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 $p=7$?
 A: Let $\alpha = \zeta_7 + \zeta_7^{2} + \zeta_7^{4} \in \mathbb{Q}(\zeta_7)$.
What is $\alpha$? Well, observe $\alpha^2 + \alpha + 2 = 0$. 
By the quadratic equation, we find:
$$\alpha = \frac{-1 \pm \sqrt{-7}}{2}$$
whence $\sqrt{-7} \in \mathbb{Q}(\zeta_7)$ as desired. QED.
(In particular, the $\pm$ should be $+$ although either case yields $\sqrt{-7} \in \mathbb{Q}(\zeta_7)$ as desired.)
A: 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.
A: 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 $\mathbb{Q}(\sqrt{-7})\subseteq \mathbb{Q}(\zeta_7)$.
A: 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 $x^2+n$.
A: The method for showing that CZ7 (the cyclotomic numbers on the heptagon, satisfying $x^7+1=0$ and the span of powers of x) contains the $\sqrt{-7}$ is to realise that it is a sixth-order system, and one can dissect an order-two unit by merging the order-3 units out of it, for example, cis2/14 + cis 4/14 + cis 8/14 is invariant to the heptagonal isomorphism, and contains a form in the type x+y sqrt(-7), where these are integers or integer-halfs.
Z7 represents the span of chords of the heptagon.  CZ7 represents the cyclotomic numbers form by dividing the half-circle into seven parts, and taking the span over Z.  This represents the verticies of the tiling {7, 14/5}.  It has a construction in terms of a general carryless base, modulo by setting a polynomial to zero, as $Pm(x^7+1=0)$.  $P$ here represents $\sum z_j x^j$, where z,j are in the integers Z and $m(\mbox{eqn})$ represents an equity equation.
Some elementry number theory can prove that if $p$ is a prime of the form $4x+3$, then the cyclotomic numbers CZp contain $Z(1, \frac 12(1+\sqrt{-p}))$.
The rule of isomorphism says that there are several solutions to this equation, and putting eg x + x' + x" = y gives a value automorphic to this value.  In the case of CZ7, an automorphic solution can be had by replacing $x$ by $x^9$ by $x^{11}$ by $x$.  So the triplets $x+x^9+x^{11}$, and $x^2+x^4+x^8$ etc, are automorphic to the $x$ -> $x^9$ transform.  
