As the title says, I'm looking for $\alpha \in \mathbb{\overline{Q}}$ such that $\operatorname{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong \mathbb{Z}_5$.

My idea was to look for a Galois extension with group $\mathbb{Z}_n$ such that $5 \mid n$. it has $\mathbb{Z}_5$ as a subgroup, so I'll find its generator and look at the field it fixes, it should be an extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}_5$, so now I only need to find a generator for this extension.

Obviously my solution isn't straightforward but I tried it because I know $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong \mathbb{Z}_n^\times$, (where $\zeta_n$ is an n-th primitive root of unity) so I quickly found that $\operatorname{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q}) \cong \mathbb{Z}_{11}^\times \cong \mathbb{Z}_{10}$ and the automorphism denoted by $\sigma$ and defined by $\zeta_{11} \mapsto \zeta_{11}^4$ is of order 5.

but the problem now is that looking for the field fixed by $\sigma$, I get:
$$ \zeta_{11}^j=\sigma(\zeta_{11}^j)=\zeta_{11}^{4j} \Rightarrow \zeta_{11}^{3j}=1 \Rightarrow 11 \mid 3j \Rightarrow 11 \mid j $$ Which means, as far as a I understand, that the field fixed by $\sigma$ is only $\mathbb{Q}$.

I'd like to know what part I got wrong, and also I'd like to hear any other solutions to the original problem, not using my idea.

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    $\begingroup$ I think you meant to write $\alpha \in \overline{\mathbb Q}$ in the title and again in the first line of the question. $\endgroup$ – Ben Aug 14 at 9:24
  • $\begingroup$ @Ben yes of course, thanks for correcting $\endgroup$ – GuySa Aug 14 at 10:03

None of the $\zeta_{11}^j$ are fixed by $\sigma$, as you have shown. However, that makes sense, as any one of them generates all of $\Bbb Q(\zeta_{11})$. Also, if you want an extension of degree $5$, you need a subgroup of index $5$, not order $5$. Which is to say you want a subgroup of order $2$.

$\zeta\mapsto \zeta^{10}$ has order $2$. And the field that is fixed by the generated subgroup is $\Bbb Q(\zeta+\zeta^{10})$.

  • $\begingroup$ thanks for your answer. by your logic in my attempt I was supposed to get a field with Galois group $\mathbb{Z}_2$, why didn't it happen? $\endgroup$ – GuySa Aug 14 at 10:25
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    $\begingroup$ @GuySa You did! You just didn't find the fixed field of your $\sigma$. You didn't find an element of $\Bbb Q(\zeta_{11})$ that is fixed by your sigma. Now that you have seen my fixed element, can you try to guess an element that is fixed by $\sigma$? $\endgroup$ – Arthur Aug 14 at 10:37
  • $\begingroup$ ok, my mistake was thinking that it's enough to look for a $j$ such that $\zeta^j = \sigma(\zeta^j)$, but I need to take a general element of $\mathbb{Q}(\zeta)$ $\endgroup$ – GuySa Aug 14 at 10:47

You should instead look for the subgroup of Gal$(\mathbb{Q}(\zeta_{11})/\mathbb{Q})$ isomorphic to $\mathbb{Z}_2$.

If $K$ is the fixed field of $\mathbb{Z}_2$, then the restriction of the Galois group to $K$ induces an isomorphism $\textrm{Gal}(K/\mathbb{Q})\cong \textrm{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})/\mathbb{Z}_2$. This gives us $\textrm{Gal}(K/\mathbb{Q})\cong \mathbb{Z}_{10}/\mathbb{Z}_2\cong \mathbb{Z}_5$.


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