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I want to construct a Galois extension of $\mathbb{Q}$ that has Galois group $\mathbb{Z}_3$ (the group with 3 elements). I know that it is possible (Galois group that is isomorphic to $\mathbb{Z}_3?$) but I would like to see an application of Artin's theorem as stated in Lang's algebra (it's theorem 1.8 in the first section dedicated to Galois theory). Artin's theorem states

Given a field $K$ and a finite group $G$ of automorphisms of $K$, then $K/K^F$, where $K^F$ is field fixed by $G$ (i.e. the subfield of $K$ such that for all $\sigma \in G$ we have $\sigma(x)=x$ for all $x\in K^F$) is a Galois extension with group $G$.

Consider $K=\mathbb{Q}(\alpha,\alpha\omega,\alpha\omega^2)$ where $\alpha=\sqrt[3]{2}$ and $\omega$ is the primitive third root of unity (this is in fact the splitting field of $X^3-2$) and $\mathbb{Z}_3$ as the subgroup of $S_3$, $G=\langle(1,2,3)\rangle=\{e,(1,2,3),(1,3,2) \}$. $G$ is a group of automorphisms of $K$ acting by permutation of the roots and the fixed field is $\mathbb{Q}$, therefore $K/\mathbb{Q}$ is a Galois extension with permutation group $G$. However this doesn't seem correct since the Galois group of $K/\mathbb{Q}$ is $S_3$. How can I fix this?

If we instead consider $G=\mathbb{Z}_2=\langle(1,2)\rangle$ as a subfield of $S_3$ then by the same argument we would get that $K/\mathbb{Q}(\alpha\omega^2)$ is a Galois extension with Galois group $G$, which I would say is correct. Am I making sense or am I mistaken?

Thanks in advance.

PunkZebra
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  • Not really since I would like to see an application of Artin's theorem, I just picked $\mathbb{Z}_3$ because it was an easy group – PunkZebra Sep 13 '22 at 12:18
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    Your construction with the field extension $K/\Bbb Q$ cannot have Galois group $C_3$, since the degree is $6$ and not $3$. So as you said, you need to compute $K^G$ correctly. See also here. So $G$ then is the Galois group of $K/K^G$. – Dietrich Burde Sep 13 '22 at 12:32

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Maybe I figured out something. By working on what @Dietrich Burde said in the comments I think that I have the wrong construction of the fixed field. Suppose that $\sigma=(1,2,3)$, then by Vieta's formulas we have that the sum of the roots is $0$, so that they are not linearly independent. The map $\sigma$ acts by permuting the roots: $$ \sigma(\alpha)\mapsto\alpha\omega \\ \sigma(\alpha\omega)\mapsto \alpha\omega^2 \\ \sigma(\alpha\omega^2)\mapsto \alpha $$ however we have also the following relations: $$ \alpha\omega^2=\sigma(\alpha\omega)=\sigma(\alpha)\sigma(\omega)=\alpha\omega\sigma(\omega) $$ which means that $\sigma(\omega)=\omega$. Moreover this implies that $\sigma(\alpha\omega^2)=\alpha$ but more importantly we have that the fixed field is generated by $\omega$ and is therefore $\mathbb{Q}(\omega)$.

PunkZebra
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  • Almost right. As you showed, you have $\sigma(\omega)=\omega$, however it's not the case that $\sigma(\alpha)=\alpha$. In fact, the associated fixed field is $\mathbb{Q}(\omega)$. – quasi Sep 13 '22 at 13:25
  • @quasi I don't even know why I wrote that, thank you very much! – PunkZebra Sep 13 '22 at 14:14
  • Perhaps edit your answer to incorporate the relevant corrections, and then I'll delete my comments. – quasi Sep 13 '22 at 15:45