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Dummit & Foot pg 593 Cor 22.

If $K$ is Galois over $F$, $G= \operatorname{Gal}(K/F)=G_1 \times G_2$ direct product. Then $K$ is the compositum of two Galois extensions $K_1,K_2$ of $F$ with $K_1 \cap K_2= F$.

The proof goes as follows:

$K_i:=K^{G_i}$ be the corresponding fixed subfields of the subgroups $G_i$. Then $K_1 K_2 = K^{G_1 \cap G_2}$ and $K_1 \cap K_2=K^{G_1G_2}$.

I can only see one inclusion in both these arguments. That is,

$K_1K_2 \subseteq K^{G_1 \cap G_2}$: If $\sigma \in G_1 \cap G_2$ then $\sigma(K_1), \sigma(K_2)$ are identity.

$K_1\cap K_2 \supseteq K^{G_1G_2}$: If $\alpha \in LHS$, then for any $\sigma \in G_1G_2$, $\sigma(\alpha)=\alpha$.

How do the other directions go?

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  • $\begingroup$ Yes, sorry, thanks a lot! $\endgroup$ – Bryan Shih Jun 3 '18 at 19:52
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As for the first equality, you can notice that $$\mathrm{Gal}(K / K_1K_2) \subset G_1 \cap G_2.$$ For instance, if $\sigma \in \mathrm{Gal}(K/F)$ fixes $K_1 K_2$ (i.e. $\sigma \in \mathrm{Gal}(K / K_1K_2)$), then $\sigma \in G_1 = \mathrm{Gal}(K / K^{G_1}) = \mathrm{Gal}(K / K_1)$.

Therefore $$K^{\mathrm{Gal}(K / K_1K_2)} = K_1K_2 \supset K^{G_1 \cap G_2}$$ gives the inclusion you needed to complete your proof.


For the second equality, you already proved $K_1\cap K_2 \subseteq K^{G_1G_2}$. Conversely, if $x \in K$ is fixed by $G_1 G_2$, then $x \in K^{G_1} = K_1$ and $x \in K^{G_2} = K_2$, i.e. $x \in K_1 \cap K_2$ as desired.

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