# compositum of galois extensions

Let $E/K$ and $F/K$ be Galois extensions and $E\cap F=K$. Show that $EF/K$ is Galois and $Gal(EF/K) \cong Gal(E/K)\times Gal(F/K)$. Here $EF$ denotes the smallest field containing both $E$ and $F$.

I know how to prove that $EF/F$ is Galois. And we can define a map from $Gal(EF/K)$ to $Gal(E/K)\times Gal(F/K)$ by $\sigma \rightarrow (\sigma|_{E}, \sigma|_{F})$. In order to show that this is an injection, we have to show that any automorphism in $Gal(EF/F)$ fixing $E$ and $F$ must be identity. For surjectivity, we have to show that for any $\sigma|_{1}\in Gal(E/K)$ and $\sigma|_{2}\in Gal(F/K)$, we can find a $\sigma$ s.t. $\sigma|_{E}=\sigma|_{1}$ and $\sigma|_{F}=\sigma_{2}$.I'm stuck in these two parts.

• Have you already covered the Fundamental Theorem of Galois Theory or, at least, portions of it? – DonAntonio May 17 '17 at 17:44
• @DonAntonio Yes，I've learned all fundamantal theorem. – mike May 17 '17 at 18:44

1. Let $N/K$ be a normal closure of $EF/K$. We then know that $N/K$ is Galois. This is immediate in char zero, and follows from separability of $E/K$ and $F/K$ otherwise (that is a bit non-trivial, but may also be a non-issue). Let $G=Gal(N/K)$, $H_1=Gal(N/E)\le G$, $H_2=Gal(N/F)\le G$.
2. Because $EF$ is the smallest subfield of $N$ containing both $E$ and $F$ we have $Gal(N/EF)=H_1\cap H_2$.
3. Because $E/K$ and $F/K$ are Galois, $H_1\unlhd G$ and $H_2\unlhd G$. Therefore $H_1\cap H_2\unlhd G$. By Galois correspondence $EF/K$ is Galois. Therefore $N=EF$ and $H_1\cap H_2=\{1_G\}$.
4. Because $H_1\unlhd G$, $H_2\unlhd G$ and $H_1\cap H_2=\{1_G\}$ a standard exercise in group theory shows that $h_1h_2=h_2h_1$ for all $h_1\in H_1$ and $h_2\in H_2$.
5. The subgroup $H_1H_2$ is the smallest subgroup of $G$ containing both $H_1$ and $H_2$. Therefore it corresponds with the intermediate field $E\cap F$, and $Gal(N/(E\cap F))=H_1H_2$.
6. Because $E\cap F=K$ we have $Gal(N/(E\cap F))=G$.
7. Therefore $G$ is the (inner) direct product $H_1\times H_2$.
All the above implies that every automorphism $\sigma\in Gal(EF/F)$ can be writtern uniquely in the form $\sigma=\tau_1\tau_2=\tau_2\tau_1$, where $\tau_1\in Gal(EF/E)$ and $\tau_2\in Gal(EF/F)$. The rest should be easy. Recall that $Gal(E/K)\simeq Gal(EF/K)/Gal(EF/E)$ and similarly $Gal(F/K)\simeq Gal(EF/K)/Gal(EF/F)$.