$ exactly when $σ \in D (P / p)$ and $σ (a) - a \in P$ for every $a \in O_{L}$ . It is a normal subgroup of D.

[F2]

Chinese remainder theorem for pairwise comaximal ideals: Let $R$ be a commutative ring and let $I_{1}, \dots, I_{r}$ be pairwise comaximal ideals, where $r \geq 1$ . Then the canonical map $R ⟶ \prod_{i = 1}^{r} R / I_{i}, x ⟼ (x + I_{1}, \dots, x + I_{r})$ is surjective, its kernel is $⋂_{i = 1}^{r} I_{i}$ , and $⋂_{i = 1}^{r} I_{i} = \prod_{i = 1}^{r} I_{i} .$ Equivalently, $R / \prod_{i = 1}^{r} I_{i} ≅ \prod_{i = 1}^{r} R / I_{i} .$

[F3]

The multiplicative group $F_{q}^{\times}$ of a finite field is cyclic: The multiplicative group $F^{\times} = F ∖ {0}$ of every finite field $F$ is cyclic.

Proof

1.1F2F3

Choose a generator u of the finite cyclic group $κ (P)^{\times}$ . CRT gives $α \in O_{L}$ reducing to u at P and to zero at every other prime above p. This also works when $κ (P) = F_{2}$ , with u=1, or when P is the only prime.

2.1step 1.1

The orbit polynomial $H (T) = \prod_{τ \in G} (T - τ α)$ has integral G-invariant coefficients, hence belongs to $O_{K} [T]$ . Its reduction has coefficients in $κ (p)$ , so $H (u^{q}) = H (u)^{q} = 0$ in $κ (P)$ . Its displayed linear factorization implies $u^{q} = \overline{σ α}$ for some $σ \in G$ .

3.1F1step 1.1step 2.1∎

If $σ^{- 1} P \neq P$ , alpha is zero at $σ^{- 1} P$ , giving $\overline{σ α} = 0$ , contrary to $u^{q} \neq 0$ . Thus $σ \in D$ . On every nonzero residue $u^{j}$ it acts by $(u^{q})^{j} = (u^{j})^{q}$ ; it fixes zero as well. This is the asserted residue action.

Depends on

Used by

Decomposition inertia exact sequence Theorem

Dependency tree · two levels

15 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

Chapter 8, Frobenius element, footnote 1 on p.141; Stein Theorem 9.3.5 (standard reference, not scraped)