# Is $G(\mathbb{Q}(\sqrt[3]{2}), i\sqrt{3})/\mathbb{Q}(i\sqrt{3}))\simeq \left \langle \mathbb{Z_3}, + \right \rangle$?

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.

• Why do you guess $\sigma_2(\alpha_2)=\alpha_1$ instead of $\alpha_3$? Feb 16, 2021 at 4:30
• Note that $\frac{-1+i\sqrt{3}}{2} \in \mathbb Q(i\sqrt3)$; so if $\sigma_2(\alpha_1) = \alpha_2$, then $\sigma_2(\alpha_2) = \sigma_2(\sqrt[3]2) \sigma_2(\frac{-1+i\sqrt{3}}{2}) = \alpha_2 \frac{-1+i\sqrt{3}}{2} = \alpha_3$ since $\sigma_2$ fixes $\mathbb Q(i\sqrt3)$. Feb 16, 2021 at 4:35
• @awllower I thought $\sigma_2$ should fix $i\sqrt{3}$, but did missed $\sigma_2$ also map $\alpha_1$ to other conjugate. Thx Feb 16, 2021 at 4:45
• @azif00 Thanks. now I know what did I missed. Feb 16, 2021 at 4:47

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 $$G(\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q}(i\sqrt{3})) \cong \mathbb Z_3$$.