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

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.

• I think you meant to write $\alpha \in \overline{\mathbb Q}$ in the title and again in the first line of the question. – Ben Aug 14 at 9:24
• @Ben yes of course, thanks for correcting – 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 $$\Bbb Q(\zeta+\zeta^{10})$$.
• 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? – GuySa Aug 14 at 10:25
• @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 $\sigma$? – Arthur Aug 14 at 10:37
• 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 $\mathbb{Q}(\zeta)$ – 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 $$\textrm{Gal}(K/\mathbb{Q})\cong \mathbb{Z}_{10}/\mathbb{Z}_2\cong \mathbb{Z}_5$$.