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 $p=7$?

I'm sure there are more elegant ways to observe this, but here's one possible way:


*

*First, show that the extension is Galois (not too hard) and that $Gal(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong Z_6$. This has subgroups of order 2 and 3. 

*Now find the fixed field of the subgroup of order 3. This is a degree 2 extension. To find the generator, find a generator of the Galois group and write the subgroup in terms of powers of the generator, and find a generator of the subgroup. Look at the orbit of the generator of that subgroup on $\zeta_7$. We guess that this will generate a degree 2 extension.

*To show that, take powers of that generator and try to find the minimal polynomial. It should be degree 2. The discriminant of the quadratic will be $\sqrt{-7}$, and you will be done.
