Exercise 12.2.9

Question

\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}}

In the situation of Theorem 12.2.5, suppose that $F\subset K$ is an extension of prime degree $p$. Prove that $\Gal(KL/K)$ is isomorphic to either $\Gal(L/F)$ or a subgroup of index $p$ 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{center} \begin{tikzpicture} ode (KL) at (3,6) {$KL$}; ode (K) at (1,4) {$K$}; ode (L) at (5,4) {$L $}; ode (KinterL) at (3,2) {$K \cap L$}; ode (F) at (3,0) {$F$}; \draw[<-] (KL) edge (K) edge (L); \draw[->] (KinterL) edge (K) edge (L); \draw[->] (F) edge (KinterL); \end{tikzpicture} \end{center} \begin{proof} Since $ p = [K:F] = [K:K \cap L]\cdot [K\cap L:F]$ is prime, the factor $ [K\cap L:F]$ of $p$ is $1$ or $p$. If $[K\cap L:F] = 1$, then $F = K\cap L$ and so $\Gal(KL/K) \simeq \Gal(L/F)$. If $[K \cap L:F] = p$, then by the Galois correspondence (Theorem 7.3.1), $\Gal(L/K\cap L)$ corresponds to $K\cap L$, and $$[K \cap L : F] = p = (\Gal(L/F) : \Gal(L/K\cap L) ).$$ Therefore $\Gal(KL/K) \simeq \Gal(L / K\cap L)$ is isomorphic to a subgroup of index $p$ in $\Gal(L/F)$. \begin{center} \begin{tikzpicture} ode (L) at (1,4) {$L$}; ode (KinterL) at (1,2) {$K \cap L$}; ode (F) at (1,0) {$F $}; \draw[->] (F) edge node[right]{$p$}(KinterL); \draw[->] (KinterL) edge (L); ode (e) at (5,4) {$\{e\}$}; ode (GKinterL) at (5,2) {$\Gal(L/K\cap L)$}; ode (GK) at (5,0) {$\Gal(L/F) $}; \draw[<-] (GKinterL) edge (e); \draw[<-] (GK) edge node[right]{$p$} (GKinterL); \end{tikzpicture} \end{center} \end{proof}

Exercise 12.2.9

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Exercise 12.2.9

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