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Have a field $\Bbb F = k(\alpha_1, \ldots, \alpha_n)$. $k^*$ - multiplicative group of k.
Consider $\xi \in k^*, O(\xi) = n$.
Let $\alpha_i^n = a_i \in k, a_i \neq a_j$ if $i \neq j$; let also $k^{*^n} = \{q^n \; | \; q \in k^*\}$.

How to show that $Gal(\Bbb F/k) \cong \langle k^{*^{n}}, a_1, \ldots, a_n \rangle / \langle k^{*^{n}} \rangle$?

Well, ideologically, since $\alpha_i \mapsto a_i$ is injective, we will have at least the proper order of $Gal(\Bbb F/k)$. More than that, perhaps we just get an automorphism of our Galois group. Quotient by $\langle k^{*^{n}} \rangle$ is needed to get rid of other $q^n$ elements of $k^{*^{n}}$. Though this is just some thoughts.

Evgeny
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  • @reuns can I ask, what is $\zeta_i$? – Evgeny Nov 19 '17 at 20:35
  • @reuns well, ok. But if we take mapping $a_i \mapsto <..>a_i$ then $<..>$ must be identity since any automorphism of extension fixes the elements of $k$ ($a_i \in k$). – Evgeny Nov 20 '17 at 08:11
  • @reuns I found a simplier way to prove this. Just consider the mapping $\beta: gk^* \mapsto g^nk^{^n}$. This mapping is surely surjective. Then one can see that $\ker \beta$ is $k^{^n}$, thus a quotient by $k^{*^n}$ gives us the result. But nevertheless we should use knowledge about nth root of unity anyway. – Evgeny Nov 20 '17 at 18:05
  • I don't understand what $\beta, g k^$ is supposed to be. We need an homomorphism (injective and surjective) from the Galois group to the group generated by the $a_i {k^}^n$. – reuns Nov 20 '17 at 18:14
  • Send $[\alpha_i \mapsto \zeta_n^{b_i\frac{n}{O(\alpha_i)}} \alpha_i] \in Gal(F/k)$ to $(a_1^{b_1}{k^}^n,\ldots, a_m^{b_m}{k^}^n)$ then to $(\prod_{i=1}^m a_i^{b_i}){k^}^n$. Show that under certain assumptions (abelian Galois extension..) the composition of both is an injective homomorphism (the question reduces essentially to the kernel of the second map). Then look at the embedding $(\prod_{i=1}^n a_i^{b_i}){k^}^n \to [ \alpha_i \mapsto \zeta_n^{b_i \frac{n}{\text{ord}(a_i (k^*)^n)}} \alpha_i] \in Gal(F/k)$. – reuns Nov 20 '17 at 18:16
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    See https://math.stackexchange.com/a/2529029/300700 – nguyen quang do Nov 20 '17 at 18:43
  • @nguyenquangdo oh, this question belongs to my friend :) this is the second question I have,.too. thanks for the link – Evgeny Nov 20 '17 at 20:02

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