ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Over $Q (ω)$ , the splitting field of $x^{3} - 2$ is a cyclic cubic extension

Example

Let $ω$ be a primitive cube root of unity and put $F = Q (ω)$ . Then $x^{3} - 2$ is irreducible over $F$ , and its splitting field is

$F (\sqrt[3]{2}),$

which is a cyclic extension of degree $3$ over $F$ .

Facts & Assumptions

Given: The field $F = Q (ω)$ and the polynomial $x^{3} - 2$ .

[L1]

Norm and trace are given by the embedding formulas (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

[L2]

Over a base containing $μ_{3}$ , a degree-three extension is cyclic exactly when it is generated by a root of an irreducible cubic $x^{3} - a$ (If $μ_{n} \subseteq F$ and $char F ∤ n$ , then a degree- $n$ extension is cyclic exactly when it is $F (α)$ with $α^{n} \in F$ and $x^{n} - α^{n}$ irreducible).

Verification

technique · direct

1.1

The field $F / Q$ has degree $2$ . If $2 = β^{3}$ for some $β \in F$ , then taking norms from $F$ to $Q$ gives $4 = N_{F / Q} (2) = N_{F / Q} (β)^{3},$ which is impossible because no rational cube equals $4$ . So $2$ is not a cube in $F$ . Hence $x^{3} - 2$ has no root in $F$ and is irreducible there.

L1algebra

2.1

Since $ω \in F$ , all roots of $x^{3} - 2$ are $\sqrt[3]{2}, ω \sqrt[3]{2}, ω^{2} \sqrt[3]{2},$ so they all lie in $F (\sqrt[3]{2})$ . The irreducibility from step 1.1 and [L2] therefore make $F (\sqrt[3]{2}) / F$ cyclic of degree $3$ .

L2step 1.1∎

Depends on

Used by

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Sources

J. S. Milne, Fields and Galois Theory, v5.10, Proposition 5.27 (standard reference, not scraped)
S. R. Ghorpade, Lectures on Field Theory and Ramification Theory, Section 1.3 (standard reference, not scraped)

Over Q(ω)Q(ω), the splitting field of x3−2x3−2 is a cyclic cubic extension