= Solution
Put $\omega=\zeta^5$, $\tau=\zeta^3$, $a=1+2\omega$, and $b=1+2(\tau+\tau^{-1})$. The primitive cubic and quintic root identities give $a^2=-3$, $b^2=5$. These elements therefore provide the following six generators:
* $\mathbb Q(a)=\mathbb Q(\sqrt{-3})$, degree two.
* $\mathbb Q(b)=\mathbb Q(\sqrt5)$, degree two.
* $\mathbb Q(ab)=\mathbb Q(\sqrt{-15})$, degree two.
* $\mathbb Q(\tau)$, degree four.
* $\mathbb Q(a+b)=\mathbb Q(\sqrt{-3},\sqrt5)$, degree four.
* $\mathbb Q(\zeta+\zeta^{-1})$, degree four.
The three quadratic fields are distinct because $-3,5,-15$ are distinct rational square classes. To justify the fifth entry, the two quadratic fields $\mathbb Q(a)$ and $\mathbb Q(b)$ are distinct, so their <compositum> has degree four. Moreover $a+b\ne0$ and
$$
(a+b)(b-a)=8,\qquad a=\frac{a+b-8/(a+b)}2,\qquad b=\frac{a+b+8/(a+b)}2.
$$
Thus their sum alone generates the compositum.
The field $\mathbb Q(\tau)$ has degree four by the quintic <cyclotomic polynomial>. Its <Galois group> is $C_4$ and its unique quadratic subfield is $\mathbb Q(b)$, so it cannot contain $a$. It is therefore different from $\mathbb Q(a,b)$. The final field has degree four by part (b) and is real; the preceding two quartic fields contain nonreal $\tau$ and $a$, respectively. Thus all three quartics are distinct, and the count in part (i) proves the list exhaustive. This gives the <subfields of the fifteenth cyclotomic field> without relying on a choice of square-root signs.
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