Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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A base-field embedding carries an algebraic element to a conjugate

Statement

Let σ:KL be an F-embedding and let αK be algebraic over F. Then σ(α) is conjugate to α over F. In particular, an F-endomorphism of a splitting field permutes the distinct roots of every base polynomial that splits there.

Facts & Assumptions

Given: An F-embedding σ:KL and an element αK algebraic over F.

[L1]

A field isomorphism transports polynomial evaluation and carries roots to roots after applying the induced coefficient map (A field isomorphism transports polynomials coefficientwise and carries roots, factorizations, and splitting to roots, factorizations, and splitting).

[L2]

An endomorphism of a splitting field fixing the base permutes the finite set of distinct roots of the defining polynomial (Every F-endomorphism of a splitting field permutes the distinct roots and is an automorphism).

[L3]

Conjugate elements are the roots of the same minimal polynomial over the base (Conjugate algebraic elements over a field).

Proof

technique · direct
1.1

Regard σ as an isomorphism Kσ(K)L. Let mαF[x] be the minimal polynomial of α. Since σ fixes F, [L1] gives mα(σ(α))=σ(mα(α))=0.

L1
2.1

Thus σ(α) is a root of mα and is conjugate to α by [L3].

step 1.1L3
3.1

When K=L is a splitting field, [L2] strengthens this root preservation to a permutation of the distinct roots.

L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources