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I will very briefly explain the computation. Note that $\phi(31) = 2\cdot 3\cdot 5$ and in $(\Bbb Z/31\Bbb Z)^\times$, one could compute the order of subgroups $\left|\left<5\right>\right|=3$ ad $\left|\left<2\right>\right|=5$. Their corresponding intermediate fields are given by $$\operatorname{Fix}\left<\sigma_5\right> = \Bbb Q[\zeta+\zeta^5+\zeta^{25}]$$ $$\operatorname{Fix}\left<\sigma_2\right> = \Bbb Q[\zeta+\zeta^2+\zeta^{4}+\zeta^8+\zeta^{16}]$$

But I have trouble computing $\operatorname{Fix}\left<\sigma_2,\sigma_5\right>$, which corresponds to the (internel) direct sum of the two subgroups $\left<\sigma_2\right>\oplus\left<\sigma_5\right>$.

I know that $$\operatorname{Fix}\left<\sigma_2,\sigma_5\right>=\Bbb Q[\zeta+\zeta^5+\zeta^{25}]\cap \Bbb Q[\zeta+\zeta^2+\zeta^{4}+\zeta^8+\zeta^{16}]$$ but how could I compute the right hand side to get a single primitive element? I am looking for a general approach that could be applied to all cyclotomic extension of $\Bbb Q$.

Some attempts: $(\Bbb Z/31\Bbb Z)^\times=\left<3\right>$ while $5 = 3^{20}\pmod {31}$ and $2=3^{24}\pmod{31}$. Rewrite above intersection as $$\Bbb Q[\alpha] = \Bbb Q[\zeta+\zeta^{3^{20}}+\zeta^{3^{9}}]\cap \Bbb Q[\zeta+\zeta^{3^{24}}+\zeta^{3^{18}}+\zeta^{3^{6}}+\zeta^{3^{12}}]$$ for some $\alpha\in\Bbb Q[\zeta]$. We need $\operatorname{Gal}(\Bbb Q[\zeta]/\Bbb Q[\alpha]) = \left<\sigma_2,\sigma_5\right>=\left<\sigma_{3^{24}},\sigma_{3^{20}}\right>=\left<\sigma_3^3\right>=\left<\sigma_{9}\right>.$ But then this means we could just take $\alpha = \zeta+\zeta^{12}+\zeta^{12^2}+...$. We have used the generator here. In the general case, what if we do not have primitive roots？

For example, let $\zeta$ be the $63^{th}$ root of unity. There is no element of order $9$ in this case. How do we find the intermediate field corresponding to the subgroup of order $9$ in $\Bbb Q[\zeta]/\Bbb Q$?

Update: For this non-cyclic case above, i.e. subgroup of order $9$ in $\Bbb Q[\zeta_{63}]/\Bbb Q$ I have asked a new question here.

William Sun
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1 Answers1

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The general approach you're looking for is based on the structure of the Galois group of the (cyclic) extension you're considering. In your case, $ \mathrm{Gal}(\mathbb{Q}(\zeta_{31})/\mathbb{Q}) \cong (\mathbb{Z}/31\mathbb{Z})^{\times} \cong C_{30}$. Indeed since 31 is prime, $\mathbb{Z}/31\mathbb{Z}$ is a finite field and thus its multiplicative group is a cyclic group (of order 30). You can identify that $3$ is a generator for $(\mathbb{Z}/31\mathbb{Z})^{\times}$.

Now, you can apply Galois correspondence: the intermediate extensions are in one-to-one correspondence with the subgroups of Galois group. For a cyclic group $C_n$, its subgroups are in correspondence with divisors of $n$ and if $g$ is a generator and $d$ a divisor of $n$, are cyclic subgroups generated by $g^d$. The integer $30$ has eight divisors, namely $\{1,2,3,5,6,10,15,30\}$. So the intermediate extensions are the subfields fixed by $\sigma_3^d$ for $d|n$ and thus the subfields fixed by $\sigma_3$, $\sigma_9$, ... , $\sigma_{25}$, $\sigma_{30}$ and $\sigma_1$. This leads to the list of intermediate extensions (in reverse order):

$ \mathbb{Q}[\zeta]$

$ \mathbb{Q}[\zeta+\zeta^{30}]$

$ \mathbb{Q}[\zeta+\zeta^{5}+\zeta^{25}]$

$ \mathbb{Q}[\zeta+\zeta^{2}+\zeta^{4}+\zeta^{8}+\zeta^{16}]$

$ \mathbb{Q}[\zeta+\zeta^{5}+\zeta^{6}+\zeta^{25}+\zeta^{26}+\zeta^{30}]$

$ \mathbb{Q}[\zeta+\zeta^{2}+\zeta^{4}+\zeta^{8}+\zeta^{15}+\zeta^{16} +\zeta^{23}+\zeta^{27}+\zeta^{29}+\zeta^{30}]$

$ \mathbb{Q}[\zeta+\zeta^{2}+\zeta^{4}+\zeta^{5}+\zeta^{7} +...+\zeta^{25}+\zeta^{28}]$

$ \mathbb{Q}[\zeta+\zeta^{2}+\zeta^{3}+\zeta^{4}+\zeta^{5} +...+\zeta^{29}+\zeta^{30}]$

Note that you can simplify the last field that is simplfy equal to $\mathbb{Q}$, because it is the subfield fixed by the whole Galois group.

If your goal is to list intermediate field extensions, I would not advocate for using techniques such as intersections of subfield but rather rely on a systematic approach by divisors. However, I'll answer your specific question about $\mathrm{Fix}<\sigma_2,\sigma_5>$. Elements that are fixed by $\sigma_2$ and $\sigma_5$ are fixed by their combinations, i.e. are fixed by $\sigma_{2^a.5^b}$ for any integers $a$ and $b$. Doing the computation (modulo 31), one gets that elements need to be fixed by $\sigma_n$ for $n\in\{1,2,4,5,7,...,20,25,28\}$. So $\mathrm{Fix}<\sigma_2,\sigma_5>$ is the penultimate intermediate extension listed above, that you can also express (if you prefer) as $\mathrm{Fix}<\sigma_9>$.

Eric Brier
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  • Thank you. I will accept this answer as it answered my original question before my edit. I have asked a new question for the non-cyclic case. See https://math.stackexchange.com/questions/4748527/intermediate-extensions-of-bbb-q-zeta-63-bbb-q?noredirect=1#comment10075904_4748527 – William Sun Aug 06 '23 at 17:12
  • Just as a "reality check", the last field, $\mathbb Q(\sum_{i=1}^{30} \zeta^i) =\mathbb Q$ as it must be by the general theory because it is the subfield fixed by the entire Galois group, and concretely because $\sum_{i=1}^{30} \zeta^i = -1$ (since $0=\zeta^{31}-1 = (\zeta-1)(\sum_{i=0}^{30} \zeta^i)$). – krm2233 Aug 06 '23 at 17:32
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    @krm2233 : In the same vein, one can also show (e.g. using Gauss sums) that the penultimate intermediate extension listed in my answer above is in fact $\mathbb{Q}(i \sqrt{31})$. – Eric Brier Aug 07 '23 at 07:07
  • This method works here (+1). But I want to comment that it does not work for finding subfields of $\Bbb{Q}(\zeta_n)$ for all $n$. The reason is that the set $\zeta^k, \gcd(k,n)=1$ is linearly independent if and only if $n$ is square-free. For example with $n=8$ we have the proper subgroup ${\overline{1},\overline{5}}\le\Bbb{Z}_8^$, but $\zeta+\zeta^5=0$ does not generate the fixed field. Similarly with $n=9$ we have the proper subgroup ${\overline{1},\overline{4},\overline{7}}\le\Bbb{Z}_9^$ and again $\zeta+\zeta^4+\zeta^7=0$. – Jyrki Lahtonen Sep 16 '23 at 03:33
  • See here for a detailed discussion. – Jyrki Lahtonen Sep 16 '23 at 03:43