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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 $$\alpha_H=\sum_{\sigma\in H}\sigma\zeta,$$ 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 $$\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\simeq \text{Gal}(\mathbb{Q}(\zeta_{p^s})/\mathbb{Q})\times\text{Gal}(\mathbb{Q}(\zeta_{q^t})/\mathbb{Q}).$$

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 $s\ge 1$? 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.

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"Then the Galois extension is cyclic of order $p$": if it were the case, there would be no intermediate extensions. The extension actually has degree $\varphi(p)=p-1$ – M Turgeon Apr 15 '12 at 18:49
thanks for the catch, fixed – user21725 Apr 15 '12 at 18:50
And thus the $\mathbb{Q}$-basis would be $1,\zeta,\cdots, \zeta^{p-2}$ – Jing Zhang Oct 27 '13 at 9:18

1 Answer 1

up vote 2 down vote accepted

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.

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Thanks for the observation, I didn't see this. I think I have the kind of answer I'm looking for, I'll try to write it up soon. Maybe my real question is for a certain kind of presentation of a general solution. – user21725 Apr 17 '12 at 1:31

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