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.

  • $\begingroup$ Why do you guess $\sigma_2(\alpha_2)=\alpha_1$ instead of $\alpha_3$? $\endgroup$
    – awllower
    Feb 16, 2021 at 4:30
  • $\begingroup$ 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 $\mathbb Q(i\sqrt3)$. $\endgroup$
    – azif00
    Feb 16, 2021 at 4:35
  • 1
    $\begingroup$ @awllower I thought $\sigma_2$ should fix $i\sqrt{3}$, but did missed $\sigma_2$ also map $\alpha_1$ to other conjugate. Thx $\endgroup$
    – JJLEE
    Feb 16, 2021 at 4:45
  • $\begingroup$ @azif00 Thanks. now I know what did I missed. $\endgroup$
    – JJLEE
    Feb 16, 2021 at 4:47

1 Answer 1


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

  • $\begingroup$ Thank you! now I know what have I mistaken, by the comment you posted. $\endgroup$
    – JJLEE
    Feb 16, 2021 at 4:48

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