I'm having some serious confusion regarding this problem. My professor is saying that $\mathbb{Q}(\sqrt[6]{2},\omega)_{\mathbb{Q}}$ is a degree $12$ Galois extension with Galois group isomorphic to $D_6$, where $\omega^2+\omega+1=0$.

I initially thought I understood, until I computed $\omega = e^{\frac{\pm 2 \pi i}{3}}$, so $\omega = \zeta^2_6$ or $\zeta^4_6$ where $\zeta_6$ is the standard $6$th root of unity. Now, I completely see how the extension has degree $12$, but I don't see how it's Galois group is $D_6$, since my professor said the order $6$ automorphism (corresponding to the order $6$ generator of $D_6$) is given by $\tau(\sqrt[6]{2}) = \omega \sqrt[6]{2}$ and $\tau(\omega) = \omega$.

How is this possible? Since $\omega$ isn't a primitive $6$th root of unity, it's pretty clear to me that $\tau^3$ is the identity automorphism as $\omega^3=1$.

Furthermore, I don't see how $\mathbb{Q}(\sqrt[6]{2},\omega)$ contains $\zeta_6 \sqrt[6]{2}$ or $\zeta^5_6 \sqrt[6]{2}$ (which are roots of $x^6-2$) since they certainly aren't contained in the field field extension $\mathbb{Q}\sqrt[6]{2}$ (which isn't Galois over $\mathbb{Q}$) and they aren't contained in $\mathbb{Q}(\omega)$ since again, $\omega$ isn't a primitive $6$th root of unity.

Now, IF $\omega$ was a primitive $6$th root of unity, then I agree, we get $D_6$. But in this case, I don't see how this is possible.

I'd be very appreciative if someone could explain what's going on here, thank you.


1 Answer 1


While $\omega$ is not a primitive $6$th root of unity,$-\omega$ is, so $\zeta_6\in \mathbb Q(\omega)$. This is no surprise, since $\phi(2n)=\phi(n)$ for any odd $n$, that is $\mathbb Q(\zeta_{2n})=\mathbb Q(\zeta_n)$ for any odd $n$, or in another way, it's easy to check $-\zeta_n$ is a primitive $2n$-th root of unity.

The order $6$ element in $D_6$ should be $\tau(\sqrt[6]{2})=-\omega\sqrt[6]{2}$ though.

To give more details, we can easily see $x^6-2$ is irreducible by Eisenstein. Since $\omega$ is imaginary, hence not in $\mathbb Q(\sqrt[3]{2})$, and we have $[\mathbb Q(\sqrt[6]{2},\omega):\mathbb Q]=[\mathbb Q(\sqrt[6]{2},\omega):\mathbb Q(\sqrt[6]{2})][\mathbb Q(\sqrt[6]{2}):\mathbb Q]=12$. From here, we also have $[\mathbb Q(\sqrt[6]{2},\omega):\mathbb Q(\omega)]=6$.

By the above, $\mathbb Q(\sqrt[6]{2},\omega)$ contains all the six roots of $x^6-2$, hence Galois over $\mathbb Q(\omega)$, therefore there must be a $\tau\in \text{Gal}(\mathbb Q(\sqrt[6]{2},\omega) | \mathbb Q(\omega))$ that sends $\sqrt[6]{2}$ to $-\omega\sqrt[6]{2}$. It's easy to check the element has order $6$, then we know the Galois group $\text{Gal}(\mathbb Q(\sqrt[6]{2},\omega) | \mathbb Q(\omega))$ is generated by this single element $\tau$.

  • $\begingroup$ I see, so $-\omega = -e^{2 \pi i / 3}$ is a primitive $6$th root of unity. How does that get $\zeta_6 \in \mathbb{Q}(\omega)$? $\mathbb{Q}(\omega) = \{p+\omega q \: | \: p,q \in \mathbb{Q}\}$ since the degree of the extension is $2$, so for what $p,q$ do we get $p+q\omega = \zeta_6$? $\endgroup$
    – Isochron
    Feb 16, 2023 at 4:25
  • $\begingroup$ Algebraically, all the primitive $n$-th roots of unity are equivalent to each other, so it doesn't matter which exact complex number is $\zeta_6$. But if you have to know, then $\zeta_6=e^{\frac{2\pi i}{6}}=e^{\frac{\pi i}{3}}=-e^{-\frac{2\pi i}{3}}$ as by Euler $e^{\pi i}=-1$. And $e^{-\frac{2\pi i}{3}}=\omega^2=-1-\omega$, hence $\zeta_6=1+\omega$. $\endgroup$ Feb 16, 2023 at 7:00
  • $\begingroup$ This is also quite intuitive, as if we draw the unit circle, $\omega$ has argument angle $360/3=120^{\circ}$, and from either trignometry or tri functions, we can easily see the real part of $\omega$ is $-1/2$, hence by adding $1$ to it, we get the symmetric point around the $y$-axis on the complex plane, whose argument angle is exactly $60^{\circ}$. $\endgroup$ Feb 16, 2023 at 7:04

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