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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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F-embeddings of F(α) into an algebraically closed field correspond to the distinct roots of mα

Statement

Let α be algebraic over F, and let Ω be an algebraically closed field containing F. Sending an F-embedding σ:F(α)→Ω to σ(α) is a bijection from the set of such embeddings to the set of distinct roots in Ω of the minimal polynomial mα. Consequently the number of embeddings is the number of distinct roots of mα, not the sum of their multiplicities.

Facts & Assumptions

Given: An algebraic element α over F, its minimal polynomial mα, and an algebraically closed overfield Ω of F.

[L1]

An F-embedding carries an algebraic element to a conjugate root of its minimal polynomial (A base-field embedding carries an algebraic element to a conjugate).

[L2]

For a monic irreducible polynomial, every chosen root in an extension induces a unique homomorphism from the quotient adjoining that root (Universal property of adjoining a root of an irreducible polynomial).

[L3]

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

Proof

technique · direct
1.1L1

By [L1], the image σ(α) of every F-embedding is a root of mα in Ω.

1.2L2

Conversely, if β∈Ω is a root of mα, [L2] applied to the two realizations of F[x]/(mα) gives a unique F-embedding F(α)→Ω with α↦β.

2.1step 1.1step 1.2

The constructions in steps 1.1 and 1.2 are inverse because an F-homomorphism on F(α) is determined by the image of α.

3.1step 2.1L3∎

The polynomial mα splits in Ω by repeated use of [L3], and the bijection indexes embeddings by its distinct roots, so repeated roots are counted once.

Depends on

Used by

Dependency tree · two levels

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Sources