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

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The trace map of a finite separable extension is surjective

Statement

If $K / F$ is a finite separable field extension, then the trace map

${Tr}_{K / F} ⁣ : K \to F$

is surjective.

Facts & Assumptions

Given: A finite separable extension $K / F$ .

[L1]

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

Proof

technique · direct

1.1

By [L1], the trace form of $K / F$ is nondegenerate. If the trace map itself were zero, then ${Tr}_{K / F} (x y) = 0$ for every $x, y \in K$ , so the vector $1 \in K$ would lie in the radical, contradicting nondegeneracy. Therefore the trace map is not the zero linear functional.

L1algebra

2.1

The image of an $F$ -linear map $K \to F$ is an $F$ -subspace of the one-dimensional vector space $F$ . A nonzero subspace of $F$ is all of $F$ , so the trace map is surjective.

step 1.1algebra∎

Depends on

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

Used by

Additive Hilbert 90: trace zero is the image of α↦α-σ(α) Theorem

Dependency tree · two levels

6 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, Theorem 2.5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Theorem 5.47 (standard reference, not scraped)