Exercise 12.2.15

Proof.

(a)

The Exercise 8.2.7 proves that $F\subset \mathit{KL}$ is Galois. Let $\sigma \in \mathrm{Gal}(\mathit{KL}∕F)$. By Theorem 7.2.5, since $F\subset L$ is normal, $\mathit{\sigma L}=L$, and $\sigma$ fixes the elements of $F$, so $\sigma {\text{|}}_L\in \mathrm{Gal}(L∕F)$. Similarly $\sigma {\text{|}}_K\in \mathrm{Gal}(K∕F)$ (there is a misprint in the statement).

(b)

Let $$\phi :\{\begin{array}{ccc}\mathrm{Gal}(\mathit{KL}∕F)\hfill & \to \hfill & \mathrm{Gal}(L∕F)\times \mathrm{Gal}(K∕F)\hfill \\ \sigma \hfill & \mapsto \hfill & (\sigma {\text{|}}_L,\sigma {\text{|}}_K)\hfill \end{array}$$

Then $\phi$ is a group homomorphism. Moreover, if $(\sigma {\text{|}}_L,\sigma {\text{|}}_K)=({\mathrm{id}}_L,{\mathrm{id}}_K)$, then $\sigma$ is the identity on both $K$ and $L$. By Exercise 6(d), $\sigma$ is the identity on $\mathit{KL}$, so $\phi$ is injective.

(c)

By the Tower Theorem $$[\mathit{KL}:F]=[\mathit{KL}:K][K:F].$$

Moreover, Theorem 12.2.5 shows that $\mathrm{Gal}(\mathit{KL}:K)\simeq \mathrm{Gal}(L∕K\cap L)=\mathrm{Gal}(L∕F)$, thus $[\mathit{KL}:K]=[L:F]$. The conclusion is

$$[\mathit{KL}:F]=[K:F][L:F].$$

(d)

So the finite sets $\mathrm{Gal}(\mathit{KL}∕F),\mathrm{Gal}(L∕F)\times \mathrm{Gal}(K∕F)$ have same cardinality, and $\phi$ is injective, therefore $\phi$ is bijective , so $\phi$ is a group isomorphism.