TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.

A monic irreducible separable cubic in characteristic not two has Galois group $A_{3}$ or $S_{3}$ according to its discriminant

Statement

Let $f \in F [x]$ be a monic irreducible separable cubic over a field of characteristic not two. Then its Galois group is $A_{3}$ when its discriminant is a square and $S_{3}$ when its discriminant is not a square.

Facts & Assumptions

Given: The orbit-stabilizer cardinality formula (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ), Lagrange's divisibility theorem (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ), and the alternating group of The alternating group $A_{n} = \ker (sgn)$ of even permutations.

[L1]

A positive-degree separable polynomial is irreducible if and only if its Galois group acts transitively on its roots (A positive-degree separable polynomial is irreducible exactly when its Galois group is transitive on the roots).

[L2]

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

Proof

technique · direct

1.1

By [L1], the Galois group $G \leq S_{3}$ acts transitively on three roots. Orbit-stabilizer makes $3$ divide $∣ G ∣$ , while Lagrange makes $∣ G ∣$ divide $6$ ; hence $∣ G ∣$ is $3$ or $6$ . In the first case every nonidentity element is a three-cycle and $G = A_{3}$ , while in the second $G = S_{3}$ .

L1givenalgebra

2.1

If the discriminant is a square, [L2] gives $G \leq A_{3}$ , so step 1.1 forces $G = A_{3}$ . If it is not a square, [L2] gives $G ≰ A_{3}$ , so step 1.1 forces $G = S_{3}$ . The two square classes are exhaustive, and separability remains an explicit hypothesis in characteristic three.

step 1.1L2∎

Depends on

Used by

Dependency tree · two levels

26 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, Example 4.7 (standard reference, not scraped)
K. Conrad, Galois Groups of Cubics and Quartics, Theorem 2.1 (standard reference, not scraped)

A monic irreducible separable cubic in characteristic not two has Galois group A3A3​ or S3S3​ according to its discriminant