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 $L\subset K$.

$$R: Gal(KL/F)\rightarrow Gal(K/F)\times Gal(L/F)\\ \text{by}\enspace R(\sigma)=(\sigma |_{K},\sigma |_{H})$$

$\hspace{100pt}$enter image description here

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?

  • $\begingroup$ Are you assuming finite extensions here? $\endgroup$ – jgon Sep 8 at 5:49
  • $\begingroup$ If you are assuming finite extensions, why not use equivalent characterizations of Galois, like $K/F$ is Galois if it is the splitting field of a separable polynomial. Then $KL$ is the splitting field of $f_Kf_L$, where $f_K$ is the polynomial for $K$ and $f_L$ is the polynomial for $L$. $\endgroup$ – jgon Sep 8 at 5:52
  • $\begingroup$ @jgon The extensions $K$ and $L$ are not necessarily finite. Btw thanks for your answer for finite extensions. $\endgroup$ – Kumar Sep 8 at 6:29
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    $\begingroup$ $R$ cannot be an isomorphism unless $K\cap L=F$. For each $\sigma$ the restrictions $\sigma\vert_L$ and $\sigma\vert_K$ agree on $K\cap L$, but this is not true for all pairs of automorphisms $(\alpha,\beta)\in Gal(L/F)\times Gal(K/F)$. $\endgroup$ – Jyrki Lahtonen Sep 9 at 9:55

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.

  • $\begingroup$ Why it is necessary to consider algebraic closure of $F$? $\endgroup$ – Kumar Sep 8 at 11:19
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    $\begingroup$ @Kumar You can pick a Galois extension of $F$ that contains both $L$ and $K$, and I just picked $\overline{F}$ - it's just a way of having a Galois group to keep track of the fields. $\endgroup$ – YumekuiMath Sep 8 at 11:32
  • $\begingroup$ Do you know if there is a way to complete my "Case $2$"? $\endgroup$ – Kumar Sep 8 at 11:59
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    $\begingroup$ @Kumar I think a counter-example is $K=\mathbb{Q}(\zeta_{5})$ and $L=\mathbb{Q}(\sqrt[4]{5},i)$. Then both fields are Galois over $\mathbb{Q}$, but $L\cap K=\mathbb{Q}(\sqrt{5})$ because $\sqrt{5}$ can be shown to be in $K$. So your map cannot be an isomorphism (the kernel of the map is of course the automorphisms that agree on the intersection of the two field extensions). So unless you suppose that the intersection of the fields is $\mathbb{Q}$ then you can't show that the map above is an isomorphism in general. $\endgroup$ – YumekuiMath Sep 8 at 15:17

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