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

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.

For $F_{p} (t) / F_{p} (t^{p})$ , the trace is identically zero

Example

Let $F = F_{p} (t^{p})$ and $K = F_{p} (t)$ . Then $K / F$ is purely inseparable of degree $p$ , its only $F$ -embedding into an algebraic closure is the inclusion, and

${Tr}_{K / F} (x) = 0 for every x \in K .$

In particular,

$N_{K / F} (t) = t^{p}, {Tr}_{K / F} (t) = 0.$

Facts & Assumptions

Given: The fields $F = F_{p} (t^{p})$ and $K = F_{p} (t)$ .

[L1]

Rational function fields are fraction fields of polynomial rings (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ ).

[L2]

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

[L3]

A simple algebraic extension has degree equal to the degree of the minimal polynomial of its generator (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L4]

In a finite extension, norm and trace are given by the embedding formulas with the inseparable exponent, and the trace vanishes when that exponent is greater than one (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

Verification

technique · direct

1.1

If $t \in F$ , then one could write $t = u (t^{p}) / v (t^{p})$ with coprime polynomials $u, v \in F_{p} [s]$ . Clearing denominators gives the polynomial identity $t v (t^{p}) - u (t^{p}) = 0.$ But the first term has degree congruent to $1$ modulo $p$ , while every term of $u (t^{p})$ has degree divisible by $p$ , impossible. Hence $t \notin F$ .

L1algebra

2.1

Since $t^{p} \in F$ and $t \notin F$ , the polynomial $x^{p} - t^{p}$ is irreducible over $F$ by [L2]. Therefore $K = F (t)$ has degree $p$ over $F$ by [L3], and it is purely inseparable because $t^{p} \in F$ . The only $F$ -embedding of $K$ into an algebraic closure is the inclusion.

step 1.1L2L3algebra

3.1

Now [L4] gives $N_{K / F} (t) = t^{p}, {Tr}_{K / F} (t) = p t = 0,$ and the same embedding formula shows ${Tr}_{K / F} (x) = 0$ for every $x \in K$ .

step 2.1L4algebra∎

Depends on

Used by

FALSE: for every finite extension, the norm is just the product over the embeddings False statement
FALSE: the trace map of every finite extension is surjective False statement

Dependency tree · two levels

27 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

B. Conrad, Norm and trace, Section 2 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Remark 5.47 (standard reference, not scraped)

For Fp(t)/Fp(tp)Fp​(t)/Fp​(tp), the trace is identically zero