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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 $K = F (α_{1}, \dots, α_{r})$ , it is the splitting field in $Ω$ of the product of the minimal polynomials of the $α_{i}$ .

Facts & Assumptions

Given: A finite extension $F \subseteq K \subseteq Ω$ 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]

An algebraic splitting extension is normal (An algebraic extension that is a splitting field of a polynomial is normal).

[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 $[K : F] = \dim_{F} K$ of a finite field extension).

Proof

technique · direct

1.1L2L5choose

Choose a finite $F$ -basis of $K$ using [L5]; it is also a finite generating family $α_{1}, \dots, α_{r}$ . Let $f_{i}$ be the minimal polynomial of $α_{i}$ over $F$ , and inside $Ω$ let $E$ be the field generated by all roots of $f_{1} \dots f_{r}$ .

2.1step 1.1L3L4

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 $α_{i}$ , hence $K$ .

2.2step 1.1algebra

If $H / F$ is any normal intermediate extension in $Ω$ containing $K$ , then each $f_{i}$ , having the root $α_{i} \in H$ , splits in $H$ . Thus $H$ contains all generators of $E$ and $E \subseteq H$ .

3.1step 2.1step 2.2L1∎

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.

Depends on

Used by

Finite separable extensions have finite minimal Galois closures Theorem

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Section 9.15: Normal extensions (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Chapter 6 (standard reference, not scraped)