Let $Q[d]$ be a cyclotomic field over $Q$ by adjoining $d-$th primitive root of unity. Let $K\subset Q[d]$ be a subfield. Then $K=Q[d]^G$ where $G$ is some $G\leq (Z_d)^\star$. Given any character $\psi$ of $(Z_m)^\star$, one can realize a kernel field by looking at $Q[d]^{Ker(\psi)}$.
$\textbf{Q:}$ Every kernel field is a subfield. Is every subfield of cyclotomic field a kernel field?
$\textbf{Q':}$ If not, how to realize any subfield via characters?
Best Answer
The answer is yes, every subfield is a kernel field, but you are not going to like the argument. This sheds no light on any important insight, it is just a very basic property of kernels of characters and their connection with normal subgroups of a group. So cyclotomic fields, or in fact fields in general are completely irrelevant in the problem.
Given a finite group $G$, every normal subgroup $N\triangleleft G$ can be written as the intersection of kernels of some irreducible characters of $G$. If these irreducible characters are $\chi_1, \ldots, \chi_k$, so $N=\cap Ker(\chi_i)$, then $N= Ker(\chi)$ where $\chi= \sum \chi_i$.
As $\mathbb{Z}_d^\star$ is commutative, every subgroup is normal in it, so every subgroup is the kernel of some character. As the cyclotomic field is a Galois extension of $\mathbb{Q}$, every subfield is the fixed field of some subgroup, so the assertion follows trivially.