The intermediate fields of $15$th cyclotomic field For the ease of typing, write $a$ to be the primitive $15$th root of unity. Then we know that $Q(a) = Z/2Z \times Z/4Z$. To be more precise it is generated by $f: a \to a^{11}$ and $g: a \to a^8$. The next thing I want to do is to write down its intermediate fields explicitly. 
It has following subgroups:
$H_1 = \{1, 2, 4, 8\}$, $H_2 = \{1, 4, 14, 11\}$, $H_3 = \{1, 4, 7, 13\}$, $H_4 = \{1, 4\}$, $H_5 = \{1, 11\}$, $H_6 = \{1, 14\}$.
My question is：
a. It seems that the fixed field of $H_1$ should be $\mathbb Q(a+a^2+a^4+a^8$). It has to be a simple extension because it is of degree $2$, so I only need to argue that $a+a^2+a^4+a^8$ is not in $\mathbb Q$. It is not so obviuos why this should be true. What is the general strategy?
b. It seems that the fixed field of $H_4$ should be $\mathbb Q(a+a^4)$, but then I need to argue $a + a^4$ is of degree $4$. What is the general strategy?
 A: A "general strategy" for the explicit determination of the subfields of a cyclotomic field is provided by the theory of "Gaussian periods", which are sums of roots of unity of certain type (perhaps the "Gauss sums" are better known). Wikipedia's article on this subject being not very informative, I refer to the bachelor thesis of D.Tijsma, chap.5, https://dspace.library.uu.nl/.../1874/.../Gaussian%20periods.pdf. 
The Gaussian periods form a basis for the ring of integers of a subfield of Q($\zeta_n$) for square-free n. This is stronger than what you need. Your special case can be treated directly. Let us first determine the 3 quadratic subfields : an obvious one is Q($\zeta_3$) = Q($\sqrt {-3}$); a second one is the unique quadratic subfield of Q($\zeta_5$), which is classically known to be Q($\sqrt 5$), see e.g. https://math.stackexchange.com/a/3255952/300700 ; so the third one is Q($\sqrt {-15}$). As for the 3 subextensions of degree 4 , they are obviously Q($\zeta_5$), Q($\zeta_{15} + \zeta_{15}^{-1}$), and  the biquadratic field Q($\sqrt 5,\sqrt {-3}$).
NB. Concerning Q($\sqrt 5$), note that the "quadratic Gauss sum" $\tau=\sum (b/5) \zeta_{5}^b$, where b runs through $(\mathbf Z/5)^*$ and $(./.)$ denotes the 5-Legendre symbol, verifies $\tau^2=(-1/5)5=5$.
