Unique subfield of order $2$ of $\mathbb{Q(\zeta_{7}})$ over $\mathbb{Q}$

Question

Actualy this subfield is $\mathbb{Q}(i\sqrt{7})$ since $X^{2}+7$ is irreducible over $\mathbb{Q}$ and has $i\sqrt{7}$ as root.

My problem here is to show unicity, I tried something using the tower law since $[\mathbb{Q(\zeta_{7})},\mathbb{Q}]=6$ but i got nowhere.

Answer 1

Use the fundamental theorem of Galois theory, namely that subfields of a Galois extension correspond to subgroups of the associated Galois group. Then you just have to show that there is a unique subgroup of order $\frac{6}{2} = 3$ in $\textrm{Gal}\left(\mathbb{Q}\left(\zeta_7\right) / \mathbb{Q}\right) = \left(\mathbb{Z}/\left(7\right)\right)^\times$.

For the sake of completeness: see here for a proof that this is the right Galois group (note that its order is indeed $6 = \varphi\left(7\right)$) and see here for an explicit description of the correspondence.