Exercise 8.2.7

Question

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Suppose that we have extensions $F \subset K_1 \subset L$ and $F\subset K_2 \subset L$ such that $F\subset K_1$ and $F \subset K_2$ are Galois. Prove that $F \subset K_1K_2$ is Galois. This will show that the compositum of two Galois extensions is again Galois.

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}
$F\subset K_1$ and $F \subset K_2$ are Galois extensions, so are separable extensions. By the Theorem of the Primitive Element, $K_1 = F(\alpha), K_2=F(\beta), \ \alpha \in K_1,\beta \in K_2$, with  $\alpha, \beta$ separable over $F$, therefore $K_1K_2 = F(\alpha,\beta)$ is separable (Proposition 7.1.6).

By Proposition 7.1.7, the Galois closure exists: let $M$ be a Galois closure of $F \subset K_1K_2$, and $\sigma$ be any element of $\Gal(M/F)$.

Let  $\gamma \in K_1 K_2 = F(\alpha,\beta)$. Then $\sigma(\gamma)\in F(\sigma(\alpha),\sigma(\beta))$, and $\sigma(\alpha) \in K_1, \sigma(\beta)\in K_2$ since $F\subset K_1$ and $F \subset K_2$ are normal extensions. Therefore $\sigma(\gamma) \in K_1K_2$.
 
Consequently $$\sigma (K_1 K_2) \subset  K_1K_2.$$

Applying this result to $\sigma^{-1}$, we obtain $\sigma^{-1}(K_1K_2) \subset K_1K_2$, therefore $K_1K_2 \subset \sigma(K_1K_2)$, and so
$$\forall \sigma \in \Gal(M/F),\ \sigma (K_1K_2) = K_1 K_2.$$
By Theorem 7.2.5, we conclude that $K_1K_2$ is a Galois extension of $F$.
\end{proof}

Exercise 8.2.7

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Exercise 8.2.7

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