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

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

Statement

False claim. For every extension $E / F_{q}$ of finite fields, every $F_{q}$ -basis of $E$ is a normal basis (Normal bases of a finite Galois extension).

Facts & Assumptions

Given: The ring $L : = F_{2} [t] / (t^{2} + t + 1)$ with $α$ the class of $t$ , so that $α^{2} = α + 1$ because $α^{2} + α + 1 = 0$ and $- 1 = 1$ in characteristic two.

[L1]

A polynomial of degree $2$ or $3$ over a field is irreducible if and only if it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field); and $F [x] / (p)$ is a field exactly when $p$ is irreducible (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible).

[L2]

A monic irreducible vanishing at $a$ is the minimal polynomial of $a$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element), and then $F (a)$ has power basis $1, a, \dots, a^{n - 1}$ with $[F (a) : F] = n$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L3]

An extension of finite fields of degree $n$ over $F_{q}$ is Galois with cyclic Galois group generated by $σ_{q} (x) = x^{q}$ , of order $n$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ , The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields, For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L4]

A normal basis of $K / F$ is an ordered $F$ -basis of the form $(σ_{1} γ, \dots, σ_{n} γ)$ for a single $γ \in K$ , indexed by $Gal (K / F)$ (Normal bases of a finite Galois extension).

[L5]

Every finite Galois extension has a normal basis (Every finite Galois extension has a normal basis).

Refutation

technique · direct

1.1

$t^{2} + t + 1$ has no root in $F_{2}$ , its values at $0$ and $1$ both being $1$ , so it is irreducible and $L$ is a field by [L1]; it is the minimal polynomial of $α$ , so $[L : F_{2}] = 2$ with ordered basis $(1, α)$ by [L2], and $L$ has four elements $0, 1, α, α + 1$ .

L1L2given

2.1

By [L3] the extension $L / F_{2}$ is Galois with $Gal (L / F_{2}) = {i d, σ}$ , where $σ (x) = x^{2}$ .

step 1.1L3

3.1

The conjugate lists of the four elements are $(0, 0)$ , $(1, 1)$ , $(α, α + 1)$ and $(α + 1, α)$ , using $α^{2} = α + 1$ and $(α + 1)^{2} = α^{2} + 1 = α$ . Their underlying sets are ${0}$ , ${1}$ and ${α, α + 1}$ .

step 1.1step 2.1given

4.1

The list $(1, α)$ is an $F_{2}$ -basis of $L$ by step 1.1, but its underlying set ${1, α}$ is none of the three sets in step 3.1, so it is not the conjugate list of any element and hence is not a normal basis by [L4]. The false claim therefore fails already for $L / F_{2}$ .

step 1.1step 3.1L4

5.1

What is true is the existential statement: some element of $L$ generates a normal basis, and $α$ does, since $(α, α + 1)$ is a list of two distinct elements whose only vanishing $F_{2}$ -combinations are trivial, as $α \neq 0$ , $α + 1 \neq 0$ and $α + (α + 1) = 1 \neq 0$ . That is the content of [L5], which asserts existence and never universality.

step 3.1step 4.1L4L5∎

Remarks

Where the false claim comes from. The normal basis theorem is an existence statement, and its proofs single out an element by a nonvanishing condition — a determinant in the infinite case, a cyclic vector in the finite case. Both conditions genuinely exclude some elements, as A normal basis of $F_{8}$ over $F_{2}$ shows over $F_{8}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

54 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

K. Conrad, Linear Independence of Characters (expository blurb), Example 3.1 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Definition 5.17 and the normal basis theorem (standard reference, not scraped)