TheoremStatement: Literature-sourcedProof: 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.

The trace form of a finite extension is nondegenerate exactly when the extension is separable

Statement

Let $K / F$ be a finite field extension and let

$T_{K / F} (x, y) = {Tr}_{K / F} (x y)$

be its trace form (The trace form $(x, y) \mapsto {Tr}_{K / F} (x y)$ of a finite extension). Then $T_{K / F}$ is nondegenerate (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space) if and only if $K / F$ is separable (Separable algebraic elements and separable extensions).

Facts & Assumptions

Given: A finite extension $K / F$ , its trace form $T_{K / F}$ , and an $F$ -basis $e_{1}, \dots, e_{n}$ of $K$ .

[F1]

The trace form is the symmetric bilinear form $(x, y) \mapsto {Tr}_{K / F} (x y)$ (The trace form $(x, y) \mapsto {Tr}_{K / F} (x y)$ of a finite extension).

[F2]

A bilinear form on a finite-dimensional space is nondegenerate exactly when its matrix in one, hence every, basis is invertible (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[L1]

The trace is the sum of the conjugates in the separable case and is identically zero in the inseparable case (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

[L2]

Distinct embeddings are linearly independent, so their evaluation matrix on a suitable basis is invertible (Dedekind's linear independence theorem for distinct characters).

Proof

technique · direct

1.1F1L1

For the forward implication from inseparability to degeneracy, suppose $K / F$ is not separable. Then [L1] makes ${Tr}_{K / F}$ identically zero, so $T_{K / F} (x, y) = {Tr}_{K / F} (x y) = 0$ for every $x, y \in K$ . Thus every vector lies in both radicals, and the form is degenerate.

1.2L2choose

For the converse direction, suppose $K / F$ is separable and let $σ_{1}, \dots, σ_{n} ⁣ : K \to Ω$ be its distinct $F$ -embeddings into an algebraic closure. Form the evaluation matrix $A = (σ_{i} (e_{j}))_{i, j}$ . By [L2], $A$ is invertible.

2.1F1L1step 1.2algebra

The matrix of $T_{K / F}$ in the basis $e_{1}, \dots, e_{n}$ is $B = ({Tr}_{K / F} (e_{i} e_{j}))_{i, j}$ . Because $K / F$ is separable, [L1] gives ${Tr}_{K / F} (e_{i} e_{j}) = \sum_{r = 1}^{n} σ_{r} (e_{i}) σ_{r} (e_{j}),$ so $B = A^{T} A .$

3.1F2step 2.1algebra

Since $A$ is invertible, the matrix $B = A^{T} A$ is invertible. Therefore [F2] makes the trace form nondegenerate.

4.1step 1.1step 3.1∎

Steps 1.1 and 3.1 prove the equivalence.

Remarks

The inseparable case is not a small defect but a total collapse. The trace itself vanishes, so the whole bilinear form vanishes.

The trace form of a finite extension is nondegenerate exactly when the extension is separable

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources