proof that the compositum of a Galois extension and another extension is Galois

Proof.

The diagram of the situation of the theorem is:

To see that $EF/F$ is Galois, note that since $E/K$ is Galois, $E$ is a splitting field of a set of polynomials over $K$; clearly $EF$ is a splitting field of the same set of polynomials over $F$. Also, if $f\in K[x]$ is separable over $K$, then also $f$ is separable over $F$. Thus $EF$ is normal and separable over $F$, so is Galois. $E$ is obviously Galois over $E\cap F$ since $E\cap F\supset K$.

Let $r$ be the restriction map

$$r:H=\mathrm{Gal}(EF/F)\to \mathrm{Gal}(E/K):\sigma \mapsto {\sigma |}_E$$

$r$ is clearly a group homomorphism, and since $E$ is normal over $K$, $r$ is well-defined.

Claim $r$ is injective. For suppose $\sigma \in \mathrm{Gal}(EF/F)$ and ${\sigma |}_E$ is the identity. Then $\sigma$ is fixed on $F$ (since it is in $\mathrm{Gal}(EF/F)$ and on $E$ (since its restriction to $E$ is the identity), so is fixed on $EF$ and thus is itself the identity.

Now, the image of $r$ is a subgroup of $\mathrm{Gal}(E/K)$ with fixed field $L$, and thus the image of $r$ is $\mathrm{Gal}(E/L)$. Claim $E\cap F=L$. $\subset$ is obvious: any element $x\in E\cap F$ is fixed by ${\sigma |}_E$ for each $\sigma \in H$ since $\sigma$ fixes $F$. Thus $E\cap F\subset L$. To see the reverse inclusion, choose $x\in L$; then $x$ is fixed by each $r(\sigma )$ for $\sigma \in H$. But $x\in L\subset E$, so that (as an element of $E$), $x$ is fixed by each $\sigma \in H$. Thus $x\in F$ so that $x\in E\cap F$.

Thus $L=E\cap F$, and $r$ is then an isomorphism $\mathrm{Gal}(EF/F)\cong \mathrm{Gal}(E/E\cap F)$.∎