... ... @@ -119,7 +119,7 @@ compatible. We have seen that every subextension $E/K/F$ of a Galois extension is Galois, but what about the extension $K/F$? We claim that \begin{theorem} \begin{theorem}\label{thm:ftgt_normal} For a Galois extension $E/F$ with $G = \Gal(E/F)$ and a subextension $E/K/F$ with $S = \Aut(E/K)$, $K/F$ is a Galois extension if and only if $S$ is normal in $G$. ... ...
 ... ... @@ -434,29 +434,24 @@ $H_{i+1}$ and that each subquotient is abelian. \begin{proof}[Proof that $H_i$ is normal in $H_{i+1}$] As we know that subextensions of Galois extensions are Galois, each $E/K_i$ is Galois. This allows us to focus on on a smaller tower $E/K_i/K_{i+1}$. We now remind ourselves of a theorem from F.T.G.T \begin{siderules} \begin{theorem*}[Reminder F.T.G.T] For a Galois extension $E/F$ with a subextension $E/K/F$, $K/F$ is Galois if and only if $\mathrm{Aut}(E/K)$ is normal in $\mathrm{Gal}(E/F)$. \end{theorem*} \end{siderules} \noindent For us, this means that remind ourselves of Theorem~\ref{thm:ftgt_normal} from F.T.G.T (normal subgroups of a Galois group are in bijection to Galois extensions over the base field contained in the original extension) and yield that % \begin{align*} \mathrm{Aut}(E/K_{i}) = \mathrm{Gal}(E/K_{i}) = H_i \triangleleft H_{i+1} = \mathrm{Gal}(E/K_{i+1}). \end{align*} % if $K_{i}/K_{i+1}$ is Galois. To exploit this theorem, we claim that $K_i$ is the splitting field of $(z^r - \gamma)$ over $K_{i+1}$, making $K_i/K_{i+1}$ Galois. To show this we write down $\{\alpha, \alpha \omega, \alpha \omega ^2, \ldots, \alpha \omega ^{r-1}\}$, with $\omega$ being the $r$th root of unity, and note that these are $r$ distinct roots of $(z^r - \gamma) \in K_{i+1}[z]$ in $K_i$: % \begin{align*} {(\alpha \omega ^k)}^r - \gamma = \alpha ^r \omega ^{rk} - \gamma ... ...