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.

Could you please help me with this problem? Thank you in advance!

  • $\begingroup$ 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 $Gal(K(a)/K) = G/(H\times 1) = 1 \times J, Gal(K(b)/K)= H \times 1$ $\endgroup$
    – reuns
    Oct 10, 2018 at 20:48

1 Answer 1


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 $[F^H:K]=p_i^{n_i}$.

More generally, the entire point of Galois theory is that it lets you identify subfields by instead identifying subgroups of the Galois group, which is usually easier. In this case the Galois group is cyclic so it is very easy to understand its subgroups.


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