I’ve been making my way through A Course in Galois Theory by Garling.
For those who have the book, I’m trying to justify a step in the middle of p. 157 for a proof of a theorem that leads to an algorithm for calculating Galois groups. As Garling points out, the algorithm is too computational, and it is more theoretical than practical. Anyway, I hope I can get help justifying the step, described below.
For those who don’t have the book, I'll try to distill out the exact part I can’t get through:
Let
$\beta = \alpha_1 x_1 + \alpha_2 x_2 + … + \alpha_n x_n$
For $\sigma \in \Sigma_n$, define $\sigma_x$ to permute the $x_i$ and define $\sigma_\alpha$ to permute the $\alpha_i$.
So
$\sigma_x(\beta) = \alpha_1 x_{\sigma(1)} + \alpha_2 x_{\sigma(2)} + … + \alpha_n x_{\sigma(n)}$
and
$\sigma_\alpha(\beta) = \alpha_{\sigma(1)} x_1 + \alpha_{\sigma(2)} x_2 + … + \alpha_{\sigma(n)} x_n$
Here is the step to justify:
$\prod_{\sigma \in A_1} (y - \tau_x \sigma_x(\beta)) = \prod_{\sigma \in A_1} (y - (\tau^{-1})_\alpha \sigma_x(\beta))$
where $A_1$ is a subgroup of $\Sigma_n$, and $\tau \in \Sigma_n$.
I see that:
$\tau_x \sigma_x(\beta) = \alpha_1 x_{\tau \sigma (1)} + \alpha_2 x_{\tau \sigma (2)} + … + \alpha_n x_{\tau \sigma (n)}$
and
$\begin{align} (\tau^{-1})_\alpha \sigma_x(\beta) & = \alpha_{\tau^{-1}(1)} x_{\sigma(1)} + \alpha_{\tau^{-1}(2)} x_{\sigma(2)} + … + \alpha_{\tau^{-1}(n)} x_{\sigma(n)} \\ & = \alpha_{\tau^{-1} (\tau (1))} x_{\sigma (\tau (1))} + \alpha_{\tau^{-1} (\tau (2))} x_{\sigma (\tau (2))} + … + \alpha_{\tau^{-1} (\tau (n))} x_{\sigma (\tau (n))} \\ & = \alpha_1 x_{\sigma \tau (1)} + \alpha_2 x_{\sigma \tau (2)} + … + \alpha_n x_{\sigma \tau (n)} \\ & = \sigma_x \tau_x(\beta) \end{align}$
But these do not seem to be equal.
Where is my mistake?
Well, I'm going to post the following as an answer because, if correct, it'll get me through to complete the proof, as an alternative approach. Please check my answer. Also, maybe with the extra details, someone can post a direct answer that justifies the step.
In the book, the first part of the proof showed that $A_1 \subseteq \Gamma_K(L)$. The second part, which includes the step I couldn't justify, shows $\Gamma_K(L) \subseteq A_1$. Here now is my attempt to show the second part.
To start, let me provide the following two details, and then I'll proceed from there.
$L = K(\alpha_1, \alpha_2, … , \alpha_n)$, where $\alpha_1, \alpha_2, … , \alpha_n$ distinct roots
and
$\prod_{\sigma \in A_1} (y - \sigma_x(\beta)) = F_1 \in K[y, x_1, x_2, … , x_n]$
Note that
$ F_1 = \prod_{\sigma \in A_1} (y - (\sigma^{-1})_\alpha (\beta)) = \prod_{\sigma \in A_1} (y - \sigma_\alpha (\beta))$
Consider
$\tau \in \Gamma_K(L)$
Then
$F_1 = \tau(F_1) = \tau_\alpha(\prod_{\sigma \in A_1} (y - \sigma_\alpha (\beta))) = \prod_{\sigma \in A_1} (y - \tau_\alpha \sigma_\alpha (\beta))$
Matching up the above factorizations for $F_1$, we must have
$\tau_\alpha \sigma_\alpha' = e_\alpha$ for some $\sigma' \in A_1$
So
$\tau = \sigma'^{-1} \in A_1$