Exercise 7.2.7

Question

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ewcommand{\ssi} {si et seulement si }

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ewcommand{\F}{\mathbb{F}}

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Suppose that $F \subset K \subset L$, where $L$ is Galois over $F$, and let $\sigma \in \Gal(L/F)$. Show that
$$K = \sigma K \iff \Gal(L/K) = \sigma \Gal(L/K) \sigma^{-1},\ \sigma\ \mathrm{ in }\  \Gal(L/F).$$

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
If $\sigma \in \Gal(L/F)$ satisfies $K = \sigma K$, then by Lemma 7.2.4,
$$\sigma \Gal(L/K)\sigma^{-1} = \Gal(L/\sigma K) = \Gal(L/K).$$
Conversely, if $\sigma \in \Gal(L/F)$ satisfies $\sigma \Gal(L/K)\sigma^{-1} = \Gal(L/K)$, then by the same Lemma, $\Gal(L/K) = \Gal(L/\sigma K)$.
As $F\subset L$ is a Galois extension, so are $K \subset L$ and $\sigma K \subset L$, the fixed field of  $\Gal(L/K)$ is $K$, and the fixed field of $\Gal(L/\sigma K)$ is $\sigma K$. As these two groups are identical, $K = \sigma K$.

$$\forall \sigma \in \Gal(L/F),\ (K = \sigma K \iff \Gal(L/K) = \sigma \Gal(L/K) \sigma^{-1}).$$
(Consequently
$$(\forall \sigma \in \Gal(L/F), \sigma K = K) \iff \Gal(L/K) \lhd \Gal(L/F)).$$
\end{proof}

Exercise 7.2.7

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Exercise 7.2.7

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