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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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.

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$F_{p} (t) / F_{p} (t^{p})$ is normal and inseparable with trivial automorphism group

Statement refuted

The assertion that every finite normal extension is separable, or that every nontrivial finite normal extension has a nontrivial relative automorphism, is false. For every prime $p$ , $F_{p} (t) / F_{p} (t^{p})$ is normal and inseparable of degree $p$ with trivial automorphism group.

Facts & Assumptions

Given: A prime $p$ , a transcendental element $t$ , the purely inseparable extension definition (Purely inseparable algebraic extensions), and the criterion that $x^{p} - u$ is irreducible when $u$ is not a $p$ -th power (If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$ ).

[L1]

Every purely inseparable algebraic extension is normal (Every purely inseparable algebraic extension is normal).

Counterexample

technique · direct

1.1

Put $u = t^{p}$ . The element $u$ is not a $p$ -th power in the rational-function field $F_{p} (u)$ , as the valuation at the prime $u$ of a $p$ -th power is divisible by $p$ . Hence $x^{p} - u$ is irreducible over $F_{p} (u) = F_{p} (t^{p})$ , while in $F_{p} (t) [x]$ it equals $(x - t)^{p}$ .

givenalgebra

2.1

Step 1.1 gives degree $p$ and shows that every element of the extension has a power in the base field, so the extension is purely inseparable and not separable. By [L1] it is normal. This includes the smallest prime $p = 2$ .

step 1.1L1

3.1

A base-field automorphism must send $t$ to another root of its minimal polynomial, but step 1.1 shows that $t$ is the unique root. Thus every such automorphism fixes $t$ and is the identity on $F_{p} (t)$ .

step 1.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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 3.8 (standard reference, not scraped)
K. Conrad, The Galois Correspondence, positive-characteristic examples (standard reference, not scraped)

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