TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For a monic separable polynomial in characteristic not two, the Galois group lies in $A_{n}$ exactly when the discriminant is a square

Statement

Let $f \in F [x]$ be a monic separable polynomial of degree $n$ , where $char F \neq 2$ . The Galois group lies in $A_{n}$ exactly when the discriminant is a square in the base field.

Facts & Assumptions

Given: A splitting field $L / F$ , an ordered root list, the discriminant definition of The discriminant of a monic polynomial as the coefficient expression of $Δ_{n}^{2}$ , the root formula $Disc (f) = δ^{2}$ and the fact that separability makes it nonzero (The discriminant is $\prod_{i < j} (α_{i} - α_{j})^{2}$ and vanishes exactly when a monic polynomial has a repeated root), the definition $A_{n} = \ker (sgn)$ (The alternating group $A_{n} = \ker (sgn)$ of even permutations), and the finite Galois correspondence, which gives $L^{Gal (L / F)} = F$ (The fundamental theorem of finite Galois theory).

[L1]

For the Vandermonde product, $σ (δ) = sgn (σ) δ$ for every Galois automorphism (The Vandermonde product transforms by the sign of the root permutation).

Proof

technique · direct

1.1

For the forward direction, suppose the Galois group lies in $A_{n}$ . Then every sign is $1$ , so [L1] shows that every automorphism fixes $δ$ . The fixed field is $F$ , hence $δ \in F$ and $Disc (f) = δ^{2}$ is a square in $F$ . This also covers $n = 0$ and $n = 1$ , when $δ = 1$ .

L1given

2.1

For the reverse direction, suppose $Disc (f) = d^{2}$ for some $d \in F$ . Since $δ^{2} = d^{2}$ , the field law gives $δ = d$ or $δ = - d$ , so $δ \in F$ . Thus every automorphism fixes $δ$ , and [L1] gives $sgn (σ) δ = δ$ . Separability gives $δ \neq 0$ , so cancellation and $1 \neq - 1$ in characteristic not two force $sgn (σ) = 1$ . Therefore every Galois permutation lies in $A_{n}$ .

L1givenalgebra∎

Remarks

The characteristic hypothesis is essential to this argument: in characteristic two the two signs have the same scalar action, so the Vandermonde equation cannot detect parity.

Depends on

Used by

Dependency tree · two levels

25 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

J. S. Milne, Fields and Galois Theory, v5.10, Corollary 4.2 (standard reference, not scraped)
K. Conrad, Galois Groups of Cubics and Quartics, Theorem 1.3 (standard reference, not scraped)

For a monic separable polynomial in characteristic not two, the Galois group lies in AnAn​ exactly when the discriminant is a square