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?
