TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Every finite Galois extension has a normal basis

Statement

Every finite Galois extension $K / F$ has a normal basis (Normal bases of a finite Galois extension): there is $α \in K$ whose family of conjugates $(σ_{1} α, \dots, σ_{n} α)$ , indexed by $Gal (K / F) = {σ_{1}, \dots, σ_{n}}$ , is an ordered $F$ -basis of $K$ .

Facts & Assumptions

Given: A finite Galois extension $K / F$ of degree $n$ , so that $K$ is an $F$ -vector space of dimension $n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension, Equivalent characterizations of a finite Galois extension).

[L1]

Every finite Galois extension of an infinite field has a normal basis (Every finite Galois extension of an infinite field has a normal basis).

[L2]

Every finite Galois extension whose Galois group is cyclic has a normal basis (Every finite cyclic extension has a normal basis).

[L3]

An extension $E / F_{q}$ of finite fields of degree $m$ is Galois with $Gal (E / F_{q})$ cyclic of order $m$ , generated by $x \mapsto x^{q}$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ ).

[L4]

A finite list $v$ of length $n$ is an ordered basis of $V$ if and only if every $x \in V$ has exactly one coordinate list $λ : n \to F$ with $x = \sum_{i < n} λ_{i} v_{i}$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L5]

A finite field is a field whose underlying set is finite, and its order is that cardinality (Finite fields and their order).

Proof

technique · cases

1.1

In the case that $F$ is infinite, [L1] applies directly and $K / F$ has a normal basis.

assume-case infL1

1.2

In the case that $F$ is finite, fix an ordered $F$ -basis $v$ of $K$ of length $n$ ; by [L4] the map sending a coordinate list $λ : n \to F$ to $\sum_{i < n} λ_{i} v_{i}$ is a bijection onto $K$ , so $∣ K ∣ = ∣ F ∣^{n}$ is finite and $K$ is a finite field.

assume-case finL4L5given

2.1

In that same finite case, $K / ⁣ F$ is therefore an extension of finite fields of degree $n$ , so $Gal (K / F)$ is cyclic by [L3], and [L2] gives a normal basis.

step 1.2L2L3

3.1

The two cases are exhaustive, a field being finite or infinite and not both, so a normal basis exists in either case.

step 1.1step 2.1cases-exhaustive∎

Remarks

Two genuinely different proofs, not one proof with a case split. The infinite case runs on a determinant that is a nonzero polynomial (Every finite Galois extension of an infinite field has a normal basis); the finite case runs on a cyclic vector for the Frobenius acting linearly (Every finite cyclic extension has a normal basis). Neither argument covers the other case: the first fails because a polynomial can vanish on all of a finite field, the second because a Galois group need not be cyclic.
The finite case is not a hypothesis on the group. It is a hypothesis on the base field, which forces the group to be cyclic through [L3]. That is the whole reason the split is by the base field rather than by the group.

Depends on

Used by

FALSE: every basis of a finite field over a subfield is a normal basis False statement

Dependency tree · two levels

62 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

J. S. Milne, Fields and Galois Theory, v5.10, Theorem 5.18 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Theorem 8.25 (standard reference, not scraped)