A question regarding the Galois extension of the cyclotomic polynomial $\Phi_{15}$ Given the cyclotomic polynomial $\Phi_{15}$, I am trying to :
i) Determine the isomorphism type of the Galois group of $\Phi_{15}$ over $\Bbb Q$.
ii)Letting ω be a primitive 15-th root of unity in $\Bbb C$. Find all quadratic extensions of $\Bbb Q$ which are contained in $\Bbb Q(ω)$ (in the form $\Bbb Q(α)$ for explicit $α$).
Here is what I did, if anyone could verify this it would be great:
i) We can write $\Phi_{15}=(x-w)(x-w^2)(x-w^4)(x-w^7)(x-w^8)(x-w^{11})(x-w^{13})(x-w^{14})$, where $w$ is the primitive $15^{th}$ root of unity. 
Clearly a splitting field for this extension is $\Bbb Q(w)$, it is a Galois extension and the degree of the extension is given by the euler totient function giving the degree to be 8 . I think one could also say that as the above polynomial is irreducible over $\Bbb Q$ and is the minimum polynomial of $w$ so it gives the degree of the extension as 8 . 
I know I'm missing some reasoning that links to what I'm about to say , I can't see clearly how to explain it, any help would be appreciated. I think it has something to do with the following:
Given that the roots are $w$ it seems that if we could send this to any other power of $w$ that we would have the group $\Bbb Z_{15}$, but we can only send it to the other roots of the above polynomial. Comparing the powers of the $w$ in that expression to the order of the elements in $\Bbb Z_{15}$ we see that they correspond to the elements in $\Bbb U_{15}=\{1,2,4,7,8,11,13,14\}$, the group of units modulo 15.
So $Gal(\Bbb Q(w)/\Bbb Q) \cong \Bbb U_{15}=\Bbb U_{5}\times \Bbb U_{3} $
ii) Given that we now know that Galois group we can look at its subgroups and use Galois correspondence to figure out the quadratic extensions.
$\Bbb U_{15}$ has 3 subgroups of order 2 
$\langle 4 \rangle=\{1,4\}$
$\langle 11 \rangle = \{1,11\}$
$\langle 14 \rangle = \{1,14\}$
So then the quadratic extensions are just :
$\Bbb Q(w^4),\Bbb Q(w^{11}),\Bbb Q(w^{14})$.
Then writing the $\alpha$ explicitly is just a matter of writing $w^4$, etc. as an exponential ?
 A: The Galois group is $(\mathbb Z/15\mathbb Z)^\times$ (which is in structure, isomorphic to $C_2 \times C_4$) this is generated by two elements $-1$ and $2$ since it's not cyclic but a product of 2 cyclic groups.
The elements $2$ and $-1 \cdot 2$ both generate half the Galois group. So if we sum their orbit on $\zeta_{15}$ we create periods, elements that live in quadratic subfields:

*

*$G_1 = \zeta_{15}^1 + \zeta_{15}^2 + \zeta_{15}^4 + \zeta_{15}^8$

*$G_2 = \zeta_{15}^1 + \zeta_{15}^{-2} + \zeta_{15}^4 + \zeta_{15}^{-8}$
$G_1$ satisfies $x^2 - x + 4$ so we have $\mathbb Q(\sqrt{-15})$
$G_2$ satisfies $x^2 - x + 1$ so we have $\mathbb Q(\sqrt{-3})$
and we will also have the subfield that comes from combining them $\mathbb Q(\sqrt{5})$.
This 3rd one comes from the 5th roots of unity, note that you can decompose $\mathbb Q(\zeta_{15}) = \mathbb Q(\zeta_{5}, \zeta_{3})$
and from the period $\zeta_5 + \zeta_5^4$ we can find
$G_3 = \zeta_5^3 + \zeta_5^{12}$ satisfies $x^2 + x - 1$ giving $\mathbb Q(\sqrt{5})$.

The diagram from Dummit and Foote is really helpful in understanding this stuff

