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 $p=7$?

The cyclotomic field $\mathbb{Q}(\zeta_n)$ is defined by adjoining a primitive $n$-th root of unity, and we have $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\phi(n)$.
In particular, it is different from $\mathbb{Q}(\sqrt{-n})$ for $n>3$. However, every cyclotomic field $\mathbb{Q}(\zeta_p)$ for $p$ prime contains a unique quadratic subfield $\mathbb{Q}(\sqrt{p})$ when $p \equiv 1 \pmod{4}$, or respectively $\mathbb{Q}(\sqrt{-p})$ when $p \equiv 3 \pmod{4}$, see here. Since $7\equiv 3 \bmod 4$ the claim follows, i.e., we have $\mathbb{Q}(\sqrt{-7})\subseteq \mathbb{Q}(\zeta_7)$.
