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I'm trying to show that $\mathbb{Q}(\sqrt[6]{-3})/\mathbb{Q}$ is a Galois extension. I would like to show that $\mathbb{Q}(\sqrt[6]{-3})/\mathbb{Q}$ is the splitting field of $f(x) = x^6 - 3$ which is irreducible by Eisenstein. If $\zeta_6 = e^{\pi i/3}$, then the roots of $f(x)$ are $\zeta_6^k\sqrt[6]{3}$ for $0 \leq k < 6$. I was able to show that $\zeta_6 \in \mathbb{Q}(\sqrt[6]{-3})$ because $\zeta_6 = \frac{1}{2} \pm \frac{\sqrt{-3}}{2}$ (depending on the choice of root of $-1$ chosen). How would I show that $\sqrt[6]{3} \in \mathbb{Q}(\sqrt[6]{-3})$ using this? From there, I would be able to deduce that all the roots of $f(x)$ are in $\mathbb{Q}(\sqrt[6]{-3})$.

As for the Galois group, because $f(x)$ is irreducible, we must have that this is a field extension of degree 6 so the Galois group is either isomorphic to $S_3$ or $\mathbb{Z}/6\mathbb{Z}$. How would I determine which group it is? I would appreciate any help!

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    $\begingroup$ It should be $x^6+3$ $\endgroup$
    – Evariste
    Jun 10 '21 at 15:25
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I'm not quite sure why you are looking at $x^6 - 3$ instead of $x^6 + 3$. I shall look at the latter.


For concreteness, fix $F = \Bbb Q(\sqrt[6]{3}\iota)$, where $\iota$ is an imaginary root of $-1$. (Note that $(\sqrt[6]{3}\iota)^6$ is indeed $-3$.)

Let $\alpha = \sqrt[6]{3}\iota$. Then, $\alpha^3 = -\iota\sqrt{2}$ and thus, $\zeta_6 := \exp(2\pi\iota/6) \in F$.

Note that $\zeta_6^k\alpha$ are roots of $x^6 \color{red}{+} 3$. Thus, $F$ contains all the roots of $x^6 + 3$ and hence, is its splitting field (being generated by one of them).

In particular, it is Galois.


Now, to compute the Galois group, we see what are the isomorphisms $F \to F$ (which fix $\Bbb Q$, but this happens automatically).

Consider $\sigma_i : F \to F$ defined by $\alpha \mapsto \zeta_6^i \alpha$ for $i = 0, \ldots, 5$.
(Does it make sense why we can do this?)

Now, it is easy to see that $\sigma_0, \ldots, \sigma_5$ are distinct (and thus, we have found all) and the composition makes it clear that the Galois group is isomorphic to $\Bbb Z/6 \Bbb Z$.

Looking at the image of $\alpha$ under the above maps shows that $\sigma_0, \ldots, \sigma_5$ are distinct. However, the group is not isomorphic to $\Bbb Z/6\Bbb Z$. It may be tempting (it certainly was, for me) to think that $$\color{red}{\sigma_i(\sigma_j(\alpha)) = \zeta_6^{i + j} \alpha}$$ but the equation above is $\color{red}{\text{not true}}$ in general since $\sigma_i$ need not fix $\zeta_6$. (Thanks to leoli1 for pointing this out!)


Let us show that there are two elements of order $2$.
$\sigma_3$ clearly has order $2$ since $$\sigma_3(\sigma_3(\alpha)) = \sigma_3(\zeta_6^3\alpha) = \sigma_3(-\alpha) = -\zeta_6^3\alpha = \alpha.$$

We now show that $\sigma_1$ also has order $2$. For this, we do some calculations. First, note that $\zeta_6 = \frac{1}{2}(1 - \alpha^3).$ Thus, we have $$\sigma_1(\alpha) = \zeta_6\alpha = \frac{\alpha}{2} - \frac{\alpha^4}{2}.$$ Applying $\sigma_1$ again gives \begin{align} \sigma_1(\sigma_1(\alpha)) &= \frac{\zeta_6\alpha}{2} - \frac{\zeta_6^4\alpha^4}{2} \\ &= \frac{\zeta_6\alpha}{2} + \frac{\zeta_6\alpha^4}{2} \\ &= \frac{1}{2} \alpha \zeta_6 (1 + \alpha^3) \\ &= \frac{1}{2} \alpha \frac{1}{2}(1 - \alpha^3)(1 + \alpha^3) \\ &= \frac{\alpha}{4}(1 - \alpha^6) = \alpha. \end{align}

Thus, the Galois group cannot be $\Bbb Z/6 \Bbb Z$. This leaves us only with $S_3$.

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  • $\begingroup$ You're right, I should have been looking at $x^6 + 3$! As for the $\sigma_i$, this can be done because the Galois group acts transitively on the roots of $x^6 + 3$, right? And lastly, why is it easy to see that the five automorphisms you have written down are distinct? Can you give me a hint as to how I can see this? $\endgroup$
    – Nick
    Jun 10 '21 at 15:51
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    $\begingroup$ 1. It can be done because of the following general phenomenon: Suppose $E/F$ is an extension of fields and $\alpha \in E$ is algebraic over $F$. Then, you can define a field homomorphism $F(\alpha) \to E$ which fixes $F$ by mapping $\alpha$ to any root of its irreducible polynomial. (No assumption of Galois-ness anywhere.) In our case, we have that $x^6 + 3$ is irreducible (by Eisenstein) and thus, what I've given actually defines a map. $$$$ 2. It is indeed easy to see why they are distinct. Simply look at where $\alpha$ goes under them! :-) $\endgroup$ Jun 10 '21 at 16:00
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    $\begingroup$ Both your points makes sense, thanks so much! :) $\endgroup$
    – Nick
    Jun 10 '21 at 16:09
  • $\begingroup$ The Galois group is not isomorphic to $\Bbb Z/6\Bbb Z$ (note that not all $\sigma_i$ fix $\zeta_6$) $\endgroup$
    – leoli1
    Jun 10 '21 at 16:23
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    $\begingroup$ @AryamanMaithani I see, thank you for clarifying that for me! :) $\endgroup$
    – Nick
    Jun 11 '21 at 21:21

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