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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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Assuming Choice, a base-field embedding extends across every algebraic extension

Statement

Assume the Axiom of Choice. Let K/F be algebraic, let Ω be algebraically closed, and let σ:FΩ be a field embedding. Then σ extends to a field embedding σ~:KΩ. The proof uses Zorn's lemma.

Facts & Assumptions

Given: The Axiom of Choice, an algebraic extension K/F, an algebraically closed field Ω, and an embedding σ:FΩ.

[L1]

Assuming Choice, a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).

[L2]

A chosen root of an irreducible polynomial induces the unique embedding of the corresponding simple root extension (Universal property of adjoining a root of an irreducible polynomial).

[L4]

Relative embeddings are field embeddings agreeing with the specified map on the base (F-homomorphisms and F-embeddings of field extensions).

[L5]

Every nonconstant polynomial over an algebraically closed field has a root (An algebraically closed field: every nonconstant polynomial has a root in the field).

Proof

technique · direct
1.1

Let P consist of pairs (E,τ) where FEK is an intermediate field and τ:EΩ is an embedding extending σ, ordered by extension of the domain and map. The pair (F,σ) shows that P is nonempty.

L4construct
2.1

The union of a chain of such pairs has the union of the domains as an intermediate field and the union of the compatible maps as an embedding into Ω, so every chain has an upper bound.

step 1.1L4
3.1

By Zorn's lemma [L1], choose a maximal pair (M,τ) in P.

step 1.1step 2.1L1choose
4.1

If MK, choose αKM. It is algebraic over M. Transport its minimal polynomial through the isomorphism Mτ(M) using [L3], choose a root in Ω by [L5], and use [L2] to extend τ to an embedding of M(α).

step 3.1L2L3L5choose
5.1

Step 4.1 contradicts maximality unless M=K. Thus τ on the maximal domain is the required extension of σ; the use of Choice is precisely [L1].

step 3.1step 4.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 6 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