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.
Best Answer
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 $\textrm{Gal}(K/\mathbb{Q})\cong \mathbb{Z}_{10}/\mathbb{Z}_2\cong \mathbb{Z}_5$.