# Showing the extension $\mathbb{Q}(\sqrt[6]{-3})/\mathbb{Q}$ is Galois and determining its Galois group.

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!

• It should be $x^6+3$ Jun 10 '21 at 15:25

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 $$S_3$$.

• 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?
– Nick
Jun 10 '21 at 15:51
• 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! :-) Jun 10 '21 at 16:00
• Both your points makes sense, thanks so much! :)
– Nick
Jun 10 '21 at 16:09
• The Galois group is not isomorphic to $\Bbb Z/6\Bbb Z$ (note that not all $\sigma_i$ fix $\zeta_6$) Jun 10 '21 at 16:23
• @AryamanMaithani I see, thank you for clarifying that for me! :)
– Nick
Jun 11 '21 at 21:21