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

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Good polynomial reduction kills inertia

Statement

Let $F \in Z [T]$ be monic separable with splitting field L. If a rational prime p does not divide $Disc (F)$ , the integral roots of F have distinct reductions at every $P ∣ p$ . The inertia group I(P/p) is trivial, so p is unramified in L.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

The discriminant of a monic polynomial as the coefficient expression of $Δ_{n}^{2}$ : By prop-vandermonde-square-is-symmetric and thm-fundamental-theorem-of-symmetric-polynomials, there is a unique polynomial $D_{n} \in Z [T_{1}, \dots, T_{n}]$ such that $Δ_{n} (x_{1}, \dots, x_{n})^{2} = D_{n} (e_{1}, \dots, e_{n}) .$ For a monic polynomial $f (t) = t^{n} + a_{1} t^{n - 1} + \dots + a_{n}$ over a commutative ring, its discriminant is $Disc (f) : = D_{n} (- a_{1}, a_{2}, \dots, (- 1)^{n} a_{n}) .$ Equivalently, in any algebra in which $f$ splits with roots $α_{1}, \dots, α_{n}$ , this coefficient expression evaluates to $Δ_{n} (α_{1}, \dots, α_{n})^{2}$ . The definition therefore depends only on the coefficients and not on a choice or ordering of roots. For a monic constant polynomial, $Disc (1) = 1$ .

[F2]

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.

[F3]

Orders of decomposition and inertia groups: For finite Galois L/K and nonzero $P ∣ p$ , writing e and f for its ramification index and residue degree, $∣ D (P / p) ∣ = e f, ∣ I (P / p) ∣ = e, ∣ D (P / p) / I (P / p) ∣ = f .$ The prime P is unramified over p if and only if its inertia group is trivial.

Proof

1.1

List the distinct roots $α_{1}, \dots, α_{n}$ in L. They are integral because F is monic. The discriminant is $\prod_{i < j} (α_{i} - α_{j})^{2}$ . Its integer value is not in P, since its contraction is (p). Therefore no difference is in P, giving distinct reductions. The constant and linear cases have an empty product equal to one.

2.1

Every inertia element permutes the roots and fixes each of their residue classes. Distinctness of those classes forces it to fix each root itself. Since these roots generate L over the rationals, the element is identity. Thus I=1 and the inertia order formula gives e=1, at every prime above p.

F2F3step 1.1∎

Depends on

Used by

Frobenius cycle type and prime splitting Theorem

Dependency tree · two levels

11 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, Proposition 8.21 and Theorem 8.23 proof, pp.144–145 (standard reference, not scraped)