Let $\mathbb{Q}$ be the field of rational number, then the splitting field of $x^{3}-2$ over $\mathbb{Q}$ is $\mathbb{Q}[\sqrt[3]{2}, \zeta]$ where $\zeta\neq 1$ be the third root of unity.

The element of $\mathbb{Q}[\sqrt[3]{2}, \zeta]$ are reprensented by $a+b\sqrt[3]{2}+c\sqrt[3]{4}+d\zeta+e\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}$.

Denote \begin{cases} \sqrt[3]{2} \mapsto \sqrt[3]{2}\zeta\\\zeta \mapsto \zeta \end{cases} \begin{cases} \sqrt[3]{2} \mapsto \sqrt[3]{2}\\ \zeta \mapsto \zeta^{2}=-1-\zeta \end{cases}

by $\sigma$ and $\tau$ respectively.

Under the action of $\sigma\tau$, $a+b\sqrt[3]{2}+c\sqrt[3]{4}+d\zeta+e\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}$ maps to $(a-d)+e\sqrt[3]{2}-c\sqrt[3]{4}-d\zeta+b\zeta\sqrt[3]{2}+(f-c)\zeta\sqrt[3]{4}$.

Now if $a+b\sqrt[3]{2}+c\sqrt[3]{4}+d\zeta+e\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}$ is fixed by $\sigma\tau$ then we must have $c=d=0, b=e$.

And then $a+b\sqrt[3]{2}+c\sqrt[3]{4}+d\zeta+e\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}=a+b\sqrt[3]{2}+b\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}$.

I do not know how to find the fixed subfield of $\mathbb{Q}[\sqrt[3]{2},\zeta]$ under the action of $\sigma\tau$. As the book of Dummit and Foote about Abstract Algebra, that fixed field is $\mathbb{Q}[\sqrt[3]{2}\zeta]$, but I do not know how to claim that.

An arbitrary element of $\mathbb{Q}[\sqrt[3]{2}\zeta]$ is of the form $a+b\sqrt[3]{2}\zeta +c\sqrt[3]{4}\zeta^2$ which is generally different from $a+b\sqrt[3]{2}+b\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}$.

I do not know what and where am I wrong. Please show me, and give me some hint or strategy to find the fixed field in genenral.


Important edit: The fixed field as in the book of Dummit Foote is $\mathbb{Q}[\zeta^2\sqrt[3]{2}]$, not $\mathbb{Q}[\zeta\sqrt[3]{2}]$ as I wrote above and so it ís easy to deduce from my argument above. Sorry for my stupid mistake. Moderator please delete this question.

  • 1
    $\begingroup$ And I think $\zeta \sqrt[3]{2}$ is not in the fixed field, as $\sigma \tau $ maps it to $\sqrt[3]{2}$ if I haven't miscalculated. So the fixed field would be $\Bbb Q(\zeta \sqrt[3]{4})$. $\endgroup$
    – benh
    Jan 3, 2014 at 12:49
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    $\begingroup$ I agree with benh. We have $\sigma(\tau(\zeta^2\root3\of2))=\zeta^2\root3\of2$, so that is in the fixed field. $\endgroup$ Jan 3, 2014 at 12:52
  • 1
    $\begingroup$ And $(\zeta^2 \sqrt[3]{2})^2 = \zeta \sqrt[3]{4}$ gives the same fixed field. $\endgroup$
    – benh
    Jan 3, 2014 at 12:55
  • $\begingroup$ If you take the opposite convention on the meaning of the product of Galois group elements (i.e. apply $\sigma$ then $\tau$), then $\zeta \sqrt[3]{2}$ is fixed. $\endgroup$
    – user14972
    Jan 3, 2014 at 15:00

1 Answer 1


Use Galois-theory:

You know that the Galois-group $G$ of $x^3-2$ has order $6 = [\Bbb Q(\sqrt[3]{2},\zeta_3) :\Bbb Q]$. As the intermediate extension $\Bbb Q(\sqrt[3]{2}) / \Bbb Q$ is not normal, $G$ cannot be abelian. Thus $G = S_3$ is the only remaining group. The group $S_3$ has three subgroups of index $3$ and one subgroup of index $2$ which by the main theorem of Galois-theory correspond to the four intermediate extensions of $\Bbb Q(\sqrt[3]{2},\zeta_3) / \Bbb Q$.

So let's analyze the fixed field of $\sigma \tau$: As $S_3$ is not cyclic, the group $0 \neq \langle \sigma \tau \rangle \subset S_3$ is a proper subgroup corresponding to an intermediate extension that is either of degree $2$ or of degree $3$.

Now we verify that $\sqrt[3]{4} \zeta_3$ is fixed under $\sigma \tau$, then we know that $\Bbb Q(\sqrt[3]{4} \zeta_3)$ is a sub-field of the fixed field of $\sigma \tau$. As it is already of degree $3$, it must be the whole fixed field.

  • $\begingroup$ Are there any easier or less complicated ways? As in my argument, I found that $a+b\sqrt[3]{2}+b\zeta\sqrt[3]{2}+f\zeta\sqrt[3]{4}$ is fixed under $\sigma\tau$'s action. Does this help? Can we use it to find the fixed field? $\endgroup$
    – Arsenaler
    Jan 3, 2014 at 18:05
  • $\begingroup$ Ok, here is something else: We know that $\sqrt[3]{4} \zeta_3$ has degree $<6$. Hence, by considering the degree we find that there cannot be a field between $\Bbb Q(\sqrt[3]{4})$ and $\Bbb Q(\sqrt[3]{2},\zeta_3) / \Bbb Q$. So $\Bbb Q(\sqrt[3]{4})$ is already the whole fixed field. That's maybe less complicated. $\endgroup$
    – benh
    Jan 3, 2014 at 20:47

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