TITLE: The Galois group of a composite of Galois extensions
QUESTION [12 upvotes]: Morandi's Field and Galois Theory, exercise 5.19b
Let $K$ and $L$ be Galois extensions of $F$. The restriction of function map, namely, $\sigma\mapsto(\sigma\vert_K,\sigma\vert_L)$ induces an injective group homomorphism $\varphi\colon\operatorname{Gal}(KL/F)\to\operatorname{Gal}(K/F)\times\operatorname{Gal}(L/F)$. Show that $\varphi$ is surjective if and only if $K\cap L=F$.

It's not hard to show that $\varphi$ is a monomorphism. If it's surjective, it's not hard to show that $K\cap L=F$ as follow:
Fix $\alpha\in K\cap L$, let $\beta$ be a root of the minimal polynomial of $\alpha$ over $F$. Since $K,L$ are normal, $\beta\in K\cap L$. By isomorphism extension theorem, we can choose $\tau_1\in\operatorname{Gal}(K/F)$ such that $\tau_1(\alpha)=\beta$. For surjectivity of the map, there's $\sigma$ such that $\sigma\vert_K=\tau_1$ and $\sigma_L=\mathrm{id}$, which forces $\alpha=\beta$, therefore $\alpha\in F$, since $K,L$ are separable over $F$.
The converse seems hard. I cannot show that when $K,L$ are arbitrary Galois extensions. If they are both finite dimensional, the statement follows from natural irrationality: $\operatorname{Gal}(KL/L)\cong\operatorname{Gal}(K/K\cap L)$, which implies that $[KL:L]=[K:K\cap L]=[K:F]$, therefore $[KL:F]=[K:F][L:F]$, and note that $\varphi$ is injective, thus surjective.
Any help? Thanks!

REPLY [2 votes]: I think this solution will give more general intuition.
Let $\varphi : Gal(KL/F) \to Gal(K/F) \times Gal(L/F)$ which sends $\sigma \to (\left.\sigma\right|_{K}, \left.\sigma\right|_{L})$.
Lemma: $Gal(KL/F)$ is isomorphic to the subgroup $H = \{(\sigma, \tau) \}$ of $Gal(K/F) \times Gal(L/F)$ where $\left.\sigma\right|_{K \cap L} = \left.\tau\right|_{K \cap L}$. So $\varphi$ is surjective if and only if $H = Gal(K/F) \times Gal(L/F)$, and this happens if and only if $K \cap L = F$.
Proof. $\varphi$ is clearly injective (any element in kernel must fix $K$ and $L$, therefore, it fixes $KF$ which implies the identity). Let $A = img(\varphi)$. It's obvious that $A \subset H$. To prove the reverse, not that $KL$ is Galois over $F$ and $K \cap L$ is a subfield containing $F$. Then any embedding from $K \cap L$ to $\bar{F}$ is of the form  $\left.\phi\right|_{K \cap L}$ for some $\phi \in Gal(KL/F)$ (generally, given field $F$ with Galois extension $K$, if $E$ is subfield of $K$ containing $F$, then any embedding of $E$ which fixes $F$ is of the form $\left.\phi\right|_{E}$ for some $\phi \in Gal(K/F)$. This is a nice property, and I'll let you play with it!). If $(\phi, \tau) \in H$, then $\left.\sigma\right|_{K \cap L} = \left.\tau\right|_{K \cap L}$ and $\left.\sigma\right|_{K \cap L} =\left.\tau\right|_{K \cap L}$ is actually of them form $\left.\phi\right|_{K \cap L}$ for some $\phi \in Gal(KL/F)$ which implies $H \subset A$, hence $H=A$.
Hope this helps!
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