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Unverified Commit ab4f2b23 authored by rrueger's avatar rrueger
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Algebra II: Lecture 8. Correctly reference FTGT theorem from lecture 5

parent 9b6a51b9
......@@ -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
......
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