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

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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.
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FALSE: the degree of a polynomial determines its Galois group

Statement

False claim. Any two separable irreducible polynomials over one field having the same degree have isomorphic Galois groups.

Facts & Assumptions

Given: Two explicit irreducible cubics over $Q$ .

[L1]

$x^{3} - 3 x + 1$ has Galois group $A_{3}$ over $Q$ ( $x^{3} - 3 x + 1$ has discriminant $81$ and Galois group $A_{3}$ over $Q$ ).

[L2]

$x^{3} - 2$ has Galois group $S_{3}$ over $Q$ ( $x^{3} - 2$ has discriminant $- 108$ and Galois group $S_{3}$ over $Q$ ).

Refutation

technique · direct

1.1

The polynomials in [L1] and [L2] both have degree three and are separable and irreducible, but their Galois groups have orders $3$ and $6$ .

L1L2

2.1

Groups of different finite orders are not isomorphic, so the common polynomial degree does not determine the Galois group.

step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

K. Conrad, Galois Groups of Cubics and Quartics, Section 2 (standard reference, not scraped)