Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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 Fp(t)/Fp(tp), the trace is identically zero

Example

Let F=Fp(tp) and K=Fp(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)=0for every x∈K.

In particular,

NK/F(t)=tp,Tr⁡K/F(t)=0.

Facts & Assumptions

Given: The fields F=Fp(tp) and K=Fp(t).

[L2]
[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,…,an−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
1.1L1algebra

If t∈F, then one could write t=u(tp)/v(tp) with coprime polynomials u,v∈Fp[s]. Clearing denominators gives the polynomial identity t v(tp)−u(tp)=0. But the first term has degree congruent to 1 modulo p, while every term of u(tp) has degree divisible by p, impossible. Hence t∉F.

2.1step 1.1L2L3algebra

Since tp∈F and t∉F, the polynomial xp−tp is irreducible over F by [L2]. Therefore K=F(t) has degree p over F by [L3], and it is purely inseparable because tp∈F. The only F-embedding of K into an algebraic closure is the inclusion.

3.1step 2.1L4algebra∎

Now [L4] gives NK/F(t)=tp,Tr⁡K/F(t)=p t=0, and the same embedding formula shows Tr⁡K/F(x)=0 for every x∈K.

Depends on

Used by

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