A few questions on subfields of $Q(\zeta_n)$ We know by standard theorems that $Gal(Q(\zeta_n)/Q)\approx (Z_n)^*$. Thus the order of the extension is $\phi(n)$.
Now if $n=p$ then we know that $(Z_p)^*\approx Z_{p-1}$ and that given a subgroup of $Gal(Q(\zeta_p)/Q)$ H, the fixed field of it is $$Q(\Sigma_{\sigma \in H}\sigma(\zeta_p))$$ which is easily computable. Are there similar "theorems" for $n=p^n$ or $n=2p^n$ when $(Z_n)^*$ is cyclic?
Furthermore, I was wondering, when can we determine for which $a,b$ do these occur: $Q(\zeta_n^a)=Q(\zeta_n^b)$ or $Q(\zeta_n^a)\subseteq Q(\zeta_n^b)$.
It is clear that when $gcd(a,n)=1$ then $Q(\zeta_n^a)=Q(\zeta_n)$ that is because then $\zeta_n^a$ is the generator of the group $<\zeta_n>$. Can we extend this reasoning to other cases? For example, $Q(\zeta_n^a)=Q(\zeta_n^b)$ $\iff$ they generate the same subgroup in $<\zeta_n>$? Also $Q(\zeta_n^a)\subseteq Q(\zeta_n^b)$ $\iff$ $<\zeta_n^a>\subseteq \zeta_n^b>$?
For my last question, consider ,$Q(\zeta_n)$ the degree of the extension is $\phi(n)$. Now let us say we find all the automorphisms and so the Galois group elements. Any element is of the form $Q(\zeta_n)=\{a_1+a_2\zeta_n+...+a_{\phi(n)-1}\zeta_n^{\phi(n)-1}\}$. Thus everything is determined by where we map $\zeta_n$. Now when we apply some automorphism on an arbitary element $a_1+a_2\zeta_n+...+a_{\phi(n)}\zeta_n^{\phi(n)}$ then the powers of $\zeta_n$ can exceed $\phi(n)$, how would we now find the decomposition of $\zeta_n^a$ in terms of the basis (we can assume $a<n$ as we can always reduce mod $n$.)? So that we can see what becomes fixed and what does not?
 A: For your third question, you may simply divide $X^a$ by the cyclotomic polynomial $\Phi_n(X)$.
Example: $n = 9$.
We have $\Phi_9(X) = x^6 + x^3 + 1$. Therefore:
$$\zeta^6 = -\zeta^3 - 1;$$
$$\zeta^7 = -\zeta^4 - \zeta;$$
$$\zeta^8 = -\zeta^5 - \zeta^2.$$
In general, for $a \geq \phi(n)$, the polynomials expressing $\zeta^a$ in terms of $\zeta$ are quite complicated.
When $n$ is squarefree, it's much easier to work with the normal basis $(\zeta^r)_{r\in (\Bbb Z/n\Bbb Z)^\times}$.

In terms of the normal basis, it's quite easy to tell when an element is fixed by an automorphism.
Let's say the automorphism is $\sigma_k:\zeta\mapsto \zeta^k$, with $k\in(\Bbb Z/n\Bbb Z)^\times$. The element $k$ generates a subgroup $\langle k\rangle$ of $(\Bbb Z/n\Bbb Z)^\times$.
Now if $x = \sum_{r \in(\Bbb Z/n\Bbb Z)^\times}a_r\zeta^r$ is an element of $\Bbb Q(\zeta)$, then the element $\sigma_k(x)$ is simply $\sum\limits_{r \in(\Bbb Z/n\Bbb Z)^\times}a_r\zeta^{kr} = \sum\limits_{r \in(\Bbb Z/n\Bbb Z)^\times}a_{rk^{-1}}\zeta^r$.
Since the elements $(\zeta^r)_{r\in(\Bbb Z/n\Bbb Z)^\times}$ form a basis, we have $\sigma_k(x) = x$ if and only if $a_r = a_{rk^{-1}}$ for all $r$.
This is again equivalent to saying that there is a function $a:(\Bbb Z/n\Bbb Z)^\times/\langle k\rangle \rightarrow \Bbb Q$ such that $a_r = a(r\mod \langle k\rangle)$.
In other words, an element of $\Bbb Q(\zeta)$ is fixed by $\sigma_k$ if and only if it is a $\Bbb Q$-linear combination of elements of the form $\zeta^r + \zeta^{rk} + \zeta^{rk^2} + \dotsc + \zeta^{rk^{m - 1}}$, where $r \in (\Bbb Z/n\Bbb Z)^\times$ and $m$ is the order of $k$ in $(\Bbb Z/n\Bbb Z)^\times$.
