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I know that the splitting field of $x^6-3$ is $\mathbb{Q}(\sqrt[6]{3},i)$, $\mathbb{Q}(\sqrt[6]{3},i)/\mathbb{Q}$ is Galois, and $[\mathbb{Q}(\sqrt[6]{3},i):\mathbb{Q}]=12$.

However I couldn't find which automorphisms generate the Galois group.

$\sqrt[6]{3}$ can be mapped only to $\pm\sqrt[6]{3}$, and $i$ con be mapped only to $\pm i$, but this gives only 4 possible cases of automorphisms.

To summarize, I want to know the generator automorphisms of $\text{Gal}(\mathbb{Q}(\sqrt[6]{3},i)/\mathbb{Q}))$.

user5876164
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    Does the $6$-th root of unity ($\frac{1}{2}+i\frac{\sqrt{3}}{2}$) belong to the field? –  Nov 23 '20 at 09:31
  • The Galois group $G$ is $G\cong \mathbb{Z}/6 \rtimes (\mathbb{Z}/6)^{\times}$, so it is not cyclic. Note that $(\mathbb{Z}/6)^{\times}$ has $\phi(6)=2$ elements. – Dietrich Burde Nov 23 '20 at 09:45
  • @StinkingBishop If you are telling me that the splitting field of $x^6-3$ is not $\mathbb{Q}(\sqrt[6]{3},i)$, now I really don't get what's going on. I really want to know the right splitting field of $x^6-3$, and since I know it is Galois, I also want to know the generator automorphisms of the Galois group. – user5876164 Nov 23 '20 at 10:53
  • I am not saying that. Here is what I am thinking. Let $\varepsilon=\frac{1}{2}+i\frac{\sqrt{3}}{2}$, $\varepsilon^6=1$. Then, the conjugates of $\sqrt[6]{3}$ are not only $\pm\sqrt[6]{3}$ but all the following numbers $\sqrt[6]{3}\cdot\varepsilon^i$ with $i=0,1,2,3,4,5$, and in fact it is more helpful to think of your splitting field as being $\mathbb Q(\sqrt[6]{3},\varepsilon)$ than being $\mathbb Q(\sqrt[6]{3},i)$ (even though both representations are correct). –  Nov 23 '20 at 11:03
  • @StinkingBishop So to summarize your comment in my point of view, $\sqrt[6]{3}\mapsto\pm\sqrt[6]{3}$ and $\sqrt[6]{3}\zeta_6\mapsto\sqrt[6]{3}\zeta_6^n$ for $n=0,\cdots,5$ will give 12 automorphisms. Is that right? if $\sqrt[6]{3}\zeta_6\mapsto\sqrt[6]{3}$, then I don't that will be an automorphism. – user5876164 Nov 23 '20 at 11:10
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    No. $\sqrt[6]{3}\mapsto \sqrt[6]{3}\zeta_6^n (n=0,1,2,3,4,5)$ and $\zeta_6\mapsto\begin{cases}\zeta_6^5=\overline{\zeta_6}\\zeta_6\end{cases}$. (Or, $i\mapsto \pm i$.) Note $\sqrt[6]{3}\mapsto -\sqrt[6]{3}$ is a special case of the former with $n=3$ ($\zeta_6^3=-1$). –  Nov 23 '20 at 11:12
  • @StinkingBishop Thanks! Now I get what was going wrong. – user5876164 Nov 23 '20 at 11:15
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    (Cont'd) In fact, the automorphism mapping $\sqrt[6]{3}\mapsto\sqrt[6]{3}\zeta_6^5$ (and $\zeta_6$ to itself) will also map $\sqrt[6]{3}\zeta_6\mapsto \sqrt[6]{3}$, so there is an automorphism of the type that you did not previously believe to exist. The point is, over $\mathbb Q$, all of the roots $\sqrt[6]{3}$ are "equivalent" and $\sqrt[6]{3}$ can map into any of them simply because $x^6-3$ is irreducible over $\mathbb Q$ (Eisenstein for $p=3$) and so it is a minimal polynomial (over $\mathbb Q$) for all of them. –  Nov 23 '20 at 11:19

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