Compute Intermediate Extensions of $\Bbb Q[\zeta]/\Bbb Q$.

Question

Let $\zeta$ be a primitive $n^{\operatorname{th}}$ root of unity over $\Bbb Q$ then there is a natural isomorphism $$\operatorname{Gal}\left(\Bbb Q\left[\zeta\right]/\Bbb Q\right)\simeq \left(\Bbb Z/n\Bbb Z\right)^\times$$

and thus subgroups of $\operatorname{Gal}\left(\Bbb Q\left[\zeta\right]/\Bbb Q\right)$, by the structure of finite abelian groups, is a product of subgroups of the form $\left<\sigma_k\right>$, where $\sigma_k(\zeta) = \zeta^k$.

I tried first to calculate the intermediate extension $\operatorname{Fix}(\sigma_k)$ corresponding to a cyclic subgroup $\left<\sigma_k\right>$, and by the fundamental theorem, $$\deg(p_{\zeta}) = \left[\operatorname{Fix}\left(\sigma_k\right)\left[\zeta\right]:\operatorname{Fix}\left(\sigma_k\right)\right] = \left|\left<\sigma_k\right>\right|$$

where $p_\zeta$ is the minimal polynomial of $\zeta$ over $\operatorname{Fix}(\sigma_k)$. Note that $\sigma_k(\zeta),\sigma_k(\sigma_k(\zeta)),...$ are also roots of $p_\zeta$, but there are exactly $d=\left|\left<\sigma_k\right>\right|$ distinct roots of $p_\zeta$, so one actually has

$$p_\zeta = (x-\zeta)(x-\sigma_k(\zeta))(x-\sigma_k(\sigma_k(\zeta)))... = (x-\zeta)(x-\zeta^k)(x-\zeta^{k^2})...(x-\zeta^{k^{d-1}})$$

Let $M$ be the extension of $\Bbb Q$ by adding all coefficients of $p_\zeta$,omitting the negative sign: $$\zeta+\zeta^k+\zeta^{k^2}+...+\zeta^{k^{d-1}}$$ $$...$$ $$\zeta\cdot\zeta^k\cdot\zeta^{k^2}\cdot...\cdot\zeta^{k^{d-1}}$$

In other words, using the Fundamental Theorem of Symmetric Polynomials, $M$ is obtained by adjoining all symmetric expressions of $\zeta,\zeta^k,...,\zeta^{k^{d-1}}$. From the construction of $M$,

$$\left[\operatorname{Fix}\left(\sigma_k\right)\left[\zeta\right]:\operatorname{Fix}\left(\sigma_k\right)\right]\geq\left[M\left[\zeta\right]:M\right]$$

but symmetric expressions of $\zeta,\zeta^k,...,\zeta^{k^{d-1}}$ is fixed by $\sigma_k$ so $M\subset \operatorname{Fix}(\sigma_k)$ and by the inequality of the extension degrees one actually has $M = \operatorname{Fix}(\sigma_k)$.

Now in general $$\operatorname{Fix}\left<\sigma_{k_1},\sigma_{k_2}\right>= \operatorname{Fix}\left<\sigma_{k_1}\right>\cap \operatorname{Fix}\left<\sigma_{k_2}\right>$$ for two different cyclic subgroups $\left<\sigma_{k_1}\right>$ and $\left<\sigma_{k_2}\right>$. So we could express all intermediate extensions of $\Bbb Q[\zeta]/\Bbb Q$ using a finite intersection of such $M$.

Question . As demonstrated in a long paragraph, $M=\operatorname{Fix}(\sigma_k)$ is the extension of $\Bbb Q$ by adjoining all symmetric expressions of $\zeta,\zeta^k,...,\zeta^{(d-1)k}$. To be more specific, $M$ is generated by certain sums of powers of $\zeta$. I am trying to express $M$ in a simpler way. For example, can one always explicitly express all intermediate extensions using a primitive element of the form $\zeta^{d_1}+...+\zeta^{d_n}$? How can we compute the degrees $d_1,...,d_n$?

Answer 1

I'll do the case $n = p^k$ an odd prime power. Choose a cyclic generator $g$ for $(\mathbb{Z}/p^k\mathbb{Z})^\times$. For $d \mid |(\mathbb{Z}/p^k\mathbb{Z})^\times| = (p - 1)p^{k - 1}$, consider the sum $\alpha := \zeta + \zeta^{g^d} + \zeta^{g^{2d}} + \dots + \zeta^{g^{(p - 1)p^{k - 1}-d}}$. This element is invariant under the subgroup of $G = \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ generated by $g^d$ (note that $(\zeta^{g^{(p - 1)p^{k - 1}-d}})^{g^d} = \zeta$) so $\mathbb{Q}(\alpha)$ is contained in the fixed field of this subgroup.

We claim that $\mathbb{Q}(\alpha)$ actually equals this fixed field. One way to see this is to note that the elements $\zeta^{m}$ for $m \in (\mathbb{Z}/n\mathbb{Z})^\times$ are $\mathbb{Q}$-linearly independent (they form a basis for the field extension), so if $\zeta \mapsto \zeta^m$ fixes $\alpha$, then we conclude that $$\{m, mg^d, \dots, mg^{(p - 1)p^{d - 1} - d}\} = \{1, g^d, \dots, g^{(p - 1)p^{d - 1} - d}\}$$ as sets modulo $n$. Using the cyclic structure of $(\mathbb{Z}/n\mathbb{Z})^\times$, this is true if and only if $m$ is a power of $g^d$. Hence $\mathbb{Q}(\alpha) = \mathrm{Fix}(\langle g^d\rangle)$.

There are two particularly interesting special cases. Taking $g^d = -1$, corresponding to complex conjugation, we have $\alpha = \zeta + \zeta^{-1} = \zeta + \overline{\zeta}$. The field generated by this element is $\mathbb{Q}(\zeta) \cap \mathbb{R}$. We can also take $d = 2$, in which case $\mathbb{Q}(\alpha)$ is the unique quadratic subfield of $\mathbb{Q}(\zeta)$, necessarily equal to $\mathbb{Q}(\sqrt{D})$ for some squarefree integer $D$. Then you can use Gauss sums to show that we can take $D = p^* := \pm p$, where the sign is chosen so that $p^* \equiv 1 \bmod 4$.