CounterexampleConstruction: 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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$Q (\sqrt[3]{2}) / Q$ is separable and nonnormal with trivial automorphism group

Statement refuted

The assertion that every finite separable extension with trivial relative automorphism group is normal is false. In fact, $Q (\sqrt[3]{2}) / Q$ has degree three and trivial automorphism group, but it is separable and not normal.

Facts & Assumptions

Given: The real cube root $a = \sqrt[3]{2}$ ; Eisenstein's irreducibility criterion (Eisenstein criterion over the integers); characteristic-zero fields are perfect (Fields of characteristic zero, finite fields, and algebraically closed fields are perfect); and the definitions of normal extension and relative automorphism (A normal algebraic extension is one in which every minimal polynomial with a root in the extension splits there, Relative field automorphisms and $Aut (K / F)$ ).

[L1]

For a simple algebraic extension, embeddings into an algebraically closed field correspond bijectively to the distinct roots of the generator's minimal polynomial ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ).

Counterexample

technique · direct

1.1

Eisenstein at $2$ makes $x^{3} - 2$ irreducible, so $[Q (a) : Q] = 3$ . Characteristic zero makes the polynomial separable, and $a \neq 0$ because $a^{3} = 2$ .

given

2.1

The other roots are $ω a$ and $ω^{2} a$ for nonreal cube roots of unity $ω, ω^{2}$ , whereas $Q (a) \subseteq R$ . Thus the minimal polynomial does not split in $Q (a)$ , so the extension is not normal.

step 1.1algebra

3.1

By [L1], a $Q$ -automorphism must send $a$ to a root of $x^{3} - 2$ that lies in $Q (a)$ . Step 2.1 leaves only $a$ , and fixing the generator fixes all of $Q (a)$ . Hence the automorphism group has exactly its identity element.

step 2.1L1∎

Depends on

Used by

FALSE: every degree-n extension has exactly n automorphisms False statement

Dependency tree · two levels

20 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 3.3 (standard reference, not scraped)
K. Conrad, The Galois Correspondence, introductory examples (standard reference, not scraped)

Q(23)/QQ(32​)/Q is separable and nonnormal with trivial automorphism group