Can we write down the explicit form of the decomposition field of cyclomotic field? If we have cyclomotic field $\mathbb{Q}(\zeta_n)$ and a prime number $p\nmid n$. Suppose the order of $p (\textrm{mod }n)$ is $f$. I guess that the decomposition field is $\mathbb{Q}(\bigcup_{(k,n)=1}\sigma_k(\alpha))$, where $\alpha=\zeta_n+\zeta_n^p+\zeta_n^{p^2}+\cdots+\zeta_n^{p^{f-1}}$ and $\sigma_k\in\textrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ sending $\zeta_n$ to $\zeta_n^k$.
I haven't found complete answer to the question. The related materials on books are low-dimensional examples. I hope someone can tell me the answer.
 A: I don't know of your background in ANT, but Galois theory applied to the decomposition of primes in a normal extension of number fields (such as in chapter 3 of Marcus' book "Number Fields") gives the answer. I reprove this directly. Since $p$ does not divide $n$, $p$ is unramified in $\mathbf Q(\zeta_n)$  (by direct computation of the discriminant), so that $\phi (n) = rf$, where $r$ is the number of prime ideals above $p$ and $f$ is the inertia of $p$ (the ramification index $e$ is $1$). Introduce the Frobenius automorphism $\sigma_p$ of $\mathbf Q(\zeta_n)$ defined by raising $\zeta_n$ to the $p$-th power and compute its order. Because $p$ is unramified, it is easily seen that for any $a\in A$ (=the the ring of integers of $\mathbf Q(\zeta_n)$), $\sigma_p (a)\equiv a^p$ mod $pA$ , which means that $\sigma_p$ actually induces the usual Frobenius automorphism extension of finite fields $(A/pA)/\mathbf F_p$$A$ 
ring of integers of $\mathbf Q(\zeta_n)$. Hence the order of $\sigma_p$ is $f$, and its fixed field $D$ has degree $r$ over $\mathbf Q$. This means that $D$ is actually the decomposition field of $p$ . 
Addendum  Let us show that the decomposition field $D$ can be generated as suggested by the OP. For simplicity, write $\zeta =\zeta_n, L = \mathbf Q(\zeta), G = Gal (L/\mathbf Q)$. The element $\alpha = \zeta + \sigma_p (\zeta) + ... + \sigma_p^{f-1} (\zeta)$ is the trace of $\zeta $ in $L/D$, hence belongs to $D$. Similarly , the conjugates $\sigma_k (\zeta)$, with $(k,n)=1$, have traces $ \sigma_k (\alpha)$ because $G$ is commutative. For any finite extension $L/K$, it is known (and easily shown) that the trace map $L \to K$ is $K$-linear, and is surjective if $L/K$ is normal. This shows, since the $\sigma_k (\zeta)$ generate $L$ as a vector space over $\mathbf Q$, that the $ \sigma_k (\alpha)$ generate $D$ over $\mathbf Q$. Note that this argument is almost purely linear algebra.
