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.
Thanks.
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 Answer
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.
