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Assuming Choice, conjugates in an algebraic closure are related by a base automorphism

Statement

Assume the Axiom of Choice. Let $Ω$ be an algebraic closure of $F$ and let $α, β \in Ω$ . Then $α$ and $β$ are conjugate over $F$ if and only if some $F$ -automorphism of $Ω$ sends $α$ to $β$ .

Facts & Assumptions

Given: The Axiom of Choice, an algebraic closure $Ω / F$ , and elements $α, β \in Ω$ .

[L1]

An $F$ -embedding carries an algebraic element to a conjugate (A base-field embedding carries an algebraic element to a conjugate).

[L2]

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

[L3]

A chosen conjugate root induces an $F$ -isomorphism between the corresponding simple extensions (Universal property of adjoining a root of an irreducible polynomial).

[L4]

Assuming Choice, a base embedding extends across an algebraic extension into an algebraically closed field (Assuming Choice, a base-field embedding extends across every algebraic extension).

[L5]

An algebraic closure is an algebraic extension whose top field is algebraically closed (An algebraic closure of a field).

[L6]

Every algebraic element has a monic irreducible minimal polynomial over the base (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Proof

technique · direct

1.1L1

If an $F$ -automorphism sends $α$ to $β$ , then [L1] makes them conjugate.

1.2L2L3

Conversely, suppose they are conjugate. By [L2] and [L3], there is an $F$ -isomorphism $F (α) \to F (β)$ sending $α$ to $β$ .

2.1step 1.2L4L5L6

Apply [L4] with base $F (α)$ to extend this map to an embedding $τ : Ω \to Ω$ . Its image $E$ is algebraically closed because it is isomorphic to $Ω$ . Every $b \in Ω$ is algebraic over $F \subseteq E$ by [L5], so [L6] gives its minimal polynomial over $E$ ; this polynomial has a root in $E$ and is therefore linear. Hence $b \in E$ , so $τ$ is surjective and is an automorphism.

3.1step 1.2step 2.1∎

This automorphism sends $α$ to $β$ , proving the reverse implication.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

J. S. Milne, Fields and Galois Theory, Chapter 6 (standard reference, not scraped)
P. L. Clark, Field Theory, Chapters 3 to 5 (standard reference, not scraped)