Editing Normal basis (section)

=== Example ===
Consider the field <math>K=\mathrm{GF}(2^3)=\mathbb{F}_{8}</math> over <math>F=\mathrm{GF}(2)=\mathbb{F}_{2}</math>, with Frobenius automorphism <math>\Phi(\alpha)=\alpha^2</math>. The proof above clarifies the choice of normal bases in terms of the structure of ''K'' as a representation of ''G'' (or ''F''[''G'']-module). The irreducible factorization 
<math display=block>X^n-1 \ =\ X^3-1\ = \ (X{-}1)(X^2{+}X{+}1) \ \in\ F[X]</math> means we have a direct sum of ''F''[''G'']-modules (by the [[Chinese remainder theorem]]):<math display=block>K\ \cong\ \frac{F[X]}{(X^3{-}\,1)} \ \cong\ \frac{F[X]}{(X{+}1)} \oplus \frac{F[X]}{(X^2{+}X{+}1)}.</math>
The first component is just <math>F\subset K</math>, while the second is isomorphic as an ''F''[''G'']-module to <math>\mathbb{F}_{2^2} \cong \mathbb{F}_2[X]/(X^2{+}X{+}1)</math> under the action <math>\Phi\cdot X^i = X^{i+1}.</math> (Thus <math>K \cong \mathbb F_2\oplus \mathbb F_4</math> as ''F''[''G'']-modules, but ''not'' as ''F''-algebras.)

The elements <math>\beta\in K</math> which can be used for a normal basis are precisely those outside either of the submodules, so that <math>(\Phi{+}1)(\beta)\neq 0</math> and <math>(\Phi^2{+}\Phi{+}1)(\beta)\neq 0</math>. In terms of the ''G''-orbits of ''K'', which correspond to the irreducible factors of:
<math display="block">t^{2^3}-t \ = \ t(t{+}1)\left(t^3 + t + 1\right)\left(t^3 + t^2 + 1\right)\ \in\ F[t],</math>
the elements of <math>F=\mathbb{F}_2</math> are the roots of <math>t(t{+}1)</math>, the nonzero elements of the submodule <math>\mathbb{F}_4</math> are the roots of <math>t^3+t+1</math>, while the normal basis, which in this case is unique, is given by the roots of the remaining factor <math>t^3{+}t^2{+}1</math>.

By contrast, for the extension field <math>L = \mathrm{GF}(2^4)=\mathbb{F}_{16}</math> in which {{nowrap|1=''n'' = 4}} is divisible by {{nowrap|1=''p'' = 2}}, we have the ''F''[''G'']-module isomorphism 
<math display="block">L \ \cong\ \mathbb{F}_2[X]/(X^4{-}1)\ =\ \mathbb{F}_2[X]/(X{+}1)^4.</math>
Here the operator <math>\Phi\cong X</math> is not [[Diagonalizable matrix|diagonalizable]], the module ''L'' has nested submodules given by [[Generalized eigenvector|generalized eigenspaces]] of <math>\Phi</math>, and the normal basis elements ''β'' are those outside the largest proper generalized eigenspace, the elements with <math>(\Phi{+}1)^3(\beta)\neq 0</math>.