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!
Best Answer
We need the following lemma.
Lemma Let $K/F$ be a (not necessarily finite dimensional) Galois extension, $L/F$ an arbitrary extension. Clearly $KL/L$ is Galois. Then the restriction map, namely, $\sigma\mapsto \sigma\mid K$ induces an isomorphism $\psi\colon \mathrm{Gal}(KL/L) \rightarrow \mathrm{Gal}(K/K\cap L)$.
Proof: We regard $\mathrm{Gal}(KL/L)$ and $\operatorname{Gal}(K/K\cap L)$ as topological groups with Krull topologies. Clearly $\psi$ is continuous and injective. Let $H = \psi(\mathrm{Gal}(KL/L))$. Since $\mathrm{Gal}(KL/L)$ is compact, $H$ is also compact. Since $\mathrm{Gal}(K/K\cap L)$ is Hausdorff, $H$ is closed. Clearly the fixed subfield of $K$ by $H$ is $K \cap L$. Hence $H = \mathrm{Gal}(K/K\cap L)$ by the fundamental theorem of (not necessarily finite dimensional) Galois theory. This completes the proof.
Now we prove the following proposition with which the OP had a problem.
Proposition Let $K$ and $L$ be Galois extensions of $F$. The restriction of function map, namely, $\sigma\mapsto(\sigma\vert_K,\sigma\vert_L)$ induces a group homomorphism $\varphi\colon\operatorname{Gal}(KL/F)\to\operatorname{Gal}(K/F)\times\operatorname{Gal}(L/F)$. Suppose $K\cap L=F$. Then $\varphi$ is an isomorphism.
Proof. Since it is clear that $\varphi$ is injective, it suffices to prove that it is surjective. Let $G_1 = \mathrm{Gal}(K/F), G_2 = \mathrm{Gal}(L/F), G = \mathrm{Gal}(KL/F)$. By the lemma, given $\sigma_1 \in G_1$, there exists $\sigma \in \mathrm{Gal}(KL/L)$ such that $\sigma\mid K = \sigma_1$. Since $\sigma \in G$ and $\sigma\mid L = 1_L$, $G_1\times 1 \subset \varphi(G)$. Similarly $1\times G_2 \subset \varphi(G)$. Hence $G_1\times G_2 = \varphi(G)$. This completes the proof.