find $\alpha \in \mathbb{Q}$ such that $Gal(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong \mathbb{Z}_5$

by GuySa   Last Updated August 14, 2019 09:20 AM - source

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

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 $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong \mathbb{Z}_n^\times$, (where $\zeta_n$ is an n-th primitive root of unity) so I quickly found that $Gal(\mathbb{Q}(\zeta_{11})/\mathbb{Q}) \cong \mathbb{Z}_{11}^\times \cong \mathbb{Z}_{10}$ 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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