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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck 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→F

is surjective.

Facts & Assumptions

Given: A finite separable extension K/F.

[L1]

The trace form (x,y)↦Tr⁡K/F(xy) 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
1.1L1algebra

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

2.1step 1.1algebra∎

The image of an F-linear map K→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.

Depends on

Used by

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