Exercise 12.2.6

Proof. In the proof of Theorem 12.2.5, we cannot use Theorem 7.2.5, since we don’t know if $F\subset \mathit{KL}$ is a Galois extension, so we prefer a direct argument.

(Consider the following counterexample: $F=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$ is a Galois extension of $F$, and $K=\mathbb{Q}(\sqrt[3]{2})$. Then $F\subset \mathit{KL}=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$ is not a Galois extension.)

If $\sigma \in \mathrm{Gal}(\mathit{KL}∕K)$, then $\sigma$ fixes $F\subset K$. Let $\alpha \in L$, and let $f\in F[x]$ be the minimal polynomial of $\alpha$ over $F$. Then $0=\sigma (f(\alpha ))=f(\sigma (\alpha ))$, so $\sigma (\alpha )$ is a root of $f$. Since $F\subset L$ is a normal extension, $\sigma (\alpha )\in L$, thus $\sigma (L)\subset L$. Moreover, $\sigma$ is $F$-linear, injective, and $[L:F]<\infty$, therefore

Write $\sigma {\text{|}}_L$ the map $\sigma {\text{|}}_L:L\to L$ defined by $\alpha \mapsto \sigma (\alpha )$.

(a)

$\mathit{\sigma \tau }\in \mathrm{Gal}(\mathit{KL}∕K)$, so $(\mathit{\sigma \tau }){\text{|}}_L:L\to L$, and also $\sigma {\text{|}}_L\tau {\text{|}}_L:L\to L$.

If $\alpha \in L$, then

$$(\sigma {\text{|}}_L\circ \tau {\text{|}}_L)(\alpha )=\sigma {\text{|}}_L(\tau {\text{|}}_L(\alpha ))=\sigma (\tau (\alpha ))=(\sigma \circ \tau )(\alpha )=(\sigma \circ \tau ){\text{|}}_L(\alpha ),$$

so

$$(\mathit{\sigma \tau }){\text{|}}_L=\sigma {\text{|}}_L\tau {\text{|}}_L.$$

(b)

By part (a), $$\sigma {\text{|}}_L{\sigma }^{-1}{\text{|}}_L=(\sigma {\sigma }^{-1}){\text{|}}_L=({\mathrm{id}}_{\mathit{KL}})|L={\mathrm{id}}_L,$$

and similarly ${\sigma }^{-1}{\text{|}}_L\sigma {\text{|}}_L={\mathrm{id}}_L$. Therefore

$${\sigma }^{-1}{\text{|}}_L={(\sigma {\text{|}}_L)}^{-1}.$$

(c)

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

where $\sigma {\text{|}}_L$ lies in $\mathrm{Gal}(L∕F)$, since $\sigma {\text{|}}_L:L\to L$ is a field automorphism, and for every $\alpha \in F\subset K\subset \mathit{KL}$, $\sigma {\text{|}}_L(\alpha )=\sigma (\alpha )=\alpha$. By part (a), if $\sigma ,\tau \in \mathrm{Gal}(\mathit{KL}∕K)$,

$$\phi (\mathit{\sigma \tau })=(\mathit{\sigma \tau }){\text{|}}_L=\sigma {\text{|}}_L\tau {\text{|}}_L=\phi (\sigma )\phi (\tau ),$$

so $\phi$ is a group homomorphism.

(d)

Let $\sigma$ be an automorphism of $\mathit{KL}$ that is the identity on both $K$ and $L$. Let $M=\{\alpha \in \mathit{KL}|\sigma (\alpha )=\alpha \}$. Then $M$ is a field, the fixed field of $\sigma$ in $\mathit{KL}$. By hypothesis, $K\subset M$ and $L\subset M$. By definition of the compositum $\mathit{KL}$, $\mathit{KL}\subset M$. Since $M\subset \mathit{KL}$ by definition, $\mathit{KL}=M$, so $\sigma$ is the identity on $\mathit{KL}$.

This prove that $\phi$ is injective.