Show $\mathbb{Q}(\zeta)$ does not contain $\sqrt{7}$

Question

Let $\zeta$ be a prmitive $14^{th}$ root of unity in $\mathbb{C}$. We are given that $\sqrt{-7}\in \mathbb{Q}(\zeta)$.

Show that $\mathbb{Q}(\zeta)$ does not contain $\sqrt{7}$.

Answer 1

The degree of the extension is $\phi(14) = 6$. If it contains $\sqrt{7}$ it contains the subfield $\mathbb{Q}(\sqrt{7},\sqrt{-7})$. The degree of this extension over $\mathbb{Q}$ is $4$, contradicting the tower law since $4\not|6$.

If it's not clear why the degree of $\mathbb{Q}(\sqrt{7},\sqrt{-7})$ over $\mathbb{Q}$ is $4$, it is because of the following:

$[\mathbb{Q}(\sqrt{-7},\sqrt{7}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{-7},\sqrt{7}):\mathbb{Q}(\sqrt{7})][\mathbb{Q}(\sqrt{7}):\mathbb{Q}], [\mathbb{Q}(\sqrt{7}):\mathbb{Q}]=2$ clearly and $[\mathbb{Q}(\sqrt{-7},\sqrt{7}):\mathbb{Q}(\sqrt{7})]=1$ or $2$, but $\mathbb{Q}(\sqrt{7})$ is a real subfield so doesn't contain $\sqrt{-7}$.