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A normal basis of $F_{8}$ over $F_{2}$

Example

Let $K : = F_{2} [t] / (t^{3} + t + 1)$ , a field of order $8$ , let $α$ be the class of $t$ , and let $σ (x) = x^{2}$ generate $Gal (K / F_{2})$ . Put

$β : = α + 1.$

Then the conjugate list

$(β, β^{2}, β^{4}) = (α + 1, α^{2} + 1, α^{2} + α + 1)$

is a normal basis of $K$ over $F_{2}$ (Normal bases of a finite Galois extension).

Not every element works. The generator $α$ itself does not: its conjugate list is $(α, α^{2}, α^{2} + α)$ , whose three members sum to $0$ , so they are linearly dependent over $F_{2}$ and are not a basis.

Facts & Assumptions

Given: The field $K = F_{2} [t] / (t^{3} + t + 1)$ with $α$ the class of $t$ , so $α^{3} = α + 1$ and $α^{4} = α^{2} + α$ ; squaring is additive in characteristic two.

[L1]

$t^{3} + t + 1$ has no root in $F_{2}$ , so it is irreducible (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field) and $K$ is a field (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible); it is the minimal polynomial of $α$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element), so $[K : F_{2}] = 3$ with power basis $1, α, α^{2}$ (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).

[L2]

$K / F_{2}$ is Galois with $Gal (K / F_{2}) = {i d, σ, σ^{2}}$ cyclic of order three, $σ (x) = x^{2}$ , and $∣ K ∣ = 8$ (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 $n$ ).

[L3]

A list $v$ of length $n$ is an ordered basis of $V$ if and only if every $x \in V$ has exactly one coordinate list $λ$ 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).

[L4]

For a linear map $T : V \to W$ with $V$ finite-dimensional, $\dim V = \dim \ker T + \dim im T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L5]

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

Verification

technique · direct

1.1

By [L1] and [L2] the space $K$ is a three-dimensional $F_{2}$ -vector space with basis $1, α, α^{2}$ , and $Gal (K / F_{2}) = {i d, σ, σ^{2}}$ acts by $x \mapsto x, x^{2}, x^{4}$ .

L1L2

2.1

The conjugate list of $α$ is $(α, α^{2}, α^{4})$ with $α^{4} = α \cdot α^{3} = α (α + 1) = α^{2} + α$ ; hence $α + α^{2} + α^{4} = α + α^{2} + α^{2} + α = 0$ . A vanishing combination with all coefficients $1$ is nontrivial, so this list is linearly dependent over $F_{2}$ and is not a basis.

step 1.1L3given

3.1

The conjugates of $β = α + 1$ are $β$ , $β^{2} = (α + 1)^{2} = α^{2} + 1$ and $β^{4} = (β^{2})^{2} = (α^{2} + 1)^{2} = α^{4} + 1 = α^{2} + α + 1$ .

step 1.1step 2.1given

4.1

The seven nonzero $F_{2}$ -combinations of $β, β^{2}, β^{4}$ are nonzero: the three single terms are $α + 1$ , $α^{2} + 1$ and $α^{2} + α + 1$ ; the three pairwise sums are $β + β^{2} = α^{2} + α$ , $β + β^{4} = α^{2}$ and $β^{2} + β^{4} = α$ ; and the total sum is $β + β^{2} + β^{4} = 1$ . None of these seven is $0$ , as each has a nonzero coordinate list in the basis $1, α, α^{2}$ .

step 1.1step 3.1given

5.1

So the $F_{2}$ -linear map $T ⁣ : F_{2}^{3} \to K$ sending $(c_{1}, c_{2}, c_{3})$ to $c_{1} β + c_{2} β^{2} + c_{3} β^{4}$ has trivial kernel by step 4.1; both spaces have dimension three by step 1.1, so [L4] makes $T$ surjective as well, hence bijective, and [L3] makes $(β, β^{2}, β^{4})$ an ordered basis of $K$ over $F_{2}$ .

step 1.1step 4.1L3L4

6.1

That list is the family of conjugates of $β$ under $Gal (K / F_{2})$ by step 3.1, so it is a normal basis, as [L5] guarantees exists for this cyclic extension.

step 3.1step 5.1L2L5∎

Remarks

A conjugate family of the right size can still fail. The list $(α, α^{2}, α^{4})$ has three distinct members and is a single Galois orbit, yet it is not a basis; what fails is independence, not the orbit condition. The normal basis theorem asserts that some element works, never that every element does (FALSE: every basis of a finite field over a subfield is a normal basis).

Depends on

Used by

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Sources

K. Conrad, Linear Independence of Characters (expository blurb), Section 3 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, the normal basis theorem (standard reference, not scraped)

A normal basis of F8F8​ over F2F2​