I wonder I have some misunderstanding of a concept of automorphism that leaving some field fixed. \
The Problem in the text(Fraleigh, p.402, 7th) is:
Referring to Example 50.9, show that
$$G(\mathbb{Q}(\sqrt[3]{2}), i\sqrt{3})/\mathbb{Q}(i\sqrt{3}))\simeq \left \langle \mathbb{Z_3}, + \right \rangle$$
I thought the elements of $G(\mathbb{Q}(\sqrt[3]{2})$ leaving $\mathbb{Q}(i\sqrt{3})$ fixed so that for the three zeros $$\alpha_{1}=\sqrt[3]{2}, \ \alpha_2=\sqrt[3]{2}\frac{-1+i\sqrt{3}}{2},\ \alpha_3=\sqrt[3]{2}\frac{-1-i\sqrt{3}}{2}$$ of $x^3-2$, I guessed the group consist of
$\sigma_1;\ \sigma_1=\iota$ (the identity map)
$\sigma_2;\ \sigma_2(\alpha_1)=\alpha_2$, $\sigma_2(\alpha_2)=\alpha_1$, $\sigma(\alpha_3)=\alpha_3$
$\sigma_3;\ \sigma_3(\alpha_1)=\alpha_3$, $\sigma_2(\alpha_2)=\alpha_2$,
$\sigma(\alpha_3)=\alpha_1$.
But in this case, the three automorphisms does not even form a group under function composition. What may I have missed? Thanks.
If $F := \mathbb{Q}(i\sqrt{3})$, then $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q}(i\sqrt{3}) = F(\sqrt[3]{2})/F$ and so $$[\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3}) : \mathbb{Q}(i\sqrt{3})] = [F(\sqrt[3]2) : F] = \deg (\min(F,\sqrt[3]2)).$$ Now $x^3-2 \in \mathbb Q[x] \subseteq F[x]$ clearly annihilates $\sqrt[3]2$, and since $$x^3-2 = (x-\sqrt[3]2)(x-\alpha_2)(x-\alpha_3)$$ we have that $x^3-2$ is irreducible over $F$ (since $\sqrt[3]2 \notin F$). Thus $\min(F,\sqrt[3]2) = x^3-2$ and then $[\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3}) : \mathbb{Q}(i\sqrt{3})] = 3$, which of course implies that $G(\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q}(i\sqrt{3})) \cong \mathbb Z_3$.