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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
1.1

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

L1
1.2

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

L2
2.1

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.

step 1.1step 1.2
3.1

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.

step 2.1L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 31 results over 7 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