Normal bases of a finite Galois extension

Let $K\mathrm{/}F$ be a finite Galois extension (Finite Galois extensions and $\mathrm{Gal}(K\mathrm{/}F)$) of degree $n=[K:F]$ (The degree $[K:F]={\dim }_{F}K$ of a finite field extension), and list its Galois group as

$\mathrm{Gal}(K\mathrm{/}F)=\{{\sigma }_1,\dots ,{\sigma }_n\},$

which has exactly $n$ elements because $\mathrm{\mid }\mathrm{Gal}(K\mathrm{/}F)\mathrm{\mid }=[K:F]$ for a finite Galois extension (Equivalent characterizations of a finite Galois extension). Scalar multiplication by $F$ makes $K$ an $F$-vector space of dimension $n$.

An element $\alpha \in K$ is a normal basis generator for $K\mathrm{/}F$ when the list

$({\sigma }_1\alpha ,{\sigma }_2\alpha ,\dots ,{\sigma }_n\alpha )$

is an ordered basis of $K$ as an $F$-vector space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Such a list is called a normal basis of $K$ over $F$: a basis that is a single orbit of the Galois group.

Two conditions, not one. A list is an ordered basis when it is injective and its image is a basis, so a normal basis generator must in particular have $n$ distinct conjugates ${\sigma }_i\alpha$. Neither half implies the other: a basis of $K$ over $F$ need not be a Galois orbit, and a Galois orbit of size $n$ need not be a basis.

The list is indexed by the group, not ordered by it. Reordering ${\sigma }_1,\dots ,{\sigma }_n$ permutes the list and leaves the property of being a basis unchanged, since a basis is a property of the underlying set together with injectivity of the list (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). The automorphisms are those of Relative field automorphisms and $\mathrm{Aut}(K\mathrm{/}F)$.