Question:
Let $K$ and $L$ be extensions of $F$. Show that $KL$ is Galois over $F$ if both $K$ and $L$ are Galois over $F$.
This question has been already asked here. But People provided incomplete solution to the problem.
I have tried to attempt the problem:
Case $1$: Either $K\subset L$ or $L\subset K$.
Then $KL$ is trivially Galois.
Case $2$: Neither $K\subset L$ nor $L\subset K$.
Consider,
$$R: Gal(KL/F)\rightarrow Gal(K/F)\times Gal(L/F)\\ \text{by}\enspace R(\sigma)=(\sigma |_{K},\sigma |_{H})$$
$\hspace{100pt}$
where $E=L\cap K$
I want to show that the map $R$ is an isomorphism. But I am unable to get started with it.
Can anyone help me, please?
Let $G=\operatorname{Gal}(\overline{F}/F)$, where $\overline{F}$ is an algebraic closure of $F$. Let $H_{L}$ be the subgroup of $G$ that corresponds to the extension $L/F$ and let $H_{K}$ be the subgroup of $G$ that corresponds to the extension $K/F$. The extension $LK/F$ corresponds to the subgroup $H_{L}\cap H_{K}$ of $G$ by the fundamental theorem of Galois theory. To show that $LK/F$ is Galois it suffices to show that $H_{L}\cap H_{K}$ is closed and normal in $G$ by the fundamental theorem of Galois theory. But $H_{L}$ is closed and normal in $G$ because $L/F$ is Galois (by the fundamental theorem of Galois theory) and likewise is $H_{K}$. By basic topology $H_{K}\cap H_{L}$ is closed in $G$ and by basic group theory $H_{K}\cap H_{L}$ is a normal subgroup of $G$. Therefore $LK/F$ is Galois by the fundamental theorem of Galois theory.