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Let $n \in \mathbb{N}$, then the order of the Galois Group of $x^n-2$ coincide with $n \phi(n)$ for $n\in \{ 1 , \dots , 36 \}$ except for $n=\{ 8, 16, 24, 32 \}$ where this order is $\frac{ n \phi (n)}{2}$ and is easy to prove that for $p$ prime we have that the order of the Galois Group of $x^p-2$ is $p(p-1)$.

What is the order of the Galois Group of $x^n-2 : n\in \mathbb{N}$ is general?

Thanks in advance.

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    $\begingroup$ Perhaps those voting to close could let the author in on the secret. $\endgroup$ – Gerry Myerson Oct 1 '13 at 23:02
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    $\begingroup$ PS: that 25 should be a 24. $\endgroup$ – Steve D Oct 2 '13 at 0:03
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    $\begingroup$ Yes, when $n$ is odd, $m=1$ and the Galois group is the full semidirect product coming from $K$ and $L$. More generally, $m$ must be a power of 2. See this article: link.springer.com/article/10.1007%2FBF02568433 $\endgroup$ – dke Oct 2 '13 at 0:10
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    $\begingroup$ @SteveD: actually I meant to link to this earlier (and open access!) article which actually contains the details: projecteuclid.org/… $\endgroup$ – dke Oct 2 '13 at 0:23
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    $\begingroup$ So to summarize all that has gone on: the degree is $n\phi(n)$, except when $8$ divides $n$, in which case the degree is $n\phi(n)/2$. $\endgroup$ – Steve D Oct 2 '13 at 0:28
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The splitting field of $x^n-2$ over $\mathbb{Q}$ is $K.L$ where $K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{2})$, so the order of the Galois group is $$ [K.L:\mathbb{Q}] = \frac{[K:\mathbb{Q}]\cdot[L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]} = \frac{n \phi(n)}{[K\cap L:\mathbb{Q}]}. $$ It remains to compute $m:=[K\cap L:\mathbb{Q}]$.

First show that $K\cap L = \mathbb{Q}(\sqrt[m]{2})$. For this, note that the norm $N_{L/(K\cap L)}(\sqrt[n]{2})$ is in $K\cap L$. This norm is the product of the $n/m$ conjugates of $\sqrt[n]{2}$ over $K$, so it is the product of $n/m$ of the conjugates of $\sqrt[n]{2}$ over $\mathbb{Q}$, and each of these conjugates has the form $\zeta_n^i \sqrt[n]{2}$. Hence the norm has the form $\zeta_n^i \sqrt[n]{2}^{n/m} = \zeta_n^i \sqrt[m]{2}$. Since this is in $K\cap L$, and $\zeta_n^i\in K$, it follows that $\sqrt[m]{2}\in K$, so $\sqrt[m]{2}\in K\cap L$. But $[\mathbb{Q}(\sqrt[m]{2}):\mathbb{Q}]=m=[K\cap L:\mathbb{Q}]$, so indeed $K\cap L=\mathbb{Q}(\sqrt[m]{2})$.

Next, since $\sqrt[m]{2}\in K$, and $K/\mathbb{Q}$ is abelian, it follows that $\mathbb{Q}(\sqrt[m]{2})/\mathbb{Q}$ is abelian and hence is Galois. Since $\zeta_m\sqrt[m]{2}$ is a conjugate of $\sqrt[m]{2}$ over $\mathbb{Q}$, it follows that $\zeta_m\in\mathbb{Q}(\sqrt[m]{2})$, so in particular $\zeta_m\in\mathbb{R}$, whence $m\le 2$.

The final step is to determine when $\sqrt{2}\in\mathbb{Q}(\zeta_n)$. For this, note that $\sqrt{2}\in\mathbb{Q}(\zeta_8)$ but $\sqrt{2}\notin\mathbb{Q}(\zeta_4)$, and that $\mathbb{Q}(\zeta_r)\cap\mathbb{Q}(\zeta_s)=\mathbb{Q}(\zeta_{(r,s)})$. Thus $\sqrt{2}\in\mathbb{Q}(\zeta_n)$ if and only if $8\mid n$. So the conclusion is that the splitting field of $x^n-2$ over $\mathbb{Q}$ has degree $n\phi(n)$ if $8\nmid n$, and has degree $n\phi(n)/2$ if $8\mid n$.

Question: is there a way to do this without showing that $K\cap L=\mathbb{Q}(\sqrt[m]{2})$, by using from the start that $K\cap L$ is a subfield of $\mathbb{Q}(\sqrt[n]{2})$ which is Galois over $\mathbb{Q}$, and then somehow going directly to the final step?

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See this stackexchange question., or this worked exercise.

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    $\begingroup$ Neither of those references address the nontrivial aspect of the question, namely the computation of $\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\sqrt[n]{2})$. The worked exercise (for $x^8-3$) simply asserts with no justification that the Galois group is as big as possible ("these moves can be done independently of each other"). The stackexchange question is based on the flawed premise that it's easy to compute the Galois group so long as the minimal polynomials of $\zeta_n$ and $\sqrt[n]{2}$ aren't the same. $\endgroup$ – Michael Zieve Oct 2 '13 at 5:30
  • $\begingroup$ I agree..... :( $\endgroup$ – Igor Rivin Oct 2 '13 at 14:56
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    $\begingroup$ The worked exercise has vanished now. $\endgroup$ – Dietrich Burde Dec 23 '16 at 15:11

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