# Can one find intermediate subextensions of a finite cyclic field extension whose degrees are prime powers?

Let $$F/K$$ be a finite cyclic extension, i.e. $$F/K$$ is finite, Galois and its Galois group is cyclic. Let us say the prime decomposition of the degree of the extension is $$[F:K] = p_1^{n_1} \cdots p_r^{n_r}$$ where $$p_i$$ are primes and $$n_i$$ are positive integers. I would like to know if you can write $$F=K_1 K_2 \dots K_r$$ where $$K_i$$ are intermediate subextensions of $$F/K$$ and $$[K_i:K] = p_i^{n_i}$$. By using the tower law, we can reduce this problem to the following

Question: Can one find intermediate fields $$K_i$$ of $$F/K$$ where $$[K_i:K]$$ is a prime power?

Ideas and Thoughts:

I know that for every intermediate subextension $$K_i$$ of $$F/K$$ the Galois group of $$K_i/K$$ because $$\operatorname{Gal}(K_i/K) = \operatorname{Gal}(F/K)/\operatorname{Gal}(F/K_i)$$, so a generator of $$\operatorname{Gal}(K_i/K)$$ would be the restriction of the generator of $$\operatorname{Gal}(F/K)$$. But I have no idea how to find this $$K_i$$. I am also not sure how to tackle the problem with the prime power degrees.

• The "optimal" decomposition of finite abelian extensions is based on this : If $K = F^G= \{ a \in F, \forall g \in G, g(a) = a\}$ then $K/F$ is a Galois with $Gal(F/K) = G$. Thus when $G = H \times J$ and $K = F^G$ then $F = K(a,b)$ where $K(a) = F^{H \times 1}, K(b ) =F^{1 \times J }$ ($a,b$ are sets of elements generating the extension) and $Gal(K(a)/K) = G/(H\times 1) = 1 \times J, Gal(K(b)/K)= H \times 1$ – reuns Oct 10 '18 at 20:48
This is immediate from Galois theory. Indeed, there exists a subgroup $$H\subseteq Gal(F/K)$$ of index $$p_i^{n_i}$$ (if $$Gal(F/K)$$ is generated by $$g$$, take the subgroup generated by $$g^{p_i^{n_i}}$$) and then the fixed field $$F^H$$ will satisfy $$[F^H:K]=p_i^{n_i}$$.