Normal Basis Theorem Proof

Letting $\equiv$ denote congruence modulo $f(x)$, we have $$ \det\left(\left(\sigma_i\sigma_kg(x)\right)_{ik}\right)^2 $$ $$ =\det\left(\left(\sum_j(\sigma_j\sigma_ig(x))(\sigma_j\sigma_kg(x))\right)_{ik}\right) $$ $$ =\det\left(\left(\sum_j\sigma_j(g_i(x)g_k(x))\right)_{ik}\right) $$ $$ \equiv\det\left(\left(\sum_j\sigma_j(\delta_{ik}g_i(x))\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\sum_j\sigma_j\sigma_ig(x)\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\sum_j\sigma_jg(x)\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\sum_jg_j(x)\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\right)_{ik}\right) $$ $$ =1. $$

The argument can be simplified as follows: the value of $\sigma_i (\sigma_j(g(x))$ at $\alpha$ is $1$ if $\sigma_i = (\sigma_j)^{-1}$ and $0$ otherwise. Therefore $D(\alpha)$ is determinant of the permutation matrix corresponding to permutation of $\sigma_1...\sigma_n$ sending $\sigma_i$ to $\sigma_i^{-1}$. It follows that $D(\alpha) = \pm 1$, so $D(x)\ne 0$. (Incidentally, $D(\alpha)=\pm 1$ follows from $D^2(x)\equiv 1 \mod f(x)$. You can prove $D^2(x) \equiv 1 \mod f(x)$ by proving that $D(\alpha)=\pm 1$ and that $D^2(x)$ lies in $F[x]$, since for any $\sigma_k$ we have $\sigma_k(D(x)) = \pm D(x)$ as $\sigma_k(D(x))$ is determinant of matrix obtained from that for $D(x)$ by a permutation of rows.)