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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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The normal closure of a finite extension exists and is finite

Statement

Let K/F be a finite extension embedded in an algebraic closure Ω of F. Its normal closure in Ω is a finite extension of F. If K=F(α1,,αr), it is the splitting field in Ω of the product of the minimal polynomials of the αi.

Facts & Assumptions

Given: A finite extension FKΩ with Ω/F an algebraic closure.

[L1]

The normal closure is the intersection of the normal intermediate extensions containing K (The normal closure of an algebraic extension inside a fixed algebraic closure).

[L2]

A finite family of nonzero polynomials has a splitting field (Every finite family of nonzero polynomials has a splitting field, obtained from their product).

[L3]
[L4]

A field generated by finitely many algebraic elements is finite over the base (An extension generated by finitely many algebraic elements is finite).

[L5]

A finite extension is finite-dimensional over its base (The degree [K:F]=dimFK of a finite field extension).

Proof

technique · direct
1.1

Choose a finite F-basis of K using [L5]; it is also a finite generating family α1,,αr. Let fi be the minimal polynomial of αi over F, and inside Ω let E be the field generated by all roots of f1fr.

L2L5choose
2.1

The field E is generated by finitely many algebraic roots, so [L4] makes E/F finite. It is a splitting field of the product and is normal by [L3], and it contains every αi, hence K.

step 1.1L3L4
2.2

If H/F is any normal intermediate extension in Ω containing K, then each fi, having the root αiH, splits in H. Thus H contains all generators of E and EH.

step 1.1algebra
3.1

Therefore E is contained in every field intersected in [L1], while step 2.1 makes E one of those fields. It equals the normal closure, which is consequently finite.

step 2.1step 2.2L1

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