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

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 $$d_1,...,d_n$$?

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 $$p^* \equiv 1 \bmod 4$$.