Prove that the $7$-th cyclotomic extension $\mathbb{Q}(\zeta_7)$ contains $\sqrt{-7}$ 
Prove that the $7$-th cyclotomic extension $\mathbb{Q}(\zeta_7)$ contains $\sqrt{-7}$

I thought that the definition of the $n$-th cyclotomic extension was: $\mathbb{Q}(\zeta_n)=\{\mathbb{Q}, \sqrt{-n}\}$. Is this correct?
How could I prove the statement? Could we consider the polynomial $X^2+7$ (which is irreducible by Eisenstein's criterion with $p=7$?

Note that $\Bbb Q(\zeta_7)$ is a totally ramified extension above $(7)$ of $\Bbb Z$ with discriminant a power of $7$. But then all subextensions must be totally ramified at $(7)$ and nowhere else. This means their discriminants must divide $7^n$. Because $\phi(7)=6$ is even and the Galois group is cyclic, we know there is a unique quadratic subfield, all of which are of the form $\Bbb Q(\sqrt{D})$ for some $D$ square free, and we know the discriminant of this field is
$$\begin{cases} D & D\equiv 1\mod 4 \\ 4D & D\equiv 2,3\mod 4\end{cases}$$
So our $D$ must be $\pm 7$, but $+7\equiv 3\mod 4$ would be ramified at $2$, hence it must be that $D=-7\equiv 1\mod 4$.

Addendum The definition of the $n^{th}$ cyclotomic field is the splitting field of $x^n-1$, your definition would be the splitting field of $x^2+n$.
