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Equivalent characterizations of a finite Galois extension

Statement

Let K/F be a finite extension and put G=Aut(K/F). The following conditions are equivalent:

  1. K/F is Galois, that is, normal and separable.
  2. K is the splitting field over F of a separable polynomial.
  3. G=[K:F].
  4. KG=F.

In particular, a finite extension is Galois if and only if it is the splitting field of a separable polynomial.

Facts & Assumptions

Given: A finite extension K/F and G=Aut(K/F); splitting fields as in Polynomials that split and splitting fields of a polynomial or a family of polynomials; the facts that endomorphisms of a splitting field permute its roots (Every F-endomorphism of a splitting field permutes the distinct roots and is an automorphism), K/F is separable exactly when [K:F]s=[K:F] (A finite extension is separable if and only if [K:F]s=[K:F]), and finite degrees multiply in towers (Tower law for finite extensions: [L:F]=[L:K][K:F]); the separable degree [K:F]s is the number of F-embeddings of K into an algebraic closure of F (The separable degree [K:F]s as a count of embeddings into an algebraic closure).

[L1]

If H is a finite group of automorphisms of a field L, then [L:LH]=H and Aut(L/LH)=H (Artin's fixed-field theorem: [K:KG]=G and Aut(K/KG)=G).

[L2]

If E/F is normal and E=F(α1,,αm), then E is the splitting field of the product of the minimal polynomials of the generators; for m=0 the product is 1 and E=F (A normal extension generated by finitely many elements is the splitting field of the product of their minimal polynomials).

[L3]

If K/F is algebraic and K=F(S) for a set S of elements separable over F, then K/F is separable (An algebraic extension generated by separable elements is separable).

[L4]

For every field extension K/F composition makes Aut(K/F) a group, and if K/F is finite then Aut(K/F)[K:F]s[K:F]; in particular the relative automorphism group is finite (Aut(K/F) is a group and Aut(K/F)[K:F]s[K:F]).

Proof

technique · direct
1.1

For the implication from condition 1 to condition 2, choose a finite generating family for K/F. By [L2], normality makes K the splitting field of the product of their distinct minimal polynomials; separability makes each factor separable, so their distinct product is separable. If K=F, the empty product 1 has splitting field F.

L2given
1.2

For the implication from condition 2 to condition 3, let K be the splitting field of the separable fF[x]. Then K is generated over F by the roots of f, and the minimal polynomial of each root divides f and therefore has no repeated root, so every generator is separable over F; by [L3], K/F is separable and the full-degree criterion in the Given gives [K:F]s=[K:F]. Every F-embedding of K into an algebraic closure permutes the roots of f and hence maps K onto itself, so the [K:F]s embeddings are precisely the elements of G and G=[K:F].

givenL3
1.3

For the implication from condition 3 to condition 4, [L1] gives [K:KG]=G=[K:F]. Since FKGK, the tower law forces [KG:F]=1, hence KG=F.

L1given
2.1

For the implication from condition 4 to condition 1, [L4] makes G finite, so [L1] applies to H=G and gives [K:KG]=G; with KG=F this reads G=[K:F]. The bound in [L4] then gives [K:F]=G[K:F]s[K:F], so [K:F]s=[K:F] and the full-degree criterion in the Given makes K/F separable. For αK, the orbit polynomial qα(x)=βGα(xβ) has distinct roots in K and coefficients fixed by G, hence in KG=F. The minimal polynomial of α divides qα, while every orbit element is one of its roots; separability makes qα divide that minimal polynomial. They are therefore equal, so every minimal polynomial over F splits in K and K/F is normal. This also covers α=0 and the degree-one extension.

L1L4givenalgebra

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