Normal Basis Theorem Proof I am a little confused by the proof of the Normal Basis Theorem in E. Artin's Galois Theory. Specifically, I am having trouble understanding why a certain squared matrix has a particular form.
The theorem and proof are as follows:

Theorem: If $E$ is a normal extension of $F$ and $\sigma_1,...,\sigma_n$ are the elements of its group $G$, there is an element $\theta$ in $E$ such that the elements $\sigma_1(\theta),...,\sigma_n(\theta)$ are linearly independent with respect to $F$.
Proof: According to Theorem 27, there exists an $\alpha$ such that $E=F(\alpha)$. Let $f(x)$ be the equation for $\alpha$, put $\sigma_i(\alpha)=\alpha_i$,
$g(x)=\frac{f(x)}{(x-\alpha)f'(\alpha)}$ and $g_i(x)=\sigma_i(g(x))=\frac{f(x)}{(x-\alpha_i)f'(\alpha_i)}$
$g_i(x)$ is a polynomial in $E$ having $\alpha_k$ as a root for $k\neq i$ and hence 
  \begin{equation}
g_i(x)g_k(x)=0\mod f(x) \text{ for } i\neq k
\end{equation}
  In the equation
  \begin{equation}
g_1(x)+g_2(x)+...+g_n(x)-1=0
\end{equation}
  the left side is of degree at most $n-1$. If $(2)$ is true for $n$ different values of $x$, the left side must be identically $0$. Such $n$ values are $\alpha_1,\alpha_2,...,\alpha_n$, since $g_i(\alpha_i)=1$ and $g_k(\alpha_i)=0$ for $k\neq i$. 
Multiplying the second equation above by $g_i(x)$ and using the first shows:
  \begin{equation}
(g_i(x))^2=g_i(x)\mod f(x)
\end{equation}
  We next compute the determinant 
  \begin{equation}
D(x)=|\sigma_i\sigma_k(g(x))|\hspace{.5cm} i,k=1,2,...,n
\end{equation}
  and prove $D(x)\neq 0$. If we square it by multiplying column by column and compute its value ($\text{mod} (f(x)$) we get from the three equations above a determinant that has $1$ in the diagonal and $0$ elsewhere.
So
  \begin{equation}
(D(x))^2=1\mod f(x).
\end{equation}
  $D(x)$ can have only a finite number of roots in $F$. Avoiding them we can find a value $a$ for $x$ such that $D(a)\neq 0$. Now set $\theta =g(a)$. Then the determinant 
  \begin{equation}
|\sigma_i\sigma_k(\theta)|\neq 0
\end{equation}
Consider any linear relation >$x_1\sigma_1(\theta)+x_2\sigma_2(\theta)+...+x_n\sigma_n(\theta)=0$ where the $x_i$ are in $F$. Applying the automorphism $\sigma_i$ to it would lead to $n$ homogeneous equations for the $n$ unknowns $x_i$. (5) shows that $x_i=0$ and our theorem is proved.

My difficulty with the above proof occurs when he squares the determinant. For instance, if $n=3$, the matrix would look like:
\begin{vmatrix}
\sigma_1\sigma_1(g(x)) & \sigma_1\sigma_2(g(x))&\sigma_1\sigma_3(g(x))\\
\sigma_2\sigma_1(g(x)) & \sigma_2\sigma_2(g(x))&\sigma_2\sigma_3(g(x))\\
\sigma_3\sigma_1(g(x)) & \sigma_3\sigma_2(g(x))& \sigma_3\sigma_3(g(x))
\end{vmatrix}.
Then the the $(2,2)$ entry in the square is:
\begin{equation}
\sigma_1\sigma_2(g(x))\sigma_2\sigma_1(g(x))+\sigma_2\sigma_2(g(x))^2+\sigma_3\sigma_2(g(x))\sigma_2\sigma_3(g(x))
\end{equation}
what I think is supposed to be happening is that each term in this summation is $\sigma_i(g(x))^2$ and thus is congruent to $\sigma_i(g(x))$, the sum of which is congruent to $1$ as $i$ goes from $1$ to $n$. However for this to be the case it seems that we need $\sigma_1\sigma_2=\sigma_2\sigma_1$. While this may be true when $n=3$, it illustrates my concern that having the math work out relies on the elements of the group commuting. So, my question is: Why does the square have the form Artin describes? 
 A: Letting $\equiv$ denote congruence modulo $f(x)$, we have 
$$
\det\left(\left(\sigma_i\sigma_kg(x)\right)_{ik}\right)^2
$$
$$
=\det\left(\left(\sum_j(\sigma_j\sigma_ig(x))(\sigma_j\sigma_kg(x))\right)_{ik}\right)
$$
$$
=\det\left(\left(\sum_j\sigma_j(g_i(x)g_k(x))\right)_{ik}\right)
$$
$$
\equiv\det\left(\left(\sum_j\sigma_j(\delta_{ik}g_i(x))\right)_{ik}\right)
$$
$$
=\det\left(\left(\delta_{ik}\sum_j\sigma_j\sigma_ig(x)\right)_{ik}\right)
$$
$$
=\det\left(\left(\delta_{ik}\sum_j\sigma_jg(x)\right)_{ik}\right)
$$
$$
=\det\left(\left(\delta_{ik}\sum_jg_j(x)\right)_{ik}\right)
$$
$$
=\det\left(\left(\delta_{ik}\right)_{ik}\right)
$$
$$
=1.
$$
A: 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.)
