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 $$ r: H = \Gal(EF/F) \to \Gal(E/K) : \sigma \mapsto \sigma |_ $$ $r$ is clearly a group homomorphism, and since $E$ is normal over $K$ , $r$ is well-defined.

Claim $r$ is injective. For suppose $\sigma\in\Gal(EF/F)$ and $\sigma|_E$ is the identity. Then $\sigma$ is fixed on $F$ (since it is in $\Gal(EF/F)$ 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 $\Gal(E/K)$ with fixed field $L$ , and thus the image of $r$ is $\Gal(E/L)$ . 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 $\Gal(EF/F)\cong \Gal(E/E\cap F)$ .