ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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.

$x^{p} - t$ is irreducible and inseparable over $F_{p} (t)$

Example

Over the rational function field $F_{p} (t)$ , the polynomial $x^{p} - t$ is irreducible and inseparable. In a field containing a $p$ th root $t^{1 / p}$ , it equals $(x - t^{1 / p})^{p}$ .

Facts & Assumptions

Given: A prime $p$ and the rational function field $F_{p} (t)$ .

[L1]

The rational function field is the fraction field of $F_{p} [t]$ (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ ).

[L2]

A polynomial ring over a field is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

[L3]

If a constant is not a $p$ th power, then $x^{p} - a$ is irreducible in characteristic $p$ (If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$ ).

[L4]

A nonzero polynomial is separable exactly when it is coprime to its derivative (A nonzero polynomial over a field is separable exactly when its gcd with its derivative is $1$ ).

Verification

technique · direct

1.1L1L2algebra

Suppose $t = (u / v)^{p}$ with coprime nonzero $u, v \in F_{p} [t]$ , using [L1]. Then $u^{p} = t v^{p}$ . In the UFD of [L2], the exponent of the irreducible factor $t$ on the left is divisible by $p$ , while on the right it is congruent to $1$ modulo $p$ , a contradiction. Thus $t$ is not a $p$ th power in $F_{p} (t)$ .

2.1step 1.1L3

By [L3], $x^{p} - t$ is irreducible.

3.1L4algebra∎

Its derivative is zero, so [L4] makes it inseparable. In an extension containing $t^{1 / p}$ , the characteristic- $p$ binomial identity gives $x^{p} - t = (x - t^{1 / p})^{p}$ .

Depends on

Used by

FALSE: every irreducible polynomial over a field is separable 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

P. L. Clark, Field Theory, Chapter 4 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Chapters 3 and 5 (standard reference, not scraped)

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