# Galois Group of $\mathbb{Q}(\sqrt[6]{2},\omega)$ where $\omega^2+\omega+1=0$?

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.

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 $$\tau$$.
• 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 $p+q\omega = \zeta_6$? Feb 16, 2023 at 4:25
• 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 $\zeta_6=1+\omega$. Feb 16, 2023 at 7:00
• 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 $60^{\circ}$. Feb 16, 2023 at 7:04