Dtyjlui

Is there a good general purpose algorithm (batch of theorems) allowing one to determine the intermediate fields between $mathbb{Q}(zeta)$ and $mathbb{Q}$, where $zeta$ is some primitive root of unity?

Let $p$ be a prime. Consider the case where $zeta=zeta_{p}$ is a primitive $p$-th root of unity. Then the Galois extension is cyclic of order $p-1$ and $1,zeta,dots,zeta^{p-1}$ is a $mathbb{Q}$-basis for the extension. In this case for any subgroup $H$ of $G=mathbb{Z}/(p-1)$, by considering the sum $$alpha_H=sum_{sigmain H}sigmazeta,$$
we can observe that $alpha_H$ lies in the fixed field for $H$, and any automorphism $tau$ not in $H$ (note automorphisms are identified with subgroups of $mathbb{Z}/(p-1)$ in the natural way), $tau$ does not fix $alpha_H$. Therefore we can conclude that $mathbb{Q}(alpha_H)$ is the fixed field of $H$.

In this way we can get all of the intermediate fields of $zeta_p$ for all odd primes $p$.

We also have a theorem that says if we have $n=p^sq^t$, then $$text{Gal}(mathbb{Q}(zeta_n)/mathbb{Q})simeq text{Gal}(mathbb{Q}(zeta_{p^s})/mathbb{Q})timestext{Gal}(mathbb{Q}(zeta_{q^t})/mathbb{Q}).$$

So what I have yet to understand is

How can one generally find the intermediate fields between $mathbb{Q}(zeta_{p^s})$ and $mathbb{Q}$ for $sge 1$? I would like to also understand the case where $p=2, s>1$ though this might turn out to be a separate case.

EDIT: Even the case $n=pq$ is a little murky to me. Even given the isomorphism given by the Chinese Remainder Theorem I don't see a priori how to get all the "product" subfields. My idea is that you can consider the separate subfields under $text{Gal}(mathbb{Q}(zeta_{p})/mathbb{Q})$ and $text{Gal}(mathbb{Q}(zeta_{q})/mathbb{Q})$ separately and then consider the pairwise product of the generators of various subfields to see if you get anything new, but my idea is too inchoate.

If $q$ is a power of an odd prime $p$, then the multiplicative group of units in the ring ${bf Z}/q{bf Z}$ is cyclic of order $q-(q/p)$, and that's also the Galois group of ${bf Q}(zeta_q)$ over the rationals, so it seems to me that your construction for the prime case works.

Let $zeta$ be a primitive $p^s$-th root of unity for a prime $p$ and a positive integer $s$. Let $G = operatorname{Gal}(mathbb{Q}(zeta)/mathbb{Q}).$ Let
$$ eta = zeta + zeta^p + ... + zeta^{p^{s-1}}.$$
One can prove by induction on $s$ that the set $mathcal{B} = {g(eta) ; | ; g in G}$ is a basis of $mathbb{Q}(zeta)/mathbb{Q}$. Now for any $alpha in mathbb{Q}(zeta)$ let $H$ be the subgroup of $G$ fixing $alpha$. Define
$$ beta = sum_{sigma in H} sigma(eta).$$
Since $tau(beta) = beta$ for all $tau in H$, $mathbb{Q}(beta)$ is a subfield of $mathbb{Q}(alpha)$. We will now show by contradiction that for any $tau in G setminus H$ that $tau(beta) neq beta$. Assume there exists a $tau in G setminus H$ such that $tau(beta) = beta$. Since $mathcal{B}$ is a basis for $mathbb{Q}(zeta)/mathbb{Q}$, there must exist a $sigma in H$ such that $tau circ sigma(eta) = iota(eta)$ where $iota$ is the identity element of $G$. Then $tau = sigma^{-1} in H$, which contradicts our assumption. We conclude that for all $tau in G setminus H$ we have $tau(beta) neq beta$. Thus $mathbb{Q}(beta)$ contains $mathbb{Q}(alpha)$. This proves $mathbb{Q}(beta) = mathbb{Q}(alpha)$.

This shows that all subfields of $mathbb{Q}(zeta)$ can be constructed as $mathbb{Q}(beta)$ where $beta = sum_{sigma in H} sigma(zeta + zeta^p + ... + zeta^{p^{s-1}})$ for a subgroup $H$ of $operatorname{Gal}(mathbb{Q}(zeta)/mathbb{Q})$.

For example, if $zeta$ is a primitive 9-th root of unity, then $eta = zeta + zeta^3$. Since 2 is a quadratic nonresidue modulo 3, the Galois group $G$ is cyclic with generator $tau$ defined by $tau(zeta) = zeta^2$. Since $varphi(9) = 6$, the subgroups of $G$ are itself, the trivial subgroup, $left< tau^2 right>$, and $left<tau^3right>$. We know what fields we'll get for the first two from basic Galois theory, so now we must now find $beta = sum_{sigma in H}sigma(eta)$ for the two proper nontrivial subgroups. When $H = left<tau^2right>$

$$beta = sum_{sigma in left< tau^2 right>} sigma(eta) = (zeta + zeta^3) + (zeta^4 + zeta^3) + (zeta^7 + zeta^3) = 3zeta^3.$$
In this case $mathbb{Q}(beta) = mathbb{Q}(zeta^3)$. When $H = left< tau^3 right>$

$$beta = sum_{sigma in left< tau^3 right>} sigma(eta) = (zeta + zeta^3) + (zeta^8 + zeta^6) = zeta + zeta^8 - 1$$
In this case $mathbb{Q}(beta) = mathbb{Q}(zeta + zeta^8) = mathbb{Q} left( cos left(frac{2 pi}{9} right) right)$.