Let $\xi_9 = e^{\frac{2\pi i}{9}}$, I want to show that there exists a unique subfield $H$ of $\mathbb{Q}(\xi_9)$ such that $[H:\mathbb{Q}]=2$.
Is this correct? In that case How one can prove this?
I know for odd prime $p$, $\exists L$ such that $[L:\mathbb{Q}]=2$ where $L$ is the unique intermediate field of $\mathbb{Q}(\zeta)/\mathbb{Q}$. In fact I know $L=\mathbb{Q}(\sqrt{\pm p})$. [Unforntunately I don't know the proof] but $9$ is not a prime, so I am not sure whether this is right or not.
If you know that the minimal polynomial of $\zeta_9$ over $\mathbb{Q}$ is $X^6+X^3+1$ (see e.g. Wikipedia article "Cyclotomic polynomial"), you know that $[\mathbb{Q}(\zeta_9):\mathbb{Q}]=6$. So the Galois group of this extension is of order 6. Therefore it can either be $\mathbb{Z}_6$ or $S_3$. Now there is a theorem by which the groups $Gal(\mathbb{Q}(\zeta_9)/\mathbb{Q})$ and the multiplicative group of $\mathbb{Z}/9\mathbb{Z}$ are isomorphic. The multiplicative group of $\mathbb{Z}/9\mathbb{Z}$ happens to be $(\mathbb{Z}/6\mathbb{Z},+)$ as $\phi(9)=6$ and $9$ is the power of an odd prime. So the Galois group of the extension must be $\mathbb{Z}_6$. But this group has a unique subgroup of index two, so the extension can only have a unique subfield of degree two.