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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17
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Restriction partitions embeddings in a finite tower into extension fibres

Statement

Let F⊆K⊆L be a finite tower and let Ω be an algebraic closure of F. Restriction defines a surjection

Hom⁡F(L,Ω)⟶Hom⁡F(K,Ω).

For every F-embedding σ:K→Ω, its fibre is nonempty and has cardinality [L:K]s after transporting the K-structure along σ.

Facts & Assumptions

Given: A finite tower F⊆K⊆L, an algebraic closure Ω/F, and an F-embedding σ:K→Ω.

[L1]

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

[L2]

A finite extension has a finite basis over its base (The degree [K:F]=dim⁡FK of a finite field extension).

[L4]

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

[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).

[L6]

For a finite extension, the number of base-field embeddings into an algebraic closure is independent of the chosen algebraic closure (The separable degree is independent of the chosen algebraic closure).

Proof

technique · direct
1.1L1

Restricting an F-embedding L→Ω to K gives an F-embedding by [L1].

1.2L2L3L4L5construct

To extend a chosen σ, take a finite K-basis a1,…,ar of L by [L2] and put Ki=K(a1,…,ai). Starting with τ0=σ, regard each τi−1 as an isomorphism onto its image, transport the minimal polynomial of ai over Ki−1 along it by [L3], choose a root in Ω by [L5], and extend τi−1 to Ki by [L4]. After finitely many steps, Kr=L, so every σ has an extension and the restriction map is surjective.

1.3L1algebra

Identify K with σ(K). The extensions of σ are exactly the σ(K)-embeddings of the scalar-transported copy of L into Ω. Since Ω is algebraically closed and algebraic over σ(K), it is an algebraic closure of that copy of K.

2.1step 1.3L6algebra

Transporting scalars and maps along the isomorphism K→σ(K) identifies the embeddings in step 1.3 with embeddings of L/K into an algebraic closure of K. By [L6], their number is the closure-independent value [L:K]s. Thus every fibre of restriction has that cardinality.

3.1step 2.1∎

In particular, transport along an isomorphism between two embedded copies of K gives a bijection between their restriction fibres.

Depends on

Used by

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Sources