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I'm pretty new to this field. I know there are $5!=120$ permutations of the roots. R. Borcherds defined the Galois Group as all the permutations preserving algebraic relations. I'm trying to reconstruct the group by this definition, to no avail. If $$\alpha_n=2^{1/5} \zeta^n, \qquad \zeta=e^{\frac{2\pi i}{5}}$$ for $n=0,1,2,3,4$ is a root, then an algebraic relation could for example be $$\alpha_m \alpha_n = \alpha_k \alpha_l$$ iff $$m+n=k+l \mod 5 .$$ Is this approach even viable, or do you never know whether you have all relations between the roots?

Diger
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  • Since an automorphism preserves addition and multiplication, it is completely determined by its action on $2^{1/5}$ and $\zeta$. The former must map to a fifth root of $2$ and the latter must map to a fifth root of unity (not $1$). – Douglas Molin Nov 17 '21 at 19:07
  • Since $n$ is odd, the Galois group is $G\cong \mathbb{Z}/5 \rtimes (\mathbb{Z}/5)^{\times}\cong \Bbb Z/20$ (see the duplicate). – Dietrich Burde Nov 17 '21 at 19:09

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First, note that $[\mathbb{Q}(\zeta); \mathbb{Q}] = 4$. And note that $[\mathbb{Q}(\sqrt[5]{2}); \mathbb{Q}] = 5$. Therefore, we have $[\mathbb{Q}(\zeta, \sqrt[5]{2}); \mathbb{Q}] = 20$.

So there are a total of 20 automorphisms in the Galois group.

Now any isomorphism must send $\zeta$ to $\zeta^n$ for some $n \in \{1, 2, 3, 4\}$, since these are the only roots of $x^4 + x^3 + x^2 + 1$. And any automorphism must send $\sqrt[5]{2}$ to $\zeta^n \sqrt[5]{2}$ for some $n \in \{0, 1, 2, 3, 4\}$, since these are the only roots of $x^5 - 2$. Picking where $\zeta$ and $\sqrt[5]{2}$ go uniquely determines the automorphism. Therefore, any choice of where $\zeta$ and $\sqrt[5]{2}$ go is a valid one.

Thomas Andrews
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Mark Saving
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  • @ThomasAndrews Wasn't paying attention and confused $\xi$ with $\zeta$ – Mark Saving Nov 17 '21 at 19:12
  • @DietrichBurde Since we know that $4$ and $5$ are coprime, it follows that the total degree must be 20. – Mark Saving Nov 17 '21 at 19:17
  • Yes, indeed it is clear (it would not be true for $x^{10}-5$, say). – Dietrich Burde Nov 17 '21 at 19:19
  • Building on your answer, is it possible to write down the field extensions for the general cubic? For example, the quintic $$x^5-x+1$$ has roots $x_1,...,x_5$ and now how do I know $[\mathbb{Q}(x_1,x_2,x_3,x_4,x_5);\mathbb{Q}]=120$? – Diger Nov 18 '21 at 15:18