# Galois group of $f := X^6 - 6 ∈ \Bbb Q[X]$

A quick sanity check:

A splitting field for $$f$$ over $$\Bbb Q$$ is $$L := \Bbb Q(\zeta_3, \sqrt[6]{6})$$. It is of degree 12 over $$\Bbb Q$$, so Gal$$(f)$$ will be a group of order 12. An automorphism of $$L$$ must send $$\sqrt[6]{6}$$ to $$± \zeta_3^k \sqrt[6]{6}$$ for some $$k ∈ \{0,1,2\}$$, but it must also send $$\zeta_3$$ to $$\zeta_3^k$$. Does this provide enough information to determine the Galois group?

• Not the same $k$. Commented Jun 13, 2020 at 19:29
• Imagine those six roots on the complex plane. They form a regular hexagon, right? If $\tau:\zeta_6\mapsto \zeta_6$, then $\tau$ must rotate the vertices of the hexagon because the image of the next zero depends on that of the previous: $\tau(x\zeta_6)=\tau(x)\zeta_6.$ On the other hand, if $\tau:\zeta_6\mapsto \overline{\zeta_6}$, then the order of rotation is reversed: $\tau(x\zeta_6)=\tau(x)\overline{\zeta_6}$. Therefore.... Commented Jun 13, 2020 at 19:35

## 1 Answer

I think rather than $$\zeta_3$$ a root of $$X^2 + X + 1$$ you need $$\zeta_6$$ a root of $$X^2 - X + 1$$.

You will have the complex conjugation automorphism $$\tau$$.

And 6 different maps $$\sigma \sqrt[6]{6} = \zeta_6^r \sqrt[6]{6}$$ for $$r$$ being $$1$$ up to $$6$$.

The complex conjugation $$\tau$$ map interacts with the $$\sigma$$ maps by reversing the cycle, similar to the dihedral group.

• That's all there is to it :-) Commented Jun 13, 2020 at 19:38
• Thanks, but why go to ζ₆? And what does $X² - X + 1$ have to do with it? Commented Jun 13, 2020 at 20:05
• Ah, I was not familiar with the $n^{\text{th}}$ cyclotomic polynomial, ($n$ not prime) Commented Jun 13, 2020 at 20:16