Normal basis theorem

Remarks

Such a basis is called a normal basis, and $\alpha$ is a normal basis generator. Equivalently, $L$ is isomorphic to the group algebra $K[G]$ as a module over $K[G]$.

When $K$ is infinite, write $G = \set{\sigma_1, \ldots, \sigma_n}$ and fix a basis $b_1, \ldots, b_n$ of $L$. The family $(\sigma_i(\alpha))$ is a basis exactly when $\det\left(\sigma_i^{-1}\sigma_j(\alpha)\right) \ne 0$; writing $\alpha = \sum x_k b_k$ makes this determinant a polynomial in $x_1, \ldots, x_n$. An argument in the spirit of Dedekind's lemma, using that $K$ is infinite, shows that this polynomial is not zero. A nonzero polynomial over an infinite field has a non-root in $K^n$. When $K$ is finite, $G$ is cyclic, generated by the Frobenius $\varphi$. Dedekind's lemma shows that the minimal polynomial of the $K$-linear map $\varphi$ is $x^n - 1$, and the theory of cyclic modules then gives the normal basis.

Example: $L = \Q(i)$ over $K = \Q$, with $G = \set{\id, \sigma}$ and $\sigma(i) = -i$. For $\alpha = 1 + i$, the family $1 + i, 1 - i$ is a basis of $\Q(i)$ over $\Q$. The choices $\alpha = 1$ and $\alpha = i$ fail: they give $1, 1$ and $i, -i$, which are linearly dependent.

Going deeper

For a student

In coordinates of a normal basis, the Galois group acts by permuting the coordinates, which makes many computations simple. Over finite fields, normal bases make the Frobenius map $x \mapsto x^q$ a cyclic shift of coordinates, which is used in computer arithmetic. The theorem says that $L$ is a free $K[G]$-module of rank $1$. Whether the ring of integers of a number field is free over $\Z[G]$ is a much subtler question of the theory of Galois modules.