Exercise 7.1.1

Question

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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}}

Given a finite extension $F \subset L$, and a subgroup $H \subset \Gal(L/F)$, prove that  $L_H = \{\alpha \in L \ | \ \forall \sigma \in H, \sigma(\alpha) = \alpha\}$ is a subfield of $L$ containing $F$.

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}
Let $H \subset \Gal(L/F)$, and $L_H =\{\alpha \in L \ \vert \ \forall \sigma \in H,\, \sigma(\alpha) = \alpha\}$.

We show that $L_H$ is a subfield of $L$ containing $F$.

$\bullet$  By definition of $\Gal(L/F)$, every element $\sigma$ of $H\subset \Gal(L/F)$ satisfies $\sigma(\alpha) = \alpha$ for all $\alpha \in F$, therefore $F \subset L_H$. In particular $1 \in F \subset L_H $, so  $L_H 
eq \emptyset$.

$\bullet$ If $\alpha, \beta \in L_H$, then 
\begin{align*}
\sigma(\alpha - \beta) &= \sigma(\alpha)- \sigma(\beta) = \alpha- \beta,\
\sigma(\alpha \beta) &= \sigma(\alpha) \sigma(\beta) = \alpha \beta,
\end{align*}
thus $\alpha-\beta, \alpha \beta \in L_H$.

$\bullet$ If $\alpha \in L_H\setminus\{0\}$, $\sigma(\alpha) = \alpha$, thus $\sigma(\alpha^{-1}) = \sigma(\alpha)^{-1} = \alpha^{-1}: \alpha^{-1} \in L_H$.

Conclusion: $L_H$ is a subfield of $L$ containing $F$.
\end{proof}

Exercise 7.1.1

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Exercise 7.1.1

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