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I'm asked to find the Galois group of irreducible polynomials of the form $X^4-k$ and $X^6-k$ for $k\in\mathbb{Q}_{>0}$.

I think I got the solution for $f=X^4-k$ but I am not entirely sure if it is correct. For starters, we have $\Omega^f=\mathbb{Q}(\sqrt[4]{k},i)$ and we have $[\Omega^f:\mathbb{Q}]=8$ because of the Tower Law and the fact that $\mathbb{Q}(\sqrt[4]{k})$ is a real field. Therefore, Gal($f$) must be a transitive subgroup of order 8 of $S_4$. The only subgroup of $S_4$ that satisfies such conditions is $D_4$, hence Gal($f$)=$D_4$. Would this be correct? I know in this case Gal($f$) can also be found identifying automorphisms directly, but I think this way is harder.

Now, for $g=X^6-k$, $\Omega^g=\mathbb{Q}(\sqrt[6]{k},\xi_6)$ and $\xi_6=e^{i\pi/3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i\implies\Omega^g=\mathbb{Q}(\sqrt[6]{k},\sqrt{3}i)\implies[\Omega^g:\mathbb{Q}]=12$ for the same reasons. Then Gal($g$) is a transitive subgroup of order $12$ of $S_6$, but I think these are much harder to characterize. How should I proceed? I have just learned about semidirect products, so it might have something to do with that.

Any hints?

simo210
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cut
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  • Should there be some restriction on $k$? What if $k=2^4$ for example? – Greg Martin Apr 27 '23 at 10:14
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    That’s what I have to find too. For the degree 4 polynomial, I can’t find any reason why I should distinguish cases, and neither for the degree 6, since you know a priori the polynomial is irreducible. – cut Apr 27 '23 at 10:31
  • ah I missed the irreducibility assumption, that's probably all you need I agree – Greg Martin Apr 27 '23 at 17:12
  • I have another question. Does $[\Omega^f:\mathbb{Q}]=8$ imply that the order of $Gal(f)$ is 8? Because if it doesn't, I couldn't simply state that $Gal(f)=D_4$. – cut Apr 29 '23 at 11:08
  • For degree $6$, there are only $2$ transitive subgroups of $S_6$ of order $12$, namely, $D_6$ and $A_4$. Can you exploit the information you have to distinguish between these two possibilities? – Travis Willse Apr 29 '23 at 20:48

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This is a special case of the classical question of the Galois Group of irreducible polynomials $X^n-k.$ See e.g. here for the general answer (see in particular the reply by Dietrich Burde): The Galois Group is $\mathbb Z/n \rtimes (\mathbb Z/n)^\times$ if and only if $k$ is odd, or $n$ is even and $\sqrt k \notin ℚ(_)$. For $n=4, n=6,$ the only exceptional cases which would need special treatment are $n=6$ and $k=-3a^2$, $a \in \mathbb Q $. In this case, $(\sqrt[6]{k})^3=$ $\sqrt k = i\sqrt 3a \in \mathbb Q(\zeta_6)$, hence $\mathbb Q(\sqrt[6]{k}) \cap \mathbb Q(\zeta_6) \ne \mathbb Q$ and the degree of the splitting field would not be equal to 12 anymore. For $n=4$ one always obtains $\mathbb Z/4 \rtimes (\mathbb Z/4)^\times$ which is isomorphic to $D_4$, as stated by the OP.

nor
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